Hairpin Heat Exchangers Explained (Livro Texto)

Prévia do material em texto

1 
 2 
 3 
Hairpin Heat 4 
Exchangers Explained 5 
Fundamentals of Double-Pipe, Multi-6 
Tube and Finned Tube Heat Exchangers 7 
 8 
Version: 1.08 9 
 10 
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Prof. Dr. Samuel Jorge Marques Cartaxo 12 
© 2015 Samuel Jorge Marques Cartaxo 13 
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Disclaimer 1 
All information provided in this document is provided "as is" without warranty of any kind, 2 
either expressed or implied. Every effort has been made to ensure accuracy and conformance to 3 
standards accepted at the time of publication. The reader is advised to research other sources of 4 
information on these topics. 5 
The material in this book is a result of many hours of work and was prepared in good faith and 6 
carefully revised and edited. However, like with any book, the author and any collaborators cannot be 7 
held responsible for errors of any sort in this content. Furthermore, since it is not possible for the 8 
author to verify the correctness and reliability of information referenced from public literature, but can 9 
only examine them for suitability for the intended purpose herein, this information cannot be 10 
warranted. Also, because the author cannot assure the experience or technical capability of the 11 
reader/user of the information and the suitability of the information for the intended purpose, the use 12 
of the contents must be based on the best judgment of the user. Therefore the final user is responsible 13 
for any damage or loss caused by the use or misuse of the information provided herein. 14 
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© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
Copyright Notice 1 
Author: Prof. Dr. Samuel Jorge Marques Cartaxo 2 
Title: Hairpin Heat Exchangers Explained: Fundamentals of Double-Pipe, Multi-Tube and 3 
Finned Tube Heat Exchangers 4 
© 2015 Samuel Jorge Marques Cartaxo 5 
 6 
ALL RIGHTS RESERVED. This book contains material protected under International and 7 
Federal Copyright Laws and Treaties. Any unauthorized reprint or use of this material is prohibited. 8 
No part of this book may be reproduced or transmitted in any form or by any means, electronic or 9 
mechanical, including photocopying, recording, or by any information storage and retrieval system 10 
without express written permission from the author / publisher. 11 
Last Revised: November 2016 12 
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Hairpin Heat Exchangers Explained: Fundamentals of 
Double-Pipe, Multi-Tube and Finned Tube Heat Exchangers 
Authored by Prof. Samuel Jorge Marques Cartaxo, Ph.D. 
ISBN-13: 9781516985401 
ISBN-10: 1516985400 
 
Contents 1 
1 An Overview of the Equipment ............................................................ 22 2 
1.1 Process applicability ........................................................................ 23 3 
1.1.1 Types of Hairpins .................................................................... 25 4 
1.1.2 First of All Concerns ............................................................... 25 5 
1.1.3 Where to Use a Double-Pipe Exchanger? ............................... 26 6 
1.1.4 About The Optimal Design...................................................... 27 7 
1.2 The Real Double-Pipe Exchanger is a Hairpin! ............................... 30 8 
1.3 Pipes and Tubes ............................................................................... 32 9 
1.3.1 Is There a Difference? ............................................................. 32 10 
1.3.2 Practical Implications of Pipe and Tube Distinction ............... 33 11 
1.3.3 Commercial Diameters and Lengths ....................................... 33 12 
2 Heat Exchanger Design Equations ....................................................... 35 13 
2.1 The Overall Heat Transfer Coefficient ............................................ 35 14 
2.1.1 “Design” and “Clean” Overall Heat Transfer Coefficients ..... 39 15 
2.1.2 Relatively Small Wall Thermal Resistance ............................. 39 16 
2.1.3 Controlling Fluid ..................................................................... 40 17 
2.2 Basic Heat Exchanger Differential Equations ................................. 40 18 
2.2.1 Parallel Flow ............................................................................ 41 19 
2.2.2 Counterflow ............................................................................. 42 20 
2.3 Log Mean Temperature Difference ................................................. 44 21 
2.3.1 Parallel Flow Double Pipe Heat Exchanger ............................ 44 22 
2.3.2 Counterflow Double Pipe Heat Exchanger ............................. 46 23 
2.3.3 The Restrictions on the Overall Heat Transfer Coefficient ..... 47 24 
2.3.4 LMTD and Fluid Flow Arrangement ...................................... 48 25 
2.3.5 Special Cases of the LMTD ..................................................... 49 26 
2.3.6 The Physical Meaning of the LMTD ....................................... 49 27 
2.3.7 Do we need the LMTD Nowadays? ........................................ 49 28 
2.3.8 Parallel flow or counterflow: what is the Difference? ............. 50 29 
2.4 Exit Temperatures of the Fluids ....................................................... 50 30 
2.4.1 Counterflow ............................................................................. 51 31 
2.4.2 Parallel Flow ............................................................................ 52 32 
Worked Example 2–1 [18] ........................................................................ 53 33 
Worked Example 2–2 [18] ........................................................................ 55 34 
3 Wall Temperatures ................................................................................ 61 35 
3.1 Cold Fluid in the Inner Pipe ............................................................. 61 36 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
3.1.1 Negligible Conductive Resistance ........................................... 63 1 
3.1.2 Thin Tube Wall ........................................................................ 63 2 
3.2 Hot Fluid in the Inner Pipe............................................................... 63 3 
3.2.1 Negligible Conductive Resistance ........................................... 64 4 
3.2.2 Thin Tube Wall ........................................................................ 64 5 
3.3 General Form of Wall Temperature Equations ................................ 65 6 
Worked Example 3–1 ................................................................................ 66 7 
Worked Example 3–2 ................................................................................ 66 8 
4 Heat Transfer Coefficients .................................................................... 69 9 
4.1 Methods of Estimation ..................................................................... 69 10 
4.1.1 Types of Correlations .............................................................. 70 11 
4.2 Using Correlations ........................................................................... 71 12 
4.3 Flow Regimes .................................................................................. 71 13 
4.4 Effects of Fluid Flow Development ................................................. 72 14 
4.4.1 Fully Developed Laminar Flow in Pipes ................................. 73 15 
Worked Example 4–1 ................................................................................ 73 16 
4.5 Heat Transfer Coefficients for Circular Pipes ................................. 74 17 
4.5.1 Exact Solutions for Laminar Fully Developed Flow ............... 74 18 
4.5.2 Turbulent ................................................................................. 75 19 
Worked Example 4–2 ................................................................................ 88 20 
4.5.3 Transition ................................................................................. 89 21 
4.5.4 Heat Transfer Coefficientsfor Water ...................................... 90 22 
4.5.5 Liquid Metals ........................................................................... 91 23 
4.6 Heat Transfer Coefficients for the Annulus ..................................... 95 24 
4.6.1 Annulus Equivalent Diameter ................................................. 95 25 
4.6.2 Petukhov-Roizen Corrector ..................................................... 96 26 
4.6.3 Gnielinski Annulus Equation [79] .......................................... 97 27 
5 Pressure Drop ...................................................................................... 100 28 
5.1 Friction Factor in Circular Tubes ................................................... 101 29 
5.1.1 Isothermal Laminar Flow ...................................................... 102 30 
5.1.2 Isothermal Turbulent Flow .................................................... 102 31 
5.1.3 All Flow Regimes .................................................................. 103 32 
5.1.4 Effect of Variable Physical Properties................................... 104 33 
5.2 Pressure Drop Calculation with the 2-K Method ........................... 104 34 
5.3 Inner Tube Pressure Drop .............................................................. 105 35 
Worked Example 5–1 .............................................................................. 108 36 
5.4 Annulus Pressure Drop .................................................................. 110 37 
5.4.1 Annulus with Bonnet-Type Return........................................ 112 38 
5.4.2 Annulus with Straight-Pipe Return........................................ 114 39 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
Worked Example 5–2 .............................................................................. 115 1 
5.5 Allowable Pressure Drop ............................................................... 116 2 
6 Series-Parallel Arrangements ............................................................. 119 3 
6.1 Mean Temperature Difference ....................................................... 121 4 
6.1.1 Hot Fluid in Series and Cold Fluid in Parallel....................... 122 5 
Worked Example 6–1 .............................................................................. 122 6 
6.1.2 Cold Fluid in Series and Hot Fluid in Parallel ....................... 123 7 
Worked Example 6–2 .............................................................................. 123 8 
6.1.3 Choice of the Fluid Division ................................................. 124 9 
6.2 Pressure Drop Calculation ............................................................. 124 10 
6.2.1 Tube Pressure Drop ............................................................... 124 11 
6.2.2 Annulus Pressure Drop .......................................................... 124 12 
Worked Example 6–3 .............................................................................. 125 13 
7 Heat Exchanger Design ...................................................................... 129 14 
7.1 Fluid Allocation ............................................................................. 129 15 
7.2 Design Margin ............................................................................... 130 16 
7.2.1 Over-design ........................................................................... 131 17 
7.2.2 Over-surface .......................................................................... 131 18 
7.2.3 Fouling over-surface .............................................................. 131 19 
7.2.4 Design Margins Ranges ......................................................... 131 20 
7.3 Physical Properties of Chemical Components ............................... 132 21 
7.4 Sizing ............................................................................................. 132 22 
7.4.1 The Design Variables ............................................................ 133 23 
7.4.2 Sizing Outline ........................................................................ 133 24 
Worked Example 7–1 .............................................................................. 134 25 
Worked Example 7–2 .............................................................................. 138 26 
7.5 Rating ............................................................................................. 150 27 
7.5.1 Rating Outline........................................................................ 151 28 
7.6 The Data Sheet ............................................................................... 152 29 
Worked Example 7–3 .............................................................................. 155 30 
Worked Example 7–4 .............................................................................. 161 31 
8 Multi-tube Heat Exchangers ............................................................... 166 32 
8.1 Tube count ..................................................................................... 166 33 
8.2 Design Method ............................................................................... 166 34 
8.2.1 Cross-flow Area ..................................................................... 166 35 
8.2.2 Shell Equivalent Diameter ..................................................... 167 36 
Worked Example 8–1 .............................................................................. 167 37 
9 Finned Tube Heat Exchangers ............................................................ 175 38 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
9.1 Fins Count ...................................................................................... 176 1 
9.2 Design Method ............................................................................... 176 2 
9.2.1 Cross-flow Area ..................................................................... 177 3 
9.2.2 Shell Equivalent Diameter ..................................................... 177 4 
9.3 Fin efficiency ................................................................................. 177 5 
9.4 Weighted fin efficiency.................................................................. 178 6 
9.5 Overall Heat Transfer Coefficient ................................................. 178 7 
Worked Example 9–1 [30] ...................................................................... 180 8 
9.6 Finned Wall Temperatures ............................................................. 181 9 
Worked Example 9–2 [28] ...................................................................... 185 10 
Worked Example 9–3 .............................................................................. 187 11 
10 References ........................................................................................... 198 12 
Appendix A Heat Exchanger Specifications ............................................ 207 13 
A.1 Piping Data - (DN, NB, NPS) ........................................................ 207 14 
A.2 B.W.G. – Birmingham Wire Gauge............................................... 215 15 
A.3 Heat Transfer Tube Data (B.W.G) ................................................. 216 16 
A.4 Finned and Multi-Tube Hairpin Design Data ................................ 218 17 
Appendix B Properties of Materials ......................................................... 222 18 
B.1 Absolute Roughness of Materials .................................................. 222 19 
B.2 Thermal Conductivity of Selected Materials ................................. 223 20 
Appendix C Physical Properties of Chemical Components ..................... 225 21 
C.1 Index .............................................................................................. 225 22 
C.1.1 List of Gases (alphabetical) ................................................... 225 23 
C.1.2 List of Vapors (alphabetical) ................................................. 22524 
C.1.3 List of Liquids (alphabetical) ................................................ 228 25 
C.2 Selected Gases – Part A ................................................................. 232 26 
C.2.1 Specific Heat .......................................................................... 233 27 
C.2.2 Thermal Conductivity ............................................................ 234 28 
C.2.3 Prandtl Number...................................................................... 235 29 
C.2.4 Specific Mass ......................................................................... 236 30 
C.2.5 Dynamic Viscosity ................................................................ 237 31 
C.3 Selected Gases – Part B ................................................................. 239 32 
C.3.1 Specific Heat .......................................................................... 240 33 
C.3.2 Thermal Conductivity ............................................................ 241 34 
C.3.3 Prandtl Number...................................................................... 242 35 
C.3.4 Specific Mass ......................................................................... 243 36 
C.3.5 Dynamic Viscosity ................................................................ 244 37 
C.4 Selected Vapors – Part A ............................................................... 246 38 
C.4.1 Heat Capacity ........................................................................ 247 39 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
C.4.2 Thermal Conductivity ............................................................ 248 1 
C.4.3 Prandtl Number...................................................................... 249 2 
C.4.4 Specific Mass ......................................................................... 250 3 
C.4.5 Dynamic Viscosity ................................................................ 251 4 
C.5 Selected Vapors – Part B ............................................................... 253 5 
C.5.1 Specific heat .......................................................................... 254 6 
C.5.2 Thermal Conductivity ............................................................ 255 7 
C.5.3 Prandtl Number...................................................................... 256 8 
C.5.4 Specific Mass ......................................................................... 257 9 
C.5.5 Dynamic Viscosity ................................................................ 258 10 
C.6 Selected Vapors – Part C ............................................................... 260 11 
C.6.1 Specific heat .......................................................................... 261 12 
C.6.2 Thermal Conductivity ............................................................ 262 13 
C.6.3 Prandtl Number...................................................................... 263 14 
C.6.4 Specific Mass ......................................................................... 264 15 
C.6.5 Dynamic Viscosity ................................................................ 265 16 
C.7 Selected Vapors – Part D ............................................................... 267 17 
C.7.1 Specific heat .......................................................................... 268 18 
C.7.2 Thermal Conductivity ............................................................ 269 19 
C.7.3 Prandtl Number...................................................................... 270 20 
C.7.4 Specific Mass ......................................................................... 271 21 
C.7.5 Dynamic Viscosity ................................................................ 272 22 
C.8 Selected Vapors – Part E................................................................ 274 23 
C.8.1 Specific heat .......................................................................... 275 24 
C.8.2 Thermal Conductivity ............................................................ 276 25 
C.8.3 Prandtl Number...................................................................... 277 26 
C.8.4 Specific Mass ......................................................................... 278 27 
C.8.5 Dynamic Viscosity ................................................................ 279 28 
C.9 Selected Liquids – Part A .............................................................. 281 29 
C.9.1 Specific Heat .......................................................................... 282 30 
C.9.2 Thermal Conductivity ............................................................ 283 31 
C.9.3 Prandtl Number...................................................................... 284 32 
C.9.4 Specific Mass ......................................................................... 285 33 
C.9.5 Dynamic Viscosity ................................................................ 286 34 
C.10 Selected Liquids – Part B ............................................................... 288 35 
C.10.1 Specific Heat .......................................................................... 289 36 
C.10.2 Thermal Conductivity ............................................................ 290 37 
C.10.3 Prandtl Number...................................................................... 291 38 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
C.10.4 Specific Mass ......................................................................... 292 1 
C.10.5 Dynamic Viscosity ................................................................ 293 2 
C.11 Selected Liquids – Part C ............................................................... 295 3 
C.11.1 Specific Heat .......................................................................... 296 4 
C.11.2 Thermal Conductivity ............................................................ 297 5 
C.11.3 Prandtl Number...................................................................... 298 6 
C.11.4 Specific Mass ......................................................................... 299 7 
C.11.5 Dynamic Viscosity ................................................................ 300 8 
C.12 Selected Liquids – Part D .............................................................. 302 9 
C.12.1 Specific heat .......................................................................... 303 10 
C.12.2 Thermal Conductivity ............................................................ 304 11 
C.12.3 Prandtl Number...................................................................... 305 12 
C.12.4 Specific Mass ......................................................................... 306 13 
C.12.5 Dynamic Viscosity ................................................................ 307 14 
C.13 Selected Liquids – Part E ............................................................... 309 15 
C.13.1 Specific heat .......................................................................... 310 16 
C.13.2 Thermal Conductivity ............................................................ 311 17 
C.13.3 Prandtl Number...................................................................... 312 18 
C.13.4 Specific Mass ......................................................................... 313 19 
C.13.5 Dynamic Viscosity ................................................................ 314 20 
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© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
List of Illustrations 1 
Figure 1: Assemble of two concentric pipes in a double-pipe heat exchanger.23 2 
Figure 2: Schematic representation of a double-pipe heat exchanger. [Adapted from Ref. [13]] 23 3 
Figure 3: A section cropped from the inner pipe of a high-pressure double-pipe heat exchanger. 4 
Source: the author. ...................................................................... 24 5 
Figure 4: Hairpin heat exchangers with integral bonnet (left end) covering the inner tubes and 6 
providingthe return chamber for the fluid in the annulus. [Adapted from Ref. [14]] 24 7 
Figure 5: Return bends for pipe and annulus of “ROD-THRU” hairpin heat exchangers used for 8 
food processing. These bends do not have a bonnet type design, being just plain pipes. 9 
[Adapted from Ref. [9]] .............................................................. 25 10 
Figure 6: Procedure for designing a heat exchanger without optimization. The process engineer is 11 
responsible for every decision in the design routine. ................. 28 12 
Figure 7: Procedure for designing a heat exchanger with the external optimization loop in place. The 13 
process engineer is responsible for setting and fine-tuning of optimizer parameters. 29 14 
Figure 8: Hairpin (double-pipe) heat exchanger. [Adapted from Ref. [29]] .. 30 15 
Figure 9: Conjugated straight multi-tube heat exchanger associated in series. [Adapted from Ref. 16 
[31]] ............................................................................................ 31 17 
Figure 10: Bank of multi-tube hairpin heat exchangers. [Adapted from Ref. [32]]32 18 
Figure 11: Bank of AISI 304 stainless steel double-pipe hairpin heat exchangers. [Adapted from 19 
Ref. [34]] .................................................................................... 33 20 
Figure 12: Thermal resistances involved in the heat transfer between a hot and a cold stream inside 21 
a heat exchanger. ........................................................................ 37 22 
Figure 13: One-dimensional representation of a double-pipe parallel flow heat exchanger. 41 23 
Figure 14: One-dimensional representation of a double-pipe counterflow heat exchanger. 43 24 
Figure 15: Schematic of the resistances in the thermal circuit in a double-pipe exchanger with the 25 
cold fluid in the inner tube. ......................................................... 61 26 
Figure 16: Nusselt numbers for liquid metals in fully developed flow and uniform heat flux 27 
boundary condition (Eq. (4.33)). ................................................ 92 28 
Figure 17: Nusselt numbers for liquid metals in fully developed flow and uniform wall temperature 29 
boundary condition (Eq. (4.34)). ................................................ 93 30 
Figure 18: Schematic representation of the annulus channel of two concentric pipes. 95 31 
Figure 19: Nusselt number correction factor for an annulus passage with the inner pipe heated or 32 
cooled (Eq. (4.42)). ..................................................................... 97 33 
Figure 20: Hairpin heat exchanger mechanical drawing. The annulus fluid returns through a bonnet. 34 
[Adapted from Ref. [15]. Courtesy of Heat Exchanger Design, Inc., Indianapolis-USA]35 
 .................................................................................................. 105 36 
Figure 21: K-factor head loss coefficient for long radius 180° return bend. Di is the internal tube 37 
diameter expressed in millimeters (Eq. (5.24)). ....................... 107 38 
Figure 22: K-factor head loss coefficient for long radius 180° return bend. Di is the internal tube 39 
diameter expressed in inches (Eq. (5.25)). ............................... 108 40 
Figure 23: Hairpin exchanger with annuli connected by a straight pipe. [Adapted from Ref. [34]]41 
 .................................................................................................. 111 42 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
Figure 24: K-factor head loss coefficient for “bonnet-type” annulus return. Dh = Ds-Do is the 1 
hydraulic diameter in millimeters (Eq. (5.36)). ........................ 112 2 
Figure 25: K-factor head loss coefficient for “bonnet-type” annulus return. Dh = Ds-Do is the 3 
hydraulic diameter in inches (Eq. (5.37)). ................................ 113 4 
Figure 26: K-factor head loss coefficient for “straight-pipe” annulus return. Dh = Ds-Do is the 5 
hydraulic diameter in millimeters (Eq. (5.40)). ........................ 113 6 
Figure 27: K-factor head loss coefficient for “straight-pipe” annulus return. Dh = Ds-Do is the 7 
hydraulic diameter in inches (Eq. (5.41)). ................................ 114 8 
Figure 28: Hairpin heat exchangers with both pipe and annulus associated in series. 119 9 
Figure 29: Hairpin heat exchangers with annulus connected in series and the pipe connected in 10 
parallel. ..................................................................................... 120 11 
Figure 30: Bank of four hairpin heat exchangers with annulus connected in series and the pipe fluid 12 
divided in two parallel streams. ................................................ 121 13 
Figure 31: Typical heat exchanger data sheet. [Adapted from Ref. [109]] .. 153 14 
Figure 32: Multi-tube hairpin heat exchanger. [Adapted from Ref. [110]] . 166 15 
Figure 33: Heat transfer tubes with longitudinal fins. [Adapted from Ref. [110]] 175 16 
Figure 34: Cross section of a double-pipe heat exchanger extended with longitudinal fins on the 17 
external surface of the internal tube. ........................................ 176 18 
Figure 35: Cross section of a multi-tube heat exchanger extended with longitudinal fins on the 19 
external surface of the internal tubes. ....................................... 177 20 
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© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
Preface 1 
This text was prepared with the student in mind. Every effort was applied in order to clarify the 2 
presented subjects, both by balancing the level of description and the material format, yet preserving 3 
the technical accuracy. Accordingly, it is targeted to facilitate the learning process and understanding 4 
of the student challenged with the subject of heat exchanger in chemical engineering, mechanical 5 
engineering, and other related engineering fields. 6 
The content is arranged in chapters organized with the purpose of promoting the understanding 7 
of the subject matter by the undergraduate student that is touching it for the first time, what makes the 8 
book not intended for use as a reference by experts in the field of heat exchangers. However, 9 
absolutely no compromise is made regarding the precision and reliability of the techniques and 10 
methods described here. Despite its pedagogical format, the knowledge presented herein envisions to 11 
be applied in professional practice, being aligned with proven information found in the scientific-12 
technical literature and reasonably up-to-date industrial practice. 13 
With regard to heat-exchangers, the Kern’s method suffices for most applications. I have seen 14 
many exchangers designed with the Kern’s method, and they performed quite satisfactorily regarding 15 
the design specifications, although it tends to overdesign the equipment in some extent. Of course, the 16 
contemporary industrial practice involves the use of closed commercial software packages that 17 
implement more sophisticated methods such as Tinker, Bell-Delaware, or derived variations. In fact, 18 
those mentioned more sophisticated methods are built upon the physical rationale of the Kern’s 19 
method, which contains the fundamental “physics” to approach the problem of designing or analyzing 20 
a heat exchanger. 21 
The knowledge of the physics supporting the heat exchanger design method is a key factor 22 
when the student learns the operation of specialized heat transfer software. Many times, the student is 23 
not aware of the underlying fundamentals behind the design method inside the software, and simply 24 
operates it in “black box” mode. In the view of the author, this move is not beneficial for someone 25 
learning engineering and involves risks when the student goes professional, since heat exchanger 26 
design software should be used uncritically in no circumstance. Usually, there are so many 27 
considerations, hypotheses and simplifications involved, which sometimes are violated,generating 28 
unpredictable errors. The engineer must have enough technical comprehension to identify such 29 
anomalies minimally. 30 
Nomenclature 31 
Subscripts 32 
0: Constant fluid properties ........................................................................ 44 33 
1: Inlet (entrance) ....................................................................................... 44 34 
2: Outlet (exit) ............................................................................................ 44 35 
eq: Equivalent ............................................................................................. 44 36 
i: Inner (internal) fluid, channel or surface ........................................... 44, 67 37 
o: Outer (external) fluid, channel or surface ........................................ 44, 67 38 
w: Evaluated at the wall ....................................................................... 44, 67 39 
Greek 40 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
µ: Viscosity at bulk temperature (Pa s or lb/ft h) ............................... 78, 104 1 
µw: Viscosity at wall temperature (Pa s or lb/ft h) .............................. 78, 104 2 
ΔP: Pressure difference (Pa or psi) ........................................................... 101 3 
ΔPdist: Distributed pressure drop (Pa or psi) ..................................... 100, 124 4 
ΔPi,dist: Distributed pressure drop in the tube (Pa or psi) .................. 105, 124 5 
ΔPi,loc: Localized pressure drop in the tube (Pa or psi) ..................... 105, 124 6 
ΔPloc: Localized pressure drop (Pa or psi) ........................................ 100, 124 7 
ΔPo,dist: Distributed pressure drop in the annulus (shell) (Pa or psi) 110, 124 8 
ΔPo,loc: Localized pressure drop in the annulus (shell) (Pa or psi) ... 110, 124 9 
ΔT1: Temperature difference in the terminal (side) 1 (°C or °F) ............... 45 10 
ΔT2: Temperature difference in the terminal (side) 2 (°C or °F) ............... 45 11 
ΔTlm,c: Logarithmic mean temperature difference for countercurrent flow (°C or °F) .................... 123 12 
ΔTlm,p: Logarithmic mean temperature difference for parallel flow (°C or °F) . 123 13 
ΔTlm: Logarithmic mean temperature difference (°C or °F) .................... 123 14 
ΔTm: Mean temperature difference (°C or °F) ................................... 35, 130 15 
ε: Absolute roughness (m or ft) ................................................................ 101 16 
ρ: Fluid specific mass (kg/m3 or lb/ft3) ..................................................... 101 17 
τw: Shear stress at the wall (Pa or lb/ft h2) ................................................ 101 18 
ϕ: Correction factor for variable physical properties .......................... 87, 104 19 
Symbols 20 
A: Accumulative heat transfer surface (m2 or ft2) ...................................... 40 21 
A: Actual (final) heat transfer area (m2 or ft2) .......................................... 130 22 
A: Heat transfer area of the heat exchanger (m2 or ft2) .............................. 36 23 
A: Reference heat transfer area of the heat exchanger (m2 or ft2) .............. 35 24 
Ac: Clean heat transfer area (m2 or ft2) ..................................................... 130 25 
Ad: Design (fouled) heat transfer area (m2 or ft2) ..................................... 130 26 
Ai: Area of the inner surface of the internal pipe (m2 or ft2) ................ 36, 67 27 
Ao: Area of the outer surface of the internal pipe (m2 or ft2) ............... 36, 67 28 
c: Specific heat of the cold fluid (J/kg K or BTU/lb °F) ............................ 42 29 
C: Specific heat of the hot fluid (J/kg K or BTU/lb °F) ............................. 41 30 
Cp1: Specific heat of the fluid 1 (J/kg K or BTU/lb °F) ............................. 51 31 
Cp2: Specific heat of the fluid 2 (J/kg K or BTU/lb °F) ............................. 51 32 
Dh: Hydraulic diameter (m or ft) .................................................. 72, 97, 110 33 
Di: Inner (internal) diameter of the internal pipe (m or ft) ................. 67, 105 34 
Di: Inner diameter of the internal pipe (m or ft) ......................................... 37 35 
Do: External diameter of the internal pipe (m or ft) ................................... 95 36 
Do: Outer (external) diameter of the internal pipe (m or ft) ............... 67, 110 37 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
Do: Outer diameter of the internal pipe (m or ft) ........................................ 37 1 
Ds: Inner (internal) diameter of the shell (m or ft) ................................... 110 2 
Ds: Internal diameter of the external pipe (shell) (m or ft) ......................... 95 3 
fD: Darcy friction factor .................................................................... 100, 101 4 
fF: Fanning friction factor ..................................................... 80, 87, 100, 101 5 
fF0: Fanning friction factor for constant fluid properties ............................ 83 6 
fFi: Fanning friction factor of the tube .............................................. 105, 124 7 
fFo: Fanning friction factor of the annulus (shell) ............................. 110, 124 8 
Fsp: Temperature difference factor for series-parallel arrangement ......... 122 9 
G: Mass flux (kg/s m2 or lb/h ft2) ....................................................... 73, 100 10 
Gi: Mass flux in the tube (kg/s m2 or lb/h ft2) .......................................... 105 11 
Go: Mass flux in the annulus (shell) (kg/s m2 or lb/h ft2) ......................... 110 12 
Gr: Grashof number .................................................................................... 71 13 
hi: Convective heat transfer coefficient of the inner surface of the internal pipe (W/m2 K or 14 
BTU/h ft2 °F) ................................................................................. 38, 67 15 
hio: Internal convective heat transfer coefficient corrected for the external (reference) heat 16 
transfer surface (W/m2 K or BTU/h ft2 °F) ......................................... 40 17 
ho: Convective heat transfer coefficient of the outer surface of the internal pipe (W/m2 K or 18 
BTU/h ft2 °F) ................................................................................. 38, 67 19 
ID: Inner (internal) diameter (m or ft) ........................................................ 73 20 
K: Friction loss coefficient ....................................................................... 100 21 
k: Thermal conductivity of the material of the internal pipe (W/m K or BTU/h ft°F) ................. 37, 67 22 
Ki: Friction loss coefficient of the tube ............................................ 106, 124 23 
Ko: Friction loss coefficient of the annulus (shell) ........................... 110, 124 24 
L: Effective length of the heat exchanger (m or ft) .................................... 36 25 
L: Pipe length (m or ft) ............................................................................... 72 26 
Lfd,T: Fully developed thermal entrance length (m or ft) ............................ 72 27 
Lfd: Fully developed entrance length (m or ft) ........................................... 72 28 
Lhp: Length of the “hairpin” heat exchanger (m or ft) ...................... 106, 124 29 
LMTD: Logarithmic mean temperature difference (°C or °F) ........... 44, 119 30 
n: Number of parallel streams in a series-parallel arrangement ............... 124 31 
nc: Number of parallel streams of the cold fluid in a series-parallel arrangement122 32 
nh: Number of parallel streams of the hot fluid in a series-parallel arrangement122 33 
Nhp: Number of “hairpin” heat exchangers ...................................... 106, 124 34 
Nu: Nusselt number based on pipe diameter ........................................ 74, 97 35 
Nu0: Nusselt number of constant fluid properties ...................................... 82 36 
Nua: Nusselt number for theannulus, based on the hydraulic diameter ..... 97 37 
Nub: Nusselt number with fluid properties at bulk temperature ................. 85 38 
Nufd: Nusselt number for fully developed flow .......................................... 89 39 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
Nut: Nusselt number for a circular tube, based on the hydraulic diameter 97 1 
Pe: Péclet number ........................................................................... 75, 93, 94 2 
Pr: Prandtl number ...................................................................................... 71 3 
Prw: Prandtl number evaluated at the wall temperature .............................. 85 4 
q: Heat transfer rate (W or BTU/h) .............................................. 35, 36, 130 5 
Rdi: Internal fouling coefficient (m2 °C/W or h ft2 °F/BTU) ...................... 37 6 
Rdo: External fouling coefficient (m2 °C/W or h ft2 °F/BTU) .................... 37 7 
Re: Reynolds number ........................................................................... 71, 72 8 
ReDh: Reynolds number based on the hydraulic diameter .......................... 97 9 
Ref: Reynolds number at fluid film temperature Tf .................................... 85 10 
Rei: Reynolds number of the tube ............................................................ 106 11 
Req: Equivalent resistance of the whole thermal circuit (°C/W or h °F/ BTU) .... 36 12 
Rew: Reynolds number evaluated with fluid properties at wall temperature103 13 
Rfi: Internal fouling resistance (°C/W or h °F/ BTU) ................................. 37 14 
Rfo: External fouling resistance (°C/W or h °F/ BTU) ............................... 37 15 
Ri: Convective resistance of the fluid flowing inside the inner pipe (°C/W or h °F/ BTU) ............... 37 16 
Rk: Conductive resistance of the wall between fluids (°C/W or h °F/ BTU)37 17 
Ro: Convective resistance for the fluid in the annular channel (°C/W or h °F/ BTU) ....................... 37 18 
S: Cross flow area (m2 or ft2) ..................................................................... 73 19 
Si: Cross flow area of the tube (m2 or ft2) ................................................. 105 20 
So: Cross flow area of the annulus (shell) (m2 or ft2) ............................... 110 21 
T: Fluid bulk temperature (°C or °F) .......................................................... 82 22 
T: Temperature (°C or °F) .......................................................................... 91 23 
t: Temperature of the cold fluid (°C or °F). .................................... 41, 42, 67 24 
T: Temperature of the hot fluid (°C or °F). .......................................... 41, 67 25 
T1,in: Entrance temperature of the fluid 1 (°C or °F) .................................. 51 26 
T1,out: Exit temperature of the fluid 1 (°C or °F)......................................... 51 27 
t1: Entrance temperature of the cold fluid (°C or °F). .............................. 119 28 
T1: Entrance temperature of the hot fluid (°C or °F). ............................... 119 29 
T2,in: Entrance temperature of the fluid 2 (°C or °F) .................................. 51 30 
T2,out: Exit temperature of the fluid 2 (°C or °F)......................................... 51 31 
t2: Exit temperature of the cold fluid (°C or °F). ...................................... 119 32 
T2: Exit temperature of the hot fluid (°C or °F). ...................................... 119 33 
Tb,in: Fluid bulk temperature at the inlet (°C or °F) .................................... 76 34 
Tb,out: Fluid bulk temperature at the outlet (°C or °F)................................. 76 35 
Tb: Fluid bulk temperature (°C or °F) ........................................................ 76 36 
ti: Temperature of the inner (internal) fluid (°C or °F). .............................. 67 37 
to Temperature of the outer (external) fluid (°C or °F). .............................. 67 38 
tw: Wall temperature (°C or °F) .................................................................. 67 39 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
Tw: Wall temperature (°C or °F) ........................................................... 76, 78 1 
twi: Temperature of the inner surface of heat transfer (°C or °F)................ 67 2 
two: Temperature of the outer surface of heat transfer (°C or °F). .............. 67 3 
U: Actual (final) overall heat transfer coefficient (W/m2 °C or BTU/h ft2 °F) .. 131 4 
Ū: Mean overall heat transfer coefficient (W/m2 K or BTU/h ft2 °F) ........ 45 5 
U: Overall heat transfer coefficient (W/m2 °C or BTU/h ft2 °F) .......... 35, 36 6 
Uc: Clean overall heat transfer coefficient (W/m2 °C or BTU/h ft2 °F) ... 130 7 
Ud: Design (fouled) overall heat transfer coefficient (W/m2 °C or BTU/h ft2 °F)130 8 
Ui: Overall heat transfer coefficient referenced to the inner surface of the internal pipe 9 
(W/m2 K or BTU/h ft2 °F) ................................................................... 36 10 
Um: Mean overall heat transfer coefficient (W/m2 °C or BTU/h ft2 °F) ... 121 11 
Uo: Overall heat transfer coefficient referenced to the outer surface of the internal pipe 12 
(W/m2 K or BTU/h ft2 °F) ................................................................... 36 13 
v: Fluid velocity (m/s or ft/s) ...................................................................... 91 14 
W: Mass flow rate (kg/s or lb/h) ................................................................. 73 15 
w: Mass flow rate of the cold fluid (kg/s or lb/h) ....................................... 42 16 
W: Mass flow rate of the hot fluid (kg/s or lb/h). ....................................... 41 17 
W1: Mass flow rate of the fluid 1 (kg/s or lb/h) ......................................... 51 18 
W2: Mass flow rate of the fluid 2 (kg/s or lb/h) ......................................... 51 19 
Wi: Mass flow rate in the tube (kg/s or lb/h). ........................................... 105 20 
Wo: Mass flow rate in the annulus (shell) (kg/s or lb/h). .......................... 110 21 
 22 
 23 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
Conversion Factors 1 
Physical Quantity Value 
Acceleration 1.0000 
𝑚
𝑠2
 =   (4.2520𝑒 + 7)  
𝑓𝑡
ℎ2
 =  3.2808 
𝑓𝑡
𝑠2
 
Area 
1.0000 𝑚2  =  10.764 𝑓𝑡2  =  1550.0 𝑖𝑛2  =  0.00024711 𝑎𝑐𝑟𝑒 
=  0.00010000 ℎ𝑎 
Density 
1.0000 
𝑘𝑔
𝑚3
 =  0.062428 
𝑙𝑏
𝑓𝑡3
 =  0.0010000 
𝑔
𝑐𝑚3
 =  0.0010000 
𝑘𝑔
𝐿
 
=  1.0000 
𝑔
𝐿
 
Dynamic viscosity 1.0000 𝑃𝑎 ⋅ 𝑠  =  1000.0 𝑐𝑃  =  10.000 𝑃  =  1.0000 
𝑘𝑔
(𝑚 ⋅ 𝑠)
 
Energy 
1.0000 𝐽  =  0.23901 𝑐𝑎𝑙  =  0.00094782 𝐵𝑇𝑈  =   (1.0000𝑒 + 7) 𝑒𝑟𝑔 
=   (6.2415𝑒 + 18) 𝑒𝑉  =  0.00027778 𝑊ℎ 
=   (2.7778𝑒 − 7) 𝑘𝑊ℎ  =   (2.3901𝑒 − 10) 𝑡𝑇𝑁𝑇 
=   (9.4782𝑒 − 9) 𝑡ℎ𝑚  =  1.0000 
𝑘𝑔 ⋅ 𝑚2
𝑠2
 
Force 
1.0000 𝑁  =  0.22481 𝑙𝑏𝑓  =   (1.0000𝑒 + 5) 𝑑𝑦𝑛  =   0.10197 𝑘𝑔𝑓 
=  3.5969 𝑜𝑧𝑓  =  0.00022481 𝑘𝑖𝑝  =  1.0000 
𝑘𝑔 ⋅ 𝑚
𝑠2
 
Fouling factor 
1.0000 
𝑚2 ⋅ 𝐾
𝑊
 =  1.0000 
𝑚2 ⋅ °𝐶
𝑊
 =  5.6783 
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
=  5.6783 
𝑓𝑡2 ⋅ ℎ ⋅ °𝑅
𝐵𝑇𝑈
 =  1.0000 
𝑠3 ⋅ °𝐶
𝑘𝑔
 
Heat flux 1.0000 
𝑊
𝑚2
 =  0.31700 
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ)
 =  1.0000 
𝑘𝑔
𝑠3
 
Heat transfer 
coefficient 
1.0000 
𝑊
(𝑚2 ⋅ 𝐾)
 =  1.0000 
𝑊
(𝑚2 ⋅ °𝐶)
 =  0.17611 
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
=  0.17611 
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝑅)
 =  1.0000 
𝑘𝑔
(𝑠3 ⋅ °𝐶)
 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
Heat volumetric rate 
1.0000 
𝑊
𝑚3
 =  0.096621 
𝐵𝑇𝑈
(𝑓𝑡3 ⋅ ℎ)
 =   (2.3901𝑒 − 7)  
𝑐𝑎𝑙
(𝑐𝑚3 ⋅ 𝑠)
 
=  1.0000 
𝑘𝑔
(𝑚 ⋅ 𝑠3)
 
Kinematic viscosity 1.0000 
𝑚2
𝑠
 =  38750.  
𝑓𝑡2
ℎ
 =  10.764 
𝑓𝑡2
𝑠
 
Latent heat 1.0000 
𝑘𝐽
𝑘𝑔
 =  0.23901 
𝑐𝑎𝑙
𝑔
 =  0.42992 
𝐵𝑇𝑈
𝑙𝑏
 =  1000.0 
𝑚2
𝑠2
 
Length 
1.0000𝑚 = 3.2808𝑓𝑡 = 39.370𝑖𝑛 = (1.0000𝑒 + 9)𝑛𝑚
= (1.0000𝑒 + 12)𝑝𝑚 = 0.00062137𝑚𝑖= 1.0936𝑦𝑑
= 0.00053996𝑛𝑚𝑖 = 236.22𝑝𝑖𝑐𝑎 
Mass 
1.0000 𝑘𝑔  =  2.2046 𝑙𝑏  =  35.274 𝑜𝑧  =  0.15747 𝑠𝑡 
=  0.023453 𝑏𝑎𝑔  =  0.068522 𝑠𝑙𝑢𝑔 
Mass flux 1.0000 
𝑘𝑔
(𝑚2 ⋅ 𝑠)
 =  737.34 
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 =  0.20482 
𝑙𝑏
(𝑓𝑡2 ⋅ 𝑠)
 
Power 1.0000 𝑊  =  3.4121 
𝐵𝑇𝑈
ℎ
 =  0.0013410 ℎ𝑝  =  1.0000 
𝑘𝑔 ⋅ 𝑚2
𝑠3
 
Pressure 
1.0000 𝑘𝑃𝑎  =  1000.0 
𝑁
𝑚2
 =  10000.  
𝑑𝑦𝑛
𝑐𝑚2
 =  20.885 
𝑙𝑏𝑓
𝑓𝑡2
 
=  0.010197 
𝑘𝑔𝑓
𝑐𝑚2
 =  1.0000 𝑘𝑃𝑎  =  0.010000 𝑏𝑎𝑟 
=  0.0098692 𝑎𝑡𝑚  =  0.14504 𝑝𝑠𝑖  =  7.5006 𝑡𝑜𝑟𝑟 
=  7.5006 𝑚𝑚𝐻𝑔  =  10.197 𝑐𝑚𝐻2𝑂  =  1000.0 
𝑘𝑔
(𝑚 ⋅ 𝑠2)
 
Surface tension 
1.0000 
𝑁
𝑚
 =  1000.0 
𝑑𝑦𝑛
𝑐𝑚
 =  0.10197 
𝑘𝑔𝑓
𝑚
 =  0.068522 
𝑙𝑏𝑓
𝑓𝑡
 
=  1.0000 
𝑘𝑔
𝑠2
 
Temperature 𝐾 = °𝐶 + 273.15 =
5
9
°𝑅 =
5
9
(°𝐹 + 459.67) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
Temperature 
difference 
1.0000 𝐾  =  1.0000 °𝐶  =  1.8000 °𝐹  =  1.8000 °𝑅 
Thermal conductivity 1.0000 
𝑊
(𝑚 ⋅ 𝐾)
 =  0.57779 
𝐵𝑇𝑈
(𝑓𝑡 ⋅ ℎ ⋅ °𝐹)
 =  1.0000 
𝑘𝑔 ⋅ 𝑚
(𝑠3 ⋅ °𝐶)
 
Thermal resistance 1.0000 
𝐾
𝑊
 =  0.52753 
ℎ ⋅ °𝐹
𝐵𝑇𝑈
 =  1.0000 
𝑠3 ⋅ °𝐶
(𝑘𝑔 ⋅ 𝑚2)
 
Volume 
1.0000 𝑚3  =  35.315 𝑓𝑡3  =  61024.  𝑖𝑛3  =  1000.0 𝐿 
=   (1.0000𝑒 + 6) 𝑚𝐿  =   (1.0000𝑒 + 6) 𝑐𝑐 
=  264.17 𝑔𝑎𝑙𝑙𝑜𝑛  =  1056.7 𝑞𝑢𝑎𝑟𝑡  =  4226.8 𝑐𝑢𝑝 
=  2113.4 𝑝𝑡  =  67628.  𝑡𝑏𝑠𝑝  =   (2.0288𝑒 + 5) 𝑡𝑠𝑝 
 1 
 2 
 3 
 4 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
Physical Constants 1 
Physical Constant Value 
Avogadro constant (6.0221𝑒 + 23)  
1
𝑚𝑜𝑙
 
Boltzmann constant 
(k) 
(1.3807𝑒 − 23) 
𝐽
𝐾
 =   (7.2700𝑒 − 27)  
𝐵𝑇𝑈
°𝑅
 =   (1.3807𝑒 − 16)  
𝑒𝑟𝑔
°𝐶
 
Molar gas constant (R) 
8.3145 
𝐽
(𝑚𝑜𝑙 ⋅ 𝐾)
 =  0.0043781 
𝐵𝑇𝑈
(𝑚𝑜𝑙 ⋅ °𝐹)
 
=   (8.2057𝑒 − 5)  
𝑚3 ⋅ 𝑎𝑡𝑚
(𝑚𝑜𝑙 ⋅ 𝐾)
 
=   (8.3145𝑒 + 7)  
𝑒𝑟𝑔
(𝑚𝑜𝑙 ⋅ °𝐶)
 =  8.3145 
𝑘𝑔 ⋅ 𝑚2
(𝑠2 ⋅ 𝑚𝑜𝑙 ⋅ °𝐶)
 
Molar mass constant 0.0010000 
𝑘𝑔
𝑚𝑜𝑙
 =  1.0000 
𝑔
𝑚𝑜𝑙
 =  0.0022046 
𝑙𝑏
𝑚𝑜𝑙
 
Molar Planck constant (3.9903𝑒 − 10)  
𝐽 ⋅ 𝑠
𝑚𝑜𝑙
 =   (3.9903𝑒 − 10)  
𝑘𝑔 ⋅ 𝑚2
(𝑠 ⋅ 𝑚𝑜𝑙)
 
Molar volume of ideal 
gas (273.15 K, 101.325 
kPa) 
0.022414 
𝑚3
𝑚𝑜𝑙
 =  0.79154 
𝑓𝑡3
𝑚𝑜𝑙
 =  22414.  
𝑐𝑚3
𝑚𝑜𝑙
 
Newtonian constant of 
gravitation 
(6.6743𝑒 − 11)  
𝑚3
(𝑘𝑔 ⋅ 𝑠2)
 =  0.013856 
𝑓𝑡3
(𝑙𝑏 ⋅ ℎ2)
 
=   (6.6743𝑒 − 8)  
𝑐𝑚3
(𝑔 ⋅ 𝑠2)
 
Standard acceleration 
of gravity 
9.8067 
𝑚
𝑠2
 =  32.174 
𝑓𝑡
𝑠2
 =   (4.1698𝑒 + 8)  
𝑓𝑡
ℎ2
 =  980.67 
𝑐𝑚
𝑠2
 
Standard atmosphere 
101.32 𝑘𝑃𝑎  =  1.0000 𝑎𝑡𝑚  =  14.696 𝑝𝑠𝑖  =  760.00 𝑚𝑚𝐻𝑔 
=  1033.2 𝑐𝑚𝐻2𝑂  =  1.0332 
𝑘𝑔𝑓
𝑐𝑚2
 =  14.696 
𝑙𝑏𝑓
𝑖𝑛2
 
=   (1.0133𝑒 + 5) 
𝑘𝑔
(𝑚 ⋅ 𝑠2)
 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
Stefan-Boltzmann 
constant 
(5.6704𝑒 − 8)
𝑊
(𝑚2 ⋅ 𝐾4)
= (1.7140𝑒 − 9)
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝑅4)
= (5.6704𝑒 − 5)
𝑒𝑟𝑔
(𝑐𝑚2 ⋅ 𝑠 ⋅ 𝐾4)
= (5.6704𝑒 − 8)
𝑘𝑔
(𝑠3 ⋅ 𝐾4)
 
 1 
 2 
 3 
 4 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
22 1 An Overview of the Equipment 
1 An Overview of the Equipment 1 
Double-pipe exchangers are the simplest type of heat transfer equipment that you might 2 
encounter in a chemical process plant. Of course, a hot fluid flowing inside a straight pipe standing in 3 
the ambient air may work as a heat exchanger device; however, this arrangement is not the most 4 
effective way to transfer a heat load. 5 
In essential terms, double-pipe heat exchangers are an assembly of two pipes mounted 6 
concentrically. A cross-section of a double-pipe is shown in Figure 1. This configuration forms two 7 
physically separated channels where the fluids involved in the heat transfer service are allocated. The 8 
central channel, simply called “pipe” or “tube”, is delimited by the inner diameter (ID) of the internal 9 
(smaller) pipe. The peripheral channel is usually referred as “annulus” and defined by the outer 10 
diameter (OD) of the internal pipe and inner diameter of the external (larger) pipe. Differently from a 11 
circle, which is specified by the single measure of a diameter, the annulus of the double-pipe 12 
exchanger requires two diameters Do and Ds to be completely defined. Generally, the outer diameter of 13 
the external pipe is not relevant for the performance of the heat exchanger, except when the heat 14 
transfer with the surroundings cannot be neglected. In such case, the impairment of the equipment 15 
performance should be minimized with a layer of thermal insulation. 16 
Pipe sections of fixed lengths varying from 4 m to 16 m can be found from manufacturers (e.g. 17 
[1]–[4]), however shorter non-standard lengths can be arranged upon special ordering. The length of 18 
the exchanger is recommended to not exceed 6 m in order to avoid bending of the inner pipe, and 19 
consequent bad flow distribution inside the annulus [5]. With relatively short pipes the unit price of 20 
the exchanger is likely to increase significantly, since the flange connections are an important part of 21 
the cost [6]. 22 
A typical device has four ports or nozzles for feeding and withdrawing the process fluids. The 23 
manner the fluid inlets are connected may have a significant impact on the heat exchanger 24 
performance. As depicted in Figure 2, when both fluids enter the equipment on the same side, the 25 
obtained arrangement is called parallel flow or concurrent. On the other hand, the insertion of the 26 
fluids by nonadjacent nozzles creates a counterflow or countercurrent flow pattern. 27 
The flow arrangement is an engineer’s decision. Usually the choice should be in favor of the 28 
countercurrent flow, because there is an advantage of the mean temperature difference produced 29 
between the streams, leading to a more efficient and less expensive design. However, there are 30 
exceptions favoring the concurrent flow, as will be discussed in Section 2.3.8. 31 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
1 An Overview of the Equipment 23 
 1 
Figure 1: Assemble of two concentric pipes in a double-pipe heat 2 
exchanger. 3 
Although “double-pipe heat exchanger” is the most common designation, this equipment is 4 
referred by several other names in the field jargon and specialized literature, including variations such 5 
as double-tube heat exchanger, concentric heat exchanger, concentric pipes heat exchanger and tube-6 
in-tube heat exchanger ([7]–[11]). The engineer should be aware of various synonyms, in order to 7 
communicate well in practical situations. 8 
1.1 Process applicability 9 
Double-pipe heat exchangers are very robust and reliable. They can operate for long periods 10 
with high fouling streams, at severe process conditions, involving extreme temperature crossings, high 11 
fluid temperatures, high pressures and large temperature ranges between the inlet and outlet of the 12 
fluids. Pressure capabilities are full vacuum to over 96,000 kPa (about 14,000 psi), limited by size, 13 
material, and design conditions [12]. These characteristics allow its use for very specialized services. I 14 
can recall of a gas phase polymerization tubular reactor that I saw in a polyethylene production plant, 15 
which was designed as a concentric heat exchanger in order to handle the high temperature and 16 
pressure required for performing the chemical reaction. Figure 3 shows a pipe section of this reactor. 17 
This is an example of process requirements that other type of exchanger could hardly match. 18 
 19 
Figure 2: Schematic representation of a double-pipe heat exchanger. 20 
[Adapted from Ref. [13]] 21 
Double-pipe exchangers are very suited for corrosive, extremely fouling services, such as high 22 
viscosity fluids or slurries. There is construction adaptations available on the market with scrapping23 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
24 1 An Overview of the Equipment 
mechanisms capable of removing even crusts deposited on the heat transfer surface. The usual design 1 
is to place the most corrosive, fouling, higher pressure and higher temperature fluid in the inner pipe 2 
[5]. 3 
 4 
Figure 3: A section cropped from the inner pipe of a high-pressure 5 
double-pipe heat exchanger. Source: the author. 6 
A heat exchanger may transfer sensible and/or latent heat between fluids. While in more 7 
integrated plants, heat exchangers are found to contact two process streams, the most common service 8 
is transferring energy between a process and a utility fluid, such as cooling water, chilled water, 9 
commercial thermal fluids, saturated vapor or air. The flow rate of the process fluid is determined by 10 
the production requirements of the plant, whereas the flow rate of the utility stream is usually 11 
manipulated in order to control the outlet temperature of the process stream. The flow rate of the 12 
utility stream may be regulated by a valve at the inlet or outlet of the heat exchanger, however, in the 13 
case of cooling water, it is recommended to restrict its flow at the entrance, because the higher 14 
temperature and vapor pressure may produce cavitation on the outlet valve [5]. 15 
 16 
Figure 4: Hairpin heat exchangers with integral bonnet (left end) 17 
covering the inner tubes and providing the return chamber for the 18 
fluid in the annulus. [Adapted from Ref. [14]] 19 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
1 An Overview of the Equipment 25 
1.1.1 Types of Hairpins 1 
Generally, two types of regular double-pipe or multi-tube heat exchangers are readily found in 2 
the market. The main distinction is the way the annulus fluid flows between the heat exchanger legs. 3 
Figure 4 shows a hairpin exchanger in which the annulus fluid passes through the internal 4 
chamber built with a flanged cover or bonnet attached to the heat exchanger outer pipes. In this case, 5 
the inner fluid usually flows in a single or multiple “U” shaped tubes, which are loaded inside the 6 
outer tube, or shell. 7 
For more fouling and severe services, manufacturers offer a so-called “ROD-THRU” hairpin 8 
exchanger type in which the inner fluid passes through a simple 180° return bend [15]. With this 9 
design, the internal tubes can be accessed from both heat exchanger ends for cleaning operations, what 10 
is not possible for “U” tubes. The annulus fluid flows from one leg to another through a plain straight 11 
tube; no bonnet is used in this construction, as seen in Figure 5. 12 
 13 
Figure 5: Return bends for pipe and annulus of “ROD-THRU” hairpin 14 
heat exchangers used for food processing. These bends do not have a 15 
bonnet type design, being just plain pipes. [Adapted from Ref. [9]] 16 
1.1.2 First of All Concerns 17 
When selecting and designing a heat exchanger, there are issues beyond costs. Certainly, costs 18 
do matter a lot, however, considering that, historically, major breakthroughs in chemical and 19 
petrochemical revenues are due to the development of new products and processes, rather than from 20 
more inexpensive designs, I endorse the point of view of Ref. [6] for asserting that safety and 21 
reliability comes in first place. In this regard, some parameters should be carefully considered while 22 
selecting a heat exchanger type for a given service, namely: 23 
• Process safety. 24 
• Environment protection. 25 
• Operation reliability. 26 
• Operating pressure and temperature. 27 
• Corrosivity of the fluid streams. 28 
• Fouling. 29 
Only after taking into account these aspects, the engineer should give a chance to a possible 30 
design configuration. Among those feasible designs, there will be one option that is more cost-31 
effective for performing the service. 32 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
26 1 An Overview of the Equipment 
1.1.3 Where to Use a Double-Pipe Exchanger? 1 
The answer for this question is not straightforward as it might appear. Generally, the design 2 
decision for using a double-pipe heat exchanger might be driven by one of the following approaches: 3 
1. Rigorous optimization. 4 
2. Rules of thumb (guidelines based on previous experiences). 5 
1.1.3.1 A Few Guidelines for Fast Selection of a Double-Pipe Exchanger 6 
Double-pipe exchangers are especially appropriate for small to moderate duties, typically 10 –7 
 20 m2 (about 100 – 200 ft2) [16], in extreme conditions. The services in which double-pipe 8 
exchangers outperforms other heat exchanger types are those requiring severe operating conditions, 9 
such as high temperatures and pressures above 3 MPa (400 psi) in the tube side [17]. Although double-10 
pipe heat exchangers are not indicated for very large duties, because the number of necessary hairpins 11 
may grow unmanageably, they are highly versatile, being able to work from high vacuum to fairly 12 
high pressures, typically not greater than 30 MPa (4,350 psi) inside the annulus and 140 MPa 13 
(20,000 psi) on the tube side [18]. 14 
The structural simplicity enables its use in highly corrosive and erosive services. Because of 15 
the robust design with a relatively small number of parts, hairpin exchangers are well suited for greater 16 
corrosion and erosion allowances, once the inner pipe may be specified with increased thickness in 17 
several millimeters, that could make other heat exchangers types impracticable to build. The ASME 18 
B31.3 code provides a formula for calculating the pipe wall thickness as the sum of the pressure 19 
design thickness with an added thickness provisioned for compensation of mass loss due to corrosion 20 
or erosion [19]. 21 
The existence of royalty-free, open and dependable design/analysis methods counts in favor of 22 
double-pipe exchangers. The availability of proven and reliable design methods is an important factor 23 
for selecting a given type of heat exchanger. Accordingly, the double-pipe heat exchanger is among 24 
the types with very strong publicly available design procedures, which can be used by anyone without 25 
incurring in royalties. Of course, there are satisfactory design methods for most heat exchanger types 26 
performing eventually any service, however the best methods for narrowly used models are often 27 
proprietary to manufacturers or research organizations [20]. 28 
1.1.3.2 Possible Applications 29 
In the absence of a more rigorous selection process, such as using optimization methods, a 30 
hairpin double-pipe or multi-tube heat exchanger should be more economical by the following 31 
guidelines (Refs. [21], [22]): 32 
• Heating or cooling of a process stream, when a relatively small heat duty is involved 33 
(normally not greater than 50 m2) [20]. 34 
• The process results in a temperature cross. 35 
• High-pressure tube side application. 36 
• High terminal temperature differences (300°F/149°C or greater). 37 
• A low-allowable pressure drop is required on one side. 38 
• When the exchanger is subject to thermal shock. 39 
• When flow-induced vibration may be a problem. 40 
• When solids are present in the process stream. 41 
• Cyclic service. 42 
• High flow rate ratios between shell side and tube side fluids. 43 
• When an augmentation device will enhance the heat transfer coefficient. 44 
• When heating or cooling vapors. 45 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
1 An Overview of the Equipment 27 
• When complete vaporization is required. 1 
• When the mechanical advantages of a hairpin are preferred. 2 
• Suitable for partial boiling or condensations. 3 
1.1.3.3 Advantages and disadvantages 4 
Some advantages favoring the use of double-pipe heat exchangers may be listed as: 5 
• Low capital cost. 6 
• Low installation cost. 7 
• Flexibility and versatility. 8 
• Simple piping arrangements[20]. 9 
• Ease of cleaning on both sides, provided that the end fittings are removable [20]. 10 
• The heat transfer area can be increased with relative simplicity in existing equipment 11 
[20]. 12 
• Adaptable flow distribution controlled by insertion of pumps between banks in series, if 13 
necessary [20]. 14 
• Very cost effective for services requiring 10 – 20 m2 (about 100 to 200 ft2) of heat 15 
transfer surface. 16 
• Do not have temperature crosses between streams, since they are inherently true 17 
countercurrent devices. 18 
• For shell-and-tube units with shell-side heat transfer coefficient lower than half the tube 19 
side, the use of double-pipes may equate the heat transfer coefficients, avoiding an 20 
undesirable controlling fluid design [5]. 21 
• Straightforward and very reliable design and rating methods are widely available. 22 
Another heat exchanger type might be considered if the following disadvantages are important 23 
issues in a given service: 24 
• A large heat duty may require an unacceptable number of double-pipe exchangers. 25 
• A bank with many exchangers is more susceptible to leakages, due to the large number 26 
of connections. 27 
• Not practical for services involving low heat transfer coefficients. 28 
• The plant space occupied by the installation may be considerable, what is especially 29 
undesirable in offshore applications. 30 
• The time required for disassembling and cleaning a large exchanger bank may be 31 
unacceptable for certain plant operations. 32 
1.1.4 About The Optimal Design 33 
When specifying a heat exchanger, the designer may achieve a satisfactory configuration 34 
through the recursive application of the design method, guided by the use of rules of thumb and 35 
recommended best practices. Such situation is illustrated in Figure 6, in which case each possible 36 
configuration is selected and tested manually by the designer. Though this manual approach can lead 37 
to a reasonable design, in practice, it has limited range in the search for an optimal solution, which is 38 
more feasible using an automated optimization method. 39 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
28 1 An Overview of the Equipment 
 1 
Figure 6: Procedure for designing a heat exchanger without 2 
optimization. The process engineer is responsible for every decision in 3 
the design routine. 4 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
1 An Overview of the Equipment 29 
 1 
Figure 7: Procedure for designing a heat exchanger with the external 2 
optimization loop in place. The process engineer is responsible for 3 
setting and fine-tuning of optimizer parameters. 4 
Although not being the first choice for “old school” process engineers, if you are given enough 5 
time and resources (knowledge and tools), optimization is the way to go. The precise assessment for 6 
using or not a particular type of heat exchanger should be based on a comparative analysis among all 7 
possible types, considering an “overall” quantitative criterion, for example: 8 
• Capital cost. 9 
• Total cost (capital and operational cost). 10 
• Available ground area for installation. 11 
• Spatial constraints like height and width of the equipment. 12 
• Total weight of the equipment, which impact on transportation and cargo costs, along 13 
with support structure costs. 14 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
30 1 An Overview of the Equipment 
• Reduction of fluid holdup inside the exchanger, because the process fluid may be 1 
hazardous for people or for the environment, and its retained volume should be kept 2 
small. 3 
To achieve the finest design founded on a quantitative decision, the use of optimization 4 
algorithms is required. Essentially, what an optimization method does is an “intelligent” 5 
search−and−evaluation of alternatives in a set of possible designs. Considering the possibilities of 6 
equipment types, geometric features for each type and operating conditions, the number of design 7 
options capable of accomplishing a given heat transfer service may be enormous, and not manageable 8 
through engineer’s direct insight or judgment. Of course, previous experience, some proven heuristics 9 
and rules of thumb should lead to a feasible design of the heat exchanger; however, no guarantees are 10 
warranted that it will be the most effective for the proposed service. 11 
In summary, within an automatic optimization scheme, the quantitative criterion is expressed 12 
as a mathematical relationship called objective function, which is evaluated for each attempted design 13 
configuration. The optimization method is responsible for the rational selection of successive 14 
configurations that should converge to an optimum heat exchanger design. As represented 15 
schematically in Figure 7, the optimization method assumes the role played by the engineer engaged in 16 
a “handmade” design. The subject of optimization techniques is vast, and out of the scope of this text. 17 
If you are interested or feel the need for going deeper in this field, there are very good sources of 18 
information in the literature, and some of them are Refs. [23]–[27]. 19 
1.2 The Real Double-Pipe Exchanger is a Hairpin! 20 
The most used type of double-pipe exchangers is not simply a pipe inside another. Actually, a 21 
“U-shape” device is mounted with two straight legs connected by a 180° return bend. Each leg is a set 22 
of concentric pipes (Figure 8). Due to its characteristic shape, this kind of double-pipe exchanger is 23 
normally entitled “Hairpin”. 24 
Large batteries of bare tubes double-pipes typically are needed in order to meet heat transfer 25 
areas exceeding 100 m2 (about 1000 ft2). It is possible to attenuate this limitation by using finned 26 
tubes. In fact, there is the commercial availability of multi-tube hairpins with area as large as 1100 m2 27 
(about 12,000 ft2) [28]. 28 
 29 
Figure 8: Hairpin (double-pipe) heat exchanger. [Adapted from Ref. 30 
[29]] 31 
The hairpin double-pipe exchanger has some practical advantages. Frequently, a single 32 
exchanger is not sufficient to perform a given heat duty. Therefore, a number of hairpins are usually 33 
associated, forming a bank or battery of exchangers, in which the U-shape is convenient, because the 34 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
1 An Overview of the Equipment 31 
outlet nozzles of one exchanger are spatially close to the inlet nozzles of the next exchanger in the 1 
bank. 2 
The nozzles proximity implies that minimum additional pipeline is necessary to connect the 3 
units. Also, there is the marginal consequence that not every hairpin in the bank needs to be of the 4 
same length, which may be a key advantage for maintenance purposes, since individual exchangers in 5 
the bank may be temporarily replaced with units of different lengths, possibly available in warehouse. 6 
In case the exchanger bank is critical equipment in the chemical process, a procedure such as that may 7 
reduce significantly the downtime of the plant, meaning sometimes a fair amount of saved revenue 8 
[30]. 9 
The structure with few connections accommodates differential thermal expansion that creates 10 
mechanical tension between the internal and external pipes. The double-pipe exchanger maintenance 11 
usually involves less time and reduced costs, because in comparison to other models, it can be easily 12 
dismounted for inspection and cleaning. If a process is discontinued, the flexibility of the double-pipe 13 
exchanger favors its use in another process of the same plant. Hairpins may be readily encountered 14 
commercially with external pipe diameter varying from 50 to 200 mm, while the internal pipe is 15 
normally in the range of 20 to 150 mm nominal diameter ([6], [28]). 16 
 17 
Figure 9: Conjugated straight multi-tube heat exchanger associated in 18 
series. [Adaptedfrom Ref. [31]] 19 
Nowadays, some manufacturers offer a bank of double-pipe exchangers composed of 20 
conjugated straight concentric pipes. The bank shown in Figure 9 is not strictly an association of 21 
hairpins, being just a sequence of several straight double-pipes connected in series by welded 22 
connections. The particular feature that distinguishes this exchanger from a hairpin is the installation 23 
of the pipes and annuli ports at opposite ends of any couple of consecutive pipes. In contrast, Figure 24 
10 shows a battery of hairpin heat exchangers from another manufacturer, with the particularity that 25 
the connections between the annuli of neighboring hairpins are welded, not flanged like in Figure 8. 26 
Roughly speaking, the capital cost of a designed double-pipe exchanger may range from 80% to 140% 27 
of the cost of a shell-and-tube fixed tube sheet unit with the same capacity [17]. 28 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
32 1 An Overview of the Equipment 
 1 
Figure 10: Bank of multi-tube hairpin heat exchangers. [Adapted from 2 
Ref. [32]] 3 
1.3 Pipes and Tubes 4 
1.3.1 Is There a Difference? 5 
Currently, the difference between pipes and tubes is just a matter of classification and the way 6 
in which they are specified and ordered from the supplier. In plain terms, a tube is a simply a hollow 7 
cylinder. Decades ago, “pipes” and “tubes” were not made by the same process. “Tubes” had more 8 
careful finish and, accordingly, their surfaces were smoother. At that time, “pipes” were less finished 9 
and their internal surface had a greater relative roughness in comparison with tubes. 10 
Any hollow cylinder produced by modern techniques may be regarded as a tube [33]. 11 
However, for classification purposes, both terms “tubes” and “pipes” are still used in industry. 12 
“Tubes” are used for mechanical and structural applications (for example, welded together in a tubular 13 
structure) and typically specified by an outside diameter given in inches or millimeters. The wall 14 
thickness of tubes is determined by standard “scales” such as BWG, SWG, among others. A 15 
representation of the BWG scale is shown in Table 2 of Appendix A.2 , and a full dataset for 16 
specification of heat transfer tubes is found in Appendix A.2, Table 3. 17 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
1 An Overview of the Equipment 33 
 1 
Figure 11: Bank of AISI 304 stainless steel double-pipe hairpin heat 2 
exchangers. [Adapted from Ref. [34]] 3 
 “Pipes” are used for fluid transportation, usually in the form of pipelines. The key dimension 4 
of pipes is the inner diameter, since it defines the cross flow area. Accordingly, pipe sizes are 5 
designated by a “Nominal Pipe Size” (NPS) or “Diameter Nominal” (DN), normally measured in 6 
inches and millimeters, respectively. Nominal Pipe Size (NPS) and Diameter Nominal (DN) tables can 7 
be found in Appendix A.1 . For nominal sizes below 12 in, the designated size is approximately equal 8 
to the internal pipe diameter. For sizes from 14 in to 36 in, the nominal size designates exactly the 9 
outer diameter. Distinctly from “tubes”, whose thickness comes directly from the BWG scale, the pipe 10 
thickness is defined by “schedules” (e.g. 40, 60, 80, 120, etc.), which for certain values are equivalent 11 
to mass classes such as STD (standard weight), XS (extra strong), XXS (double extra strong), etc. 12 
1.3.2 Practical Implications of Pipe and Tube Distinction 13 
Beside the fact that “tubes” and “pipes” are essentially the same thing today, the exchanger 14 
designer should be aware of the former distinction when using equations for estimating heat transfer 15 
coefficients and friction factors. Some traditional and reliable correlations are based on experimental 16 
data measured from rough pipes and smooth tubes, separately. Good examples of this situation are the 17 
Sieder-Tate equations. As a result, the equations derived from rough pipe data give overestimated heat 18 
transfer coefficients and friction factors when applied to modern tubes. The engineer must be aware of 19 
these errors, because they penalize the heat transfer area, increasing the risk of a failed heat exchanger 20 
design. However, in ordinary practical cases, such deviations are balanced by the conservative 21 
approximations and overdesign performed on the stages of the heat exchanger sizing procedure. 22 
1.3.3 Commercial Diameters and Lengths 23 
Heat transfer tubes 24 
Tubes specially manufactured for heat transfer services can be acquired in a variety of metals 25 
and alloys from major suppliers in several standardized sizes, e.g. from 16 mm to 51 mm according to 26 
DIN 28180, and ranging from ½ in to 3 in by the ASTM A179 standard [33]. Usually, it is possible to 27 
arrange the specification in any size standard upon previous agreement with the supplier. The offered 28 
lengths of heat transfer tubes vary according to the manufacturer, however the buyer is normally 29 
allowed to request the tubes cut at any length smaller than the full length, defined by the 30 
manufacturing process. Heat transfer tubes in about 18 m is easily found in the market [35]. 31 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
34 1 An Overview of the Equipment 
Ordinary pipes 1 
The inner and outer tubes of double pipe heat exchangers, like the outer tube (shell) of shell 2 
and tube heat exchangers are commonly built from general purpose “regular” pipes, specified under 3 
NPS or DN standards, however ordering in accordance with other size standard is possible upon 4 
agreement with the manufacturer. Typical commercial lengths are pipe continuous sections of 5 to 5 
10 m (16 to 32 ft) [36]. 6 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 35 
2 Heat Exchanger Design Equations 1 
Designing a heat exchanger means essentially a two-step procedure: (1) to select a type or 2 
model of the equipment and (2) determining its size. The selection of the heat exchanger type is 3 
usually a qualitative step, where the process engineer should consider several elements such as safety, 4 
environment, operating conditions, corrosion, fouling, risks, etc. For the specific case of a double-pipe 5 
heat exchanger selection, this issue is discussed in more detail in Section 1.1.3. The second step, 6 
namely the sizing of the exchanger heat transfer surface, is strictly quantitative, and achieved via any 7 
design method, which should be appropriate to the model type chosen in step one. 8 
The most used design methods for sizing a heat exchanger are based on a very simple 9 
integrated equation. The foundations for this method were put in place by a few researchers, pioneered 10 
by Nagle ([37], [38]), Colburn [39] and Bowman ([37], [40]). The referred design equation may be 11 
written as: 12 
𝑞 = 𝑈𝐴𝛥𝑇𝑚 (2.1) 
Where: 13 
• q is the heat transfer rate (W or BTU/h), i.e. the heat load that the equipment must match. 14 
• U is the overall heat transfer coefficient (W/m2 °C or BTU/h ft2 °F), which is a combination of 15 
all the thermal resistances between the energy source (the hot fluid) and the energy target (the 16 
cold fluid). Typically, there are three thermal resistances inside a heat exchanger, being two of 17 
them convective resistances and the third is the conductive thermal resistance of the wall 18 
separating the fluid streams. 19 
• ΔTm is a kind of mean temperature difference taken over all the local temperature differences 20 
between the two fluids exchanging energy (°C or °F). 21 
• A is the reference heat transfer area of the heat exchanger (m2 or ft2). 22 
Given that the calculation of the overall heat transfer coefficient (U) and mean temperature 23 
difference (ΔTm) is performed appropriately, Eq. (2.1) may be used for the design of several types of 24 
heat exchangers. In fact, these lasttwo parameters are the only information that restricts the Eq. (2.1) 25 
for a specific type of heat exchanger. 26 
2.1 The Overall Heat Transfer Coefficient 27 
The overall heat transfer coefficient is a very useful concept because, in a practical heat 28 
exchanger, the heat transfer process involves multiple thermal resistances. Although, in some 29 
situations, one or more resistances may be disregarded with respect to another, generally there are five 30 
thermal resistances to deal with, namely: 31 
1. Convective resistance on the hot fluid side. 32 
2. Fouling resistance on the hot fluid side. 33 
3. Conductive resistance on the separation wall. 34 
4. Fouling resistance on the cold fluid side 35 
5. Convective resistance on the cold fluid side. 36 
In a double-pipe heat exchanger, the geometry is tubular, i.e. cylindrical; therefore the 37 
separation wall is actually the wall of the internal pipe. This can be seen in Figure 12, which shows a 38 
zoomed cut of a counterflow double pipe heat exchanger. The thermal resistances are identified as 39 
forming a thermal circuit of five resistances in series. 40 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
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36 2 Heat Exchanger Design Equations 
Flow regimes in industrial process situations are commonly turbulent. Accordingly, due to the 1 
mixing effect of turbulence, the temperature gradient in the turbulent core regions of the flow field is 2 
greatly reduced. Consequently, most of the temperature variation, starting from the wall and going into 3 
the hot or cold fluids, is located in a quite thin region referred as the “thermal boundary layer”. This is 4 
the place where almost all convective thermal resistance resides. In fact, the very definition of a 5 
thermal boundary layer states that it ends when 99% of the temperature variation has undergone ([41]–6 
[43]), departing from the solid wall. 7 
The concept of overall heat transfer coefficient comes from the definition of the equivalent 8 
thermal resistance for a given heat exchange circuit. Taking advantage from a direct analogy with the 9 
equivalent resistance in electrical circuits taught in fundamental physics courses, we could rewrite the 10 
design equation (Eq. (2.1)) as: 11 
𝑞 =
𝛥𝑇𝑚
1/𝑈𝐴
=
𝛥𝑇𝑚
𝑅𝑒𝑞
 (2.2) 
At this point, you should be a little confused… For a defined heat load q in Eq. (2.2), the 12 
numeric value of the equivalent resistance must be unique; consequently, the value of (1/UA) must be 13 
unique also, but, for a cylindrical wall, i.e. a pipe, there two heat transfer surfaces crossed by the radial 14 
heat flux. Therefore, to obtain the same heat load q, the conclusion is that the overall heat transfer 15 
coefficient is necessarily tied to a given reference heat transfer area A; if the reference area changes, 16 
the overall heat transfer coefficient changes to compensate. The inside surface Ai and the outside 17 
surface Ao of the internal tube in a heat exchanger are possible choices for the reference heat transfer 18 
area, then the equivalent thermal resistance may be expressed as: 19 
𝑅𝑒𝑞 =
1
𝑈𝐴
=
1
𝑈𝑖𝐴𝑖
=
1
𝑈𝑜𝐴𝑜
 (2.3) 
Where: 20 
U: Mean overall heat transfer coefficient (W/m2 K or BTU/h ft2 °F). 21 
Ui, Uo: Overall heat transfer coefficients referenced to the inner and outer surface of the 22 
internal pipe, respectively (W/m2 K or BTU/h ft2 °F). 23 
A: Area of heat transfer surface (m2 or ft2). 24 
Req: Mean equivalent thermal resistance for the whole heat transfer surface A. 25 
Ai and Ao are the internal and external heat transfer surfaces of the inner tube with effective 26 
length L, respectively: 27 
𝐴𝑖 = 𝜋𝐷𝑖𝐿 (2.4) 
𝐴𝑜 = 𝜋𝐷𝑜𝐿 (2.5) 
Thermal resistances are, in general terms, local quantities, since they depend on physical 28 
properties that have different values from point to point of the heat transfer surface. Therefore, we 29 
should notice that the equivalent thermal resistance in Eq. (2.3) is a mean value for the total area of 30 
heat transfer. This is a direct consequence from the fact that Eq. (2.2) correlates extensive physical 31 
quantities, which change with the size of the equipment, such as A, ΔTm and U. 32 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 37 
Considering the thermal circuit shown in Figure 12, there are five resistances associated in 1 
series. The equivalent resistance for this circuit may be expressed as: 2 
𝑅𝑒𝑞 = 𝑅𝑖 + 𝑅𝑓𝑖 + 𝑅𝑘 + 𝑅𝑓𝑜 + 𝑅𝑜 (2.6) 
Where: 3 
Ri: Convective resistance of the fluid flowing inside the inner pipe. 4 
Rk: Conductive resistance of the wall between fluids. 5 
Ro: Convective resistance of the fluid in the annular channel. 6 
Rfi: Internal fouling resistance (°C/W or h °F/ BTU). 7 
Rfo: External fouling resistance (°C/W or h °F/ BTU). 8 
 9 
 Figure 12: Thermal resistances involved in the heat transfer between 10 
a hot and a cold stream inside a heat exchanger. 11 
Conventionally, fouling thermal resistances Rfi and Rfo are expressed using required values of 12 
“fouling coefficients” or “fouling factors” Rdi and Rdo, which do not depend on the size of the heat 13 
transfer surface and correspond conceptually to the inverse of convective heat transfer coefficients. 14 
Therefore, the fouling resistances are given by: 15 
𝑅𝑓𝑖 =
𝑅𝑑𝑖
𝐴𝑖
 (2.7) 
𝑅𝑓𝑜 =
𝑅𝑑𝑜
𝐴𝑜
 (2.8) 
The wall conductive resistance for a cylindrical duct with thermal conductivity k, inner 16 
diameter Di and outer diameter Do can be expressed as: 17 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
38 2 Heat Exchanger Design Equations 
𝑅𝑘 =
𝐷𝑖 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
2𝐴𝑖𝑘
 (2.9) 
With the heat transfer coefficients hi and ho for the tube and annulus fluids, respectively, the 1 
convective thermal resistances are: 2 
𝑅𝑖 =
1
ℎ𝑖𝐴𝑖
 (2.10) 
𝑅𝑜 =
1
ℎ𝑜𝐴𝑜
 (2.11) 
With Eq. (2.3) and substituting Eqs (2.7) − (2.11) into (2.6), the equivalent thermal resistance 3 
becomes: 4 
𝑅𝑒𝑞 =
1
𝑈𝑖𝐴𝑖
=
1
𝑈𝑜𝐴𝑜
=
𝑅𝑑𝑖
𝐴𝑖
+
1
ℎ𝑖𝐴𝑖
+
𝐷𝑖 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
2𝐴𝑖𝑘
+
1
ℎ𝑜𝐴𝑜
+
𝑅𝑑𝑜
𝐴𝑜
 (2.12) 
A widely accepted (although arbitrary) convention in the technical literature and industry is to 5 
define the tube external surface as the reference for the design and analysis of tubular heat exchangers. 6 
Following such convention, the design equation Eq. (2.1) is rewritten as: 7 
𝑞 = 𝑈𝑜𝐴𝑜𝛥𝑇𝑚 (2.13) 
And the thermal conductance based on the outside area is UoAo = 1/Req, where: 8 
𝑈𝑜𝐴𝑜 = �
𝑅𝑑𝑖
𝐴𝑖
+
1
ℎ𝑖𝐴𝑖
+
𝐷𝑖 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
2𝐴𝑖𝑘
+
1
ℎ𝑜𝐴𝑜
+
𝑅𝑑𝑜
𝐴𝑜
�
−1
 (2.14) 
Substituting Eqs. (2.4) and (2.5) into Eq. (2.14), solving for Uo and canceling terms, we have 9 
the overall heat transfer coefficient referenced to the outer heat transfer area as: 10 
𝑈𝑜 = �
1
ℎ𝑖
�
𝐷𝑜
𝐷𝑖
� +
𝐷𝑜
2𝑘
𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� +
1
ℎ𝑜
+ 𝑅𝑑𝑖 �
𝐷𝑜
𝐷𝑖
� + 𝑅𝑑𝑜�
−1
 (2.15) 
Keep in mind that, hereafter, the overall heat transfer coefficient refers to the outer heat 11 
transfer surface (the outer surface convention), and then let us drop the subscript “o” from Eqs. (2.13) 12 
and (2.15) to obtain the final equations: 13 
𝑞 = 𝑈𝐴𝛥𝑇𝑚 (2.16) 
𝑈 = �
1
ℎ𝑖
�
𝐷𝑜
𝐷𝑖
� +
𝐷𝑜
2𝑘
𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� +
1
ℎ𝑜
+ 𝑅𝑑𝑖 �
𝐷𝑜
𝐷𝑖
� + 𝑅𝑑𝑜�
−1
 (2.17) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 39 
Notice that, although each term in the right-hand side of Eq. (2.17) is related to a type of 1 
thermal resistance, they are not thermal resistances at all, at least by its strict definition. Thermal 2 
resistances have dimensions of K/W or °F h/BTU, as we may see from Eq. (2.3); therefore it is not the 3 
case here, since all summed terms are in m2K/Wor h ft2 °F/BTU. 4 
2.1.1 “Design” and “Clean” Overall Heat Transfer Coefficients 5 
Heat exchangers are generally sized with the provision of some additional transfer area to 6 
compensate the performance loss due to fouling along their operation cycle. The fouling coefficients 7 
Rdi and Rdo are responsible for this additional heat transfer area; hence their values must be judiciously 8 
specified. 9 
The actual U value used in the evaluation of the heat transfer area by Eq. (2.16) is called the 10 
“design” overall heat transfer coefficient (Ud), which considers the equipment fouled condition in the 11 
form: 12 
𝑈𝑑 = �
1
ℎ𝑖
�
𝐷𝑜
𝐷𝑖
� +
𝐷𝑜
2𝑘
𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� +
1
ℎ𝑜
+ 𝑅𝑑𝑖 �
𝐷𝑜
𝐷𝑖
� + 𝑅𝑑𝑜�
−1
 (2.18) 
The “clean” overall heat transfer coefficient (Uc) ignores the fouling effect, being obtained 13 
simply by setting Rdi = Rdo = 0 in Eq. (2.18): 14 
𝑈𝑐 = �
1
ℎ𝑖
�
𝐷𝑜
𝐷𝑖
� +
𝐷𝑜
2𝑘
𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� +
1
ℎ𝑜
�
−1
 (2.19) 
With Uc from Eq. (2.19), Ud can also be evaluated more conveniently as: 15 
𝑈𝑑 = �
1
𝑈𝑐
+ 𝑅𝑑𝑖 �
𝐷𝑜
𝐷𝑖
� + 𝑅𝑑𝑜�
−1
 (2.20) 
2.1.2 Relatively Small Wall Thermal Resistance 16 
Eq. (2.19) is a general form for Uc, and it can be simplified in many ways. According to the 17 
physical conditions appropriate for a given heat exchanger and chemical process, some reduced 18 
versions of the equivalent thermal resistance may be used. For example, in such a case where the pipe 19 
material is highly conducive − which is usually the case, because metals are the most common 20 
material for constructing heat exchangers − the thermal resistance of the wall is small in comparison to 21 
the sum of the convective resistances, i.e.: 22 
𝐷𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
2𝑘
<<
1
ℎ𝑖
�
𝐷𝑜
𝐷𝑖
� +
1
ℎ𝑜
 (2.21) 
This approximation is widely used by Kern [44] in his classic book, in which almost all designs 23 
disregard the wall thermal resistance, and the resulting overall heat transfer coefficient becomes: 24 
𝑈𝑐 =
ℎ𝑖𝑜ℎ𝑜
ℎ𝑖𝑜+ℎ𝑜
 (2.22) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
40 2 Heat Exchanger Design Equations 
Where hio is the internal (tube) convective heat transfer coefficient corrected for the external 1 
(reference) heat transfer surface: 2 
ℎ𝑖𝑜 = ℎ𝑖 �
𝐷𝑖
𝐷𝑜
� (2.23) 
Under this assumption, the design overall coefficient can be calculated from: 3 
𝑈𝑑 = �
1
𝑈𝑐
+ 𝑅𝑑𝑖 �
𝐷𝑜
𝐷𝑖
� + 𝑅𝑑𝑜�
−1
 (2.24) 
2.1.3 Controlling Fluid 4 
Sometimes one of the fluids rules the heat transfer rate inside the heat exchanger. When there 5 
is one fluid in the heat exchanger for which the convective heat transfer coefficient is remarkably 6 
small in comparison with the other fluid, the former is usually said to be the “controlling fluid” or 7 
“controlling stream” of the heat transfer operation. The reason is that a change in the convective heat 8 
transfer coefficient of the controlling fluid has a direct impact in the overall heat transfer coefficient of 9 
the whole exchanger, as we can see from Eq. (2.22) by taking for example hio >> ho (the annulus fluid 10 
is assumed controlling), giving: 11 
𝑈 ≈ ℎ𝑜 (2.25) 
2.2 Basic Heat Exchanger Differential Equations 12 
When two fluids, with distinct temperatures, exchange energy on a heat exchanger, their 13 
temperatures are modified along the fluid path and a spatial temperature distribution develops, as the 14 
heat moves from the hot to the cold fluid. A few physical laws, namely the first and second laws of 15 
Thermodynamics associated with a heat transfer rate relation, govern this heat exchange. 16 
Considering a differential control volume, with the form of a cross section of infinitesimal 17 
thickness, in each stream of a double-pipe heat exchanger, we may apply an energy balance for the hot 18 
and cold fluids in order to obtain the differential equations governing the heat transfer process. A 19 
relatively simple, yet representative, mathematical description of this heat exchanger can be obtained 20 
by imposing the following assumptions: 21 
1. One-dimensional axial flow path1. 22 
2. Steady-state operation. 23 
3. Only the heat transfer between the fluids changes their enthalpy, i.e. there is no heat loss to 24 
the surroundings and kinetic and potential energies are neglected [45]. 25 
4. Constant specific heats and flow rates. 26 
5. Heat diffusion in the direction of the flow is ignored [45]. 27 
With previously stated simplifications, a one-dimensional representation of a double-pipe heat 28 
exchanger is defined using the “accumulative” heat transfer area (A), starting from the left-end and 29 
increasing in the direction of the right-end of the heat exchanger (Figure 13). Since the inflow and 30 
outflow of energy accounted for in the energy balance is dependent on the fluid flow direction for each 31 
1 Some authors (e.g. Ref. [123]) add an explicit assumption for the uniform thermal history of the fluid particles in 
each stream. Herein, such condition is automatically achieved by the combined assumptions of steady-state and one-
dimensional axial flow path. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
2 Heat Exchanger Design Equations 41 
stream, we need to deal separately with the exchanger operating in parallel flow or counterflow 1 
arrangements. 2 
Hot 
fluid
Cold 
fluid
T(A)
t(A)
WC
wc
U
T1
T2
t1
t2
A
1 2
A A+dA
dA
AWCT dAAWCT +
Awct dAAwct +
dq
 3 
Figure 13: One-dimensional representation of a double-pipe parallel 4 
flow heat exchanger. 5 
2.2.1 Parallel Flow 6 
The differential equations for the parallel flow heat exchanger are more straightforward to 7 
develop, since both fluids flow in the same positive direction of the accumulated heat transfer surface 8 
(A). This helps to avoid possible pitfalls involving the signs of the convective terms of the counterflow 9 
differential equations, as we will see in Section 2.2.2. 10 
2.2.1.1 Energy Balance for Hot Fluid 11 
From the energy balance on the heat transfer area element identified in Figure 13, evaluating 12 
WCT and wct at positions A and A+dA with a Taylor series expansion truncated at the first order 13 
term, we have: 14 
𝑊𝐶𝑇���
𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 
𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 
𝐼𝑁𝑇𝑂 𝑡ℎ𝑒 
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑎𝑡 𝑓𝑎𝑐𝑒 𝐴 
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
− �𝑊𝐶𝑇 +
𝑑(𝑊𝐶𝑇)
𝑑𝐴
𝑑𝐴�
���������������
𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 
𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦
𝑂𝑈𝑇 𝑜𝑓 𝑡ℎ𝑒 
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑎𝑡 𝑓𝑎𝑐𝑒 𝐴+𝑑𝐴
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
− 𝑈𝑑𝐴(𝑇 − 𝑡)���������
ℎ𝑒𝑎𝑡 𝑅𝐸𝑀𝑂𝑉𝐸𝐷 
𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 
ℎ𝑜𝑡 𝑓𝑙𝑢𝑖𝑑 
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
= 0 
(2.26) 
Where: 15 
W: Mass flow rate of the hot fluid (kg/s or lb/h). 16 
C: Specific heat of the hot fluid (J/kg K or BTU/lb °F). 17 
T: Temperature of the hot fluid (°C or °F). 18 
t: Temperature of the cold fluid (°C or °F). 19 
Notice that the term of heat removal must be carefully defined with the proper sign, which is 20 
forced by the second law of thermodynamics. Since the energy is subtracted from the hot fluid, it is 21 
preceded by the negative sign, therefore the positivity of the expression dq = U dA (T − t) must be 22 
assured by taking the temperature difference from the hot to the cold fluid. 23 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
42 2 Heat Exchanger Design Equations 
Considering that WC does not change with accumulated area A along the exchanger, canceling 1 
terms and rearranging, we may take out from the derivative to get the differential equation for T(A) as: 2 
𝑑𝑇
𝑑𝐴
+
𝑈
𝑊𝐶
(𝑇 − 𝑡) = 0 (2.27) 
2.2.1.2 Energy Balance for Cold Fluid 3 
As for the hot fluid, doing the same energy balance, to the cold fluid, we get: 4 
𝑤𝑐𝑡�
𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 
𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 
𝐼𝑁𝑇𝑂 𝑡ℎ𝑒 
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑎𝑡 𝑓𝑎𝑐𝑒 𝐴 
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
− �𝑤𝑐𝑡 +
𝑑(𝑤𝑐𝑡)
𝑑𝐴
𝑑𝐴�
�������������
𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 𝑜𝑓 
𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦
𝑂𝑈𝑇 𝑜𝑓 𝑡ℎ𝑒 
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑎𝑡 𝑓𝑎𝑐𝑒 𝐴+𝑑𝐴
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
+ 𝑈𝑑𝐴(𝑇 − 𝑡)���������
ℎ𝑒𝑎𝑡 𝐴𝐷𝐷𝐸𝐷 
𝑡𝑜 𝑡ℎ𝑒 
𝑐𝑜𝑙𝑑 𝑓𝑙𝑢𝑖𝑑
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
= 0 
(2.28) 
Where: 5 
w: Mass flow rate of the cold fluid (kg/s or lb/h). 6 
c: Specific heat of the cold fluid (J/kg K or BTU/lb °F). 7 
t: Temperature of the cold fluid (°C or °F). 8 
The term for the heat added to the cold fluid control volume is defined accordingly to the 9 
second law of thermodynamics. It should be led by a positive sign, provided the expression 10 
dq = U dA (T − t) is guaranteed to be positive also, since T > t. 11 
Subsequently, under the consideration that specific heat and mass flow rate (wc) may be 12 
assumed constant, then: 13 
𝑑𝑡
𝑑𝐴
−
𝑈
𝑤𝑐
(𝑇 − 𝑡) = 0 (2.29) 
Eq. (2.27) and (2.29) can be solved in order to allow the prediction of the fluid temperature in 14 
each position of the heat transfer surface. 15 
2.2.2 Counterflow 16 
To develop the differential equations for a counterflow exchanger, let’s start with the schematic 17 
representation in Figure 14. The independent variable accumulated heat transfer area (A) has its axis 18 
oriented from left- to right-hand side as indicated. For the one-dimensional description done here, this 19 
simple “A-axis” defines a reference frame to which the developed differential equations are tied. 20 
In this counterflow arrangement, the flow of the hot fluid is oriented in the positive A-axis 21 
direction, but the cold fluid has the opposite A-axis direction. Such fluid orientations, with respect to 22 
which fluid has the same axis orientation, are in fact totally arbitrary! Provided the energy balances are 23 
performed correctly, i.e. in strict accordance with the chosen orientations, the final forms of the 24 
generated differential equations will be the same, for the defined reference frame. 25 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 43 
Hot 
fluid
Cold 
fluid
T(A)
t(A)
WC
wc
U
T1 T2
t2 t1
1 2
A A A+dA
dA
AWCT dAAWCT +
Awct dAAwct +
dq
 1 
Figure 14: One-dimensional representation of a double-pipe 2 
counterflow heat exchanger. 3 
2.2.2.1 Energy Balance for Hot Fluid 4 
With a truncated Taylor series expansion for the enthalpy function WCT, we can draw the 5 
following balance in Eq. (2.30). 6 
𝑊𝐶𝑇���
𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 
𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 
𝐼𝑁𝑇𝑂 𝑡ℎ𝑒 
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑎𝑡 𝑓𝑎𝑐𝑒 𝐴 
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
− �𝑊𝐶𝑇 +
𝑑(𝑊𝐶𝑇)
𝑑𝐴
𝑑𝐴�
���������������
𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 
𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦
𝑂𝑈𝑇 𝑜𝑓 𝑡ℎ𝑒 
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑎𝑡 𝑓𝑎𝑐𝑒 𝐴+𝑑𝐴
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
− 𝑈𝑑𝐴(𝑇 − 𝑡)���������
ℎ𝑒𝑎𝑡 𝑅𝐸𝑀𝑂𝑉𝐸𝐷 
𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 
ℎ𝑜𝑡 𝑓𝑙𝑢𝑖𝑑 
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
= 0 
(2.30) 
Assuming the hypothesis that WC does not change with A significantly, we may write: 7 
𝑑𝑇
𝑑𝐴
+
𝑈
𝑊𝐶
(𝑇 − 𝑡) = 0 (2.31) 
2.2.2.2 Energy Balance for Cold Fluid 8 
�𝑤𝑐𝑡 +
𝑑(𝑤𝑐𝑡)
𝑑𝐴
𝑑𝐴�
�������������
𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 
𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦
𝐼𝑁𝑇𝑂 𝑡ℎ𝑒 
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑎𝑡 𝑓𝑎𝑐𝑒 𝐴+𝑑𝐴
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
− 𝑤𝑐𝑡�
𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 
𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 
𝑂𝑈𝑇 𝑜𝑓 𝑡ℎ𝑒 
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑎𝑡 𝑓𝑎𝑐𝑒 𝐴 
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
+ 𝑈𝑑𝐴(𝑇 − 𝑡)���������
ℎ𝑒𝑎𝑡 𝐴𝐷𝐷𝐸𝐷 
𝑡𝑜 𝑡ℎ𝑒 
𝑐𝑜𝑙𝑑 𝑓𝑙𝑢𝑖𝑑
(𝑊 𝑜𝑟 𝐵𝑇𝑈/ℎ)
= 0 
(2.32) 
Taking specific heat and flow rate of the cold fluid as constants, the resulting differential 9 
equation is: 10 
𝑑𝑡
𝑑𝐴
+
𝑈
𝑤𝑐
(𝑇 − 𝑡) = 0 (2.33) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
44 2 Heat Exchanger Design Equations 
2.3 Log Mean Temperature Difference 1 
The design equation (Eq. (2.1)) requires the determination of a kind of mean temperature 2 
difference that is representative for the heat transfer service, specifically for the temperature 3 
distributions inside the heat exchanger. A complication arises from the fact that such mean 4 
temperature difference is a result of the temperature profiles of both fluids, and, in turn, these profiles 5 
depend upon the equipment design and performance. In addition, the geometry and fluid dynamics 6 
inside the heat exchanger have direct impact on the value of the overall heat transfer coefficient U, 7 
which determine the equipment size (i.e. its geometry). Therefore, the sizing process is inherently 8 
iterative. 9 
A detailed derivation of the logarithmic mean of the temperature difference (LMTD) in a 10 
double-pipe heat exchanger may be found in Kern’s text [44] and will not be repeated here. Instead, let 11 
us take another way to this important definition. Let us start from the descriptive equations for steady 12 
state double-pipe heat exchanger and undertake a mathematical procedure in order to develop the 13 
formula for the mean temperature difference corresponding to the area of heat transfer as a whole. 14 
2.3.1 Parallel Flow Double Pipe Heat Exchanger 15 
In Section 2.2.1, we obtained the differential equations for a double-pipe heat exchanger 16 
operating with parallel flow streams as Eqs. (2.27) and (2.29), which are rewritten here for 17 
convenience: 18 
𝑑𝑇
𝑑𝐴
+
𝑈
𝑊𝐶
(𝑇 − 𝑡) = 0 (2.34) 
𝑑𝑡
𝑑𝐴
−
𝑈
𝑤𝑐
(𝑇 − 𝑡) = 0 (2.35) 
A new differential equation may be obtained by subtracting Eq. (2.34) and (2.35), 19 
consequently: 20 
𝑑(𝑇 − 𝑡)
𝑑𝐴
+ �
𝑈
𝑊𝐶
+
𝑈
𝑤𝑐
� (𝑇 − 𝑡) = 0 (2.36) 
Let us define the temperature difference between both fluids as a new variable: 21 
𝛥𝑇 = 𝑇 − 𝑡 (2.37) 
And, substituting in Eq. (2.36), we get: 22 
𝑑𝛥𝑇
𝑑𝐴
+ �
𝑈
𝑊𝐶
+
𝑈
𝑤𝑐
�𝛥𝑇 = 0 (2.38) 
Eq. (2.38) may be readily integrated by separation of variables from the terminal (1) to the 23 
terminal (2) of the heat exchanger, and using the hypothesis of constant specific heat and flow rates, 24 
follows: 25 
�
𝑑𝛥𝑇
𝛥𝑇
=
𝛥𝑇2
𝛥𝑇1
− �
1
𝑊𝐶
+
1
𝑤𝑐
�� 𝑈𝑑𝐴
𝐴
0
 (2.39) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 45 
Notice that the assumptions of constant specific heat and flow rates for both fluids are already 1 
embedded in the descriptive differential equations (2.34) and (2.35). Thus, we are not placing an 2 
additional restriction in this derivation of the logarithmic mean temperature difference. 3 
One should notice that in the general case, the physical properties affecting the overall heat 4 
transfer coefficient U are dependent on the fluid temperature, therefore the remaining integral on the 5 
right-hand side of Eq. (2.39) may be rigorously evaluated only if the distribution of U(A) over the heat 6 
transfer area is known. At this point, let us use the definition of a mean overall heat transfer coefficient 7 
(Ū) for the whole heat transfer surface: 8 
Ū =
1
𝐴
� 𝑈𝑑𝐴
𝐴
0
 (2.40) 
From Eq. (2.40), the Eq. (2.39) may be rewritten as: 9 
𝑙𝑛 �
𝛥𝑇2
𝛥𝑇1
� = −�
1
𝑊𝐶
+
1
𝑤𝑐
�𝑈�𝐴 (2.41) 
Assuming that all energy lost by the hot fluid is gained by the cold fluid, an energy balance for 10 
both streams yields: 11 
𝑞 = 𝑊� 𝐶𝑑𝑇 =
𝑇1
𝑇2
𝑤� 𝑐𝑑𝑡
𝑡2
𝑡1
 (2.42) 
And, using mean values for the specific heats, one may write: 12 
𝑞 = 𝑊𝐶(𝑇1 − 𝑇2) = 𝑤𝑐(𝑡2 − 𝑡1) (2.43) 
Thus: 13 
1
𝑊𝐶
=
(𝑇1 − 𝑇2)
𝑞
 (2.44) 
1
𝑤𝑐
=
(𝑡2 − 𝑡1)
𝑞
 (2.45) 
𝑞 𝑙𝑛 �
𝛥𝑇2
𝛥𝑇1
� = −[(𝑇1 − 𝑡1) − (𝑇2 − 𝑡2)]𝑈�𝐴 (2.46) 
For a parallel flow heat exchanger, the temperature differences at the terminals are: 14 
𝛥𝑇1 = 𝑇1 − 𝑡1 (2.47) 
𝛥𝑇2 = 𝑇2 − 𝑡2 (2.48) 
Using Eqs. (2.47) and (2.48) in Eq. (2.46), and rearranging, results in: 15 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
46 2 Heat Exchanger Design Equations 
𝑞 = 𝑈�𝐴
(𝛥𝑇1 − 𝛥𝑇2)
𝑙𝑛 �𝛥𝑇1𝛥𝑇2
�
 (2.49) 
Comparing Eqs. (2.49) and (2.1), we may conclude that the mean temperature difference (ΔTm) 1 
takes the formbelow: 2 
𝛥𝑇𝑚 =
(𝛥𝑇1 − 𝛥𝑇2)
𝑙𝑛 �𝛥𝑇1𝛥𝑇2
�
 (2.50) 
This result is an important type of mean temperature difference designated as the Logarithmic 3 
Mean Temperature Difference (LMTD), hence let us restate its definition: 4 
𝐿𝑀𝑇𝐷 =
(𝛥𝑇1 − 𝛥𝑇2)
𝑙𝑛 �𝛥𝑇1𝛥𝑇2
�
 (2.51) 
2.3.2 Counterflow Double Pipe Heat Exchanger 5 
The descriptive equations for the stream temperature profiles of a counterflow double pipe heat 6 
exchanger were developed in Section 2.2.2, and are reprinted here for convenience: 7 
𝑑𝑇
𝑑𝐴
+
𝑈
𝑊𝐶
(𝑇 − 𝑡) = 0 (2.52) 
𝑑𝑡
𝑑𝐴
+
𝑈
𝑤𝑐
(𝑇 − 𝑡) = 0 (2.53) 
A new differential equation may be obtained by subtracting Eqs. (2.52) and (2.53), 8 
consequently: 9 
𝑑(𝑇 − 𝑡)
𝑑𝐴
+ �
𝑈
𝑊𝐶
−
𝑈
𝑤𝑐
� (𝑇 − 𝑡) = 0 (2.54) 
Using the local temperature difference (Eq. (2.37)) as new variable results in: 10 
𝑑𝛥𝑇
𝑑𝐴
+ �
𝑈
𝑊𝐶
−
𝑈
𝑤𝑐
�𝛥𝑇 = 0 (2.55) 
Eq. (2.55) may be readily integrated by separation of variables from the terminal (1) to the 11 
terminal (2) of the heat exchanger, and reusing the hypothesis of constant specific heats and flow rates, 12 
follows: 13 
�
𝑑𝛥𝑇
𝛥𝑇
=
𝛥𝑇2
𝛥𝑇1
− �
1
𝑊𝐶
−
1
𝑤𝑐
�� 𝑈𝑑𝐴
𝐴
0
 (2.56) 
With the definition of the average overall heat transfer coefficient (Ū), Eq. (2.40), we can 14 
rewrite Eq. (2.56) in the form: 15 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 47 
𝑙𝑛 �
𝛥𝑇2
𝛥𝑇1
� = −�
1
𝑊𝐶
−
1
𝑤𝑐
�𝑈�𝐴 (2.57) 
Considering that the hot fluid does not lose energy to the surroundings, and using mean values 1 
for the specific heats, the integral energy balance for both streams is stated as: 2 
𝑞 = 𝑊𝐶(𝑇1 − 𝑇2) = 𝑤𝑐(𝑡2 − 𝑡1) (2.58) 
Thus: 3 
1
𝑊𝐶
=
(𝑇1 − 𝑇2)
𝑞
 (2.59) 
1
𝑤𝑐
=
(𝑡2 − 𝑡1)
𝑞
 (2.60) 
Using Eqs. (2.59) and (2.60) in Eq. (2.57), and rearranging the temperature differences in the 4 
right-hand side, yields: 5 
𝑞 𝑙𝑛 �
𝛥𝑇2
𝛥𝑇1
� = −[(𝑇1 − 𝑡2) − (𝑇2 − 𝑡1)]𝑈�𝐴 (2.61) 
Noticing that, for a counterflow heat exchanger, the temperature differences at the terminals 6 
are: 7 
𝛥𝑇1 = 𝑇1 − 𝑡2 (2.62) 
𝛥𝑇2 = 𝑇2 − 𝑡1 (2.63) 
Therefore, with Eqs. (2.62) and (2.63), Eq. (2.61) may be rearranged as: 8 
𝑞 = 𝑈�𝐴
(𝛥𝑇1 − 𝛥𝑇2)
𝑙𝑛 �𝛥𝑇1𝛥𝑇2
�
 (2.64) 
The comparison of this result with Eq. (2.1), allow us to conclude that the expression for the 9 
mean temperature difference along the heat exchanger has the form: 10 
𝐿𝑀𝑇𝐷 =
(𝛥𝑇1 − 𝛥𝑇2)
𝑙𝑛 �𝛥𝑇1𝛥𝑇2
�
 (2.65) 
2.3.3 The Restrictions on the Overall Heat Transfer Coefficient 11 
The derivation of the logarithmic mean temperature difference does not require a constant 12 
overall heat transfer coefficient. It is a common misconception that the use of the LMTD for heat 13 
exchanger design imposes the restriction of constant overall heat transfer coefficient (U) for the whole 14 
heat transfer surface, which is not true! As was shown in previous Sections 2.3.1 and 2.3.2, the 15 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
48 2 Heat Exchanger Design Equations 
derivation of the LMTD for parallel flow and counterflow arrangements could be performed directly 1 
from the descriptive differential equations of the one-dimensional heat exchanger, developed under the 2 
assumptions (1) to (5) of Section 2.2, which do not require a constant U over the area. Therefore, Eqs. 3 
(2.27) and (2.29) (for parallel flow) and Eqs. (2.31) and (2.33) (for counterflow) do not imply a 4 
constant U. Additionally, the use of an integrated definition of the averaged U, given by Eq. (2.40), 5 
may account for any functional variation of the overall heat transfer coefficient over the heat transfer 6 
surface U(A), as a result, a uniform heat transfer coefficient is just the special case of a constant 7 
function U(A) = Uconst. 8 
2.3.4 LMTD and Fluid Flow Arrangement 9 
The mean temperature for a heat exchanger is the driving force for the heat transfer, when 10 
evaluated using the integrated design equation. In general, the greater the mean temperature 11 
difference, the more efficient is the heat exchanger, and a smaller heat transfer surface will suffice to 12 
perform a specified service. The only exceptions are found when the chosen flow arrangement 13 
penalizes the mean temperature difference, but this effect is compensated by a greater increase in the 14 
overall heat transfer coefficient. 15 
In the case of a double pipe heat exchanger, the mean temperature difference, under certain 16 
assumptions, was proven to be the logarithmic mean temperature difference (LMTD). Surprisingly, 17 
the expressions for the LMTD have the same form, regardless of the flow orientation of the exchanger, 18 
be it in parallel flow or counterflow, as seen on Eqs. (2.51) and (2.65). Such verification may lead 19 
mistakenly to the conclusion that two heat exchangers operating in parallel flow and counterflow have 20 
the same thermal efficiency, which is not true! This equality is only apparent when the LTMD is 21 
written using the two temperature differences at the heat exchanger terminals ΔT1 and ΔT2, however, 22 
we should notice that the definition of the temperature differences at the exchanger terminals is tied to 23 
the flow arrangement, and in terms of the four process temperatures T1, T2, t1 and t2, the LMTD for 24 
both situations becomes distinct. 25 
With Eqs. (2.47) and (2.48) substituted into Eq. (2.51), we get the LMTD for a parallel heat 26 
exchanger as: 27 
𝐿𝑀𝑇𝐷(𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑓𝑙𝑜𝑤) =
(𝑇1 − 𝑡1) − (𝑇2 − 𝑡2)
𝑙𝑛 �𝑇1 − 𝑡1𝑇2 − 𝑡2
�
 (2.66) 
The temperature differences for a counterflow heat exchanger are Eqs. (2.62) and (2.63), 28 
therefore, substituting in the LMTD definition, we have: 29 
𝐿𝑀𝑇𝐷(𝑐𝑜𝑢𝑛𝑡𝑒𝑟𝑓𝑙𝑜𝑤) =
(𝑇1 − 𝑡2) − (𝑇2 − 𝑡1)
𝑙𝑛 �𝑇1 − 𝑡2𝑇2 − 𝑡1
�
 (2.67) 
Considering defined process temperatures, a counterflow heat exchanger will be normally more 30 
efficient than a device with the same streams contacted in parallel flow. The reason for this outcome is 31 
that the LMTD value is larger for counterflow than for the parallel flow arrangement. Although we are 32 
not going to demonstrate here, it can be proved [46] that Eq. (2.67) always produces a greater mean 33 
temperature difference than Eq. (2.66), except when one of the streams undergoes an isothermal 34 
condensation or vaporization. In such case, there is no distinction between parallel flow and 35 
counterflow, because both LMTD’s are identical. 36 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 49 
2.3.5 Special Cases of the LMTD 1 
The LMTD as given by Eq. (2.51) can be undetermined in some special cases. For example, 2 
consider a heat exchanger where the temperature differences are equal to each other at both terminals. 3 
The Eq. (2.43) shows that such situation is found when the heat capacities of both streams equals each 4 
other, i.e. WC = wc. Consequently, Eq. (2.51) cannot be directly evaluated; however, applying the 5 
L’Hopital’s rule to Eq. (2.51), we find that: 6 
𝐿𝑀𝑇𝐷 = 𝛥𝑇1 = 𝛥𝑇2 =
𝛥𝑇1 + 𝛥𝑇2
2
 (2.68) 
2.3.6 The Physical Meaning of the LMTD 7 
At this point, you might be questioning: why the logarithmic mean temperature difference 8 
(LMTD) is distinct from others definitions of mathematical means such as arithmetic or geometric? 9 
What advantages it has over other means? 10 
The answer for this question arises from the very derivation of the LMTD. Notice that our 11 
starting point for the development of the LMTD formula was the one-dimensional differential 12 
equations (2.27) and (2.29), describing the fluid temperature profiles along the heat transfer area of a 13 
double-pipe heat exchanger. In their turn, those descriptive equations resulted from the application of 14 
physical principles such as the first and second laws of thermodynamics. For thatreason, we may say 15 
that when such physical principles are applied to a double-pipe heat exchanger, aiming the derivation 16 
of a mean temperature difference expression, the single one expression that arises naturally is the log 17 
mean temperature difference, the LMTD, and no other! Consequently, the LMTD contains physics of 18 
heat transfer phenomena (e.g. conservation of energy) that other mathematical means definitions 19 
simply does not. 20 
2.3.7 Do we need the LMTD Nowadays? 21 
No, we do not! This is the short answer. However, for a longer, more elaborated answer; it can 22 
become a sound yes! The choice is all about the engineering task you are doing, what kind of design 23 
information is needed, and the required level of detail. Additionally, the available computational tools 24 
play an important role. 25 
The development of the LMTD provides a very simple, yet effective design method, however, 26 
at the cost of accuracy loss in some situations. As we could see from Eq. (2.49), the expression of the 27 
LMTD arises naturally inside a form of the heat exchanger design equation (Eq. (2.1)), suggesting a 28 
direct method for sizing or rating a heat exchanger device. The simplifying assumptions embedded in 29 
the LMTD are a key element to produce this straightforward design method, comprising only 30 
algebraic expressions, which are especially suited for manual calculations. 31 
Besides the clean and elegant design method founded on the LMTD concept, we should remark 32 
that the simplifications used to derive the LMTD are not imperative anymore. The LMTD concept was 33 
developed at ages where nowadays computers were out of reach of the most forward-looking dreams, 34 
when numerical calculations were mostly performed by hand and integrals and differential equations 35 
were solved by graphical methods. Today, we have mature numerical methods and personal computers 36 
able to do thousands of mathematical operations in less than a fraction of a second, which allow the 37 
direct solution of the descriptive differential equations for a given exchanger type and flow 38 
arrangement. 39 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
50 2 Heat Exchanger Design Equations 
2.3.7.1 Algorithm for Area Sizing Using the Descriptive Differential Equations 1 
1. Choose a plausible initial estimate of the required heat transfer area. 2 
2. Solve the differential equations over the estimated area to obtain the temperature profile of 3 
the fluid streams. 4 
3. Integrate the total heat transfer load from the local temperature differences between each 5 
pair of streams. 6 
4. Compare the calculated heat transfer load with the heat load required by the process. If the 7 
calculated heat load is: 8 
a. Smaller than the process requirement, then increase heat transfer area and solve the 9 
differential equations again. 10 
b. Greater than the process requirement, then decrease the heat transfer area and solve 11 
the differential equations again. 12 
c. Approximately equal to the process requirement, a feasible heat exchanger size was 13 
achieved. 14 
2.3.8 Parallel flow or counterflow: what is the Difference? 15 
Given the task of designing a single or a battery of hairpins to fulfill a specified service, it is up 16 
to the designer to decide how the fluids are to be thermally contacted: in parallel or counterflow. 17 
Depending on the process fluids and operating conditions, one arrangement or another will yield the 18 
most effective heat exchanger. 19 
In almost all cases, the counterflow operation should be the right choice, because it provides 20 
the highest mean temperature difference driving the heat transfer process. Consequently, for a 21 
specified heat load, the designed heat exchanger will be smaller and less expensive. However, a 22 
parallel flow arrangement may be favored if the cold fluid is highly viscous. In such situation, the feed 23 
of the cold fluid at the same hot fluid terminal allows an early heating, which lowers its viscosity more 24 
quickly along the flow path. As a result, the increase of the overall heat transfer coefficient may be 25 
significant to the point of compensating for the lower mean temperature difference inherent to the 26 
parallel flow. 27 
2.4 Exit Temperatures of the Fluids 28 
There are two common occasions in which the heat exchanger specifications (type, size, 29 
material, etc.) are known and the exit temperatures of both fluids must be determined. 30 
The first case is for a freshly designed heat exchanger with a large built-in design margin. 31 
Typically, a new heat exchanger placed into service delivers a heat load greater than the nominal value 32 
used in its sizing calculations. This is a consequence of the various conservative approximations done 33 
during the calculations, of the fouling factors included in the overall heat transfer coefficient and of a 34 
final safety factor that might be incorporated in the design. The excess in the heat load makes the exit 35 
temperatures of the hot and cold fluids become lower and higher than the design levels, respectively. 36 
An exchanger with a large design margin may shift expressively the operating exit temperatures from 37 
the reference values applied in its design. If the actual exit temperatures yielded by the exchanger are 38 
considerably distinct from the design temperatures, it is advisable to evaluate the impact on the 39 
operation of the connected equipment downstream. 40 
The second situation is when a spare heat exchanger is reused to perform a given service, 41 
which has been carried out by a primary heat exchanger. Sometimes, using a standby heat exchanger 42 
for replacing another one that is shut down for cleaning or repairing is a valuable way of decreasing 43 
the process downtime. Beside other process requirements that the fallback exchanger must match, 44 
such as operating pressure, corrosion resistance, space restrictions, etc., it has to be able to achieve the 45 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 51 
exit temperatures demanded for the cold and hot streams, i.e. being able to supply the at least the 1 
design heat load. 2 
The estimation of the exit temperatures is inherently an iterative procedure, because their 3 
calculation requires the knowledge of the mean overall heat transfer coefficient for the exchanger, 4 
which is itself a function of the entrance and exit temperatures of the fluids. 5 
2.4.1 Counterflow 6 
To develop more general equations, let us label the fluids exchanging heat simply as fluid 1 7 
and 2. With this approach, the resulting equations remain valid independently of what stream is the hot 8 
or cold fluid. Using the definition of the logarithmic mean of the temperature difference (LMTD) in a 9 
counterflow heat exchanger (Eq. (2.66)) of area A and average overall heat transfer coefficient U, the 10 
heat transfer rate can be written as: 11 
𝑞 =
𝐴𝑈�𝑇1,𝑖𝑛 − 𝑇1,𝑜𝑢𝑡 + 𝑇2,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡�
𝑙𝑛 �
𝑇1,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡
𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛
�
 (2.69) 
Under the same assumptions made in the development of the LMTD, the enthalpy variation of 12 
each fluid is numerically equal to the respective heat transfer rates q1 and q2, respectively: 13 
𝑞1 = 𝐶𝑝1𝑊1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� (2.70) 
𝑞2 = 𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� (2.71) 
Therefore, the summation of the enthalpy changes must be zero: 14 
𝐶𝑝1𝑊1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� + 𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� = 0 (2.72) 
It can be observed that the LMTD term in Eq. (2.69) and the right-hand side of Eq. (2.71) both 15 
assume a positive sign when fluid 1 is the hot fluid, or a negative sign if the fluid 1 is the cold fluid. 16 
Hence, the following equality is valid: 17 
𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� =
𝐴𝑈�𝑇1,𝑖𝑛 − 𝑇1,𝑜𝑢𝑡 + 𝑇2,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡�
𝑙𝑛 �
𝑇1,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡
𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛
�
 (2.73) 
The system composed byEqs. (2.70)-(2.73) can be solved for the exit temperature of the fluids 18 
(T1,out and T2,out) and the heat transfer rates (q1 and q2). After some algebraic manipulation, the exit 19 
temperatures are found to be weighted averages of the entrance temperatures in the form: 20 
𝑇1,𝑜𝑢𝑡 =
𝑅2(𝑅 − 1)𝑇1,𝑖𝑛 + (𝑅2 − 𝑅1)𝑇2,𝑖𝑛
𝑅𝑅2 − 𝑅1
 (2.74) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
52 2 Heat Exchanger Design Equations 
𝑇2,𝑜𝑢𝑡 =
𝑅(𝑅2 − 𝑅1)𝑇1,𝑖𝑛 + 𝑅1(𝑅 − 1)𝑇2,𝑖𝑛
𝑅𝑅2 − 𝑅1
 (2.75) 
𝑞1 =
𝐶𝑝1𝑊1(𝑅1 − 𝑅2)�𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛�
𝑅𝑅2 − 𝑅1
 (2.76) 
𝑞2 =
𝐶𝑝1𝑊1(𝑅2 − 𝑅1)�𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛�
𝑅𝑅2 − 𝑅1
 (2.77) 
Where: 1 
𝑅 =
𝐶𝑝1𝑊1
𝐶𝑝2𝑊2
 (2.78) 
𝑅1 = 𝑒
𝐴𝑈
𝐶𝑝1𝑊1 (2.79) 
𝑅2 = 𝑒
𝐴𝑈
𝐶𝑝2𝑊2 (2.80) 
Notice that, alternatively to Eqs. (2.76) and (2.77), the heat transfer rate can be calculated from 2 
Eqs., (2.70) and (2.71). As expected, if the hot stream is the fluid 1, Eqs. (2.70) and (2.76) yield a 3 
negative heat transfer rate. 4 
2.4.2 Parallel Flow 5 
Following the same procedure from the previous section, using the LMTD for a parallel flow 6 
exchanger, its heat load is given by: 7 
𝑞 =
𝐴𝑈�𝑇1,𝑖𝑛 − 𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛 + 𝑇2,𝑜𝑢𝑡�
𝑙𝑛 �
𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛
𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑜𝑢𝑡
�
 (2.81) 
Again, the right-hand side of Eq. (2.81) follows the same sign of Eq. (2.71), therefore it is valid 8 
to write: 9 
𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� =
𝐴𝑈�𝑇1,𝑖𝑛 − 𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛 + 𝑇2,𝑜𝑢𝑡�
𝑙𝑛 �
𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛
𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑜𝑢𝑡
�
 (2.82) 
The combination of Eqs. (2.70), (2.71), (2.72) and (2.82) results in a set, whose solutions for 10 
T1,out, T2,out, q1 and q2 can be reduced to the following form: 11 
𝑇1,𝑜𝑢𝑡 =
(𝑅𝑅1𝑅2 + 1)𝑇1,𝑖𝑛 + (𝑅1𝑅2 − 1)𝑇2,𝑖𝑛
𝑅1𝑅2(𝑅 + 1)
 (2.83) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 53 
𝑇2,𝑜𝑢𝑡 =
𝑅(𝑅1𝑅2 − 1)𝑇1,𝑖𝑛 + (𝑅 + 𝑅1𝑅2)𝑇2,𝑖𝑛
𝑅1𝑅2(𝑅 + 1)
 (2.84) 
𝑞1 =
𝐶𝑝1𝑊1(1 − 𝑅1𝑅2)�𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛�
𝑅1𝑅2(𝑅 + 1)
 (2.85) 
𝑞2 =
𝐶𝑝1𝑊1(𝑅1𝑅2 − 1)�𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛�
𝑅1𝑅2(𝑅 + 1)
 (2.86) 
Where: 1 
𝑅 =
𝐶𝑝1𝑊1
𝐶𝑝2𝑊2
 (2.87) 
𝑅1 = 𝑒
𝐴𝑈
𝐶𝑝1𝑊1 (2.88) 
𝑅2 = 𝑒
𝐴𝑈
𝐶𝑝2𝑊2 (2.89) 
Worked Example 2–1 [18] 2 
In a double-pipe heat exchanger with 10 m2 of effective area, the hot fluid enters at 100°C with 3 
a flow rate of 1.97 kg/s. The entrance temperature of the cold fluid is 15°C and its flow rate is 0.6 kg/s. 4 
The average overall heat transfer coefficient can be taken as U = 500 W/m2K. The specific heats of the 5 
hot and cold fluids are 1.88 J/gK and 4.19 J/gK, respectively. Determine the heat load and exit 6 
temperatures if the streams are arranged in counterflow and parallel flow. 7 
Solution 8 
Setting the hot and cold fluids as fluid 1 and 2, respectively, the data of the heat exchanger 9 
streams is summarized as: 10 
Fluid 1 (hot) Fluid 2 (cold) 
𝑇1,𝑖𝑛 = 100.0 °𝐶 
𝑊1 = 1.97 
𝑘𝑔
𝑠
 
𝐶𝑝1 = 1880.0 
𝐽
(𝑘𝑔 ⋅ 𝐾)
 
𝑇2,𝑖𝑛 = 15.0 °𝐶 
𝑊2 = 0.6 
𝑘𝑔
𝑠
 
𝐶𝑝2 = 4190.0 
𝐽
(𝑘𝑔 ⋅ 𝐾)
 
With Eqs. (2.78) to (2.80) we can evaluate the terms R, R1 and R2 as: 11 
𝑅 =
𝐶𝑝1𝑊1
𝐶𝑝2𝑊2
=
1.9700 ⋅ 1880.0
0.60000 ⋅ 4190.0
= 1.4732 
𝑅1 = 𝑒
𝐴𝑈
𝐶𝑝1𝑊1 = 𝑒
10.000⋅500.00
1.9700⋅1880.0 = 3.8576 
𝑅2 = 𝑒
𝐴𝑈
𝐶𝑝2𝑊2 = 𝑒
10.000⋅500.00
0.60000⋅4190.0 = 7.3072 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
54 2 Heat Exchanger Design Equations 
For the counterflow arrangement, using Eq. (2.74) and (2.75), the exit temperatures of the hot 1 
and cold fluids are: 2 
𝑇1,𝑜𝑢𝑡 =
𝑅2𝑇1,𝑖𝑛(𝑅 − 1) + 𝑇2,𝑖𝑛(−𝑅1 + 𝑅2)
𝑅𝑅2 − 𝑅1
=
100.00 ⋅ 7.3072(1.4732 − 1) + 15.000(−3.8576 + 7.3072)
1.4732 ⋅ 7.3072 − 3.8576
= 57.550 °𝐶 
𝑇2,𝑜𝑢𝑡 =
𝑅𝑇1,𝑖𝑛(−𝑅1 + 𝑅2) + 𝑅1𝑇2,𝑖𝑛(𝑅 − 1)
𝑅𝑅2 − 𝑅1
=
1.4732 ⋅ 100.00(−3.8576 + 7.3072) + 15.000 ⋅ 3.8576(1.4732 − 1)
1.4732 ⋅ 7.3072 − 3.8576
= 77.537 °𝐶 
The Eq. (2.85) allows the calculation of the heat transfer rate from the hot fluid as: 3 
𝑞1 =
𝐶𝑝1𝑊1(𝑅1 − 𝑅2)�𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛�
𝑅𝑅2 − 𝑅1
=
1.9700 ⋅ 1880.0(100.00 − 15.000)(3.8576 − 7.3072)
1.4732 ⋅ 7.3072 − 3.8576
= (−1.5722𝑒 + 5) 𝑊 
Since fluid 1 is the hot stream. The negative sign just indicates that it loses energy. 4 
For the parallel flow arrangement, R, R1 and R2 are the same as evaluated previously, and then 5 
the exit temperatures are obtained from Eq. (2.83) and (2.84): 6 
𝑇1,𝑜𝑢𝑡 =
𝑇1,𝑖𝑛(𝑅𝑅1𝑅2 + 1) + 𝑇2,𝑖𝑛(𝑅1𝑅2 − 1)
𝑅1𝑅2(𝑅 + 1)
=
100.00(1.4732 ⋅ 3.8576 ⋅ 7.3072 + 1) + 15.000(3.8576 ⋅ 7.3072 − 1)
3.8576 ⋅ 7.3072(1.4732 + 1)
= 66.851 °𝐶 
𝑇2,𝑜𝑢𝑡 =
𝑅𝑇1,𝑖𝑛(𝑅1𝑅2 − 1) + 𝑇2,𝑖𝑛(𝑅 + 𝑅1𝑅2)
𝑅1𝑅2(𝑅 + 1)
=
1.4732 ⋅ 100.00(3.8576 ⋅ 7.3072− 1) + 15.000(1.4732 + 3.8576 ⋅ 7.3072)
3.8576 ⋅ 7.3072(1.4732 + 1)
= 63.835 °𝐶 
In addition, Eq. (2.85) gives the heat load transferred from the hot fluid in the exchanger: 7 
𝑞1 =
𝐶𝑝1𝑊1�𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛�(−𝑅1𝑅2 + 1)
𝑅1𝑅2(𝑅 + 1)
=
1.9700 ⋅ 1880.0(100.00 − 15.000)(−3.8576 ⋅ 7.3072 + 1)
3.8576 ⋅ 7.3072(1.4732 + 1)
= (−1.2277𝑒 + 5) 𝑊 
Alternatively, using heat balances of the fluids (Eq. (2.70) and (2.71)), along with the 8 
calculated exit temperatures, the heat transfer rates for both streams are checked below: 9 
𝑞1 = 𝐶𝑝1𝑊1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 1.9700 ⋅ 1880.0(66.851 − 100.00) = (−1.2277𝑒 + 5) 𝑊 
𝑞2 = 𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� = 0.60000 ⋅ 4190.0(63.835 − 15.000) = (1.2277𝑒 + 5) 𝑊 
Conclusion 10 
From the previous calculations, we verify that by simply choosing the counterflow 11 
arrangement, the heat transfer rate increases in about 28%. 12 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 55 
Worked Example 2–2 [18] 1 
A 3000kg/h stream of hot water is cooled from 50°C to 30°C in the tube of a double-pipe heat 2 
exchanger, using 6000kg/h of water at a temperature of 10°C. The fluids are contacted in parallel flow. 3 
The heat transfer coefficients for both sides were estimated as 5000W/m2K. The fouling and tube wall 4 
conductive resistances are negligible. (a) What is the required heat transfer area to perform this 5 
service? Assuming the heat transfer coefficients obey the Sieder-Tate equation for turbulent flow, 6 
determine (b) the exit temperatures if the hot water flow rate doubles, and (c) if the flow rate of both 7 
fluids is doubled. 8 
Solution 9 
a) Case 1: original flow rates. 10 
Let the hot water be named fluid 1, then its average bulk temperature is: 11 
𝑇1𝑚 =
𝑇1,𝑖𝑛
2
+
𝑇1,𝑜𝑢𝑡
2
=
50.000
2
+
30.000
2
= 40.000°𝐶  
From Appendix C.13 , its specific heat at 40°C is estimated as Cp1 = 4080.0 J/(kg⋅K). 12 
Therefore, we have the heat transfer rate of the hot fluid as: 13 
𝑞1 = 𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 0.83333 ⋅ 4080.0(30.000 − 50.000) = −68000.  𝑊 
As a first estimate, the specific heat of the cold water (fluid 2) is evaluated at the entrance 14 
temperature as Cp2 = 4070.0 J/((kg⋅K)), then the exit temperature can be calculated from the heat 15 
balance equation: 16 
𝑇2,𝑜𝑢𝑡 = 𝑇2,𝑖𝑛 +
𝑞2
𝐶𝑝2𝑊2
= 10.000 +
68000.
1.6667 ⋅ 4070.0
= 20.020 °𝐶 
The log mean temperature difference for parallel flow is: 17 
𝛥𝑇1 = 𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛 = −10.000 + 50.000 = 40.000 °𝐶 
𝛥𝑇2 = 𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑜𝑢𝑡 = −20.020 + 30.000 = 9.9800 °𝐶 
𝛥𝑇𝑙𝑚,𝑝 =
𝛥𝑇1 − 𝛥𝑇2
𝑙𝑛 �𝛥𝑇1𝛥𝑇2
�
=
40.000 − 9.9800
𝑙𝑛 �40.0009.9800�
= 21.624 °𝐶 
With the initial values of the heat transfer coefficients, the overall heat transfer coefficient for 18 
the whole exchanger area is: 19 
𝑈 = �
1
ℎ𝑜
+
1
ℎ𝑖
�
−1
=
5000.0
2
= 2500.0 
𝑊
(𝑚2 ⋅ 𝐾)
 
And the heat transfer surface can be calculated from Eq. (2.1): 20 
𝐴 =
𝑞
𝑈𝛥𝑇𝑙𝑚,𝑝
=
68000.
21.624 ⋅ 2500.0
= 1.2579 𝑚2 
b) Case 2: hot water flow rate doubled. 21 
The fluid flow rates are: 22 
© 2015 Samuel Jorge Marques Cartaxo Hairpin HeatExchangers Explained 
56 2 Heat Exchanger Design Equations 
𝑊1 = 𝑊𝑛𝑒𝑤 = 2 ⋅ 𝑊𝑜𝑙𝑑 = 2 ⋅ 0.83333 = 1.6667 
𝑘𝑔
𝑠
 
𝑊2 = 1.6667 
𝑘𝑔
𝑠
 
The Sieder-Tate correlation for turbulent flow gives that h ∝ Re0.8 ∝ W0.8, therefore the internal 1 
heat transfer coefficient (hi) for the new flow rate is: 2 
ℎ𝑖 = ℎ𝑛𝑒𝑤 = ℎ𝑜𝑙𝑑 �
𝑊𝑛𝑒𝑤
𝑊𝑜𝑙𝑑
�
0.8
= 5000.0 �
1.6667
0.83333
�
0.8
= 8705.7 
𝑊
(𝑚2 ⋅ 𝐾)
 
The overall heat transfer coefficient is: 3 
𝑈 = �
1
ℎ𝑜
+
1
ℎ𝑖
�
−1
= �
1
8705.7
+
1
5000.0
�
−1
= 3175.9 
𝑊
(𝑚2 ⋅ 𝐾)
 
From Eqs. (2.87)-(2.89), the parameters R, R1 and R2 are evaluated: 4 
𝑅 =
𝐶𝑝1𝑊1
𝐶𝑝2𝑊2
=
4080.0
4070.0
= 1.0025 
𝑅1 = 𝑒
𝐴𝑈
𝐶𝑝1𝑊1 = 𝑒
1.2579⋅3175.9
1.6667⋅4080.0 = 1.7995 
𝑅2 = 𝑒
𝐴𝑈
𝐶𝑝2𝑊2 = 𝑒
1.2579⋅3175.9
1.6667⋅4070.0 = 1.8021 
Eqs. (2.83) and (2.84) give the exit temperatures for parallel flow: 5 
𝑇1,𝑜𝑢𝑡 =
𝑇1,𝑖𝑛(𝑅𝑅1𝑅2 + 1) + 𝑇2,𝑖𝑛(𝑅1𝑅2 − 1)
𝑅1𝑅2(𝑅 + 1)
=
10.000(1.7995 ⋅ 1.8021 − 1) + 50.000(1.0025 ⋅ 1.7995 ⋅ 1.8021 + 1)
1.7995 ⋅ 1.8021(1.0025 + 1)
= 36.185 °𝐶 
𝑇2,𝑜𝑢𝑡 =
𝑅𝑇1,𝑖𝑛(𝑅1𝑅2 − 1) + 𝑇2,𝑖𝑛(𝑅 + 𝑅1𝑅2)
𝑅1𝑅2(𝑅 + 1)
=
1.0025 ⋅ 50.000(1.7995 ⋅ 1.8021− 1) + 10.000(1.0025 + 1.7995 ⋅ 1.8021)
1.7995 ⋅ 1.8021(1.0025 + 1)
= 23.850 °𝐶 
Using the fluids entrance temperatures T1,in and T2,in, the rate of heat loss from the hot fluid can 6 
be evaluated from Eq. (2.85): 7 
𝑞1 =
𝐶𝑝1𝑊1�𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛�(−𝑅1𝑅2 + 1)
𝑅1𝑅2(𝑅 + 1)
=
1.6667 ⋅ 4080.0(−10.000 + 50.000)(−1.7995 ⋅ 1.8021 + 1)
1.7995 ⋅ 1.8021(1.0025 + 1)
= −93946.  𝑊 
Moreover, the heat gained by the cold (Eq. (2.86)) fluid is: 8 
𝑞2 =
𝐶𝑝1𝑊1�𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛�(𝑅1𝑅2 − 1)
𝑅1𝑅2(𝑅 + 1)
=
1.6667 ⋅ 4080.0(−10.000 + 50.000)(1.7995 ⋅ 1.8021 − 1)
1.7995 ⋅ 1.8021(1.0025 + 1)
= 93946.  𝑊 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 57 
We can double-check the streams heat loads by making the calculation using the four process 1 
temperatures in the heat balances Eqs. (2.70) and (2.71): 2 
𝑞1 = 𝐶𝑝1𝑊1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 1.6667 ⋅ 4080.0(36.185 − 50.000) = −93944.  𝑊 
𝑞2 = 𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� = 1.6667 ⋅ 4070.0(−10.000 + 23.850) = 93951.  𝑊 
c) Case 3: both streams flow rates doubled. 3 
In the present case, both streams have their flow rate doubled: 4 
𝑊1 = 𝑊𝑛𝑒𝑤 = 2 ⋅ 𝑊𝑜𝑙𝑑 = 2 ⋅ 0.83333 = 1.6667 
𝑘𝑔
𝑠
 
𝑊2 = 𝑊𝑛𝑒𝑤 = 2 ⋅ 𝑊𝑜𝑙𝑑 = 2 ⋅ 1.6667 = 3.3334 
𝑘𝑔
𝑠
 
Updating the heat transfer coefficients with the flow rate changes, we have: 5 
ℎ𝑖 = ℎ𝑛𝑒𝑤 = ℎ𝑜𝑙𝑑 �
𝑊𝑛𝑒𝑤
𝑊𝑜𝑙𝑑
�
0.8
= 5000.0 �
1.6667
0.83333
�
0.8
= 8705.7 
𝑊
(𝑚2 ⋅ 𝐾)
 
ℎ𝑜 = ℎ𝑛𝑒𝑤 = ℎ𝑜𝑙𝑑 �
𝑊𝑛𝑒𝑤
𝑊𝑜𝑙𝑑
�
0.8
= 5000.0 �
3.3334
1.6667
�
0.8
= 8705.5 
𝑊
(𝑚2 ⋅ 𝐾)
 
Overall heat transfer coefficient: 6 
𝑈 = �
1
ℎ𝑜
+
1
ℎ𝑖
�
−1
= �
1
8705.7
+
1
8705.5
�
−1
= 4352.8 
𝑊
(𝑚2 ⋅ 𝐾)
 
Then, the new exit temperatures for this overall heat transfer coefficient are calculated: 7 
𝑅 =
𝐶𝑝1𝑊1
𝐶𝑝2𝑊2
=
1.6667 ⋅ 4080.0
3.3334 ⋅ 4070.0
= 0.50123 
𝑅1 = 𝑒
𝐴𝑈
𝐶𝑝1𝑊1 = 𝑒
1.2579⋅4352.8
1.6667⋅4080.0 = 2.2371 
𝑅2 = 𝑒
𝐴𝑈
𝐶𝑝2𝑊2 = 𝑒
1.2579⋅4352.8
3.3334⋅4070.0 = 1.4972 
𝑇1,𝑜𝑢𝑡 =
𝑇1,𝑖𝑛(𝑅𝑅1𝑅2 + 1) + 𝑇2,𝑖𝑛(𝑅1𝑅2 − 1)
𝑅1𝑅2(𝑅 + 1)
=
10.000(1.4972 ⋅ 2.2371 − 1) + 50.000(0.50123 ⋅ 1.4972 ⋅ 2.2371 + 1)
1.4972 ⋅ 2.2371(0.50123 + 1)
= 31.310 °𝐶 
𝑇2,𝑜𝑢𝑡 =
𝑅𝑇1,𝑖𝑛(𝑅1𝑅2 − 1) + 𝑇2,𝑖𝑛(𝑅 + 𝑅1𝑅2)
𝑅1𝑅2(𝑅 + 1)
=
0.50123 ⋅ 50.000(1.4972 ⋅ 2.2371− 1) + 10.000(0.50123 + 1.4972 ⋅ 2.2371)
1.4972 ⋅ 2.2371(0.50123 + 1)
= 19.368 °𝐶 
The heat load evaluated with Eqs. (2.85) or (2.86) are: 8 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
58 2 Heat Exchanger Design Equations 
𝑞1 =
𝐶𝑝1𝑊1�𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛�(−𝑅1𝑅2 + 1)
𝑅1𝑅2(𝑅 + 1)
=
1.6667 ⋅ 4080.0(50.000 − 10.000)(−1.4972 ⋅ 2.2371 + 1)
1.4972 ⋅ 2.2371(0.50123 + 1)
= (−1.2709𝑒 + 5) 𝑊 
𝑞2 =
𝐶𝑝1𝑊1�𝑇1,𝑖𝑛 − 𝑇2,𝑖𝑛�(𝑅1𝑅2 − 1)
𝑅1𝑅2(𝑅 + 1)
=
1.6667 ⋅ 4080.0(50.000 − 10.000)(1.4972 ⋅ 2.2371 − 1)
1.4972 ⋅ 2.2371(0.50123 + 1)
= (1.2709𝑒 + 5) 𝑊 
On the other hand, using the evaluated exit temperatures, from the heat balance equations, we 1 
have: 2 
𝑞1 = 𝐶𝑝1𝑊1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 1.6667 ⋅ 4080.0(31.310 − 50.000) = (−1.2709𝑒 + 5) 𝑊 
𝑞2 = 𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� = 3.3334 ⋅ 4070.0(19.368 − 10.000) = (1.2710𝑒 + 5) 𝑊 
Conclusion 3 
The following table summarizes the obtained results: 4 
 Case 1 Case 2 Case 3 
T1,in 50°C 50°C 50°C 
T2,in 10°C 10°C 10°C 
W1 0.83333 kg/s 1.6667 kg/s 1.6667 kg/s 
W2 1.6667 kg/s 1.6667 kg/s 3.3334 kg/s 
T1,out 30°C 36.185°C 31.310°C 
T2,out 20.020°C 23.850°C 19.368°C 
q 68000. W 93946. W (1.2709e+5) W 
Comparing to the initial operating conditions (case 1), the heat transfer rate increases after 5 
doubling the flow rate of any of the fluids. Notice that in the case 2, despite the increased heat load of 6 
the heat exchanger, there is a raise in the exit temperature of the fluid 1 (hot), which is a consequence 7 
of the greater flow rate. 8 
Problems 9 
2.1) Consider the carbon steel heat exchanger tube shown in Figure 3. Given the convective 10 
heat transfer coefficients of hi = 1841.8 BTU/(ft2⋅h⋅°F) and ho = 94.298 BTU/(ft2⋅h⋅°F), the pipe 11 
dimensions ID = 1 in, OD = 4 in and the material heat conductivity of k = 29.467 BTU/(ft h °F), 12 
determine: 13 
a) Clean overall heat transfer coefficient 14 
b) Simplified clean overall heat transfer coefficient ignoring the wall conductive resistance 15 
c) Design overall heat transfer coefficient 16 
d) The error introduced if the simplified clean overall heat transfer coefficient is used to 17 
design the exchanger 18 
Answer: (a) Uc = 48.503 BTU/(ft2 h °F), (b) Uc (ignoring wall)= 78.269 BTU/(ft2 h °F), (c) The 19 
same as Uc, (d) Error: 38%. 20 
2.2) A fluid is heated from t1 to t2 by cooling another fluid from T1 to T2. Their heat capacities 21 
are wc and WC, respectively. If the overall heat transfer coefficient can be assumed approximately 22 
constant, derive the expressions for the log mean temperature difference for (a) parallel and (b) 23 
countercurrent flow. 24 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
2 Heat Exchanger Design Equations 59 
Answer: (a) Δ𝑇𝑙𝑚 =
𝑇1−𝑡1+𝑡2−𝑇2
ln�𝑇1−𝑡1𝑇2−𝑡2
�
, (b) Δ𝑇𝑙𝑚 =
𝑇1−𝑡2+𝑡1−𝑇2
ln�𝑇1−𝑡2𝑇2−𝑡1
�
. 1 
2.3) Crude oil is heated from 20°C to 60°C with saturated vapor condensing at 100°C. Evaluate 2 
the logarithmic mean temperature difference (LMTD) for (a) parallel flow and (b) counterflow. 3 
Answer: (a) LMTD = 57.7°C, (b) LMTD = 57.7°C. 4 
2.4) A battery of counterflow double-pipe exchangers of effective area 14 m2 is used to heat up 5 
a 2.5 kg/s stream of water from 30°C to 80°C with a 4 kg/s flue gas discharge cooling from 280°C to 6 
157°C. Evaluate the overall heat transfer for this heat exchanger. The physical properties of the flue 7 
gas were found to be very similar to carbon dioxide at the same operating conditions. 8 
Answer: U = 226.85 W/m2⋅K. 9 
2.5) A stream of ethanol with 5.2 kg/s flow rate is heated from 30°C to 65°C using 4 kg/s of 10 
liquid hot water at 90°C. The service is performed in a counterflow carbon steel heat exchanger with 11 
26.928 m2 effective area. Determine: 12 
a) Heat transfer rate 13 
b) Exit temperature of the hot water 14 
c) Log mean temperature difference 15 
d) Overall heat transfer coefficient 16 
Answer: (a) q = 5.0960e+5 W, (b) Tout = 59.5°C, (c) LMTD = 27.2K, (d) Ud = 695.87 W/m2 K 17 
2.6) Consider the service from the Problem 2.5. Assuming the ethanol is controlling the heat 18 
transfer rate, in turbulent flow, the overall heat transfer of the exchanger can be taken as 19 
approximately proportional to its mass flow rate (i.e. U ∝ w0.8). Estimate the heat transfer rate and 20 
ethanol exit temperature if its flow rate changes to (a) 3 kg/s and (b) 7 kg/s. As a first approximation, 21 
assumethat the water exit temperature remains unchanged. 22 
Answer: (a) q = 3.1893e+5 W, Tout = 67.9°C, (b) q = 6.5925e+5 W, Tout = 63.6°C 23 
2.7) Considering the same changes in the ethanol flow rate to (a) 3 kg/s and (b) 7 kg/s (from 24 
the Problem 2.6), determine the “exact” heat transfer rate and the error introduced in the heat transfer 25 
rate by the assumed simplification of constant exit water temperature. 26 
Answer: (a) q = 3.4131e+5 W, Error = −6.5% (b) q = 6.1451e+5 W, Error = 7.3% 27 
2.8) In a chemical plant, 3.3 kg/s of liquid benzene at 90°C is cooled to 60°C by heating a 28 
phenol stream from 30°C to 50°C. Assume a clean overall heat transfer coefficient of 29 
Uc = 190 W/(m2 K). A fouling factor of Rdi = 0.00053 m2K/W is provided for the phenol stream 30 
flowing in the inner tube. Calculate the required heat transfer areas if the flow arrangement is (a) 31 
parallel and (b) counterflow. (c) What is the most cost effective flow arrangement in this case? 32 
Answer: (a) 35.9 m2 (b) 28.8 m2 (c) Countercurrent flow. 33 
2.9) An effluent of 31747 lb/h hot wastewater must be cooled from 176°F to 95°F, before 34 
being discharged to a watercourse. A process stream of aniline is available at 68°F, and the outlet 35 
temperature is designed to be 104°F. Fouling factors are specified for both streams as 36 
Rdi = 0.002 ft2 h °F/BTU and Rdo = 0.001 ft2 h °F/BTU. Carbon steel double-pipe exchangers are used, 37 
which have the following piping data: 38 
Inner pipe: ID = 2.067in, OD = 2.375in 39 
Outer pipe: ID = 7.981in, OD = 8.625in 40 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
60 2 Heat Exchanger Design Equations 
Heat conductivity: 29.467 BTU/(ft h °F) 1 
The convective heat transfer coefficient for the inner tube and annulus are 2 
1841.8 BTU/(ft2 h °F) and 94.3 BTU/(ft2 h °F), respectively. Determine (a) the design overall heat 3 
transfer coefficient and (b) required area to perform this service. 4 
Answer: (a) Ud = 66.697 BTU/(ft2 h °F), (b) A = 823.55 ft2. 5 
2.10) Given the process conditions and exchanger specifications, determine the expressions for 6 
the temperature profiles of hot and cold fluids in a parallel flow double pipe heat exchanger by solving 7 
the differential equations (2.27) and (2.29). 8 
2.11) Calculate the temperature distributions for the fluids in a counterflow double pipe heat 9 
exchanger. Determine the actual mean temperature difference by integrating the difference of the 10 
temperature distribution curves. 11 
 12 
 13 
 14 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
3 Wall Temperatures 61 
3 Wall Temperatures 1 
The Sieder-Tate correlation and most modern equations for the prediction of the heat transfer 2 
coefficient, such as Petukhov-Popov, Gnielinski and Notter-Sleicher, among others, require the 3 
evaluation of fluid physical properties in the temperature of the solid surface heating or cooling the 4 
fluid, the so-called wall temperature. 5 
Rigorously, there are two “wall temperatures”, one for the inner surface (twi) and another on the 6 
outer surface (two), which depend on the convective coefficient of both streams exchanging heat, on 7 
the thermal conductivity of the solid sheet separating the fluids, and on its thickness. Typically, a 8 
metallic cylindrical wall separates the fluids; however, there are heat exchangers made with more 9 
uncommon materials, for instance plastic or ceramics. 10 
 11 
Figure 15: Schematic of the resistances in the thermal circuit in a 12 
double-pipe exchanger with the cold fluid in the inner tube. 13 
In this section, firstly, we are going to develop general equations for calculating the internal 14 
and external wall temperatures in a tubular heat exchanger. Then, the general equations can be 15 
simplified to the most common cases of heat exchangers, such as where the conductive resistance of 16 
the pipe wall is negligible in comparison to the convective resistances of the fluids. 17 
3.1 Cold Fluid in the Inner Pipe 18 
When two fluids exchange energy with a solid wall through the heat path, there are three 19 
resistances opposing this heat flux: a convective resistance of the inner fluid, the wall conductive 20 
resistance and the convective resistance in the outer fluid. This circuit is represented in Figure 15 for 21 
the case where the cold fluid flows within the inner pipe passage. 22 
The heat transfer rate in the individual resistances may be written as: 23 
𝑞 = 𝐴𝑖ℎ𝑖(𝑡𝑤𝑖 − 𝑡) (𝑎) 
𝑞 =
2𝐴𝑜𝑘(𝑡𝑤𝑜 − 𝑡𝑤𝑖)
𝐷𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
(𝑏) 
(3.1) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
62 3 Wall Temperatures 
𝑞 = 𝐴𝑜ℎ𝑜(𝑇 − 𝑡𝑤𝑜) (𝑐) 
𝑞 =
𝑇 − 𝑡
𝐷𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
2𝐴𝑜𝑘
+ 1𝐴𝑜ℎ𝑜
+ 1𝐴𝑖ℎ𝑖
(𝑑)
 
Eqs. (3.1)-(a) to (3.1)-(c) can be combined to form the system of equations: 1 
𝐴𝑖ℎ𝑖(𝑡𝑤𝑖 − 𝑡) =
2𝐴𝑜𝑘(𝑡𝑤𝑜 − 𝑡𝑤𝑖)
𝐷𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
(𝑎) 
𝐴𝑜ℎ𝑜(𝑇 − 𝑡𝑤𝑜) =
2𝐴𝑜𝑘(𝑡𝑤𝑜 − 𝑡𝑤𝑖)
𝐷𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
(𝑏) 
(3.2) 
Where the internal and external heat transfer surfaces of the inner tube may be expressed as: 2 
𝐴𝑖 = 𝜋𝐷𝑖𝐿 (3.3) 
𝐴𝑜 = 𝜋𝐷𝑜𝐿 
Solving Eq. (3.2)-(a) and (3.2)-(b) for the inner wall (twi) and outer wall (two), temperatures, 3 
and using Eq. (3.3), we have: 4 
𝑡𝑤𝑖 =
𝐷𝑖ℎ𝑖 �𝐷𝑜ℎ𝑜𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� 𝑡 + 2𝐷𝑜ℎ𝑜𝑘𝑇
𝐷𝑖ℎ𝑖 �𝐷𝑜ℎ𝑜𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� + 2𝐷𝑜ℎ𝑜𝑘
 (3.4) 
𝑡𝑤𝑜 =
𝐷𝑜ℎ𝑜 �𝐷𝑖ℎ𝑖𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� 𝑇 + 2𝐷𝑖ℎ𝑖𝑘𝑡
𝐷𝑖ℎ𝑖 �𝐷𝑜ℎ𝑜𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� + 2𝐷𝑜ℎ𝑜𝑘
 (3.5) 
In the limiting case of a very thin pipe wall, or highly conductive material, or where the 5 
conductive thermal resistance is small compared with the convective resistances, the inner and outer 6 
wall temperatures become nearly the same, i.e. twi = two = tw. 7 
Noticeably, these equations may be quite cumbersome for convenient use; however, we may 8 
verify that the terms in parenthesis have the same units of the tube thermal conductivity, assuming the 9 
meaning of a corrected thermal conductivity for the effects of pipe curvature. Then, let us define the 10 
factors Ki, Ko and Kio as below: 11 
𝐾𝑖 = 𝐷𝑖ℎ𝑖 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘 (3.6) 
𝐾𝑜 = 𝐷𝑜ℎ𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘 (3.7) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
3 Wall Temperatures 63 
𝐾𝑖𝑜 = 𝐷𝑖ℎ𝑖𝐾𝑜 + 2𝐷𝑜ℎ𝑜𝑘 (3.8) 
With Eqs. (3.6)−(3.8), the wall temperature equations can be reduced to the more compact 1 
forms of Eqs. (3.9) and (3.10). 2 
𝑡𝑤𝑖 =
𝐷𝑖ℎ𝑖𝐾𝑜𝑡 + 2𝐷𝑜ℎ𝑜𝑘𝑇
𝐾𝑖𝑜
 (3.9) 
𝑡𝑤𝑜 =
𝐷𝑜ℎ𝑜𝐾𝑖𝑇 + 2𝐷𝑖ℎ𝑖𝑘𝑡
𝐾𝑖𝑜
 (3.10) 
3.1.1 Negligible Conductive Resistance 3 
If the wall conductive resistance is very small when compared with the convective resistances 4 
of the fluids, Eq. (3.4) and (3.5) reduces to the form of Eq. (3.11). 5 
𝑡𝑤 =
𝐷𝑖ℎ𝑖𝑡 + 𝐷𝑜ℎ𝑜𝑇
𝐷𝑖ℎ𝑖 + 𝐷𝑜ℎ𝑜
 (3.11) 
Notice that, although the wall conductive resistance vanished in Eq. (3.11), the effect of pipe 6 
curvature on the heat flux is still there, being taken account by the presence of the inner and outer 7 
diameters 8 
3.1.2 Thin Tube Wall 9 
For a relatively thin tube wall, we have that Di ≈ Do, therefore Eq. (3.4) and (3.5) take the form 10 
of Eq. (3.12). 11 
𝑡𝑤 =
ℎ𝑜𝑇 + ℎ𝑖𝑡
ℎ𝑖 + ℎ𝑜
 (3.12) 
A comparison of the Eq. (3.11) with (3.12) reveals that the thin wall approximation is a 12 
stronger simplification than small conductive wall resistance, since that later result ignores more 13 
system descriptive information, namely the internal and external diameters of the inner pipe of the heat 14 
exchanger. 15 
3.2 Hot Fluid in the Inner Pipe 16 
With the hot fluid passing through the inner tube, the heat transfer equations for the thermal 17 
circuit are similar, as seen in Eq. (3.13). 18 
𝑞 = 𝐴𝑖ℎ𝑖(𝑇 − 𝑡𝑤𝑖) (𝑎) 
𝑞 =
2𝐴𝑜𝑘(𝑡𝑤𝑖 − 𝑡𝑤𝑜)
𝐷𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
(𝑏) 
𝑞 = 𝐴𝑜ℎ𝑜(𝑡𝑤𝑜 − 𝑡) (𝑐) 
(3.13)© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
64 3 Wall Temperatures 
𝑞 =
𝑇 − 𝑡
𝐷𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
2𝐴𝑜𝑘
+ 1𝐴𝑜ℎ𝑜
+ 1𝐴𝑖ℎ𝑖
(𝑑)
 
Then, the system of equations for obtaining the wall temperatures is given by: 1 
𝐴𝑖ℎ𝑖(𝑇 − 𝑡𝑤𝑖) =
2𝐴𝑜𝑘(𝑡𝑤𝑖 − 𝑡𝑤𝑜)
𝐷𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
(𝑎) 
𝐴𝑜ℎ𝑜(𝑡𝑤𝑜 − 𝑡) =
2𝐴𝑜𝑘(𝑡𝑤𝑖 − 𝑡𝑤𝑜)
𝐷𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
�
(𝑏) 
(3.14) 
Accordingly, the solutions of Eq. (3.14)-(a) and (3.14)-(b) are: 2 
𝑡𝑤𝑖 =
𝐷𝑖ℎ𝑖 �𝐷𝑜ℎ𝑜𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� 𝑇 + 2𝐷𝑜ℎ𝑜𝑘𝑡
𝐷𝑖ℎ𝑖 �𝐷𝑜ℎ𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� + 2𝐷𝑜ℎ𝑜𝑘
 (3.15) 
𝑡𝑤𝑜 =
𝐷𝑜ℎ𝑜 �𝐷𝑖ℎ𝑖𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� 𝑡 + 2𝐷𝑖ℎ𝑖𝑘𝑇
𝐷𝑖ℎ𝑖 �𝐷𝑜ℎ𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� + 2𝐷𝑜ℎ𝑜𝑘
 (3.16) 
And using the K factors from Eqs. (3.6) − (3.8), we have: 3 
𝑡𝑤𝑖 =
𝐷𝑖ℎ𝑖𝐾𝑜𝑇 + 2𝐷𝑜ℎ𝑜𝑘𝑡
𝐾𝑖𝑜
 (3.17) 
𝑡𝑤𝑜 =
𝐷𝑜ℎ𝑜𝐾𝑖𝑡 + 2𝐷𝑖ℎ𝑖𝑘𝑇
𝐾𝑖𝑜
 (3.18) 
3.2.1 Negligible Conductive Resistance 4 
For a high thermally conductive material, the convective resistances dominate the heat transfer 5 
process, and both Eqs (3.15) and (3.16) takes the same form of Eq. (3.19): 6 
𝑡𝑤 =
𝐷𝑖ℎ𝑖𝑇 + 𝐷𝑜ℎ𝑜𝑡
𝐷𝑖ℎ𝑖 + 𝐷𝑜ℎ𝑜
 (3.19) 
3.2.2 Thin Tube Wall 7 
A simplified version of Eq. (3.19) is obtained for the approximation of a tube wall with small 8 
thickness, i.e. Di ≈ Do: 9 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
3 Wall Temperatures 65 
𝑡𝑤 =
ℎ𝑖𝑇 + ℎ𝑜𝑡
ℎ𝑖 + ℎ𝑜
 (3.20) 
3.3 General Form of Wall Temperature Equations 1 
Defining a notation wherein the temperatures of the fluids in the inner tube and annulus are ti 2 
and to, respectively, disregarding which is the hot or cold fluid, a single set of equations may be 3 
written as described below. 4 
Significant Conductive and Convective Resistances 5 
The full equations including the effects of curvature on the conductive and convective thermal 6 
resistances are: 7 
𝑡𝑤𝑖 =
𝐷𝑖ℎ𝑖 �𝐷𝑜ℎ𝑜𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� 𝑡𝑖 + 2𝐷𝑜ℎ𝑜𝑘𝑡𝑜
𝐷𝑖ℎ𝑖 �𝐷𝑜ℎ𝑜𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� + 2𝐷𝑜ℎ𝑜𝑘
 (3.21) 
𝑡𝑤𝑜 =
𝐷𝑜ℎ𝑜 �𝐷𝑖ℎ𝑖𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� 𝑡𝑜 + 2𝐷𝑖ℎ𝑖𝑘𝑡𝑖
𝐷𝑖ℎ𝑖 �𝐷𝑜ℎ𝑜𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘� + 2𝐷𝑜ℎ𝑜𝑘
 (3.22) 
Or in the reduced forms of Eq. (3.23) and (3.24): 8 
𝑡𝑤𝑖 =
𝐷𝑖ℎ𝑖𝐾𝑜𝑡𝑖 + 2𝐷𝑜ℎ𝑜𝑘𝑡𝑜
𝐾𝑖𝑜
 (3.23) 
𝑡𝑤𝑜 =
𝐷𝑜ℎ𝑜𝐾𝑖𝑡𝑜 + 2𝐷𝑖ℎ𝑖𝑘𝑡𝑖
𝐾𝑖𝑜
 (3.24) 
Where: 9 
𝐾𝑖 = 𝐷𝑖ℎ𝑖 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘 (3.25) 
𝐾𝑜 = 𝐷𝑜ℎ𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘 (3.26) 
𝐾𝑖𝑜 = 𝐷𝑖ℎ𝑖𝐾𝑜 + 2𝐷𝑜ℎ𝑜𝑘 (3.27) 
Negligible Conductive Wall Resistance 10 
𝑡𝑤 =
𝐷𝑖ℎ𝑖𝑡𝑖 + 𝐷𝑜ℎ𝑜𝑡𝑜
𝐷𝑖ℎ𝑖 + 𝐷𝑜ℎ𝑜
 (3.28) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
66 3 Wall Temperatures 
Thin Tube Wall 1 
𝑡𝑤 =
ℎ𝑖𝑡𝑖 + ℎ𝑜𝑡𝑜
ℎ𝑖 + ℎ𝑜
 (3.29) 
It is important to notice that the formulas given in Eq. (3.21) and (3.22) are recommended for 2 
more accurate estimations of the internal and external wall temperatures. However, in ordinary 3 
situations, the incurred error from using the simplified forms (Eqs. (3.28) and (3.29)) are likely to be 4 
admissible, and, indeed, these are the expressions typically offered by some reliable technical sources 5 
(e.g. Refs. [44], [47], [19] and [28]). 6 
Worked Example 3–1 7 
Consider a double-pipe exchanger with the hot fluid allocated in the inner tube with an average 8 
bulk temperature of 82 °C and convective heat transfer coefficient 500 W/m2K. The cold fluid flows in 9 
the annulus and has an average temperature of 55 °C and an estimated heat transfer coefficient of 10 
300 W/m2K. The high pressure inner pipe (schedule XXS) has internal and external diameters of 11 
22.8 mm and 42.2 mm, respectively. (a) Estimate the inner tube wall temperature assuming a 12 
negligible conductive resistance. (b) What is the wall temperature for the thin wall approximation? 13 
Solution 14 
(a) If the conductive resistance of the wall is disregarded, Eq. (3.28) may be used with inner 15 
fluid temperature ti = 82 °C and outer temperature to = 55 °C. Substituting the numeric values, we 16 
have: 17 
𝑡𝑤 =
𝐷𝑖ℎ𝑖𝑡𝑖 + 𝐷𝑜ℎ𝑜𝑡𝑜
𝐷𝑖ℎ𝑖 + 𝐷𝑜ℎ𝑜
=
0.02280 ⋅ 500.0 ⋅ 82.00 + 0.04220 ⋅ 300.0 ⋅ 55.00
0.02280 ⋅ 500.0 + 0.04220 ⋅ 300.0
= 67.79°𝐶 
(b) The thin wall approximation is a stronger simplification, under which the wall temperature 18 
is given by Eq. (3.29) as: 19 
𝑡𝑤 =
ℎ𝑖𝑡𝑖 + ℎ𝑜𝑡𝑜
ℎ𝑖 + ℎ𝑜
=
(300.0 ⋅ 55.00 + 500.0 ⋅ 82.00)
300.0 + 500.0
= 71.88°𝐶 
Notice that the consideration of the pipe curvature brings the average wall temperature closer 20 
to the outer fluid temperature, which, in this case is the cold fluid. The difference between both tw 21 
values is about 4.09 °C. 22 
Worked Example 3–2 23 
The exchanger analyzed in Worked Example 3–1 has the inner pipe built in stainless steel 430, 24 
with thermal conductivity k = 8.1 W/mK. How do the “exact” inner and outer walls temperatures 25 
compare with the one estimated with the thin wall simplification? 26 
Solution 27 
Using Eqs. (3.25) to (3.27) the K’s can be evaluated as: 28 
𝐾𝑖 = 𝐷𝑖ℎ𝑖 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘 = 0.02280 ⋅ 500.0 𝑙𝑛 �
0.04220
0.02280
� + 2 ⋅ 8.100 = 23.22
𝑊
(𝑚 ⋅ 𝐾)
 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
3 Wall Temperatures 67 
𝐾𝑜 = 𝐷𝑜ℎ𝑜 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 2𝑘 = 0.04220 ⋅ 300.0 𝑙𝑛 �
0.04220
0.02280
� + 2 ⋅ 8.100 = 23.99
𝑊
(𝑚 ⋅ 𝐾)
 
𝐾𝑖𝑜 = 𝐷𝑖𝐾𝑜ℎ𝑖 + 2𝐷𝑜ℎ𝑜𝑘 = 0.02280 ⋅ 23.99 ⋅ 500.0 + 2 ⋅ 0.04220 ⋅ 300.0 ⋅ 8.100
= 478.6
𝑊2
(𝑚2 ⋅ 𝐾2)
 
Then, the internal and external wall temperatures are obtained from Eqs. (3.23) and (3.24): 1 
𝑡𝑤𝑖 =
1
𝐾𝑖𝑜
(𝐷𝑖𝐾𝑜ℎ𝑖𝑡𝑖 + 2𝐷𝑜ℎ𝑜𝑘𝑡𝑜)
=
1
478.6
(0.02280 ⋅ 23.99 ⋅ 500.0 ⋅ 82.00 + 2 ⋅ 0.04220 ⋅ 300.0 ⋅ 55.00 ⋅ 8.100)
= 70.43°𝐶 
𝑡𝑤𝑜 =
1
𝐾𝑖𝑜
(2𝐷𝑖ℎ𝑖𝑘𝑡𝑖 + 𝐷𝑜𝐾𝑖ℎ𝑜𝑡𝑜)
=
1
478.6
(2 ⋅ 0.02280 ⋅ 500.0 ⋅ 8.100 ⋅ 82.00 + 0.04220 ⋅ 23.22 ⋅ 300.0 ⋅ 55.00)
= 65.42°𝐶 
In this case, the use of the common assumption of thin wall estimates the tube temperature as 2 
tw = 71.88 °C. This result would outcome a wall temperature error of: 3 
• tw – twi = 1.45 °C for the pipe (hot) fluid 4 
• tw – two = 6.46 °C for the annulus (cold) fluid. 5 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Problems 6 
3.1) The heat transfer coefficients for the cold and hot fluids in a double-pipe heat exchanger 7 
are hi = 2667 W/m2K and ho = 940 W/m2K, respectively. The bulk mean temperature is 47.5°C for the 8 
cold fluid and 81.2°C for the hot fluid. What is the average wall temperature in this heat exchanger? 9 
Answer: tw = 56.3°C. 10 
3.2) Triethylene glycol is heated from 30°C to 65°C, using hot water varying from 90°C to 11 
80°C in a counterflow heat exchanger. The overall heat transfer coefficient was evaluated as 12 
U = 160 W/m2K. In this service, the water flows through the annulus and develops a convective heat 13 
transfer of ho = 777 W/m2K. Estimate (a) the convective heat transfer of the triethylene glycol, (b) the 14 
wall temperature at the cold terminal, (c) at the hot terminal, and (d) the mean wall temperature. 15 
Answer: (a) hi = 201.49 W/m2K (b) tw(cold) = 69.7°C (c) tw(hot) = 84.8°C (d) tw(mean) = 77.3°C. 16 
3.3) The heat exchanger used in Problem 3.2 has an internal pipe of DN 50 mm schedule 160. 17 
What are the wall temperatures if the effect of the pipe curvature is taken into account (a) at the cold 18 
terminal, (b) at the hot terminal, and (c) the average wall temperature? 19 
Answer: (a) tw(cold) = 72.2°C (b) tw(hot) = 86.1°C (c) tw(mean) = 79.2°C. 20 
3.4) A high pressure carbon steel heat exchanger with an NPS 2-1/2 − sch XXS internal pipe 21 
cools an ethylene glycol stream using regular cooling water. The bulk temperatures are 149°F and 22 
78°F for ethylene glycol and water, respectively. Ethylene glycol flows inside the inner pipe with 23 
convective heat transfer hi = 136 BTU/(ft2 h °F). The water convective heat transfer is ho = 97424 
BTU/(ft2 h °F). Calculate the wall temperature assuming the approximations of (a) thin wall and (b) 25 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
68 3 Wall Temperatures 
negligible wall conductive resistance. Additionally, determine (c) the temperature difference between 1 
both sides of the inner tube wall and the temperature deviations between the common thin-wall 2 
approximation and the more exact estimations given by (d) Eq. (3.23) and (e) Eq. (3.24). 3 
Answer: (a) tw = 86.7 °F (b) tw = 83.6 °F (c) (twi − two) = 9.4 °F (d) (tw − twi) = −5.5 °F 4 
(e) (tw – two) = 3.8 °F. 5 
3.5) A chemical plant uses 39683.0 lb/h of cooling water from 77.0 °F to 94.7 °F to cool down 6 
an explosive stream of 15873.0 lb/h of 2,4,6-trinitrotoluene from 336.2 °F to 212.0 °F in a carbon steel 7 
(k = 29.467 BTU/(h ft°F)) double-tube heat exchanger. Its inner pipe is NPS 2 in – schedule STD and 8 
carries the 2,4,6-trinitrotoluene stream. The internal and external heat transfer coefficients are 9 
hi = 47 BTU/(h ft2°F) and ho = 427 BTU/(h ft2°F), respectively The trinitrotoluene normal melting 10 
point is about 179.3 °F. 11 
a) What is the wall thickness of the inner tube? 12 
b) Do you anticipate the fouling formation during the operation of this exchanger? Why? 13 
c) Calculate the average wall temperature in this exchanger assuming thin wall. 14 
d) Do you think that the approximation of thin wall yields a significant error? Why? 15 
e) What is the temperature difference between tw from the negligible conductive resistance 16 
and internal and external wall temperatures twi and two? 17 
Answer: (a) 0.154 in (b) Yes (justify!). (c) tw(thin wall) = 104.5 °F (d) No (justify!) (e) (tw –18 
 twi) = −2.94 °F and (tw – two) = 0.28 °F. 19 
3.6) Consider that the trinitrotoluene/water exchanger from Problem 3.5 has an effective length 20 
of 192 ft. 21 
a) Evaluate the heat fluxes and 22 
b) the heat transfer rates for the internal and external fluids using the wall temperature 23 
estimated through the thin wall approximation (Problem 3.5, item c). 24 
c) Does the previously calculated heat transfer rate apparently violate the principle of 25 
conservation of energy? Explain this result. 26 
d) Determine the heat fluxes and 27 
e) the heat transfer rates over the inner an outer surface of the internal pipe using their specific 28 
wall temperatures twi and two, respectively. 29 
Answer: (a) (q/Ai) = −7970.3 BTU/(ft2⋅h), (q/Ao) = 7972.1 BTU/(ft2⋅h) 30 
(b) qi = −8.28e+5 BTU/h, qo = 9.52e+5 BTU/h (c) Yes (justify!) 31 
(d) (q/Ai) = −7936.0 BTU/(ft2⋅h), (q/Ao) = 6908.9 BTU/(ft2⋅h) (e) qi = −8.24e+5 BTU/h, 32 
qo = 8.24e+5 BTU/h. 33 
 34 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
4 Heat Transfer Coefficients 69 
4 Heat Transfer Coefficients 1 
The calculation of the overall heat transfer coefficient depends on the determination of the 2 
pipe-wall conductive resistance and on the convective heat transfer coefficients for both surfaces of 3 
the internal pipe. The conductive resistance of the pipe is not such a problem, because it can be 4 
straightforwardly calculated from the geometric properties, such as the inner diameter and wall 5 
thickness, along with the materials selected to build the heat exchanger. On the other hand, the 6 
convective heat transfer coefficients for the fluids are key information for evaluating the heat 7 
exchanger performance or size, and normally impose greater uncertainty on the design than the one 8 
embedded in the conductive resistance. 9 
4.1 Methods of Estimation 10 
The evaluation of the heat transfer coefficients for the fluids in the heat exchanger can be 11 
performed by several approaches: 12 
1. Analytical solutions of the conservation equations. 13 
2. Numerical solutions of the conservation equations. 14 
3. Correlations: 15 
a. Empirical: derived from measured data. 16 
b. Theoretical: derived from the analytical or numerical solution of a mathematical 17 
model. 18 
c. Semi-empirical: the fitted solution of the mathematical model is achieved using 19 
experimental data. 20 
Relatively simple geometries such as flat plates, spheres, cylindrical ducts, etc. admit analytical 21 
treatment for the determination of the convective heat transfer coefficients. In these cases, simplifying 22 
hypotheses (e.g. one-dimensional and laminar flow) are applied in order to reduce the mathematical 23 
complexity of the equations to a level achievable by the known analytical methods of solution of 24 
differential equations. Unfortunately, current real world problems are commonly characterized by 25 
elaborated geometric features and turbulent flow, making the problem solution usually unattainable by 26 
analytical methods. 27 
In a few cases, the heat transfer coefficients may be obtained with some accuracy from direct 28 
computation, however, they are typically estimated by the use of empirical correlations involving 29 
nondimensional groups. Nowadays, convection coefficients may be calculated numerically by solving 30 
the differential equations expressing the conservation of mass, momentum and energy. The tools and 31 
methods involved in these solutions comprise a whole branch of engineering named computational 32 
fluid dynamics (CFD), which are out of the scope of this text. The main difficulty encountered with 33 
numerical methods is still the phenomenon of turbulence, which is particularly hard to simulate with 34 
reasonable accuracy and within a practical time frame necessary to design heat exchangers. Another 35 
challenge faced by CFD is related to the geometry of a heat exchanger, which may be fairly complex 36 
for various types of industrial importance. Intricate geometries require refined discretization meshes, 37 
rising significantly the time span required to solve the equations. As a result, even with the great 38 
computational power at rather low cost available today, the predominant approach for estimating the 39 
heat transfer coefficients is derived from experimental data posed in the form of mathematical 40 
correlations. 41 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
70 4 Heat Transfer Coefficients 
4.1.1 Types of Correlations 1 
4.1.1.1 Empirical Correlations 2 
In practice, empirical correlations are the predominant source for calculation of the convective 3 
heat transfer coefficients used in heat exchanger design and analysis. Empirical correlations are 4 
essentially mathematical relations among sets of nondimensional numbers obtained from experimental 5 
data. Therefore, an empirical correlation is as good and reliable as is the accuracy of the original 6 
measured data, provided the conditions of use are respected. The engineer must select an adequate 7 
correlation for the studied system. A very accurate correlation may give poor results if applied to a 8 
wrong geometry, for which it was not developed. 9 
4.1.1.2 Theoretical Correlations 10 
A number of heat transfer correlations are not simply the statistical fitting of experimental data 11 
points. In some cases, the correlation represents the results obtained by means of a mathematical 12 
model solved analytically or numerically, for example, the solution of the differential equations of 13 
conservation of mass, energy and momentum. In such situation, the correlation can be named a 14 
“theoretical correlation”, because its data source has no component of experimental data. 15 
The validity of theoretical correlations must be confirmed by experimental data. Although a 16 
theoretical correlation is based entirely on calculated (not measured) results, it should always be 17 
validated against reliable experimental data before using in practical situations. The reason for this is 18 
very straightforward: there is no such thing as a “perfect” mathematical model. Nature is complex and 19 
every single model has assumptions and simplifications in order to be solvable. Themanner to confirm 20 
if a model produces good results is “asking the Nature” if the mathematical model predictions are 21 
correct, and we ask questions directly to Nature by doing experiments. 22 
At this point, you may be questioning right now: why someone would need a correlation if the 23 
exact solution is at hand? Well, the answer has to do with practicality, appropriateness and 24 
convenience. Engineers like practical things, methods, procedures, etc. Practicality reverts to 25 
efficiency, saved time and resources after all. Typically, the data source of a theoretical correlation is 26 
the fluid temperature distribution in the form of a complicated function, involving summation, 27 
integrals or even differential equations for extraction of eigenvalues2. Therefore, the translation of the 28 
temperature profile into the heat transfer coefficient required for equipment design may not be a trivial 29 
task. On the other hand, a simple and straightforward equation, with the necessary accuracy, is 30 
probably the adequate tool for the designer. 31 
4.1.1.3 Semi-empirical Correlations 32 
There is also what is commonly called semi-empirical correlations. It is a blend of theory and 33 
experimentation. A semi-empirical correlation fits the solution of the mathematical model, however, in 34 
order to solve accurately the model, some piece of experimental data may be required. This is the 35 
usual for turbulent flows, for example, while solving the energy equation for a flowing fluid, we have 36 
to know the turbulent velocity distribution. There are two ways to get this information: (1) solving the 37 
momentum equation coupled with the energy equation, along with some turbulence model; (2) make 38 
use of an experimental measured velocity profile. The second option was the choice for several heat 39 
transfer correlations available in the technical literature. 40 
2 The Notter-Sleicher equations presented in Section 4.5.2 match this case. See [69]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
4 Heat Transfer Coefficients 71 
4.2 Using Correlations 1 
Correlations for convective heat transfer give the value of the Nusselt (Nu) or Stanton (St) 2 
numbers as a function of Reynolds (Re), Prandtl (Pr), Grashof (Gr) and other relevant nondimensional 3 
numbers. With the value of the Nusselt or Stanton numbers, we can evaluate the heat transfer 4 
coefficient for the studied problem. Correlations for prediction of forced convection heat transfer 5 
coefficients usually involve only the Reynolds number, however there are exceptions. For example, 6 
heat transfer correlations combining the effects of forced and natural convection relate the Nusselt or 7 
Stanton numbers as a function of both Reynolds and Grashof numbers. 8 
Keep in mind that every correlation has its “user manual” and you must stick to the rules! In 9 
general, an empirical correlation is tied to a specific geometry and domain of validity, usually defined 10 
by the minimum and maximum values of the nondimensional groups measured in the experiments. 11 
The domain of validity for an empirical correlation may be given as ranges for the nondimensional 12 
numbers involved, or sometimes, as ranges for products of primary nondimensional numbers. If the 13 
value of the nondimensional numbers for the case studied are not within the valid ranges of a given 14 
heat transfer correlation, you should look for another more adequate correlation. 15 
Never put all the eggs (estimations) in a single basket (correlation). Every correlation you 16 
found in a reference book or scientific paper is susceptible to errors, be they experimental, statistical or 17 
typos. Therefore, if you are designing a heat exchanger and evaluate heat transfer coefficients from a 18 
mistaken correlation, you may be in serious trouble. I have already found typos in widely used and 19 
referenced books, which generate errors of some orders of magnitude. It is recommended to evaluate 20 
heat transfer coefficients from a few sources before picking the final value to apply in a project. Use at 21 
least two or three sources for a given task. 22 
Considering a set of correlations claiming comparable accuracies, usually, the older and more 23 
reproduced is a heat transfer correlation, the better. From the time of its publication, the fact that a heat 24 
transfer correlation is often reprinted and referenced in textbooks, handbooks, technical manuals, etc. 25 
is an indication of higher reliability. An early correlation is more probable to have been used in several 26 
projects, providing a greater chance for problems to be found. Avoid the use of a recent or obscure 27 
heat transfer correlation as the single source. 28 
4.3 Flow Regimes 29 
Convective heat transfer in internal flows is largely dependent on the flow regime, which can 30 
be laminar, turbulent or assume some transition behavior in between the two. The heat transfer rate 31 
between a fluid in contact with a heated or cooled solid surface is significantly increased by the 32 
macroscopic mixing present in turbulent flow, which renews continuously the fluid particles adjacent 33 
to the heat transfer surface. The replacement of the fluid near the wall increases the local temperature 34 
difference, preventing the boundary layer from reaching the thermal equilibrium with the heat transfer 35 
surface. 36 
The fundamental dimensionless number for classifying the flow regime is the Reynolds 37 
number. The shift from laminar to turbulent flow depends on boundary factors affecting the fluid 38 
motion, such as the wall roughness, mechanical vibration or sudden deviations in the flow direction. 39 
For internal flow in circular pipes, the Reynolds number below 2000 indicates a sustained laminar 40 
regime in most practical situations; however this laminar threshold is often stretched to 2300 [47]. In 41 
the range ~2000 < Re < 4000 the flow behavior swings unsteadily between laminar and turbulent, 42 
corresponding to the transition regime. 43 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
72 4 Heat Transfer Coefficients 
In non-isothermal flows, a significant viscous layer may persist near the wall even at 1 
4000 < Re < 10000, impairing the convective heat transfer3. Because of this, when heat transfer is 2 
involved, the transition regime may be considered to occur in the range of ~2000 < Re < 10000. A 3 
Reynolds number above 10000 characterizes the fully turbulent regime, where only a very thin viscous 4 
sublayer remains adjacent to the solid wall. 5 
4.4 Effects of Fluid Flow Development 6 
The flow of a fluid inside a pipe may be classified as developing or fully developed, 7 
hydrodynamically and/or thermally. In developing flows, the velocities and temperature profiles of the 8 
fluid change for each cross section of the pipe. This implies that the conditions for momentum and 9 
heat transfer inside the fluid vary from the entrance to a certain length of the pipe, which is commonly 10 
called “entrance length” or “entrance region”. 11 
The entrance length of laminar flow is significantly greater than for turbulent flow. For this 12 
reason, the common practice is to use results for fully developed flows in design calculations when the 13 
flow regime is turbulent, ignoring completely the hydrodynamic and thermal entrance lengths [47]. On 14 
the other hand, the entrance length should be taken into consideration if the flow field is laminar, 15 
accordingly the more accurate correlations for forced laminar convection evaluates the Nusselt or 16 
Stanton numbers as a function of the dimensionless number (D/L), where L is the pipe length. 17 
In any case, to achieve higher accuracy, the engineer is recommended to identify if the flow is 18 
fully developed hydrodynamically and thermally for the approached design, in order to apply the 19 
proper methods for evaluating the frictionfactors and heat transfer coefficients. 20 
For the laminar regime of a Newtonian fluid, the entrance length − i.e. the pipe length counted 21 
from the fluid inlet, in which the velocity profile is still deforming – may be estimated by Eq. (4.1) 22 
[48]. 23 
𝐿𝑓𝑑
𝐷ℎ
= 0.05𝑅𝑒 𝑅𝑒 ≤ 2100 (𝑙𝑎𝑚𝑖𝑛𝑎𝑟) (4.1) 
Dh is the hydraulic diameter, which is the internal diameter D for a circular tube, Re is based on 24 
the on Dh, and Lfd is the fully developed entrance length. 25 
Some texts recommend the fine tuning of the coefficient to 0.04 (see Ref. [49]) and other 26 
suggests increasing it to 0.056 as in Ref. [18]. Since the transition from undeveloped to fully 27 
developed flow with the position along the pipe is smooth, these modifications of Eq. (4.1) are 28 
immaterial for practical purposes. 29 
The thermal entrance length for laminar flow is given by Eq. (4.2) [50], where Lfd,T is the fully 30 
developed entrance length for the thermal boundary layer. 31 
𝐿𝑓𝑑,𝑇
𝐷ℎ
= 0.05𝑅𝑒𝑃𝑟 𝑅𝑒 ≤ 2100 (𝑙𝑎𝑚𝑖𝑛𝑎𝑟) (4.2) 
In turbulent flow, the eddy diffusivity dominates the transport of energy and momentum, and 32 
the thermal entrance length is essentially not dependent on the Prandtl number. For practical 33 
situations, the turbulent thermal and hydrodynamic entrance lengths are about the same, varying from 34 
10 to 20 tube diameters. For more precise assessments, the hydrodynamic entrance length for turbulent 35 
flow can be evaluated from Eq. (4.3) [18]. 36 
3 For a detailed explanation of this phenomenon, please refer to Ref. [50]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
4 Heat Transfer Coefficients 73 
𝐿𝑓𝑑
𝐷ℎ
= 1.359𝑅𝑒0.25 𝑅𝑒 > 10000 (𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡) (4.3) 
For turbulent flow, the thermal entrance length is estimated from Eq. (4.4) [48]: 1 
𝐿𝑓𝑑,𝑇
𝐷ℎ
= 1.359𝑅𝑒0.25𝑃𝑟 𝑅𝑒 > 10000 (𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡) (4.4) 
4.4.1 Fully Developed Laminar Flow in Pipes 2 
Can we apply fully developed flow heat transfer coefficients for heat exchangers operating in 3 
the laminar regime? Well, the estimation of the heat transfer coefficient assuming fully developed 4 
flow is conservative for the design, i.e. the estimative is smaller than the real value. Therefore, it is a 5 
safer way to go, especially in situations where some degree of overdesign is allowed or even 6 
recommended. However, you should not forget that overdesign also means additional cost, which is 7 
generally undesirable. 8 
Worked Example 4–1 9 
A stream of W = 0.58 kg/s of 1-butanol at a mean bulk temperature of 93 °C is heated in a 10 
pipe with internal diameter ID = 0.0301 m. What is the pipe length necessary to achieve fluid dynamic 11 
and thermal fully developed flows? 12 
Solution 13 
From Appendix C.9 the physical properties for the fluid at the mean bulk temperature are: 14 
Physical Properties 
1-butanol, 366.15 K, 101325 Pa 
µ = 0.00063397 s Pa 
ρ = 761.9 kg/m3 
Cp = 2529.0 J/(kg K) 
k = 0.1392 W/(m K) 
Pr = 11.518 
The cross flow area is: 15 
𝑆 =
𝜋𝐷2
4
=
𝜋0.03012
4
= 0.00071158 𝑚2 
The mass flux and Reynolds number are: 16 
𝐺 =
𝑊
𝑆
=
0.58
0.00071158
= 815.09 
𝑘𝑔
𝑠 𝑚2
 
𝑅𝑒 =
𝐷𝐺
𝜇
=
0.0301
0.00063397
815.09 = 38699.32 (𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤) 
For turbulent flow, the hydrodynamic entrance length is evaluated from Eq. (4.3): 17 
𝐿𝑓𝑑 = 1.359𝐷√𝑅𝑒
4 = 1.359 ⋅ 0.0301√38699.324 = 0.57 𝑚 
And the thermal entrance length for turbulent regime is given by Eq. (4.4) as: 18 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
74 4 Heat Transfer Coefficients 
𝐿𝑓𝑑,𝑇 = 1.359𝐷𝑃𝑟√𝑅𝑒
4 = 1.359 ⋅ 0.0301 ⋅ 11.518√38699.324 = 6.61 𝑚 
Considering the previous results, we may observe that the high value of the Pr number causes a 1 
significantly slower development of the thermal boundary layer in comparison with the hydrodynamic 2 
boundary layer. 3 
4.5 Heat Transfer Coefficients for Circular Pipes 4 
The inner pipe of a double-pipe heat exchanger is just a circular duct. There is a great variety 5 
of correlations for predicting the heat transfer in forced convection for such geometry. We will discuss 6 
here the most used and proven correlations, considering their domain of validity, for the involved 7 
nondimensional numbers. The Reynolds number is commonly a reliable indicator of the fluid flow 8 
regime, which can be laminar, transitional or turbulent. There is no single correlation able to predict 9 
with high accuracy the heat transfer coefficients for all fluid dynamic regimes; therefore those 10 
correlations are typically bound to a particular fluid flow regime, delimited by a Reynolds number 11 
range. In some cases, the same mathematical formula is applied; however, different sets of parameters 12 
are used for each fluid dynamic region. 13 
4.5.1 Exact Solutions for Laminar Fully Developed Flow 14 
The convective heat transfer in laminar flow is suitable for certain analytical treatment and 15 
some important results may be derived, which provide insight for the development of correlations for 16 
other flow regimes, such as transitional and turbulent. 17 
In laminar flow, the way the heat transfer occurs at the surface of the pipe, i.e. the thermal 18 
boundary conditions, is a major issue, having great impact on the value of the mean heat transfer 19 
coefficient. The most common boundary conditions we can find are: 20 
1. Uniform heat transfer rate at the pipe wall. 21 
2. Uniform temperature of the pipe wall. 22 
4.5.1.1 Uniform wall temperature of the pipe (UWT) 23 
In this situation, the heat transfer surface of the pipe is maintained at a constant temperature, 24 
meaning that every cross section along the pipe axis has the same temperature on the wall. This 25 
boundary condition is appropriate, in good approximation, for heat exchangers where one of the fluids 26 
performs a phase change, such as evaporators and condensers. Similarly, liquid-to-gas heat exchangers 27 
[18], where the liquid flow rate is high, are well represented by this condition, since the gas typically 28 
is the controlling fluid and has a relatively low heat capacity, causing the wall temperature to stay very 29 
close to the liquid temperature. 30 
Assuming fully developed laminar flow, negligible axial conduction, and that the fluid 31 
properties can be considered constant with a uniform wall temperature boundary condition, the 32 
governing differential equations of momentum and energy for the fluid can be solved directly4 to yield 33 
a constant value for the Nusselt number based on the pipe internal diameter D: 34 
𝑁𝑢 = 3.658 (4.5) 
35 
Fact Sheet 
Fluids Gases and liquids 
4 See Ref. [124] for a detailed derivation of the solution. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
4 Heat Transfer Coefficients 75 
Pr > 0.6 
Properties Evaluated using the local fluid bulk temperature 
Conditions 
• Laminar 
• Fully developed flow 
• Uniform wall temperature 
The Péclet number (Pe) can be used to evaluate the importance of the fluid axial conduction. 1 
For Pe > 10, the axial conduction is negligible [47], however for smaller values, a more accurate Nu is 2 
obtained from [51]: 3 
𝑁𝑢 = �
4.1807(1 − 0.0439𝑃𝑒)
3.6568(1 + 1.227/𝑃𝑒2 ) 
𝑃𝑒 < 1.5
𝑃𝑒 > 5 (4.6) 
4.5.1.2 Uniform heat flux at the pipe wall (UHF) 4 
A boundary condition of constant heat transfer rate in the wall holds when the local heat 5 
transfer flux (heat transfer rate per unit of peripheral pipe surface) is approximately constant along the 6 
axial direction of the pipe. Therefore, the lateral area corresponding to each cross section transfers the 7 
same amount of heat per unit time. A direct consequence of such thermal condition is that the pipe 8 
wall temperature does not remain constant axially. As the fluid flows in contact with the heat transfer 9 
surface,it gains or loses internal energy, changing its temperature, then the wall temperature must 10 
necessarily increase or decrease, respectively, in order to keep the heat flux at the same value. 11 
𝑁𝑢 = 4.363 (4.7) 
12 
Fact Sheet 
Fluids Gases and liquids 
Pr > 0.6 
Properties Evaluated using the local fluid bulk temperature 
Conditions 
• Laminar 
• Fully developed flow 
• Uniform heat flux 
4.5.2 Turbulent 13 
4.5.2.1 Dittus-Boelter [52] 14 
Dittus-Boelter [52] gave the precursor form of his famous equation at 30’s. It is one of the 15 
earliest equations for the prediction of convective heat transfer coefficients in pipe flow, being 16 
originally developed for the design of car radiators in the automotive industry. Because it was 17 
published a long time ago, this equation is presented in almost all text and reference books, being 18 
widely known by engineers and practitioners, and probably in great use even these days. 19 
Several versions of this correlation may be found in the literature. Usually the structure of the 20 
equation is preserved and the coefficients and powers are modified. Apparently, the most famous (and 21 
current) form of the Dittus-Boelter equation is due to McAdams [53], which recalibrated its coefficient 22 
from the original value of 0.0243 to 0.023 [54]. 23 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
76 4 Heat Transfer Coefficients 
The modern formulation of the Dittus-Boelter correlation (with the McAdams’ modifications 1 
[53]) is: 2 
𝑁𝑢 = 0.023𝑅𝑒0.8𝑃𝑟𝑛 (4.8) 
𝑛 = 0.4 (ℎ𝑒𝑎𝑡𝑖𝑛𝑔,𝑇𝑤 > 𝑇𝑏)
𝑛 = 0.3 (𝑐𝑜𝑜𝑙𝑖𝑛𝑔,𝑇𝑤 < 𝑇𝑏)
 
3 
Fact Sheet 5 
Fluids Gases and liquids 
Re 104 – 107 
Pr 0.7 – 100 
L/D > 60 
Properties Evaluated using the average of the fluid bulk temperature: Tb=(Tb,in+Tb,out)/2 
Error ±25% from correlated data −10% to −25% for gases and +40% for liquids, confirmed by later studies 
Conditions 
• Turbulent 
• Fully developed flow 
• Constant fluid properties (small to moderate temperature difference Tb-Tw 
and low heat transfer rate) 
• Both uniform wall temperature and uniform heat flux (with deviation of ±25% 
of correlated experimental data) 
• Smooth pipes 
There is some disagreement about the validity domain of this relation, see, for example, Refs. 4 
[55], [42], [50], [56], [57], [48], [58] and [47]. Sometimes the discordance is on the Prandtl number, 5 
where the minimum value ranges from 0.5 to 0.7, and the maximum valid Prandtl is given as 100, 120, 6 
160 or even 170, depending on the source. The lowest recommended Reynolds number also is not 7 
settled, being reported as 2500, 6000 or 10000. The later divergence is of special importance, since it 8 
endorses the use of this equation in the transition flow regime (roughly Re = 2000 – 10000 for pipes), 9 
which is strongly NOT recommended. 10 
The correlation is appropriate for uniform or constant fluid properties, i.e. the temperature 11 
difference between the entrance and exit of the fluid should not be large. In other words, the rate of 12 
heating or cooling of the fluid must not be capable of deforming the velocity distribution from the 13 
isothermal conformation. All physical properties are to be evaluated at the fluid bulk temperature of 14 
the fluid Tb, whereas an average bulk temperature along the pipe (Tb=(Tb,in+Tb,out)/2) may be used to 15 
find an average heat transfer coefficient for the whole heat transfer surface. For rough calculations, the 16 
exponent n = 0.4 can be used both for heating and cooling [54]. 17 
The application of Dittus-Boelter correlation (Eq. (4.8)) over the years revealed some 18 
inaccuracies for certain ranges of Reynolds and Prandtl numbers. It was reported that this correlation 19 
may overestimate the heat transfer coefficient by 10 to 25% for gases, and underestimate in about 40% 20 
the heat transfer coefficient for liquids [56]. Another rigorous study has confirmed underestimations of 21 
5 to 15% for water flowing at several Prandtl values [59]. Additionally, experiments from Ref. [60] 22 
reveled underpredictions of 2 – 12% and 20% for water and oil, respectively. Ref [61] confirmed 23 
predictions with –20% deviations for water. 24 
5 The Reynolds range of applicability varies significantly in the literature, for example, Ref. [50] recommends a 
wider range of Re = 6×103 – 107 and Ref. [48] recommends a Prandtl range of Pr = 0.7 – 120. Therefore, we are going 
conservative here by recommending a narrower validity domain in order to reduce deviations and consequent design risk. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
4 Heat Transfer Coefficients 77 
Such deviations, especially the overvaluing error, pose a critical risk while sizing or analyzing 1 
a heat exchanger. In compensation for the possibility of overestimation of the heat transfer coefficient, 2 
the designer should preventively apply a generous safety factor to his/her calculations. For that reason, 3 
nowadays it is advised against the use of this equation, except with the purpose of crosschecking along 4 
with other predictions of the heat transfer coefficient in question. Of course, in the absence of more 5 
reliable correlation or experimental data, the Dittus-Boelter equation is satisfactory dependable for 6 
practical use, provided an appropriate safety factor is in place. 7 
4.5.2.2 Sieder-Tate [62] 8 
The empirical correlations proposed by Sieder and Tate are still among the most reliable 9 
available in the open technical literature. It is based on over a hundred points of experimental data, 10 
measured from several fluids while heating and cooling in a double-pipe heat exchanger with an 11 
internal smooth copper tube, positioned both in horizontal and vertical. The authors also included 12 
experimental data available from literature and other sources at the time. The fluids where mostly 13 
hydrocarbon mixtures (crude oil cuts), but also included others, such as glycerol, benzene, kerosene, 14 
spindle oil and water. 15 
The correlation was proven to be accurate for gases and liquids in general; however the heat 16 
transfer coefficient predictions for water suffer the larger deviations, once this fluid was a minor part 17 
of the experimental data set. The mean deviations of the correlation from experimental points for 18 
water were one order of magnitude greater than the deviations for oils, reaching +63% in the transition 19 
range (2100 <Re <10000). For turbulent flow (Re > 10000), the water deviations remained in a more 20 
acceptable value of ±10% [62]. In addition, apparently the equation overestimated consistently the 21 
water heat transfer coefficient for Re > 60000, what leads to a tangible risk of non-conservative heat 22 
exchanger designs. 23 
Despite the recommendations against the use of Sieder-Tate for water (Ref. [44] and [62]), a 24 
more recent experimental study [59] measured the heating of water in smooth tubes for 104 < Re <105 25 
and 6.0 < Pr < 11.6, and performed an assessment of several popular correlations, including Sieder and 26 
Tate. It has been found that the Sieder and Tate turbulent correlation (Eq. (4.10)) underestimated 27 
regularly their data in about-5% to -15%. Such result endorses the use of Sieder and Tate for water, by 28 
placing its predictions for water in the safe side for design. After all, as long as the designer avoid the 29 
flow transition range, the equation seems acceptable, even for water, since the deviation for higher 30 
Reynolds numbers is not large, and might be accounted for by the provision of a proper safety factor. 31 
Probably one major contribution of Sieder and Tate, apart from the correlations themselves, 32 
was the introduction of the dimensionless group (µ/µw) 0.14, a viscosity correction factor, to 33 
compensate the significant deformations of the fluid velocity profile when the heat transfer rate is 34 
large, and thephysical properties of the fluid cannot be regarded as constants along the pipe 35 
considerable errors. The viscosity correction factor allowed fitting Nusselt experimental data for 36 
cooling and heating with a single unified curve. Earlier correlations, e.g. the Dittus-Boelter equation 37 
(4.8), were offered as two separated equations for cooling and heating, since the heat transfer data 38 
could not be adequately represented by a single functional form. 39 
Sieder and Tate points that their correlations are able to estimate local heat transfer 40 
coefficients, provided the physical properties are evaluated using the bulk and the wall temperature of 41 
the corresponding tube cross section [62]. It is worth noting that the correlations presented herein 42 
(Eqs. (4.9) and (4.10)) are almost invariably attributed to Sieder and Tate; nevertheless they are not 43 
presented in the currently used forms in their original work [62]. 44 
The application of Sieder-Tate for gases deserves some caution. Despite the fact that there are 45 
important references which does not point out any warnings on its use for gases (e.g. [63], [44], [47], 46 
[42] , among others), there are indications of poor predictions of this equation, for gases, see for 47 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
78 4 Heat Transfer Coefficients 
example [56]. In addition, it is noticed that the original work of Sieder and Tate applied strictly 1 
experimental data from liquids (Table 2 from Ref. [62]) in the correlation development. 2 
Laminar 3 
In the laminar flow regime, the Sieder-Tate equation is Eq. (4.9), where L is the effective pipe 4 
length performing the heat transfer. All the properties should be evaluated at the fluid bulk 5 
temperature, except for the viscosity µw, which is based on the temperature of the heating or cooling 6 
surface Tw. 7 
𝑁𝑢 = 1.86 �𝑅𝑒𝑃𝑟
𝐷
𝐿
�
1/3
�
𝜇
𝜇𝑤
�
0.14
 (4.9) 
8 
Fact Sheet 
Fluids Gases, organic liquids and aqueous solutions [44] and water ([59], [28]) 
Re < 2100 
Pr 0.48 – 16700 [62] 
µ/µw 0.0044 – 9.75 [62] 
(Re Pr D/L)1/3(µ/µw)0.14 > 2 
Properties 
All properties evaluated using the fluid bulk temperature Tb, 
except for µw, which is calculated at the wall temperature Tw. The 
average bulk temperature (Tb = (Tb,in+Tb,out)/2) may be used to 
calculate the mean heat transfer coefficient for the whole surface 
Error ±12% in the range 100 < Re < 2100 [44] ±20% for Re Pr D/L > 10 [43] 
Conditions 
• Heat transfer coefficients are estimated not conservatively 
for water [44], however, there remains some controversy 
about this statement 
• Applicable for smooth and rough pipes 
• Variable fluid properties 
• Isothermal wall boundary condition 
• Ref. [63] asserts the applicability for uniform heat flux 
along pipe, although no supportive study is mentioned. 
Among the conditions of use of Eq. (4.9) is that (Re Pr D/L)1/3(µ/µw)0.14 > 2. This is necessary, 9 
because the Nusselt number given this correlation will approach zero for very long pipes (D/L → 0), 10 
what is physically incorrect. In such case, the predicted Nusselt number is expected to approach 11 
Nu = 3.658, as estimated by Eq. (4.5) for a fully developed laminar flow inside a pipe with uniform 12 
wall temperature. 13 
Note that Eq. (4.9) accounts for the entrance length of the tube, where the heat transfer 14 
coefficients reach higher values. Actually, in basic theory, the Nusselt number is infinite at the 15 
beginning of the tube and diminishes asymptotically along the tube axis to the fully developed flow 16 
value. 17 
Turbulent 18 
For fully developed turbulent flow, use Eq. (4.10), which follows the same application rules of 19 
the Sieder-Tate correlation for the laminar regime (Eq. (4.9)). 20 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
4 Heat Transfer Coefficients 79 
𝑁𝑢 = 0.027𝑅𝑒0.8𝑃𝑟1/3 �
𝜇
𝜇𝑤
�
0.14
 (4.10) 
1 
Fact Sheet 
Fluids Gases6, organic liquids and aqueous solutions [44] and water ([59], [28]) 
Re > 104 
Pr 0.48 – 16700 [62] 
L/D > 10 [43] 
µ/µw 0.0044 – 9.75 [62] 
Properties 
All properties evaluated using the fluid bulk temperature Tb, except for µw, which is 
calculated at the wall temperature Tw. The average bulk temperature 
(Tb = (Tb,in+Tb,out)/2) may be used to calculate the mean heat transfer coefficient for 
the whole surface 
Error −10% to +15% for Re > 10
4 [44] 
±20% in the range 104 < Re < 106 and 0.6 < Pr <100 [43] 
Conditions 
• Heat transfer coefficients are estimated not conservatively for water [44], 
however, there remains some controversy about this statement 
• Applicable for smooth and rough pipes 
• Fully developed flow 
• Variable fluid properties 
• Isothermal wall boundary condition 
• Ref. [63] asserts the applicability for uniform heat flux along pipe, although 
no supportive study is mentioned. 
Based on the reasoning that Eq. (4.9) is based mostly on viscous liquids, and also considering 2 
additional experimental evidence not available at the publication time of the original Sieder and Tate 3 
equations, it is proposed (Ref. [19]) a fine tuning the coefficient in the following manner: 4 
• 0.027 for more viscous liquids, 5 
• 0.023 for less viscous liquids and 6 
• 0.021 for gases. 7 
Although the change of the coefficient according to the fluid category may yield some 8 
accuracy improvement in the prediction of the Nusselt number, to some engineers, this gain seems not 9 
worth for normal practical applications. It is important to warn, however, that the small absolute 10 
difference between these coefficients is quite misleading. In fact, considering the larger range from 11 
0.021 to 0.027, we are talking about a variation of nearly 30% in the heat transfer coefficient! 12 
4.5.2.3 Petukhov-Popov [64] 13 
The Petukhov-Popov correlation is a remarkable instance of successful theoretical development 14 
in the turbulent heat transfer arena. Based on an elaborated analytical and numerical procedure, 15 
Petukhov & Popov [64] proposed a solution for the turbulent heat transfer in fully developed flow in a 16 
smooth tube, assuming constant physical properties and uniform heat flux at the wall. Their correlation 17 
is in fact an interpolation formula; however the correlated database was entirely produced by their 18 
calculations, and validated later with selected experimental data. 19 
6 There are a few contraindications against the use of this correlation with gases in the technical literature. See, for 
example, Ref. [56]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
80 4 Heat Transfer Coefficients 
The assumed hypotheses limited the results to the range 104 < Re < 5×106 and 0.5 < Pr < 2000, 1 
covering gases, non-viscous and viscous liquids, except liquid metals flowing in fully turbulent 2 
regime. Although the calculations presumed the uniform heat flux wall condition, it was shown 3 
previously (Refs. [65]–[67]) that within the assumed Pr and Re ranges, the difference between the 4 
results for uniform heat flux and uniform temperature at the pipe wall is immaterial. 5 
For the case in which the fluid properties can be taken as approximately constants along the 6 
pipe, they proposed an equation resembling the original Prandtl correlation [20] in the form: 7 
𝑁𝑢 =
(𝑓𝐹/2)𝑅𝑒𝑃𝑟
𝐾1 + 𝐾2(𝑓𝐹/2)1/2(𝑃𝑟2/3 − 1)
 (4.11) 
𝐾1 = 1 + 13.6𝑓𝐹
𝐾2 = 11.7 + 1.8/𝑃𝑟1/3
 
Where fF is the Fanning friction factor for constant fluid properties in smooth pipes given by 8 
Filonenko’s equation [68]: 9 
𝑓𝐹 =
1
4
(0.79 𝑙𝑛(𝑅𝑒) − 1.64)−2 (4.12) 
10 
Fact Sheet 
Fluids Gases and liquids, except liquid metals 
Re 104 − 5×106 
Pr 0.5 – 2000 
Properties 
All properties evaluated using the fluid bulk temperature Tb. The average bulk 
temperature (Tb = (Tb,in+Tb,out)/2) may be used to calculate the mean heat transfer 
coefficient for the whole surfaceConditions 
• Applicable for smooth and rough pipes 
• Fully developed flow 
• Constant fluid properties 
• Isothermal wall boundary condition 
• Uniform heat flux at the wall 
Error7 
Under ±1% from theoretical predictions for 104 < Re < 5×105 and 0.5 < Pr < 200 
±1-2% from theoretical predictions in the range 5×105 < Re < 5×106 and 
200 < Pr < 2000 
±5-6% from experimental data tested in the range 104 < Re < 4×105 and 
1 < Pr < 100 
Any correlation other than Eq. (4.12) may be used for the calculation of the friction factor 11 
coefficient, thus Eq. (4.11) is not restricted to smooth tubes. If an appropriate friction factor for rough 12 
tubes is used, it remains valid for estimating the heat transfer coefficient. 13 
7 From Ref. [72]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
4 Heat Transfer Coefficients 81 
Liquids and Gases with Constant Properties (Compact Form) 1 
𝑁𝑢 =
(𝑓𝐹/2)𝑅𝑒𝑃𝑟
1.07 + 12.7(𝑓𝐹/2)1/2(𝑃𝑟2/3 − 1)
 (4.13) 
In Eq. (4.13), fF is the Fanning friction factor for constant fluid properties in smooth pipes 2 
given by Filonenko’s equation [68]: 3 
𝑓𝐹 =
1
4
(0.79 𝑙𝑛(𝑅𝑒) − 1.64)−2 (4.14) 
4 
Fact Sheet 
Fluids Gases and liquids, except liquid metals 
Re 104 − 5×106 
Pr 0.5 – 2000 
Properties 
All properties evaluated using the fluid bulk temperature Tb. The average bulk 
temperature (Tb = (Tb,in+Tb,out)/2) may be used to calculate the mean heat transfer 
coefficient for the whole surface 
Conditions 
• Applicable for smooth and rough pipes 
• Fully developed flow 
• Constant fluid properties 
• Isothermal wall boundary condition 
• Uniform heat flux at the wall 
Error8 
±5-6% from experimental data tested in the range 104 < Re < 5×106 and 
0.5 < Pr < 200 
±10% from experimental data tested in the range 104 < Re < 5×106 and 
0.5 < Pr < 2000 
Alternative correlations for estimating the friction factor instead of Eq. (4.14) may be used with 5 
Eq. (4.13), therefore it is not restricted to smooth tubes. It remains valid for estimating the heat 6 
transfer coefficient using an appropriate friction factor equation for rough tubes. 7 
Variable Physical Properties 8 
Heat Transfer Coefficient in Liquids with Variable Properties 9 
The Nusselt numbers given by Eqs. (4.11) and (4.13) can be corrected to take into account the 10 
effect of varying physical properties using the factor below: 11 
𝑁𝑢 = 𝑁𝑢0 �
𝜇
𝜇𝑤
�
𝑛
 (4.15) 
𝑛 = �0.11
(ℎ𝑒𝑎𝑡𝑖𝑛𝑔)
0.25 (𝑐𝑜𝑜𝑙𝑖𝑛𝑔) 
8 From Ref. [72]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
82 4 Heat Transfer Coefficients 
Where Nu0 is the Nusselt number for the condition of constant fluid properties, µ and µw are 1 
the fluid viscosities at the bulk and wall temperatures, respectively. 2 
Fact Sheet9 
Fluids Liquids, except liquid metals 
Re 104 – 1.25×105 
Pr 2 – 140 
µw/µ 0.08 – 40 
µ/µw 0.025 – 12.5 
Properties 
µ is evaluated at bulk temperature Tb and µw at the wall temperature Tw. The average 
temperatures Tb = (Tb,in+Tb,out)/2 and Tw = (Tw,in+Tw,out)/2 may be used to calculate 
the mean heat transfer coefficient for the whole surface of the tube. 
Conditions 
• Applicable for smooth and rough pipes 
• Fully developed flow 
• Variable liquid properties 
• Isothermal wall boundary condition 
• Uniform heat flux at the wall 
Heat Transfer Coefficient in Gases with Variable Properties (Compact Form) 3 
𝑁𝑢 = 𝑁𝑢0 �
𝑇
𝑇𝑤
�
𝑛
 (4.16) 
𝑛 = �0.47
(ℎ𝑒𝑎𝑡𝑖𝑛𝑔)
0.36 (𝑐𝑜𝑜𝑙𝑖𝑛𝑔) 
Where Nu0 is the Nusselt number for the condition of constant fluid properties, T and Tw are 4 
the bulk and wall temperatures, respectively. 5 
Fact Sheet10 
Fluids Gases 
Re 104 – 106 
Pr 2 – 140 
Tw/T 0.37 – 3.1 
T/Tw 0.32 – 2.70 
Conditions 
• Smooth and rough pipes 
• Fully developed flow 
• Variable physical properties 
• Isothermal wall boundary condition 
• Uniform heat flux at the wall 
Error ±6% from analytical results 
9 From Ref. [72]. 
10 From Ref. [72]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
4 Heat Transfer Coefficients 83 
Friction Factor for Liquids with Variable Properties 1 
𝑓𝐹 = 𝑓𝐹0 
1
6
�7 −
𝜇
𝜇𝑤
�
𝜇
𝜇𝑤
> 1 (ℎ𝑒𝑎𝑡𝑖𝑛𝑔)
𝑓𝐹 = 𝑓𝐹0 �
𝜇
𝜇𝑤
�
−0.24 𝜇
𝜇𝑤
< 1 (𝑐𝑜𝑜𝑙𝑖𝑛𝑔)
 (4.17) 
Where fF0 is the Fanning friction factor for the condition of constant fluid properties. 2 
Fact Sheet11 
Fluids Liquids, except liquid metals 
Re 104 – 2.3×105 
Pr 1.3 – 10 
µ/µw 0.5 – 2.86 
Properties 
µ is evaluated at bulk temperature Tb and µw at the wall temperature Tw. The average 
temperatures Tb = (Tb,in+Tb,out)/2 and Tw = (Tw,in+Tw,out)/2 may be used to calculate 
the mean heat transfer coefficient for the whole surface of the tube. 
Conditions 
• Smooth and rough pipes 
• Fully developed flow 
• Variable liquid properties 
• Isothermal wall boundary condition 
• Uniform heat flux at the wall 
Friction Factor for Gases with Variable Properties (Compact Form) 3 
𝑓𝐹 = 𝑓𝐹0 �
𝑇
𝑇𝑤
�
𝑛
 (4.18) 
𝑛 = �0.52
(ℎ𝑒𝑎𝑡𝑖𝑛𝑔)
0.38 (𝑐𝑜𝑜𝑙𝑖𝑛𝑔) 
4 
Fact Sheet12 
Fluids Gases 
Tw/T 0.37 – 3.7 
T/Tw 0.27 – 2.7 
Conditions 
• Smooth and rough pipes 
• Fully developed flow 
• Variable physical properties 
• Isothermal wall boundary condition 
• Uniform heat flux at the wall 
Error ±7% from analytical results for heating 
11 From Ref. [72]. 
12 From Ref. [72]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
84 4 Heat Transfer Coefficients 
±4% from analytical results for cooling 
4.5.2.4 Notter-Sleicher ([69], [56]) 1 
Notter-Sleicher [69] solved numerically the energy equation for the non-isothermal fluid flow 2 
in a circular pipe with boundary conditions of uniform temperature and uniform heat flux at the wall. 3 
The determination of the two-dimensional temperature distribution of the flow field required the 4 
assumption of constant fluid properties and the use of empirical data for the turbulent velocity profile 5 
and eddy diffusivity distribution in the energy differential equation. The obtained numerical results 6 
were correlated within ±10% of deviation in the range 104 < Re < 106 and 0.1 < Pr < 104. The authors 7 
also performed validation with available experimental data. The Notter-Sleicher fitted equation was 8 
given as: 9 
𝑁𝑢 = 5 + 0.015𝑅𝑒𝑎𝑃𝑟𝑏 (4.19) 
𝑎 = 0.88 −
0.24
(4 + 𝑃𝑟)
 
𝑏 =
1
3
+ 0.5𝑒−0.6𝑃𝑟 
10 
Fact Sheet 
Fluids Gases and liquids 
Re 104 – 106 
Pr 0.1 – 104 
Properties 
Evaluated at bulk temperature Tb The average bulk temperature Tb = (Tb,in+Tb,out)/2 
may be used to calculate the mean heat transfer coefficient for the whole surface of 
the tube. 
Conditions 
• Constant fluid properties 
• Smooth pipes 
• Fully developed flow 
• Isothermal wall boundary condition 
• Uniform heat flux at the wall 
Error 
±10% from analytical results 
±10% from experimental data tested for high Prandtl numbers 
Negligible deviation from tested experimental data for gases 
Notice that Eq. (4.19) has a Pr range able to account for liquid metals. A more accurate and 11 
simplified version of Eq. (4.19) specifically for gases takes the form given in Eq. (4.20), which 12 
correlates the source data within only 4% of deviation. 13 
𝑁𝑢 = 5 + 0.012𝑅𝑒0.83(𝑃𝑟 + 0.29) (4.20) 
14 
Fact Sheet 
Fluids Gases 
Re 104 – 106 
Pr 0.6 – 0.9 
Properties Evaluated at bulk temperature Tb The average bulk temperature Tb = (Tb,in+Tb,out)/2 may be used to calculate the mean heat transfer coefficient for the whole surface of 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
4 Heat Transfer Coefficients 85 
the tube. 
Conditions 
• Constant fluid properties 
• Smooth pipes 
• Fully developed flow 
• Isothermal wall boundary condition• Uniform heat flux at the wall 
Error ±4% from analytical results 
Variable Physical Properties 1 
A few years after the publication of Notter-Sleicher equation, the Sleicher-Rouse correlation 2 
was proposed as an adaptation for use with non-uniform properties fluid flows [56]. The 3 
Sleicher-Rouse equation (Eq. (4.21)) is essentially the same as the Notter-Sleicher’s, but with a 4 
key difference: the evaluation of the physical properties at several temperatures, namely, the 5 
bulk, film and wall temperatures. 6 
Notice that Eq. (4.21) can be applied to uniform fluid properties and for variable fluid 7 
properties, however the domain of validity for Ref and Prw is further restricted when the fluid 8 
properties change significantly along the pipe. The authors reported a maximum deviation of 9 
±20% and an average deviation of about ±7% from the experimental data set. The correlated heat 10 
transfer experimental data was limited to measurements of local heat transfer coefficients, therefore 11 
the correlation is supposed to be adequate for local heat transfer evaluations. 12 
𝑁𝑢𝑏 = 5 + 0.015𝑅𝑒𝑓𝑎𝑃𝑟𝑤𝑏 (4.21) 
𝑎 = 0.88 −
0.24
(4 + 𝑃𝑟𝑤)
 
𝑏 =
1
3
+ 0.5𝑒−0.6𝑃𝑟𝑤 
Where Nub is the Nusselt number with fluid properties at bulk temperature, Ref is the Reynolds 13 
number with properties at the film temperature, and Prw is the Prandtl number evaluated at the wall 14 
temperature. 15 
Fact Sheet 
Fluids Liquids and gases 
Ref 
104 – 106 for constant properties fluids 
104 – 5×105 for variable properties fluids13 
Prw 
0.1 – 104 for constant properties fluids 
0.7 – 75 for variable properties fluids13 
Properties 
Nub, Ref and Prw are evaluated at bulk, film and wall temperatures, respectively. 
The average temperatures for the whole tube length may be used to calculate the 
mean heat transfer coefficient for the whole surface of the tube. 
Conditions • Constant fluid properties 
• Variable fluid properties 
13 This range of validity seems to be restricted by the experimental data for varying fluid properties available to 
the authors at that time. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
86 4 Heat Transfer Coefficients 
• Smooth pipes 
• Fully developed flow 
• Isothermal wall boundary condition 
• Uniform heat flux at the wall 
Error ±20% maximum deviation in some points from correlated experimental data ±6.9% average deviation from correlated experimental data 
The modified correlation for gas flow with non-uniform properties is: 1 
𝑁𝑢𝑏 = 5 + 0.012𝑅𝑒𝑓0.83(𝑃𝑟𝑤 + 0.29) (4.22) 
2 
Fact Sheet 
Fluids Gases 
Ref 
104 – 106 for constant properties gases 
104 – 5×105 for variable properties gases13 
Prw 0.6 – 0.9 
Properties 
Nub, Ref and Prw are evaluated at bulk, film and wall temperatures, respectively. 
The average temperatures for the whole tube length may be used to calculate the 
mean heat transfer coefficient for the whole surface of the tube. 
Conditions 
• Constant fluid properties 
• Variable fluid properties 
• Smooth pipes 
• Fully developed flow 
• Isothermal wall boundary condition 
• Uniform heat flux at the wall 
Error ±6.9% average deviation from correlated experimental data 
4.5.2.5 Gnielinski [70] 3 
Gnielinski ([70] in [20]) modified the theoretical equation from Petukhov-Kirillov by adding 4 
the Hausen corrector ([71] in [45]) to consider the effect of the entrance length on the heat transfer 5 
coefficient. 6 
The deformation of the fluid velocity and temperature profiles due to non-isothermal variations 7 
of the physical properties were taken into account with a factor in the same fashion proposed by 8 
Petukhov [72], however, this factor was recalibrated in order to correlate simultaneously the heat 9 
transfer data for heating and cooling. 10 
The Gnielinski correlation seems to be increasing in popularity. There is not complete 11 
consensus, but important references (e.g. [48]) consider it the most accurate correlation at the present 12 
time, while others like Ref. [47] consider the Petukhov’s equations (see Section 4.5.2.3) the best, due 13 
to its solid theoretical grounds. 14 
𝑁𝑢 =
0.5𝑓𝐹(𝑅𝑒 − 1000)𝑃𝑟
1 + 8.98𝑓𝐹
1/2(𝑃𝑟2/3 − 1)
�1 + �
𝐷
𝐿
�
2/3
�𝜙 (4.23) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
4 Heat Transfer Coefficients 87 
Where fF is the Fanning friction factor for constant fluid properties in smooth pipes given by 1 
Filonenko’s equation [68]: 2 
𝑓𝐹 =
1
4
(0.79 𝑙𝑛(𝑅𝑒) − 1.64)−2 (4.24) 
Variable Physical Properties 3 
The recommended correction factor for variable properties flow is distinct for liquids or gases, 4 
given as: 5 
𝜙 = �
(𝑃𝑟/𝑃𝑟𝑤)0.11 (𝑙𝑖𝑞𝑢𝑖𝑑𝑠)
(𝑇/𝑇𝑤)0.45 (𝑔𝑎𝑠𝑒𝑠)
 (4.25) 
In which the Prandtl numbers Pr and Prw are evaluated at the fluid bulk temperature (T) and 6 
wall temperature (Tw), respectively. 7 
Fact Sheet 
Fluids Gases and liquids 
Re 2300 − 106 [20] 
Pr 0.6 – 2000 (for constant fluid properties) 0.6 – 200 (for variable fluid properties) [50] 
D/L < 1 
Pr/Prw 0.1 – 10 [20] 
T/Tw 0 – 0.5 [20] 
Properties 
All properties evaluated using the fluid bulk temperature Tb. The average bulk 
temperature (Tb = (Tb,in+Tb,out)/2) may be used to calculate the mean heat 
transfer coefficient for the whole surface [45] 
Conditions 
• Applicable for smooth and rough pipes 
• Fully developed and developing flow 
• Constant and variable fluid properties 
• Isothermal wall boundary condition 
• Uniform heat flux at the wall 
Error14 ±20% from tested experimental data [20] 
This correlation was extensively verified against experimental data and presented a very good 8 
agreement with the Prandtl number varying from 0.6 to 1000 [70]. 9 
The Sieder-Tate correction factor for variable physical properties may also be used with Eq. 10 
(4.23) [48]: 11 
𝜙 = (𝜇/𝜇𝑤)0.14 (4.26) 
12 
Fact Sheet 
Fluids Gases and liquids 
µ/µw 1 – 40 [48] 
14 From Ref. [20]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
88 4 Heat Transfer Coefficients 
In practice, the difference between the Prandtl-based and viscosity-based correction factor is 1 
usually small, therefore if both factors are within the range of validity, the choice is merely a matter of 2 
taste. 3 
Worked Example 4–2 4 
A stream of 1.7 m/s of heptanol is heated from 30°C to 50°C in the inner pipe of a heat 5 
exchanger with 5 cm ID and 3 m long. The pipe internal surface is maintained at Tw = 90°C. Estimate 6 
the Nusselt number using the Eq. (4.23) and evaluate the error introduced if the flow is assumed to be 7 
fully developed. 8 
Solution 9 
From Appendix C.9 , the physical properties of 1-heptanol at the mean bulk (T = 40°C) and 10 
wall temperatures are: 11 
Physical Properties: 1-heptanol, 101325 Pa 
T = 313.15 K 
ρ = 805.72 kg/m3 
k = 0.15563 W/(m⋅K) 
μ = 0.0036201 s⋅Pa 
Pr = 47.706 
Tw = 363.15 K 
μw = 0.0010822 s⋅Pa 
Prw = 16.867 
The Reynolds number can be calculated as: 12 
𝑅𝑒 =
𝐷𝜌
𝜇
𝑣 =
0.050000
0.0036201
1.7000 ⋅ 805.72 = 18918.→ 𝑓𝑢𝑙𝑙𝑦 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 
This result indicates that the flow regime is fully turbulent; therefore, Eq. (4.23) is within its 13 
domain of validity and can be used for the Nusselt estimation. 14 
Using Filonenko’s correlation (Eq. (4.24)), the Fanning friction factor is: 15 
𝑓𝐹 =
0.25
(0.79 𝑙𝑛(𝑅𝑒) − 1.64)2
=
0.25
(0.79 𝑙𝑛(18918. ) − 1.64)2
= 0.0066229 
And the correction factor accounting for variation of physical properties in liquids is given by 16 
Eq. (4.25): 17 
𝜙 = �
𝑃𝑟
𝑃𝑟𝑤
�
0.11
= �
47.706
16.867
�
0.11
= 1.1212 
Combining the last results into Eq. (4.23), the Nusselt for a non-completely developed flow is: 18 
𝑁𝑢 =
0.5𝑃𝑟𝑓𝐹(𝑅𝑒 − 1000.0)
8.98𝑓𝐹0.5 �𝑃𝑟
2
3 − 1.0� + 1.0
��
𝐷
𝐿
�
2
3
+ 1.0�𝜙
=
0.5 ⋅ 0.0066229 ⋅ 1.1212 ⋅ 47.706(18918.−1000.0)��0.0500003.0000 �
2
3
+ 1.0�8.98 ⋅ 0.00662290.5 �47.706
2
3 − 1.0� + 1.0
= 342.11 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
4 Heat Transfer Coefficients 89 
The Nusselt number for fully developed flow (Nufd) may be obtained simply by setting 1 
D/L = 0 in Eq. (4.23), then: 2 
𝑁𝑢𝑓𝑑 =
0.5𝑃𝑟𝑓𝐹(𝑅𝑒 − 1000.0)𝜙
8.98𝑓𝐹0.5 �𝑃𝑟
2
3 − 1.0� + 1.0
=
0.5 ⋅ 0.0066229 ⋅ 47.706(18918.−1000.0) ⋅ 1.1212
8.98 ⋅ 0.00662290.5 �47.706
2
3 − 1.0� + 1.0
= 321.16 
The relative deviation from the Nusselt number accounting for the effect of the entrance length 3 
on the heat transfer is: 4 
𝐸𝑟𝑟𝑜𝑟 (%) =
100
𝑁𝑢
�𝑁𝑢𝑓𝑑 − 𝑁𝑢� =
100
342.11
(321.16 − 342.11) = −6.1238% 
Conclusion 5 
By ignoring the flow development length, the Nusselt number is underestimated in about 6%. 6 
4.5.3 Transition 7 
The design of a heat exchanger operating in the transition region between laminar and turbulent 8 
flow is not recommended. Do that only if it is strictly necessary. The fluctuations of the friction factor 9 
and Nusselt number in the transitional flow with very small changes in the Reynolds number are 10 
relatively higher than those for laminar or turbulent flow. Such amplified fluctuations may cause 11 
operational instabilities resulting from minor variations of the flow rate in the process stream, which of 12 
course is undesirable and poses an additional difficulty for the equipment control. 13 
The critical Reynolds number for fluid flow in rounded pipes is Re = 2300. The transition 14 
regime for circular tubes in normal process conditions is usually defined with the Reynolds number in 15 
the range 2300 < Re < 104. For Re > 104, it is generally safe to assume that the flow regime is fully 16 
turbulent. Although, with mechanical vibrations on the pipeline, it is possible to find transitional flow 17 
at 2000 < Re < 2300, if Re < 2000, the flow field returns to laminar, even under strong intermittent 18 
vibrations [73]. 19 
It is of major importance for the engineer to identify correctly the flow regime under which 20 
each stream of the heat exchanger will operate, in order to calculate appropriately the heat transfer 21 
coefficients. For example, if the flow regime used in the design is turbulent and the actual flow regime 22 
when the equipment is in production is laminar, then certainly the heat exchanger will underperform 23 
and possibly not match the heat transfer duty. The behavior of a turbulent flow, and consequently the 24 
heat transfer coefficient, depends on some aspects such as the configuration of the tube entrance15, the 25 
roughness of the tube internal surface and disturbances in the fluid velocity. These elements are of 26 
minor importance if the flow regime is laminar. 27 
There are several methods of heat transfer coefficient estimation for the transition flow regime, 28 
some of them is going to be presented following. 29 
4.5.3.1 Gnielinski Correlation 30 
The Gnielinski Eq. (4.23) combined with Eqs. (4.24) and (4.25) may be used for estimating 31 
Nusselt numbers in the transition range within ±20% of accuracy. 32 
15 Methods of estimation of Nusselt number for various entrance shapes may be found in Ref. [47]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
90 4 Heat Transfer Coefficients 
4.5.3.2 Nusselt Number Interpolation 1 
The Nusselt number for transition flow may be approximated by simple interpolation between 2 
the limiting Nusselt values for laminar and turbulent regimes. This method was verified to agree very 3 
well with experimental data [45]. 4 
The Nusselt number for any transition Reynolds is calculated from: 5 
𝑁𝑢 = (1 − 𝛾)𝑁𝑢(2300) + 𝛾𝑁𝑢(104) (4.27) 
With γ being given by: 6 
𝛾 =
𝑅𝑒 − 2300
104 − 2300
 , 0 ≤ 𝛾 ≤ 1 (4.28) 
Nu(2300) and Nu(104) are the average Nusselt numbers for Re = 2300 and Re = 104, respectively, 7 
evaluated with any appropriate equation. If the pipe is assumed at constant wall temperature, the 8 
average Nusselt number of laminar flow with Re = 2300 in a tube of length L can be calculated from 9 
Eq. (4.29) [45]. 10 
𝑁𝑢(2300) = 53.8��
𝐷𝑃𝑟
𝐿
�
3/2
�21.3 �
𝐷𝑃𝑟
𝐿
�
1/3
− 0.7�
3
�
1
22𝑃𝑟 + 1
�
1/2
�
1/3
 (4.29) 
On its turn, for turbulent flow at either constant wall temperature or constant heat flux, Nu(104) 11 
may be taken, for example, from Eq. (4.11) or Eq. (4.23). 12 
4.5.3.3 Hausen Correlation [74] 13 
The Hausen equation (Eq. (4.30)) is frequently recommended for the transition flow regime for 14 
Reynolds in the range 2100 < Re < 104, given by: 15 
𝑁𝑢 = 0.116�𝑅𝑒2/3 − 125�𝑃𝑟1/3 �
𝜇
𝜇𝑤
�
0.14
�1 + �
𝐷
𝐿
�
2/3
� (4.30) 
16 
Fact Sheet 
Fluids Liquids and gases 
Re 2100 – 104 
Properties 
All properties evaluated using the fluid bulk temperature Tb, except for µw, which is 
calculated at the wall temperature Tw. The average bulk temperature 
(Tb = (Tb,in+Tb,out)/2) may be used to calculate the mean heat transfer coefficient for 
the whole surface 
Conditions • Transitional flow 
• Smooth pipe 
4.5.4 Heat Transfer Coefficients for Water 17 
The heat transfer coefficient of water flow in tubes can be estimated within practical 18 
accurateness from general equations like Petukhov-Popov (Eqs. (4.11) and (4.13)) or Gnielinski (Eq. 19 
(4.23)). However, a more accurate estimation is attainable through specific data for water, such as 20 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
4 Heat Transfer Coefficients 91 
provided by Eagle and Ferguson [75]. Empirical correlations based on Eagle-Ferguson measurements 1 
are presented in Eqs. (4.31) or (4.32)16. Notice that those equations do not use nondimensional groups, 2 
as is usual. On the contrary, they are tied to a set of units, and the convective heat transfer coefficient 3 
(h) is a function of the water velocity (v) and temperature (T). 4 
In the SI system, we have: 5 
ℎ = 1057(1.352 + 0.02𝑇)
𝑣0.8
𝐷0.2
 (4.31) 
[ℎ] = 𝑊/𝑚2°𝐶
[𝑣] = 𝑚 𝑠⁄
[𝐷] = 𝑚
[𝑇] = °𝐶
 
Alternatively, in US customary units: 6 
ℎ = 0.13(1 + 0.011𝑇)
𝑣0.8
𝐷0.2
 (4.32) 
[ℎ] = 𝐵𝑡𝑢/ℎ𝑓𝑡2°𝐹
[𝑣] = 𝑓𝑡 𝑠⁄
[𝐷] = 𝑓𝑡
[𝑇] = °𝐹
 
7 
Fact Sheet 
Fluids Liquid water 
T 5 – 104°C (40 − 220°F) 
4.5.5 Liquid Metals 8 
For some applications, liquid metals may be the most suitable option as the cooling fluid. The 9 
low vapor pressure at high temperatures combined with the greater heat capacity usually leads to more 10 
cost effective heat exchanger, with smaller size and operating pressure. The use of liquid metals as a 11 
heat transfer fluid is possible for most types of heat exchangers, including the double-pipe type, 12 
however, some caution should be taken with welded and brazed heat exchangers, since increased 13 
chemical corrosion could take place when the liquid metal contacts some alloys [76]. 14 
Axial thermal diffusion becomes an important mechanism of heat transfer in the forced 15 
convection of liquid metals. The thermal conductivity of liquid metals is distinctly high in comparison 16 
with other fluids. In accordance, liquid metals are characterized by very low Prandtl numbers (Pr); 17 
being typical values in the range of Pr = 0.001 to 0.1. 18 
The convective heat transfer coefficient in liquid metals is strongly dependent on the Péclet 19 
number (Pe). Most correlations for forced convection in internal flows give the Nusselt number as a 20 
function of the Reynolds number (Re) and Prandtl number (Pr) separately, i.e. Nu = f(Re,Pr). For 21 
liquid metals, the Péclet number can accurately explain Nu experimental data in many cases, therefore, 22 
it is also common to correlate the Nusselt number as a sole function of the Péclet number (i.e. 23 
Nu = f(Pe)), which is a combination of both Re and Pr, defined as Pe = RePr. The Péclet number 24 
16 From Ref. [58]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained92 4 Heat Transfer Coefficients 
provides a measure of the relative importance of the heat transported in the form of internal energy 1 
while the fluid moves, compared to the heat transferred by thermal diffusion in the flow direction. The 2 
smaller the Péclet number, the greater is the role of thermal heat conduction for the convective heat 3 
transfer rate. 4 
4.5.5.1 Notter-Sleicher [69] 5 
Semi-empirical equations based on a combination of experimental data and numerical analysis 6 
were proposed by [69] for the boundary conditions of uniform heat flux and uniform temperature at 7 
the pipe wall. 8 
Uniform wall heat flux: 9 
𝑁𝑢 = 6.3 + 0.0167𝑅𝑒0.85𝑃𝑟0.93 (4.33) 
 10 
Figure 16: Nusselt numbers for liquid metals in fully developed flow 11 
and uniform heat flux boundary condition (Eq. (4.33)). 12 
Uniform wall temperature: 13 
𝑁𝑢 = 4.8 + 0.0156𝑅𝑒0.85𝑃𝑟0.93 (4.34) 
14 
Fact Sheet 
Fluids Liquid metals in general 
Re 104 – 106 
Pr 0 − 0.1 
L/D > 10 
Properties Evaluated using the average of the fluid bulk temperature: Tb=(Tb,in+Tb,out)/2 
Error ±5% in the range 0.004 < Pr < 0.1 and 10
4 < Re < 5×105 
±10% in the range 0 < Pr < 0.1 and 104 < Re < 106 
Conditions 
• Turbulent 
• Fully developed flow 
• Small value of the difference Tb-Tw 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
4 Heat Transfer Coefficients 93 
• Smooth pipe 
The accuracy of correlations (4.33) and (4.34) can be improved to a maximum deviation of 5% 1 
when applied within a narrower range of 0.004 <Pr < 0.1 and 104 < Re < 5×105. 2 
For convenient reading of Nusselt values, a graphic representation of Eqs. (4.33) and (4.34) is 3 
given in Figures 16 and 17, respectively. 4 
 5 
Figure 17: Nusselt numbers for liquid metals in fully developed flow 6 
and uniform wall temperature boundary condition (Eq. (4.34)). 7 
4.5.5.2 Skupinski, Tortel, & Vautrey [77] 8 
This is an empirical correlation based on experiments performed with an alloy of 44% sodium 9 
and 56% potassium (NaK) under fully developed turbulent flow with uniform heat flux at the wall. 10 
The Nusselt number evaluated is local with the properties evaluated at the local bulk temperature. An 11 
average Nusselt number corresponding to the entire heat transfer surface may be estimated using the 12 
mean bulk temperature for evaluating the fluid properties. 13 
𝑁𝑢 = 4.82 + 0.0185(𝑅𝑒𝑃𝑟)0.827
𝑜𝑟
𝑁𝑢 = 4.82 + 0.0185(𝑃𝑒)0.827
 (4.35) 
Where Pe is the Péclet number using the pipe diameter as the characteristic length. 14 
Fact Sheet 
Fluids Liquid metals in general 
Re 3.6×103 – 9.05×105 
Pr 0.003 – 0.05 
Pe 102 – 104 
Conditions 
• Turbulent 
• Fully developed flow 
• Constant heat flux on the wall 
• Smooth pipe 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
94 4 Heat Transfer Coefficients 
Properties Evaluated using the average of the fluid bulk temperature 
Error ±15% 
4.5.5.3 Seban & Shimazaki [65] 1 
The correlation of Seban & Shimazaki [65] is based on experiments using lead-bismuth as 2 
thermal fluid. It evaluates the local Nusselt number of fully developed turbulent flows under the 3 
boundary condition of constant wall temperature. 4 
𝑁𝑢 = 5.0 + 0.025(𝑅𝑒𝑃𝑟)0.8
𝑜𝑟
𝑁𝑢 = 5.0 + 0.025(𝑃𝑒)0.8
 (4.36) 
Where Pe is the Péclet number using the pipe diameter as the characteristic length. 5 
Fact Sheet 
Fluids Liquid metals in general 
Re 3.6×103 – 9.05×105 
Pr 0.003 – 0.05 
Pe > 100 
L/D > 30 
Properties Evaluated using the local fluid bulk temperature 
Error ±15% 
Conditions 
• Turbulent 
• Fully developed flow 
• Constant wall temperature 
• Smooth pipe 
4.5.5.4 Effect of Impurities 6 
The correlations (4.33), (4.34), (4.35) and (4.36) are valid for cooling or heating of the liquid 7 
metal. However, care should be taken to the concentration of impurity content, e.g. oxygen, nitrogen, 8 
etc., in the liquid metal, which is required to be below the oxide solubility limit at the operating 9 
temperature of the heat exchanger [20]. When there is oxide deposition in the wall, the heat transfer 10 
coefficient may be significantly decreased by the oxide layer. An estimation of the reduced Nusselt 11 
may be obtained from: 12 
𝑁𝑢 = 4.3 + 0.0021(𝑅𝑒𝑃𝑟)
𝑜𝑟
𝑁𝑢 = 4.3 + 0.0021(𝑃𝑒)
 (4.37) 
13 
Fact Sheet 
Fluids Liquid metals in general with high concentration of impurities (occurrence of oxide deposition) 
Pe 102 – 104 
L/D > 60 
Properties Evaluated using the local fluid bulk temperature 
Conditions • Fully developed flow 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
4 Heat Transfer Coefficients 95 
• Smooth pipe 
4.6 Heat Transfer Coefficients for the Annulus 1 
The annulus region formed by the assembly of two concentric tubes in a double-pipe heat 2 
exchanger is characterized geometrically by the external diameter of the internal pipe (Do) and the 3 
internal diameter of the shell (Ds), depicted in Figure 18. 4 
The annular channel of two concentric tubes exhibits a distinct hydrodynamic and thermal 5 
behavior from a circular duct. The fluid flowing in the annulus may exchange heat in three ways: 6 
1. From/to the inner surface with the outer surface insulated. 7 
2. From/to the outer surface with the inner surface insulated. 8 
3. From/to both inner and outer surfaces simultaneously. 9 
Considering the possible combinations of the heat transfer surface (inner, outer or both) and the 10 
heat transfer mode (heating or cooling), there are eight arrangements for the fluid exchanging heat, 11 
each one with slightly different velocity and thermal profiles, and consequently distinct heat transfer 12 
behavior. 13 
 14 
Figure 18: Schematic representation of the annulus channel of two 15 
concentric pipes. 16 
Typically, the outer pipe of a double-pipe heat exchanger is insulated (or the heat loss is 17 
ignored), therefore, the most commonly encountered boundary conditions for this equipment are: 18 
1. fluid heated by the inner surface and 19 
2. fluid cooled by the inner surface. 20 
There are several methods for evaluation of the heat transfer coefficient for these situations. 21 
Let us mention some of them following. 22 
4.6.1 Annulus Equivalent Diameter 23 
Perhaps, the most used technique to estimate the annulus heat transfer coefficient is using the 24 
hydraulic diameter (Dh) as a kind of equivalent diameter (De) to adapt the Nusselt number and friction 25 
factor data originally tailored for circular tubes. Such approximation works satisfactorily for several 26 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
96 4 Heat Transfer Coefficients 
noncircular ducts17, including the annulus of a double-pipe exchanger, within 15% accuracy [47], 1 
being a commonplace recommendation in several technical sources ([19], [50], [47], [57], [18] and 2 
[58]). 3 
The hydraulic diameter for a noncircular channel is defined in the form: 4 
𝐷ℎ = 4 ×
𝑓𝑙𝑜𝑤 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
 (4.38) 
Setting Do as the outer diameter of the internal tube, and the inner diameter of the outer tube 5 
(shell) as Ds (Figure 18), the flow cross section and wetted perimeter is given by: 6 
𝑓𝑙𝑜𝑤 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 =
𝜋
4
(𝐷𝑠2 − 𝐷𝑜2) (4.39) 
𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 𝜋(𝐷𝑠 + 𝐷𝑜) (4.40) 
Substituting in the hydraulic diameter expression and simplifying, we have the equivalent 7 
diameter as: 8 
𝐷𝑒 = 𝐷ℎ =
𝐷𝑠2 − 𝐷𝑜2
𝐷𝑠 + 𝐷𝑜
= 𝐷𝑠 − 𝐷𝑜 (4.41) 
In general, throughout this text, unless stated otherwise, the equivalent diameter (De) for 9 
noncircular ducts is going to be taken as equal to the hydraulic diameter (Dh). 10 
4.6.2 Petukhov-Roizen Corrector 11 
A method ([78], [20]) for diminishing the deviation from using a circular pipe equation to 12 
estimate the annulus heat transfer coefficient consists in the application of a correction factor on the 13 
Nusselt number for a circular pipe, as follows:14 
𝑁𝑢𝑎 = 𝐹𝑎(𝑖𝑛𝑛𝑒𝑟) 𝑁𝑢𝑡 (𝑖𝑛𝑛𝑒𝑟 𝑡𝑢𝑏𝑒 ℎ𝑒𝑎𝑡𝑒𝑑 𝑜𝑟 𝑐𝑜𝑜𝑙𝑒𝑑)
𝐹𝑎(𝑖𝑛𝑛𝑒𝑟) = 0.86 �
𝐷𝑜
𝐷𝑠
�
−0.16 (4.42) 
𝑁𝑢𝑎 = 𝐹𝑎(𝑜𝑢𝑡𝑒𝑟) 𝑁𝑢𝑡 (𝑜𝑢𝑡𝑒𝑟 𝑡𝑢𝑏𝑒 ℎ𝑒𝑎𝑡𝑒𝑑 𝑜𝑟 𝑐𝑜𝑜𝑙𝑒𝑑)
𝐹𝑎(𝑜𝑢𝑡𝑒𝑟) = 1 − 0.14 �
𝐷𝑜
𝐷𝑠
�
0.6 (4.43) 
17 Despite the great convenience, the use of the hydraulic diameter as the characteristic length in circular pipes 
correlations for estimating the friction factor and Nusselt number for noncircular ducts with sharp corners produces 
unacceptable deviations [47]. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
4 Heat Transfer Coefficients 97 
 1 
Figure 19: Nusselt number correction factor for an annulus passage 2 
with the inner pipe heated or cooled (Eq. (4.42)). 3 
 4 
The Nusselt number in Eqs. (4.42) and (4.43) is evaluated from any correlation targeted to 5 
circular tubes, such as Eqs. (4.11), (4.19) or (4.23), with the characteristic length as the pipe internal 6 
diameter. The Nua is the Nusselt number for the annulus. Nut is the Nusselt number calculated from 7 
equations tailored for circular tubes. The Nusselt numbers Nua and Nut and any nondimensional 8 
numbers used in their determination must be evaluated using the hydraulic diameter Dh = Ds – Do. 9 
In the case of a double-pipe heat exchanger, the equation to be used is Eq. (4.42), since 10 
typically the outer pipe wall is approximately adiabatic in practical situations. For convenience of 11 
calculation, Figure 19 presents the curve corresponding to the Nusselt correction factor for the annulus 12 
with the inner tube heated or cooled. 13 
4.6.3 Gnielinski Annulus Equation [79] 14 
A relatively recent method for predicting the heat transfer coefficient for annular ducts is a 15 
modification of the Petukhov-Kirillov equation proposed by Gnielinski after correlating a significant 16 
amount of experimental data ([45], [79]). The suggested equation is Eq. (4.44), in which Nu and Re 17 
are based on the annulus hydraulic diameter (Dh). 18 
𝑁𝑢 =
(𝑓𝐹/2)𝑅𝑒𝑃𝑟
𝑘1 + 12.7(𝑓𝐹/2)1/2(𝑃𝑟2/3 − 1)
�1 + �
𝐷ℎ
𝐿
�
2/3
�𝐹𝑎 (4.44) 
𝑘1 = 1.07 +
900
𝑅𝑒
−
0.63
(1 + 10𝑃𝑟)
 (4.45) 
The annulus Fanning friction factor fF used in Eq. (4.44) is also adapted from Filonenko’s 19 
equation for the annular geometry, using a modified Reynolds number in the form: 20 
𝑓𝐹 =
1
4
(0.79 𝑙𝑛(𝑅𝑒∗) − 1.64)−2 (4.46) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
98 4 Heat Transfer Coefficients 
𝑅𝑒∗ = 𝑅𝑒 +
(1 + 𝑎2)𝑙𝑛(𝑎) + (1 − 𝑎2)
(1 − 𝑎)2 𝑙𝑛(𝑎)
 (4.47) 
𝑎 =
𝐷𝑜
𝐷𝑠
 (4.48) 
The factor Fa is recalibrated from the Petukhov-Roizen corrector (Eq. (4.42)), and accounts for 1 
the thermal boundary condition under which the heat is transferred to or from the fluid in the annulus. 2 
When the heat is transferred through the inner pipe wall and the outer pipe is insulated, this factor 3 
becomes: 4 
𝐹𝑎(𝑖𝑛𝑛𝑒𝑟) = 0.75𝑎−0.17 (4.49) 
If the heat is transferred through the outer pipe wall and the inner pipe is adiabatic, the 5 
boundary corrector is given by Eq. (4.50). 6 
𝐹𝑎(𝑜𝑢𝑡𝑒𝑟) = 0.9 − 0.15𝑎0.6 (4.50) 
7 
Fact Sheet 
Fluids Liquids and gases 
Re 104 – 106 
Pr 0.6 – 1000 
D/L 0 – 1 
Properties Evaluated using fluid mean bulk temperature: Tb=(Tb,in+Tb,out)/2 
Conditions 
• Turbulent 
• Fully developed flow 
• Smooth pipe 
Problems 8 
4.1) 2 kg/s of benzene flows in a 10 m long pipe of ID = 52.502 mm. The benzene inlet and 9 
outlet temperatures are 52 °C and 30 °C, respectively. The pipe absolute roughness is about 0.035 mm. 10 
Determine: 11 
a) Reynolds number. 12 
b) Fluid dynamic and thermal entrance lengths. 13 
c) The mean convective heat transfer coefficient estimated by the Sieder-Tate equation. 14 
d) The mean convective heat transfer coefficient estimated by the Petukhov-Popov equation, 15 
using the friction factor from the Filonenko’s equation, 16 
e) and with the friction factor given by the Haaland’s equation, respectively. 17 
Answer: (a) Re = 98784 (b) Lfd = 1.2649m, Lfd,T = 6.9065m (c) h = 1246.5 W/(m2⋅K) 18 
(d) h = 1380.4 W/(m2⋅K) (e) h = 1523.4 W/(m2⋅K) 19 
4.2) A 3.46 kg/s stream of liquid acetone is heated from 20.0 °C to 30.0 °C in the annulus of a 20 
5 m long double-pipe hairpin. The annulus inner and outer diameters are 60.325 mm and 128.19 mm, 21 
respectively; and can be considered smooth with its inner surface maintained at 34.5 °C. Determine: 22 
a) The hydraulic diameter and Reynolds number. 23 
b) Fluid dynamic and thermal entrance lengths. 24 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
4 Heat Transfer Coefficients 99 
c) The relative deviation between the Petukhov’s and Sieder-Tate’s heat transfer coefficients, 1 
assuming constant physical properties. Take Petukhov’s estimation as the reference. 2 
d) The viscosity and Prandtl based correction factors for variable physical properties of the 3 
fluid. 4 
e) The convective heat transfer coefficient given by Petukhov’s equation, taking into account 5 
the variation of fluid properties with the Prandtl correction factor. 6 
Answer: (a) Dh = 0.067865m, Re = 76164 (b) Lfd = 1.5322m, Lfd,T = 5.7915m (c) Error: −3.3% 7 
(d) (μ/μw)0.14 = 1.0120, (Pr/Prw)0.11 = 1.0048 (e) h = 828.77 W/(m2⋅K) 8 
4.3) Cooling water at a flow rate of 2.23 kg/s and 20°C is used to chill 3 kg/s of diethanolamine 9 
from 110°C to 90°C in a double-pipe exchanger with tubes DN 50mm/40S and DN 90mm/40S. The 10 
diethanolamine is allocated in the inner tube. Evaluate (a) the heat transfer coefficients assuming 11 
constant fluid properties, (b) the wall temperature, (c) the Prandtl-based correction factors for variable 12 
fluid properties in the tube and annulus, (d) the actual (corrected) heat transfer coefficients, (e) the 13 
overall heat transfer coefficient for this heat exchanger, (f) the overall heat transfer coefficient if non 14 
corrected convective heat transfer coefficients were used and (g) the error introduced on its value. 15 
Answer: (a) hi = 470.15 W/(m2⋅K) (transition), ho = 3321.5 W/(m2⋅K) (turbulent) (b) tw = 16 
41.0°C (c) ϕi = 0.74693, ϕo = 1.0142 (d) hi = 351.17 W/(m2⋅K), ho = 3368.7 W/(m2⋅K) 17 
(e) Uc = 273.90 W/(m2⋅K) (f) Uc = 364.30W/(m2⋅K) (g) Error: +30% 18 
4.4) In an experiment to measure heat transfer coefficients, a 32.8 ft pipe, ID = 2.0661 in, has 19 
its internal surface at 68.9°F. A air stream of 2381 lb/h is fed at 230.0°F and exits at 140.0°F. 20 
Determine: 21 
a) Total heat transfer rate of the air stream. 22 
b) Mean temperature difference between the fluid and the pipe. 23 
c) Convective heat transfer coefficient. 24 
d) The pipe fraction corresponding to the hydrodynamic entrance length. 25 
e) The pipe fraction corresponding to the thermal entrance length. 26 
f) The relative deviation of the calculated convective heat transfer coefficient from the value 27 
predicted by the Sieder-Tate equation. 28 
Answer: (a) q = 51859 BTU/h (b) 110.03°F (c) h = 47.182 BTU/(ft2⋅h⋅°F) (d) 17% (e) 12% 29 
(f) h(Sieder-Tate) is +41% greater from the experimental value. 30 
4.5) As an alternative process, the air stream referred on the Problem 4.4 is cooled in a double-31 
pipe heat exchanger using 1587.3 lb/h of water passing through the annulus with inlet temperature of 32 
68.0°F. The exchanger piping is carbon steel NPS 2 in/40S and 2-1/2 in/40S. Evaluate: 33 
a) Outlet temperature of the water. 34 
b) Pipe wall temperature, assuming negligible wall thermal conductive resistance. 35 
c) Annulus heat transfer coefficient. 36 
d) The heat exchanger overall heat transfer coefficient. 37 
Answer: (a) Twater,out = 101.65°F (b) tw = 91.6°F (c) ho = 571.07 BTU/(ft2⋅h⋅°F) 38 
(d) U = 37.62 BTU/(ft2⋅h⋅°F). 39 
 40 
 41 
 42 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
100 5 Pressure Drop 
5 Pressure Drop 1 
When both fluids exchanging heat pass througha hairpin heat exchanger, the initial pressures 2 
decrease along the flow path, from the inlet to the outlet nozzles, respectively. The total pressure drop 3 
is a combination of twofold causes: 4 
1. Distributed pressure losses along the linear pipe length of inner tube and annulus. 5 
2. Localized pressure losses at the fittings, connecting the pipes and annuli to build the hairpin 6 
assembly. 7 
Then, the total pressure drop for the whole fluid path may be given by Eq. (5.1) where ΔPdist 8 
and ΔPloc are the distributed and localized pressure losses, respectively. 9 
𝛥𝑃 = 𝛥𝑃𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑙𝑜𝑐 (5.1) 
The pipe distributed losses along ducts are evaluated with the aid of the friction factor. There 10 
are several definitions of friction factor; however the most common are that of Fanning (fF) and Darcy 11 
(fD), related by fF = fD/4. Eq. (5.13) evaluates the duct distributed pressure loss based on the Fanning 12 
friction factor as: 13 
𝛥𝑃𝑑𝑖𝑠𝑡 = 2𝑓𝐹
𝐿
𝐷
𝜌𝑣2 (5.2) 
Or in terms of the mass flux G (kg/m2s), with Eq. (5.14). 14 
𝛥𝑃𝑑𝑖𝑠𝑡 = 2𝑓𝐹
𝐿
𝐷
𝐺2
𝜌
 (5.3) 
Formerly, the localized pressure drop in fittings and flow path “events” was taken as 15 
proportional to the number of velocity heads of the stream, being the proportionality constant the loss 16 
coefficient K. A velocity head expressed in pressure units is the quantity ρv2/2, i.e. the kinetic energy 17 
per unit volume of the fluid. Therefore, local losses in a pipe circuit are calculated from Eq. (5.4). 18 
𝛥𝑃𝑙𝑜𝑐 = 𝐾
𝜌𝑣2
2
 (5.4) 
 19 
Or in terms of the mass flux G (kg/m2s), with Eq. (5.5). 20 
𝛥𝑃𝑙𝑜𝑐 = 𝐾
𝐺2
2𝜌
 (5.5) 
The current knowledge is that the relationship between localized pressure drop and the fluid 21 
velocity is not that simple. Accumulated experimental evidence has shown that the loss coefficient K 22 
is actually a function of Reynolds number and pipe diameter and roughness. Nonetheless the relations 23 
given by Eq. (5.4) and (5.5) are still valid if we keep in mind that K = K(Re,D), and evaluates the loss 24 
coefficient accordingly. 25 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
5 Pressure Drop 101 
The (constant) K-factor method is still in wide use. As the uncertainties involved in Reynolds 1 
number and diameter variations are covered by generous safety factors, more advanced methods such 2 
as the New Crane K [80], the Hooper’s 2-K [81] and the Darby’s 3-K ([82], [83]) methods are in 3 
increasing acceptance. 4 
5.1 Friction Factor in Circular Tubes 5 
We need friction factors to calculate pressure losses. When a fluid flows inside a pipe, 6 
provided the nonslip condition holds, a force will be exerted on the internal surface of the wall with 7 
the same direction of the flow. This force per unit area creates a tangential stress on the wall τw, which 8 
can be made dimensionless by dividing by the volumetric specific kinetic energy of the flowing fluid 9 
ρv2/2 to give: 10 
𝑓𝐹 =
𝜏𝑤
𝜌𝑣2
2
 (5.6) 
The left-hand side of Eq. (5.6) is the Fanning friction factor, a dimensionless number. Notice 11 
that the multiplier 2 in this equation is actually arbitrary. Any other number could be used and the 12 
equation would still remain a dimensionless definition for a friction factor. In fact, there are other 13 
types of friction factors. For example, the Darcy friction factor is in common use by civil and 14 
mechanical engineers, which is defined as fD = 4fF. 15 
So, how we can evaluate pressure drop from the friction factor? A straightforward force 16 
balance for a fluid flowing with bulk velocity v in a tubular section of length L and inner diameter D 17 
shows that: 18 
𝜋𝐷𝐿𝜏𝑤 =
𝜋𝐷2
4
𝛥𝑃 (5.7) 
Solving for ΔP, we get: 19 
𝛥𝑃 = 4
𝐿
𝐷
𝜏𝑤 (5.8) 
But, from Eq. (5.6), τw is: 20 
𝜏𝑤 = 𝑓𝐹
𝜌𝑣2
2
 (5.9) 
Therefore, combining Eq. (5.9) into Eq. (5.8), the relation between the pressure drop and the 21 
Fanning friction factor is obtained as: 22 
𝛥𝑃 = 2𝑓𝐹
𝐿
𝐷
𝜌𝑣2 (5.10) 
Dimensional analysis shows that the friction factor depends on the Reynolds number of the 23 
flow and the relative roughness of the duct wall in contact with the fluid, i.e. fF = fF(Re,ε/D). 24 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
102 5 Pressure Drop 
5.1.1 Isothermal Laminar Flow 1 
The laminar regime admits accurate analytical treatment and the friction factor for Re < 2100 is 2 
calculated from Eq. (5.11). 3 
𝑓𝐹 =
16
𝑅𝑒
 (5.11) 
4 
Fact Sheet 
Fluids Liquids and gases 
Re Re < 2100 
Properties Evaluated using fluid mean bulk temperature: Tb=(Tb,in+Tb,out)/2 
Conditions 
• Laminar 
• Fully developed flow 
• Smooth and rough pipes 
5.1.2 Isothermal Turbulent Flow 5 
Smooth Tubes 6 
Possibly, the simplest friction factor correlation for turbulent flow inside a smooth tube is the 7 
Blasius’ equation [45]: 8 
 𝑓𝐹 =
0.0791
𝑅𝑒0.25
 (5.12) 
9 
Fact Sheet 
Fluids Liquids and gases 
Re 4×103 <Re < 105 
Properties Evaluated using fluid mean bulk temperature: Tb=(Tb,in+Tb,out)/2 
Conditions 
• Turbulent 
• Fully developed flow 
• Smooth pipes 
Rough Tubes 10 
For rough pipes in the turbulent flow range, the Fanning friction factor fF may be evaluated 11 
from the Haaland’s equation (Eq. (5.13)) [84]. This correlation agrees within about +1.21% [47] to the 12 
Colebrook-White [85] implicit equation, considered the golden standard for friction factors in circular 13 
pipes. 14 
𝑓𝐹 = 0.41�𝑙𝑛 �0.23 �
𝜖
𝐷
�
10
9 +
6.9
𝑅𝑒
��
−2
 (5.13) 
15 
Fact Sheet 
Fluids Liquids and gases 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
5 Pressure Drop 103 
ReD 4×103 – 108 ([43], [47]) 
ε/D 0 – 0.05 [43] 2×10-8 – 0.1 [47] 
Properties Evaluated using fluid mean bulk temperature: Tb=(Tb,in+Tb,out)/2 
Conditions 
• Turbulent 
• Fully developed flow 
• Smooth and rough pipes 
5.1.3 All Flow Regimes 1 
5.1.3.1 Churchill Equation [86] 2 
A quite general friction factor equation was developed by Churchill [86], throughout an 3 
interpolated (Ref. [87]) blend of results for the various flow regimes. The correlation in the form of 4 
Fanning friction factor is shown in Eq. (5.14). The author recommended the evaluation of the 5 
Reynolds number at the wall temperature. The Eq. (5.14) is no appropriate for very viscous flows 6 
(with Re < 7), in which case the friction factor should be calculated from the analytical result Eq. 7 
(5.11). 8 
𝑓𝐹 = 2���
8
𝑅𝑒𝑤
�
10
+ �
𝑅𝑒𝑤
36500
�
20
�
−0.5
+ �2.21𝑙𝑛 �
𝑅𝑒𝑤
7
��
10
�
−1/5
 (5.14) 
9 
Fact Sheet 
Fluids Liquids and gases 
Rew > 7 
Properties Rew is calculated at the wall temperature Tw. 
Conditions • Smooth circular ducts 
• Fully developed flow 
5.1.3.2 Bhatti & Shah Equation [88] 10 
The Bhatti-Shah [88] equation is targeted to all Reynolds regimes, and can be used to evaluate 11 
the friction factor for smooth circular ducts with fully developed flow, as given in Eq. (5.15). 12 
 𝑓𝐹 = 𝐴 +
𝐵
𝑅𝑒
1
𝑚
 (5.15) 
13 
 𝐴 = 0 𝐵 = 16 𝑚 = 1 𝑅𝑒 < 2100
𝐴 = 0.0054 𝐵 = 2.3 × 10−8 𝑚 = −
2
3
2100 < 𝑅𝑒 < 4000
𝐴 = 1.28 × 10−3 𝐵 = 0.1143 𝑚 = 3.2154 𝑅𝑒 > 4000
 
14 
Fact Sheet 
Fluids Liquids and gases 
Rew All 
Properties Evaluated using fluid mean bulk temperature: Tb=(Tb,in+Tb,out)/2 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
104 5 Pressure Drop 
Conditions • Smooth circular ducts 
• Fully developed flow 
5.1.4 Effect of Variable Physical Properties 1 
Sieder and Tate used a fair amount of experimental data to develop a corrector factor for the 2 
effect of large variations of fluid properties due to the heating or cooling of the fluid flowing in 3 
circular pipes [62]. 4 
𝑓𝐹 = 𝑓𝐹0𝜙 (5.16) 
𝜙 =
⎩
⎪
⎨
⎪
⎧1.1 �
𝜇
𝜇𝑤
�
−0.25
(𝑅𝑒 < 2100)
1.02 �
𝜇
𝜇𝑤
�
−0.14
(2100 < 𝑅𝑒 < 105)
 (5.17) 
5 
Fact Sheet 
Fluids Gases, organic liquids and aqueous solutions and water [62] 
Re18 2100 – 105µ/µw 
0.0044 – 9.75 (laminar) 
0.1 – 7 (turbulent) 
Properties 
All properties evaluated using the fluid bulk temperature Tb, except for µw, 
which is calculated at the wall temperature Tw. The average bulk temperature 
(Tb = (Tb,in+Tb,out)/2) may be used to calculate the mean heat transfer coefficient 
for the whole surface 
Conditions 
• Smooth pipes 
• Variable fluid properties 
• Isothermal wall boundary condition 
• Fully developed flow 
The friction factor for isothermal flow fF0 used in Eq. (5.16) by Sieder and Tate the Fanning 6 
friction factor; though the Darcy friction factor may also be corrected for non-isothermal flow with 7 
this approach. 8 
5.2 Pressure Drop Calculation with the 2-K Method 9 
A more comprehensive technique for evaluating localized pressure drop is the 2-K method, 10 
which was developed by Hooper (1981) [81]. The 2-K method addresses some weak points of the 11 
conventional K-factor method, by incorporating the dependences of the head loss coefficient K on the 12 
Reynolds number and size of the fitting. The method is accurate for a wide range of diameters, even 13 
for large ones (ID = 0.2 to 80 in), and for low Reynolds number flows (Re = 10 to 106). For further 14 
details about this method, please follow the Refs. [81], [16] and [89]. 15 
18 The actual experimental data for friction factor correlated by Ref. [62] spanned strictly to Re < 10000, however 
the extrapolation to greater values of Reynolds number is widely accepted. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
5 Pressure Drop 105 
 1 
Figure 20: Hairpin heat exchanger mechanical drawing. The annulus 2 
fluid returns through a bonnet. [Adapted from Ref. [15]. Courtesy of 3 
Heat Exchanger Design, Inc., Indianapolis-USA] 4 
Considering its accuracy advantages, in this section we are going to apply the 2-K method in 5 
the development of a procedure for the calculation of pressure losses in the heat exchanger pipe and 6 
the annulus. A note is worth about not using the 3-K method here. Although the 3-K method claims 7 
greater accuracy for a wider range of fitting diameters, their constants K’s were adjusted using 8 
nominal diameters, while the 2-K method is optimized for the actual internal pipe diameter [81]. Since 9 
the internal diameter is already used for several calculations, such as mass flux, Reynolds number, 10 
friction factor, etc., and because of its greater simplicity, the presented procedure will be based on the 11 
2-K method. 12 
The Figure 20 identifies the sources of distributed (A) and localized (B, C and D) pressure 13 
losses for the pipe and the annulus of a hairpin with bonnet return. 14 
5.3 Inner Tube Pressure Drop 15 
The main component of pressure drop in the inner tube is the linear pipe length. Entrance and 16 
exit losses are typically negligible in comparison with linear pipe losses [44], because the nozzles are 17 
usually aligned with the process pipeline and the cross section variation between the nozzles and the 18 
inner pipe is not large. 19 
The distributed pressure drop for the inner pipe channel may be calculated using Eq. (5.18), 20 
where ΔPi,dist is the pressure drop for a mass flux Gi of fluid with specific mass ρ flowing along a 21 
straight pipe of length L and internal diameter Di. fFi is the pipe Fanning friction factor, evaluated from 22 
any valid correlation. This equation is not tied to a unit system; therefore any consistent units set could 23 
be used. 24 
𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 =
2𝐺𝑖2𝐿𝑓𝐹𝑖
𝐷𝑖𝜌
 (5.18) 
The cross flow area of the inner pipe is given by Eq. (5.19), and for a mass flow rate Wi, the 25 
tube mass flux is calculated with Eq. (5.20). 26 
𝑆𝑖 =
𝜋𝐷𝑖2
4
 (5.19) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
106 5 Pressure Drop 
𝐺𝑖 =
𝑊𝑖
𝑆𝑖
 (5.20) 
It should be noted that for an association of Nhp hairpins in series, L comprises the full pipe 1 
length, including all exchanger legs; however the length of the return bend is ignored − it is later 2 
considered as a local source of pressure loss − i.e. if the effective length of the exchanger is Lhp, the 3 
entire fluid path is computed as L = 2NhpLhp, since a “U” hairpin exchanger has two legs. 4 
𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 =
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑖
𝐷𝑖𝜌
𝐺𝑖2 (5.21) 
The Fanning friction factor to be applied in Eq. (5.21) is evaluated from a valid equation for 5 
the flow conditions of the process. Notice that Eq. (5.21) is essentially isothermal; hence, for large 6 
variations of fluid temperature, the friction factor should be corrected with a Sieder-Tate (Eq. (5.16)) 7 
or Petukhov factor (Eq. (4.20) and (4.21)). 8 
The pipe fluid performs a 180° turn at one end of the hairpin exchanger. Moreover, because of 9 
the small specific heat transfer area, double-pipe exchangers are industrially mounted in banks (see 10 
Figures 4, 5 and 23), where the inner tube of one exchanger is connected to the next exchanger by a 11 
long radius bend, which should also be accounted as a source of pressure loss. Assuming these bends 12 
have approximately the same internal diameter of the inner tube, no fluid expansion or contraction 13 
takes place, then the loss coefficient for a flanged/welded short radius 180° bend is: 14 
𝐾(180°) = 0.35 +
1000
𝑅𝑒
+
8.89
𝐷ℎ(𝑚𝑚)
 (5.22) 
𝐾(180°) = 0.35 +
1000
𝑅𝑒
+
0.35
𝐷ℎ(𝑖𝑛)
 (5.23) 
Therefore the total number of velocity heads for the inner tube (assuming two long radius 180° 15 
flanged or welded bend) is evaluated with the 2-K method from Eq. (5.24), where Ki is the friction 16 
loss coefficient for the inner pipe of one hairpin with internal diameter Di expressed in millimeters. 17 
𝐾𝑖 = 0.7 +
2000
𝑅𝑒𝑖
+
17.78
𝐷𝑖(𝑚𝑚)
 (5.24) 
Or, for Di measured in inches, Ki is calculated by: 18 
𝐾𝑖 = 0.7 +
2000
𝑅𝑒𝑖
+
0.7
𝐷𝑖(𝑖𝑛)
 (5.25) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
5 Pressure Drop 107 
 1 
Figure 21: K-factor head loss coefficient for long radius 180° return 2 
bend. Di is the internal tube diameter expressed in millimeters (Eq. 3 
(5.24)). 4 
The Eq. (5.24) is represented in Figure 21, where it is noticeable the significant increase of the 5 
K value for laminar flows. In this regime, the application of a constant number of velocity heads to 6 
account for local head losses may dangerously underestimate the exchanger pressure drop. As 7 
predicted by the classical K-factor method, the K value becomes practically constant for higher 8 
Reynolds numbers flows. The effect of the pipe diameter is also significant for smaller diameters. 9 
However, for greater pipe sizes, an increase of diameter affects the head loss coefficient to a lesser 10 
extent. 11 
For a battery containing Nhp hairpin exchangers connected in series, with Eqs. (5.5) and (5.24), 12 
there are (Nhp−1) additional 180° return bends connecting the tubes of adjacent hairpins, and then we 13 
write the total localized pressure loss in the inner pipe as: 14 
𝛥𝑃𝑖,𝑙𝑜𝑐 =
𝐺𝑖2𝐾𝑖
2𝜌
�2𝑁ℎ𝑝 − 1� (5.26) 
Tube: long radius 180° return bend 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
108 5 Pressure Drop 
 1 
Figure 22: K-factor head loss coefficient for long radius 180° return 2 
bend. Di is the internal tube diameter expressed in inches (Eq. (5.25)). 3 
Hence, the total pressure drop for a bank of Nhp hairpin exchangers with length Lhp is given by 4 
Eq. (5.27): 5 
𝛥𝑃𝑖 =
𝐺𝑖2𝐾𝑖
2𝜌
�2𝑁ℎ𝑝 − 1� +
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑖
𝐷𝑖𝜌
𝐺𝑖2 (5.27) 
Worked Example 5–1 6 
A process stream of 1.26 kg/s of ethylenediamine should be heated from 15 °C to 50 °C 7 
through the inner pipe in a serial bank of seven 5 m length hairpin double pipe heat exchangers. The 8 
internal tube is carbon steel NPS 32 mm-Schedule 40. Determine if the inner tube pressure drop is 9 
within an allowable limit of 150 kPa. 10 
Solution 11 
a) Fluid physical properties 12The average temperature of the Ethylenediamine inside the exchanger is: 13 
𝑇𝑚 =
𝑇𝑖𝑛
2
+
𝑇𝑜𝑢𝑡
2
=
288.15
2
+
323.15
2
= 305.65 𝐾 
From Appendix C.11, the estimated physical properties of the fluid given in the following 14 
table: 15 
𝑬𝒕𝒉𝒚𝒍𝒆𝒏𝒆𝒅𝒊𝒂𝒎𝒊𝒏𝒆:𝟑𝟎𝟓.𝟔𝟓𝑲− 𝟏𝟎𝟏𝟑𝟐𝟓𝑷𝒂 
𝜇 = 0.00145 s ⋅ Pa 
𝜇 = 1.45 cP 
𝜌 = 551.0
kg
m3
 
b) Piping specifications 16 
From the Appendix A.1 the tube specifications are: 17 
Tube: long radius 180° return bend 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
5 Pressure Drop 109 
NPS 32 mm-Schedule 40 
Di = ID = 35.08 mm (inner diameter) 
Do = OD = 42.2 mm (outer diameter) 
c) Friction factor 1 
Using the Haaland’s equation (Eq. (5.13)) to calculate the friction factor for a smooth pipe, we 2 
have: 3 
- Flow cross section and mass flux 4 
𝑆𝑖 =
𝜋𝐷𝑖2
4
=
𝜋0.0350802
4
= 0.00096652 𝑚2 
𝐺𝑖 =
𝑊𝑖
𝑆𝑖
=
1.2600
0.00096652
= 1303.6
𝑘𝑔
(𝑚2 ⋅ 𝑠)
 
- Reynolds number 5 
𝑅𝑒𝑖 =
𝐷𝑖𝐺𝑖
𝜇𝑖
=
0.035080
0.0014500
1303.6 = 31538.→ 𝑓𝑢𝑙𝑙𝑦 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 
The absolute roughness of a new carbon steel pipe surface from Appendix B.1 is estimated as 6 
ε = 0.035 mm, therefore the friction factor is given by: 7 
𝑓𝐹𝑖 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑖
�
10
9 + 6.9𝑅𝑒𝑖
�
=
0.41
𝑙𝑛2 �0.23 �(3.5000𝑒 − 5)0.035080 �
10
9
+ 6.931538.�
= 0.0063571 
d) Pressure drop 8 
- Distributed pressure drop 9 
Using Eq. (5.21), the distributed pressure loss in the tube is: 10 
𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 =
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑖
𝐷𝑖𝜌𝑖
𝐺𝑖2 =
4 ⋅ 0.0063571 ⋅ 5.0000
0.035080 ⋅ 551.00
1303.62 ⋅ 7.0000 = 78246.
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 78.246 𝑘𝑃𝑎 
- Localized pressure drop 11 
With the internal diameter Di(mm) = 35.08 mm, the friction loss coefficient can be evaluated 12 
from Eq. (5.24). 13 
𝐾𝑖 = 0.7 +
2000
𝑅𝑒𝑖
+
17.78
𝐷𝑖(𝑚𝑚)
= 0.7 +
17.78
35.080
+
2000
31538.
= 1.2703 
Therefore, the local pressure loss is given by Eq. (5.26): 14 
𝛥𝑃𝑖,𝑙𝑜𝑐 =
𝐺𝑖2𝐾𝑖
2𝜌𝑖
�2𝑁ℎ𝑝 − 1� =
1.2703 ⋅ 1303.62
2 ⋅ 551.00
(2 ⋅ 7.0000 − 1) = 25466.
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 25.466 𝑘𝑃𝑎 
- Total pressure drop 15 
The full pressure loss in the pipe is the summation of distributed and local losses, then: 16 
𝛥𝑃𝑖 = 𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑖,𝑙𝑜𝑐 = 25466. +78246. = (1.0371𝑒 + 5)
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 103.71 𝑘𝑃𝑎 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
110 5 Pressure Drop 
Considering the calculated pressure loss, we conclude that this bank of seven exchangers does 1 
not exceed the available pressure drop. 2 
5.4 Annulus Pressure Drop 3 
The fluid inside the annulus of a hairpin double-pipe exchanger experiences localized pressure 4 
drop at several places, namely: the entrance and exit nozzles and in the 180° return connecting the two 5 
hairpin legs (Figure 20). Therefore, the total annulus pressure is a combination of linear or distributed 6 
friction losses along the annulus channel with the localized losses at these specified spots. 7 
The distributed pressure loss is given by Eq.(5.28), where Dh = Ds−Do is the hydraulic diameter 8 
of the annulus channel (see Section 4.6.1). The annulus Fanning friction factor fFo must be evaluated 9 
with the Reynolds number based on this hydraulic diameter. 10 
𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 =
2𝐺𝑜2𝐿𝑓𝐹𝑜
𝐷ℎ𝜌
 (5.28) 
The flow cross section and the mass flux through the annulus channel (Go) are given by: 11 
𝑆𝑜 =
𝜋
4
(𝐷𝑠2 − 𝐷𝑜2) (5.29) 
𝐺𝑜 =
𝑊𝑜
𝑆𝑜
 (5.30) 
The equation for the local pressure loss in the annulus of a single hairpin, using the annulus 12 
friction loss coefficient Ko is: 13 
𝛥𝑃𝑜,𝑙𝑜𝑐 =
𝐺𝑜2𝐾𝑜
2𝜌
 (5.31) 
Typically, the annulus channel of two adjacent hairpins composing a bank is connected with 14 
rather short pipe sections, which are taken as part of the internal friction losses of each hairpin, 15 
therefore the localized pressure loss for a series association of Nhp exchangers may be expressed as: 16 
𝛥𝑃𝑜,𝑙𝑜𝑐 =
𝐺𝑜2𝐾𝑜𝑁ℎ𝑝
2𝜌
 (5.32) 
And the total pressure loss for the annulus of a bank with Nhp exchangers of effective length 17 
Lhp comes from Eq. (5.28) and (5.32) in the form: 18 
𝛥𝑃𝑜 =
𝐺𝑜2𝐾𝑜𝑁ℎ𝑝
2𝜌
+
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑜
𝐷ℎ𝜌
𝐺𝑜2 (5.33) 
The way of estimating the annulus additional pressure drop in Eq. (5.31) is not universally 19 
settled in the technical literature. Generally, the localized pressure drop is considered by adding a 20 
certain number of velocity heads regarding the fluid returns, entrance and exit losses; however, the 21 
recommended amount of velocity heads varies significantly depending on the source. For example: 22 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
5 Pressure Drop 111 
• Kern (1950) [44] and Coker (1995) [90] recommend the addition of one velocity head 1 
per hairpin exchanger to compensate for entrance and exit losses. 2 
• Bejan & Kraus (2003) [48] account with four velocity heads for the return and localized 3 
losses in a single hairpin exchanger. 4 
• Serth (2007) [28] suggests adding 1.2 (turbulent flow) to 1.5 (laminar flow) velocity 5 
head for a long radius 180° return bend, and the addition of another 1.5 velocity heads 6 
for entrance and exit losses if the flow is turbulent or 3 velocity heads for laminar flow, 7 
according to standard formulas for head losses in nozzles found in Ref. [20]. Therefore, 8 
this method adds a total of 2.7 − 4.5 velocity heads, almost quintuplicating the value 9 
placed by Refs. [44] and [90]. 10 
 11 
Figure 23: Hairpin exchanger with annuli connected by a straight 12 
pipe. [Adapted from Ref. [34]] 13 
The aforementioned methods for estimating the localized pressure losses in the annulus are all 14 
based on the standard friction loss coefficient concept, i.e. the conventional K-factor method ([91], 15 
[92]), which represents the number of velocity heads lost due to a fitting in a pipeline circuit. The 16 
friction loss coefficient K has a fixed value for a given pipe fitting, not varying with the fitting 17 
dimension or with the flow regime. It was verified that this approach to evaluate local pressure losses 18 
can have significant accuracy limitations, since the K value is not actually constant. It may increase 19 
expressively for viscous flows under Re < 1000 and does not scale exactly with the size of the pipe 20 
fitting [16], because smaller diameter fittings are more sensitive to roughness and their K values tend 21 
to be higher in comparison with the same shape fitting in greater sizes, for example the K for a ¼ in 22 
elbow does not equal its value for a 4 in elbow. 23 
Currently, concerning the manner the annulus 180° return is assembled; heat exchanger 24 
manufactures may offer two distinct models of hairpins. This is illustrated in Figures 20 and 23. In the 25 
first type (Figure 20), the annulus fluid is reverted inside a chamber covered with a “bonnet” flanged 26 
to the end of the heat exchanger. In the second type shown in Figure 23, the fluid return is simply a 27 
straight pipe connecting both annuli in each leg of the hairpin. Since such difference in construction 28 
enables distinct spots of local friction losses, they are treated separately in the following Sections 5.4.1 29 
and 5.4.2. 30 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
112 5 Pressure Drop 
 1 
Figure 24: K-factor head loss coefficient for “bonnet-type” annulus 2 
return. Dh = Ds-Do is the hydraulic diameter in millimeters (Eq. 3 
(5.36)). 4 
5.4.1 Annulus with Bonnet-Type Return 5 
If the hairpin exchanger has one or multiple “U” shaped internal tubes. Conventionally, the 6 
annulus fluid return is built with a flanged bonnet covering the inner pipes. Referring to Figure 20, in 7 
the absence of the detailed nozzle specification, the entrance and exit regions of the annulus can be 8 
fairly approximated as a branched tee, i.e. a tee with one leg blocked, forcing the fluid to perform a 9 
90° curve. Basedon the annulus hydraulic diameter, the 2-K head loss coefficient for a standard 10 
welded or flanged tee [81] is given by Eq. (5.34). 11 
𝐾(𝑡𝑒𝑒) = 0.7 +
500
𝑅𝑒
+
17.78
𝐷ℎ(𝑚𝑚)
 (5.34) 
𝐾(𝑡𝑒𝑒) = 0.7 +
500
𝑅𝑒
+
0.7
𝐷ℎ(𝑖𝑛)
 (5.35) 
Annulus: “bonnet-type” return 
 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
5 Pressure Drop 113 
 1 
Figure 25: K-factor head loss coefficient for “bonnet-type” annulus 2 
return. Dh = Ds-Do is the hydraulic diameter in inches (Eq. (5.37)). 3 
Therefore, for this hairpin type, the flow path for the annulus fluid has its localized pressure 4 
losses approximated as two branched tees (mounted as 90° elbows) and a bonnet return of 180°, and 5 
the total number of velocity heads for such circuit is in Eqs. (5.36) and (5.37), where Dh is the 6 
hydraulic diameter of the annulus measured in millimeters and inches, respectively. The Figures 24 7 
and 25 show a graphic representation of the aforementioned equations. 8 
𝐾𝑜 = 1.75 +
2000
𝑅𝑒𝑜
+
44.45
𝐷ℎ(𝑚𝑚)
 (5.36) 
𝐾𝑜 = 1.75 +
2000
𝑅𝑒𝑜
+
1.75
𝐷ℎ(𝑖𝑛)
 (5.37) 
 9 
Figure 26: K-factor head loss coefficient for “straight-pipe” annulus 10 
return. Dh = Ds-Do is the hydraulic diameter in millimeters (Eq. 11 
(5.40)). 12 
Annulus: “bonnet-type” return 
 
Annulus: “straight-pipe” return 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
114 5 Pressure Drop 
Using Eq. (5.36), the complete equation for the exchanger pressure loss with a bonnet channel 1 
in the annulus is Eq. (5.38): 2 
𝛥𝑃𝑜 =
𝐺𝑜2𝑁ℎ𝑝
2𝜌
�1.75 +
2000
𝑅𝑒𝑜
+
44.45
𝐷ℎ(𝑚𝑚)
� +
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑜
𝐷ℎ𝜌
𝐺𝑜2 (5.38) 
𝛥𝑃𝑜 =
𝐺𝑜2𝑁ℎ𝑝
2𝜌
�1.75 +
2000
𝑅𝑒𝑜
+
1.75
𝐷ℎ(𝑖𝑛)
� +
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑜
𝐷ℎ𝜌
𝐺𝑜2 (5.39) 
5.4.2 Annulus with Straight-Pipe Return 3 
Manufactures also offer a hairpin model in which the return of the annulus fluid is performed 4 
through a simple straight pipe connection between the two annuli. In this case, there is no bonnet. An 5 
example of this construction is shown in Figure 23. 6 
The direct annuli connection enforces the flow through a couple of abrupt 90° turns, which is 7 
approximated by the pressure losses of two branched tees, as given by Eq. (5.34). Therefore, the 8 
localized total pressure loss for this hairpin type may be estimated with Eq. (5.40) or (5.41). 9 
𝐾𝑜 = 2.8 +
2000
𝑅𝑒𝑜
+
71.12
𝐷ℎ(𝑚𝑚)
 (5.40) 
𝐾𝑜 = 2.8 +
2000
𝑅𝑒𝑜
+
2.8
𝐷ℎ(𝑖𝑛)
 (5.41) 
For convenient quick calculations, the velocity heads number charts based on Eqs. (5.40) and 10 
(5.41) are plotted in Figures 26 and 27, respectively. 11 
 12 
Figure 27: K-factor head loss coefficient for “straight-pipe” annulus 13 
return. Dh = Ds-Do is the hydraulic diameter in inches (Eq. (5.41)). 14 
Annulus: “straight-pipe” return 
 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
5 Pressure Drop 115 
Worked Example 5–2 1 
Determine the pressure drop for the annulus of a bank of eight NPS 32 mm (Sch 40) × 80 mm 2 
(Sch 40) × 6 m long carbon steel hairpin double pipe heat exchangers. The processed fluid is a stream 3 
of 1.15 kg/s of heptanal, cooled from 65 °C to 38 °C. 4 
Solution 5 
a) Fluid physical properties 6 
The average temperature of the hot fluid from 65 °C to 38 °C is: 7 
𝑇𝑚 =
𝑇1,𝑖𝑛
2
+
𝑇1,𝑜𝑢𝑡
2
=
311.15
2
+
338.15
2
= 324.65 𝐾 
Using the Appendix C.11 , the estimated physical properties for the heptanal are: 8 
𝑯𝒆𝒑𝒕𝒂𝒏𝒂𝒍:𝟑𝟐𝟒.𝟔𝟓𝑲 − 𝟏𝟎𝟏𝟑𝟐𝟓𝑷𝒂 
𝜇 = 0.000642s ⋅ Pa 
𝜇 = 0.642cP 
𝜌 = 689.0
kg
m3
 
 9 
b) Piping specifications 10 
The annulus channel of the heat exchanger is delimited by the external diameter of the inner 11 
pipe Do (NPS 32 mm – schedule 40) and internal diameter of the outer pipe Ds (NPS 80 mm – 12 
schedule 40). From the Appendix A.1 the tube specifications for both pipes are: 13 
NPS 32 mm - Schedule 40 
Di = ID = 35.08 mm (inner diameter) 
Do = OD = 42.2 mm (outer diameter) 
NPS 80 mm - Schedule 40 
Ds = ID = 77.92 mm (inner diameter) 
OD = 88.9 mm (outer diameter) 
c) Friction factor 14 
The annulus cross section and fluid mass flux are evaluated from Eqs. (5.29) and (5.30) as: 15 
- Annulus mass flux 16 
𝑆𝑜 =
𝜋
4
(𝐷𝑠2 − 𝐷𝑜2) =
𝜋
4
(0.0779202 − 0.0422002) = 0.0033699 𝑚2 
𝐺𝑜 =
𝑊𝑜
𝑆𝑜
=
1.1500
0.0033699
= 341.26
𝑘𝑔
(𝑚2 ⋅ 𝑠)
 
- Hydraulic diameter (Eq. (4.41)) and Reynolds number 17 
𝐷ℎ = 𝐷𝑠 − 𝐷𝑜 = 0.077920 − 0.042200 = 0.035720 𝑚 
𝑅𝑒𝑜 =
𝐷ℎ𝐺𝑜
𝜇𝑜
=
0.035720
0.00064200
341.26 = 18987. 
The absolute roughness of the carbon steel (new) pipe surface from Appendix B.1 is estimated 18 
as ε = 0.035 mm, therefore the friction factor is estimated by the Haaland’s equation (Eq. (5.13)): 19 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
116 5 Pressure Drop 
𝑓𝐹𝑜 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷ℎ
�
10
9 + 6.9𝑅𝑒𝑜
�
=
0.41
𝑙𝑛2 �0.23 �(3.5000𝑒 − 5)0.035720 �
10
9
+ 6.918987.�
= 0.0069739 
d) Distributed pressure drop 1 
𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 =
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑜
𝐷ℎ𝜌𝑜
𝐺𝑜2 =
4 ⋅ 0.0069739 ⋅ 6.0000
0.035720 ⋅ 689.00
341.262 ⋅ 8.0000 = 6336.0
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 6.336 𝑘𝑃𝑎 
e) Localized pressure drop 2 
From Eq. (5.36), with the hydraulic diameter expressed in millimeters Dh(mm) = 35.72 mm: 3 
𝐾𝑜 = 1.75 +
2000
𝑅𝑒𝑜
+
44.45
𝐷ℎ(𝑚𝑚)
= 1.75 +
44.45
35.720
+
2000
18987.
= 3.0997 
𝛥𝑃𝑜,𝑙𝑜𝑐 =
𝐺𝑜2𝐾𝑜𝑁ℎ𝑝
2𝜌𝑜
=
3.0997 ⋅ 341.262 ⋅ 8.0000
2 ⋅ 689.00
= 2095.7
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 2.0957 𝑘𝑃𝑎 
f) Total annulus pressure drop 4 
Combining the distributed and localized contributions, we get the total annulus pressure drop 5 
as: 6 
𝛥𝑃𝑜 = 𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑜,𝑙𝑜𝑐 = 2095.7 + 6336.0 = 8431.7
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 8.4317 𝑘𝑃𝑎 
5.5 Allowable Pressure Drop 7 
Before the installation of any piece of equipment in a process line, it should assure that the line 8 
has enough head to support the new equipment in operation; otherwise, the stream will be obstructed. 9 
This premise applies to heat exchangers also. Therefore, right after the thermal design is done; the 10 
engineer is responsible for evaluating the pressure drop (or head loss) that the heat exchanger is going 11 
to produce during operation. If the process line has not enough pump (or compressor/blower) power 12 
available, then the heat exchanger must be redesigned in order to meet the allowable pressure drop 13 
requirements. 14 
The allowance for pressure drop in a heat exchanger is a balance between the cost of the 15 
equipment itself and the cost of the pumping power for its operation. In basic terms, with given 16 
process conditions and equipment type, heat exchangers designed to produce higher pressure drop are 17 
almost always smaller (i.e. cheaper) than units with lower pressure drop. In compensation, the smaller 18 
exchanger with higher pressure drop will require more energy to run the pumps moving the fluids 19 
during the whole lifetime of the equipment. Therefore, clearly we see a tradeoff in the sense that there 20 
must be a pressure drop specification that minimizes the total cost (fixed and variable costs) associated 21 
with the heat exchanger. 22 
In principle, there is no universal standard for the allowable pressure loss in a chemical plant. 23 
Each process and, more specifically, each process line may have a different pressure drop threshold. 24 
Commonly, the pressure drop for a process stream is determined by the nature of the process, or the 25 
process conditions. In vacuum lines, the pressure drop available may be quite restricted as a few Pa; in 26 
contrast, high pressure systems may use dozens of kPa to drive the fluids. As a general rule for an 27 
initial reference value, for low to medium viscosity liquids, the allowable pressure drop for liquids 28 
should be in the range of 35 – 70 kPa (5 – 10 psi) [44]. Other possible pressure drop design limits are 29 
shown in Table 1. 30 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat ExchangersExplained 
5 Pressure Drop 117 
Table 1: General reference design criteria for heat exchange allowable pressure drop. 1 
Allowable Pressure Drop (General Rule) 
Liquids 
Low viscosity: < 1 mPa-s (1 cP) 35 kPa (5 psi) [19] 
Medium viscosity: 1 – 10 mPa-s 
(1 – 10 cP) 
50 – 70 kPa (7 – 10 psi) ([93], [94], 
[18], [19]) 
High viscosity Possibly > 70 kPa (10 psi), according to the pumping system. 
Gases and vapors, pressurized19 
1 – 2 bar 
> 10 bar 
0.5 × gauge pressure (50 – 100 kPa) 
0.1× gauge pressure (> 100 kPa) [19] 
Gases and vapors, vacuum 
High vacuum 
Medium vacuum 
0.4 – 0.8 kPa (0.06 – 0.12 psi) 
0.1× absolute pressure [19] 
 2 
Problems 3 
5.1) A fluid with a flow rate of 13492.0 lb/h is cooled in a 32.8 ft long, ID = 1.049in, straight 4 
pipe. The inlet and outlet temperatures are 140.0°F and 104.0°F, respectively. The pipe can be 5 
considered smooth and its inner surface temperature is tw = 95.1°F. Determine: 6 
a) Heat transfer coefficient, assuming constant physical properties. 7 
b) Heat transfer coefficient improved with the Prandtl correction factor. 8 
c) Fluid pressure drop. 9 
Fluid properties: 10 
T = 122.0°F (582.0°R) 
μ = 3.27lb/(ft⋅h) 
μ = 1.35cP 
ρ = 52.3lb/ft3 
Cp = 0.597BTU/(lb⋅°F) 
k = 0.0868BTU/(ft⋅h⋅°F) 
Pr = 22.5 
T = 95.1°F (555.0°R) 
μ = 4.36lb/(ft⋅h) 
μ = 1.8cP 
ρ = 53.4lb/ft3 
Cp = 0.581BTU/(lb⋅°F) 
k = 0.0886BTU/(ft⋅h⋅°F) 
Pr = 28.6 
Answer: (a) h = 626.73 BTU/(ft2 h °F) (b) h = 610.41 BTU/(ft2 h °F) (c) ΔPdist = 6.39 psi 11 
5.2) A cooling water stream is heated in the annulus of a double-pipe hairpin exchanger. The 12 
water flow rate, entrance and exit temperatures are 16574.0 lb/h, 68.0°F and 86.0°F, respectively. The 13 
hairpin is Lhp = 16.4 ft long and has a straight pipe annulus return. The inner and outer annulus 14 
diameters are Do = 1.315 in and Ds = 2.469 in. Calculate the (a) distributed, (b) localized and (c) total 15 
annulus pressure drop. 16 
19 Refs. [93] and [94] recommend the range 5 – 20 kPa (1 – 3 psi) as a rough rule. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
118 5 Pressure Drop 
Answer: (a) ΔPo,dist = 0.47431psi (b) ΔPo,loc = 0.31993psi (c) ΔPo = 0.79424psi 1 
5.3) A bank of 6 double-pipe exchangers is designed to heat 7 kg/s of benzene from 30°C to 2 
50°C in the annulus, by cooling 3.94 kg/s of toluene in the tube from 85°C to 53°C. Each hairpin is 3 
built in stainless steel with 4 m length and piping of DN 50 mm and DN 90 mm, both schedule 40S. 4 
The annulus fluid return is a straight pipe. Determine the friction factor for (a) the tube and (b) the 5 
annulus, using the Haaland’s equation; (c) the localized pressure drop in the pipe. Assuming an 6 
allowable pressure drop of 70kPa for both streams, (d) verify if this bank is suitable to perform the 7 
service regarding to its developed pressure drop. 8 
Answer: (a) fF=0.0044427 (b) fF=0.0051723 (c) ΔPi,loc=23.225kPa (d) No, it is not. 9 
5.4) As an attempt to meet the service requirements indicated in the Problem 5.3 with the 10 
already designed heat exchangers bank, the fluid allocation is inverted. Calculate (a) the tube and (b) 11 
the annulus total pressure drop. 12 
Answer: (a) ΔPi=167.6kPa (b) ΔPo=49.734kPa. 13 
5.5) In a biofuel production facility, a stream of 3 kg/s of ethanol need to chill before storage 14 
from 60°C to 35°C. It is available cooling water at an initial temperature of 25°C. The supply of water 15 
is not an issue, but its outlet temperature must not surpass 50°C. The storehouse has in standby a 16 
battery of 11 carbon steel, DN 50×80mm/10S, 4m long double pipe hairpins. The annulus return is of 17 
bonnet type and the ethanol is allocated in the inner tube. Calculate the following: 18 
a) Required heat transfer rate. 19 
b) Final overall heat transfer coefficient. 20 
c) Total pressure loss in the tube. 21 
d) Total pressure loss in the annulus. 22 
Answer: (a) 2.1e+5 W (b) 1259.7 W/(m2⋅K) (c) ΔPi = 52.566kPa (d) ΔPo = 46.951kPa. 23 
 24 
 25 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
6 Series-Parallel Arrangements 119 
6 Series-Parallel Arrangements 1 
As an outcome of its simple construction and relatively small heat transfer area per volume 2 
ratio, quite rarely a single double-pipe heat exchanger is designed to match a real-world heat transfer 3 
service. Possibly, this would require a very long hairpin with large diameters non-standardized pipes, 4 
which is not practical economically. The viable solution is to design an appropriate sized double-pipe 5 
exchanger and combine a number of them in a bank or battery by connecting in sequence their tubes 6 
and annulus, what we could designate as a series-series association for obvious reasons. With the 7 
fluids flowing in countercurrent mode, the maximum value of the LMTD is achieved in this 8 
arrangement, and the heat transfer area is minimized accordingly. This common “series-series” 9 
association is demonstrated in Figure 28, in which two exchangers appear with their tubes and annulus 10 
connected in sequence. 11 
T1
T
T2
t2
t
t1
(I)
(II)
w
W
 12 
Figure 28: Hairpin heat exchangers with both pipe and annulus 13 
associated in series. 14 
 15 
If both process streams have comparable flow rates and temperature range between the inlet 16 
and outlet of the exchanger, the series-series arrangement will be the most probable to meet the service 17 
effectively, however there are situations in which this arrangement is not the best design choice. Let us 18 
consider the integral energy balance for the fluids in the form: 19 
𝐶𝑊(𝑇1 − 𝑇2) = 𝑐𝑤(𝑡2 − 𝑡1) (6.1) 
By rearranging, we have the mass flow rate ratio as: 20 
𝑊
𝑤
=
𝑐(𝑡2 − 𝑡1)
𝐶(𝑇1 − 𝑇2)
 (6.2) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
120 6 Series-Parallel Arrangements 
From Eq. (6.2), we realize that the flow rate for a given stream is inversely proportional to the 1 
product of its specific heat with the temperature range. Therefore, if one of the fluids has a relatively 2 
small specific heat or a very short temperature range, its flow rate needs to be quite large to match the 3 
heat transfer duty. This situation can raise a difficulty in the specification of the heat exchanger, 4 
because the fluid with the larger flow rate may exceed the available pressure drop in great extent, 5 
while the other fluid produces a relatively small pressure drop, impairing its heat transfer coefficient, 6 
and possibly the performance of the equipment as a whole. A conceivable solution is to divide both 7 
streams in halves and use two independent batteries of exchangers in parallel. While such design may 8 
solve the excess of pressure drop, the partition of the stream with smaller flow rate will reduce even 9 
more its heat transfer coefficient in both banks of exchangers, increasing the required total heat 10 
transfer area, the size of the individual hairpins and the equipment cost as a whole. 11 
Clearly, there are drawbacks with the previous approach, and a better design can be the sole 12 
division of the stream with larger flow rate, which is exceeding the allowable pressure loss. It is 13 
possible with the series-parallel association of double-pipe exchangers. As shown in Figure 29, the 14 
inner fluid in the tube is divided with half flow rate entering the exchanger (I), and the other half 15 
entering the exchanger (II). As a result, the flow pipe length L and the mass flux G are reduced by a 16 
factor of ½. From Eq. (4.13) for tube pressure loss, the factor (L/2)(G/2)2 = (1/8)(LG2) cuts the 17 
pressure loss to 1/8 of the original value. Readily, we see that the division of the tube fluid in three 18 
parallel streams reduces the pressure loss by 1/27, and the division in N pipe parallel streams cuts the 19 
pressure loss by a factor of 1/N3, which is a significant reduction, permitting the accommodation of 20 
nearly arbitrarilylarge flow rates as the inner fluid. 21 
T1
T
T2
t2
(I)
(II)
t1 w
w/2
w/2
 22 
Figure 29: Hairpin heat exchangers with annulus connected in series 23 
and the pipe connected in parallel. 24 
Notice that if a service requires larger heat transfer surfaces, each hairpin depicted in Figure 29 25 
may in fact be replaced by a bank of exchangers with the pipe and annuli both connected in series 26 
inside the bank. A design in which two banks of hairpins are linked in series-parallel arrangement is 27 
presented in Figure 30. In this case, the inner tube stream is divided in two parallel streams with each 28 
half fed in one of the banks. The annulus fluid flows in series through all the exchangers. 29 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
6 Series-Parallel Arrangements 121 
6.1 Mean Temperature Difference 1 
Kern (1950) [44] demonstrated that the mean temperature difference is not the log mean 2 
temperature difference (LMTD) for the series-parallel association of double pipe exchangers. In fact, 3 
the use of the LMTD for sizing could generate unacceptable deviations from the true mean 4 
temperature difference, and consequently for the calculated heat transfer area. 5 
Consider the design equation (Eq. (6.3)), where Um is the mean overall heat transfer coefficient 6 
representing the entire heat transfer area A, and ΔTm is the “exact” mean of all temperature differences 7 
along the whole exchanger. 8 
𝑞 = 𝑈𝑚𝐴𝛥𝑇𝑚 (6.3) 
T1 t2
(III)
(IV)
t1T2
(I)
(II)
w
w/2
w/2
W
 9 
Figure 30: Bank of four hairpin heat exchangers with annulus 10 
connected in series and the pipe fluid divided in two parallel streams. 11 
Starting from Eq. (6.3), Kern (1950) showed that the actual mean temperature difference for 12 
the series-parallel association has the form given in Eq. (6.4), i.e. a “series−parallel temperature 13 
factor” (Fsp), varying from zero to one, multiplied by the maximum temperature difference found in 14 
the equipment (T1-t1). 15 
𝛥𝑇𝑚 = 𝐹𝑠𝑝(𝑇1 − 𝑡1) (6.4) 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
122 6 Series-Parallel Arrangements 
6.1.1 Hot Fluid in Series and Cold Fluid in Parallel 1 
After extensive algebra, the Fsp factor for the hot fluid flowing in the annulus and the cold fluid 2 
divided into nc parallel pipe streams was evaluated in the form of Eq. (6.5). For a detailed derivation, 3 
please follow Ref. [44]. 4 
𝐹𝑠𝑝 =
𝑃1 + 𝑅1(1 − 𝑃1) − 1
𝑅1𝑛𝑐𝑙𝑛 �
1
𝑅1
�𝑅1 �
1
𝑃1
�
1
𝑛𝑐 − � 1𝑃1
�
1
𝑛𝑐 + 1��
 
(6.5) 
𝑃1 =
𝑇2 − 𝑡1
𝑇1 − 𝑡1
 (6.6) 
𝑅1 =
𝑇1 − 𝑇2
𝑛𝑐(𝑡2 − 𝑡1)
 (6.7) 
Worked Example 6–1 5 
In a service, a battery of double-pipe heat exchangers associated in series-parallel, with the hot 6 
fluid through the annulus in series and temperature ranging from 250 °C to 100 °C. The cold fluid has 7 
a very large flow rate, being partitioned in four parallel streams in the tube of the exchangers, with 8 
temperature ranging from 83 °C to 120 °C. What is the mean temperature difference in this bank? 9 
Solution 10 
Since the cold fluid is in parallel, we should use Eq. (6.5), with parameters given from Eqs. 11 
(6.6) and (6.7) as: 12 
𝑃1 =
𝑇2 − 𝑡1
𝑇1 − 𝑡1
=
100 − 83
250 − 83
= 0.1 
𝑅1 =
𝑇1 − 𝑇2
𝑛𝑐(𝑡2 − 𝑡1)
=
250 − 100
4(120 − 83)
= 1.01 
The temperature difference factor is: 13 
𝐹𝑠𝑝 =
0.10 + 1.01(−0.10 + 1) − 1
1.01 ⋅ 4 𝑙𝑛 � 11.01�1.01 �
1
0.10�
1
4
− � 10.10�
1
4
+ 1��
= 0.292 
Substituting in Eq. (6.4), the actual mean temperature difference for the entire bank is: 14 
𝛥𝑇𝑚 = 𝐹𝑠𝑝(𝑇1 − 𝑡1) = 0.292(250 − 83) = 48.8 °𝐶 
Let us compare this result with the log mean temperature difference for countercurrent flow 15 
(ΔTlm.c), calculated as: 16 
𝛥𝑇𝑙𝑚,𝑐 =
𝑇1 − 𝑇2 + 𝑡1 − 𝑡2
𝑙𝑛 �𝑇1 − 𝑡2𝑇2 − 𝑡1
�
=
−100 − 120 + 250 + 83
𝑙𝑛 �−120 + 250100 − 83 �
= 55.5 °𝐶 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
6 Series-Parallel Arrangements 123 
Considering that the logarithmic mean temperature difference is 55.5 °C and the “actual” 1 
reference value is 48.8 °C, the relative error between both results is about 13.8%. Since the actual 2 
mean temperature difference is lower, the use of the log mean temperature would lead to an 3 
underestimated heat transfer area. 4 
6.1.2 Cold Fluid in Series and Hot Fluid in Parallel 5 
When the cold fluid goes through the annulus in series and the hot fluid is divided into nh 6 
parallel tube streams, the proper equation for Fsp is Eq. (6.8). 7 
𝐹𝑠𝑝 =
𝑃2𝑅2 − 𝑃2 − 𝑅2 + 1
𝑛ℎ𝑙𝑛 �−𝑅2 �
1
𝑃2
�
1
𝑛ℎ + 𝑅2 + �
1
𝑃2
�
1
𝑛ℎ�
 
(6.8) 
𝑃2 =
𝑇1 − 𝑡2
𝑇1 − 𝑡1
 (6.9) 
𝑅2 =
𝑛ℎ(𝑇1 − 𝑇2)
𝑡2 − 𝑡1
 (6.10) 
Worked Example 6–2 8 
Consider the same service presented in the Worked Example 6–1 . At this time, however, the 9 
cold fluid is the smaller stream and should be allocated in the annulus in series with its temperature 10 
ranging from 83 °C to 120 °C. The hot fluid will be divided in four parallel streams in the tube of the 11 
exchangers, with temperature ranging from 250 °C to 100 °C. What is the mean temperature 12 
difference in this bank? 13 
Solution 14 
With Eq. (6.9) and (6.10), we have: 15 
𝑃2 =
𝑇1 − 𝑡2
𝑇1 − 𝑡1
=
250 − 120
250 − 83
= 0.78 
𝑅2 =
𝑛ℎ(𝑇1 − 𝑇2)
𝑡2 − 𝑡1
=
4(250 − 100)
120 − 83
= 16.2 
The temperature factor comes from Eq. (6.8): 16 
𝐹𝑠𝑝 =
0.78 ⋅ 16.2 − 0.78 − 16.2 + 1
4 𝑙𝑛 �−16.2 � 10.78�
1
4
+ 16.2 + � 10.78�
1
4
�
= 0.214 
Substituting in Eq. (6.4), the actual mean temperature difference for the entire bank is: 17 
𝛥𝑇𝑚 = 𝐹𝑠𝑝(𝑇1 − 𝑡1) = 0.214(250 − 83) = 35.8 °𝐶 
The log mean temperature difference in counterflow in this case is: 18 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
124 6 Series-Parallel Arrangements 
𝛥𝑇𝑙𝑚,𝑐 =
𝑇1 − 𝑇2 + 𝑡1 − 𝑡2
𝑙𝑛 �𝑇1 − 𝑡2𝑇2 − 𝑡1
�
=
−100 − 120 + 250 + 83
𝑙𝑛 �−120 + 250100 − 83 �
= 55.5 °𝐶 
Considering that the counter current logarithmic mean temperature difference is 55.5 °C and 1 
the “actual” value is only 35.8 °C, the relative error between both results is about 55.2%. Such 2 
deviation is a dangerous risk in the design, because if the ΔTlm were used to size this bank the heat 3 
transfer area would be seriously underspecified, and the energy duty would not be fulfilled at all. 4 
6.1.3 Choice of the Fluid Division 5 
Although, a series-parallel association is commonly encountered with tube in parallel and the 6 
annulus in series, as in Figures 29 and 30, this condition is not mandatory. The decision for splitting 7 
the tube or annulus stream is irrelevant for the mean temperature difference between the fluids; 8 
therefore, the designer may choose to set the annulus fluid in parallel if necessary. 9 
That said, there is a reason for the division of the inner fluid being far more common 10 
industrially, and it is related to piping and fittings. Due to the very structure of a hairpin, with the 11 
annuli nozzles placed side by side in a bank, joining the annuli in series is easier and quicker, and 12 
requires less piping and man work. 13 
6.2 Pressure Drop Calculation 14 
6.2.1 Tube Pressure Drop 15 
In a bank of Nhp hairpins of length Lhp connected in series-parallel arrangement, with the pipe 16 
fluid partitioned in n parallel streams, the pressure loss equation should be modified in the following 17 
ways: 18 
1. The actual mass flux in each individual pipe stream is Gi/n. 19 
2. The number of exchangers comprising each train is Nhp/n. 20 
With these changes, the total pressure drop for the pipe may be calculated by rewriting Eq. 21 
(5.27) in the form of Eq. (6.11), where Ki is the friction loss coefficient for the inner pipe with internal 22 
diameter Di expressed in inches, evaluated from Eq. (5.24). 23 
𝛥𝑃𝑖 =
𝐺𝑖2𝐾𝑖
2𝑛2𝜌
�
2𝑁ℎ𝑝
𝑛
− 1� +
4𝐺𝑖2𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑖
𝐷𝑖𝑛3𝜌
 (6.11) 
Notice that the inner tube friction factor (fFi) is tobe evaluated with the actual mass flux of the 24 
divided stream Gi/n. 25 
6.2.2 Annulus Pressure Drop 26 
For a series-parallel association of double-pipe heat exchangers involving only the division of 27 
the tube stream, the pressure loss calculation for the annulus passage remains unaltered, such as 28 
discussed is Section 5.4. The Eq. (5.33) may be used directly, which is reproduced below for 29 
convenience, where Nhp is the total number of exchangers with length Lhp and hydraulic diameter Dh. 30 
The friction loss factor Ko is obtained from Eq. (5.36) for bonnet type return or from Eq. (5.40) if the 31 
exchanger has a direct return for the annular fluid. 32 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
6 Series-Parallel Arrangements 125 
𝛥𝑃𝑜 =
𝐺𝑜2𝐾𝑜𝑁ℎ𝑝
2𝜌
+
4𝐺𝑜2𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑜
𝐷ℎ𝜌
 (6.12) 
Worked Example 6–3 1 
A stream of 1.8 kg/s of triethylene glycol is cooled with 2 kg/s water in a bank of 20 series-2 
parallel carbon steel double-pipe exchangers (hairpin with return bend). The triethylene glycol goes 3 
inside the inner tube divided in two parallel streams, while the water flows through the annulus in 4 
series in the whole bank. The mean bulk temperatures are 71 °C and 42.7 °C for triethylene glycol and 5 
water, respectively. The internal pipe is ID = 52.48 mm and OD = 60.3 mm. The outer tube is 6 
ID = 90.12 mm. Estimate the tube and annulus pressure drop for this bank. 7 
Solution 8 
1) Fluid properties 9 
Using the mean bulk temperatures, the physical properties are estimated from Appendix C.8 10 
and C.13 as: 11 
Tube: Triethylene glycol 
344.15K − 101325 Pa 
Annulus: Water 
305.15K − 101325 Pa 
μ = 0.00732 s⋅Pa 
μ = 7.32 cP 
ρ = 1230.0 kg/m3 
Cp = 1860.0 J/(kg⋅K) 
k = 0.196 W/(m⋅K) 
Pr = 69.2 
μ = 0.000775 s⋅Pa 
μ = 0.775 cP 
ρ = 1060.0 kg/m3 
Cp = 4070.0 J/(kg⋅K) 
K = 0.62 W/(m⋅K) 
Pr = 5.09 
2) Tube 12 
The inner fluid is divided into two streams, therefore the actual mass flow rate is: 13 
𝑊𝑖 = 0.9
𝑘𝑔
𝑠
 
For the internal diameter Di = ID = 52.48 mm, the inner pipe cross section, mass flux and 14 
Reynolds are: 15 
𝑆𝑖 =
𝜋𝐷𝑖2
4
=
𝜋0.0524802
4
= 0.0021631𝑚2 
𝐺𝑖 =
𝑊𝑖
𝑆𝑖
=
0.90000
0.0021631
= 416.07
𝑘𝑔
(𝑚2 ⋅ 𝑠)
 
𝑅𝑒𝑖 =
𝐷𝑖𝐺𝑖
𝜇𝑖
=
0.052480
0.0073200
416.07 = 2983.0 → 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 
From Appendix B.1 , let us take the absolute roughness of a carbon steel pipe surface as 16 
ε = 0.035 mm. In this case, the friction factor accounting for the tube roughness can be evaluated 17 
conservatively from the Haaland’s equation: 18 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
126 6 Series-Parallel Arrangements 
𝑓𝐹𝑖 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑖
�
10
9 + 6.9𝑅𝑒𝑖
�
=
0.41
𝑙𝑛2 �0.23 �(3.5000𝑒 − 5)0.052480 �
10
9
+ 6.92983.0�
= 0.011238 
We must also halve the total number of exchangers in the bank, then: 1 
𝑁ℎ𝑝 =
𝑁ℎ𝑝,𝑡𝑜𝑡𝑎𝑙
𝑛
=
20.000
2.0000
= 10.000 
The distributed pressure drop is calculated from: 2 
𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 =
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑖
𝐷𝑖𝜌𝑖
𝐺𝑖2 =
4 ⋅ 0.011238 ⋅ 10.000
0.052480 ⋅ 1230.0
4.0000 ⋅ 416.072 = 4822.2
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 4.8222𝑘𝑃𝑎 
The localized pressure drop for the inner fluid is given by: 3 
𝐷𝑖(𝑚𝑚) = 52.48𝑚𝑚 
𝐾𝑖 = 0.7 +
2000
𝑅𝑒𝑖
+
17.78
𝐷𝑖(𝑚𝑚)
= 0.7 +
17.78
52.480
+
2000
2983.0
= 1.7093 
𝛥𝑃𝑖,𝑙𝑜𝑐 =
𝐺𝑖2𝐾𝑖
2𝜌𝑖
�2𝑁ℎ𝑝 − 1� =
1.7093 ⋅ 416.072
2 ⋅ 1230.0
(2 ⋅ 10.000 − 1) = 2285.4
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 2.2854𝑘𝑃𝑎 
The total tube pressure drop, combining distributed and local losses, is: 4 
𝛥𝑃𝑖 = 𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑖,𝑙𝑜𝑐 = 2285.4 + 4822.2 = 7107.6
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 7.1076𝑘𝑃𝑎 
3) Annulus 5 
The heat exchangers annuli are connected in series, then the total flow rate is used in the 6 
pressure drop calculation: 7 
𝑊𝑜 = 2.0
𝑘𝑔
𝑠
 
a) Reynolds number 8 
The cross section area and mass flux for the annulus are: 9 
𝑆𝑜 =
𝜋
4
(𝐷𝑠2 − 𝐷𝑜2) =
𝜋
4
(0.0901202 − 0.0603002) = 0.0035229𝑚2 
𝐺𝑜 =
𝑊𝑜
𝑆𝑜
=
2.0000
0.0035229
= 567.71
𝑘𝑔
(𝑚2 ⋅ 𝑠)
 
Using the equivalent diameter as the hydraulic diameter, we have: 10 
𝐷𝑒 = 𝐷ℎ = 𝐷𝑠 − 𝐷𝑜 = 0.090120 − 0.060300 = 0.029820 𝑚 
And finally the Reynolds number is: 11 
𝑅𝑒𝑜 =
𝐷𝑒𝐺𝑜
𝜇𝑜
=
0.029820
0.00062400
567.71 = 27130 → 𝑓𝑢𝑙𝑙𝑦 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
6 Series-Parallel Arrangements 127 
b) Friction factor 1 
With the Haaland’s equation, the Fanning friction factor is estimated as: 2 
𝑓𝐹𝑜 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑒
�
10
9 + 6.9𝑅𝑒𝑜
�
=
0.41
𝑙𝑛2 �0.23 �(3.5000𝑒 − 5)0.029820 �
10
9
+ 6.927130.�
= 0.0066190 
c) Pressure drop 3 
The total number of hairpins in series is: 4 
𝑁ℎ𝑝 = 𝑁ℎ𝑝,𝑡𝑜𝑡𝑎𝑙 = 20.000 = 20.000 
From Eq. (5.28), the distributed pressure loss can be calculated as: 5 
𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 =
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑜
𝐷ℎ𝜌𝑜
𝐺𝑜2 =
4 ⋅ 0.0066190 ⋅ 20.000
0.029820 ⋅ 1050.0
4.0000 ⋅ 567.712 = 21802.
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 21.802𝑘𝑃𝑎 
The friction loss coefficient for a straight pipe return from Eq. (5.40) is: 6 
𝐷ℎ(𝑚𝑚) = 29.82𝑚𝑚 
𝐾𝑜 = 2.8 +
2000
𝑅𝑒𝑜
+
71.12
𝐷ℎ(𝑚𝑚)
= 2.8 +
71.12
29.820
+
2000
27130.
= 5.2587 
The local pressure drop is then obtained as: 7 
𝛥𝑃𝑜,𝑙𝑜𝑐 =
𝐺𝑜2𝐾𝑜𝑁ℎ𝑝
2𝜌𝑜
=
20.000 ⋅ 5.2587 ⋅ 567.712
2 ⋅ 1050.0
= 16141.
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 16.141𝑘𝑃𝑎 
And the total pressure loss in the annulus is the sum of distributed and local losses: 8 
𝛥𝑃𝑜 = 𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑜,𝑙𝑜𝑐 = 16141. +21802. = 37943.
𝑘𝑔
(𝑚 ⋅ 𝑠2)
 = 37.943𝑘𝑃𝑎 
Problems 9 
6.1) In a battery of double-pipe heat exchangers, the hot fluid temperature varies from 50°F to 10 
30°F, while the cold fluid temperature ranges from 22°F to 41°F. The heat load is 6.7e+5 BTU/h and 11 
the overall heat transfer coefficient was evaluated as U=103 BTU/(ft2⋅h⋅°F). Assuming the overall heat 12 
transfer coefficient is nearly constant, determine the effective mean temperature difference and the 13 
required heat transfer area for this service when a series-parallel arrangement is used with (a) the cold 14 
fluid and (b) the hot fluid divided in 3 parallel streams. 15 
Answer: (a) ΔTm = 3.8657°F, A=1682.7 ft2 (b) ΔTm = 2.8095°F; A=2315.3 ft2 16 
6.2) The benzene/toluene heat transfer service examined in the Problems 5.3 and 5.4 could not 17 
be accomplished by fluid relocation. Either way, the benzene stream exceeded the allowable the 18 
pressure loss. Analyze an alternative design in which the bank of 6 exchangers is restructured with 19 
benzene allocated in the tube divided in two parallel streams, while the exchangers’ annuli are kept 20 
connected in series. Determine: 21 
a) Tube friction factor, using Haaland’s equation. 22 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
128 6 Series-Parallel Arrangements 
b) Annulus friction factor, using Haaland’s equation. 1 
c) Distributed and localized pressure loss in the tube. 2 
d) Distributed and localized pressure loss in the annulus. 3 
e) Heat transfer coefficients for the internal and external fluids. 4 
f) Clean overall heat transfer coefficient (ignoring the wall conductive resistance). 5 
g) Clean overall heat transfer coefficient. 6 
h) Does the issue of exceeding the allowable pressure drop was solved? 7 
i) Can this series-parallel bank arrangement handle the required heat load? 8 
Answer: (a) fFi = 0.004632 (b) fFo = 0.0053007 (c) ΔPi,dist = 12.927kPa; ΔPi,loc = 8.0132kPa 9 
(d) ΔPo,dist = 25.967kPa; ΔPo,loc = 23.767kPa (e) hi = 2213.3W/(m2⋅K); ho = 1955.6W/(m2⋅K) 10 
(f) Uc = 970.41W/(m2⋅K) (g) Uc = 776.46W/(m2⋅K) (h) Yes (justify!) (i) No (justify!) 11 
6.3) A fluid is heated from 10°C to 50°C with 3.3 kg/s of another process stream varying from 12 
55°C to 40°C. The heat is transferred by a battery of 33 hairpins, where the annuli are connected in 13 
series and the hot fluid flows inside the pipe divided in three parallel streams, each one passing 14 
through 11 hairpins.15 
a) What is the mean temperature difference for this battery? 16 
b) If the divided stream is the cold fluid, what is the mean temperature difference? 17 
Answer: (a) ΔTm=11.491°C (b) Not defined 18 
6.4) Acetic acid at a flow rate of 10.0 kg/s is cooled from 65°C to 50°C using cooling water 19 
from 20°C to 35°C. The service is accomplished in a bank of 4 carbon steel double-pipe 20 
DN 50×90 mm/10S, 7 m long, exchangers with their annuli connected in series. The acetic acid flows 21 
in two parallel streams inside the tubes, while the water flows through the annulus. Determine: 22 
a) Mean temperature difference for this battery. 23 
b) Tube and annulus convective heat transfer coefficient. 24 
c) Clean overall heat transfer coefficient. 25 
d) Pressure loss in the tube and annulus. 26 
e) Percentage of the over-surface. 27 
Answer: (a) ΔTm = 29.462°C (b) hi = 2846.4W/(m2⋅K); ho = 4820.9W/(m2⋅K) 28 
(c) Uc = 1535.5W/(m2⋅K) (d) ΔPi = 35.188kPa; ΔPo = 28.838kPa (e) Over-surface = 65% 29 
6.5) Consider the service specified in the Problem 6.4. Test if a greater design margin can be 30 
attained by inverting the fluid allocation, associating the tubes in series and dividing the annulus fluid 31 
in two parallel streams. Evaluate: 32 
a) Mean temperature difference for this battery. 33 
b) Tube and annulus convective heat transfer coefficient. 34 
c) Clean overall heat transfer coefficient. 35 
d) Pressure loss in the tube and annulus. 36 
e) Percentage of over-surface. 37 
Answer: (a) ΔTm = 29.462°C (b) hi = 7526.0W/(m2⋅K); ho = 1800.0W/(m2⋅K) 38 
(c) Uc = 1317.8W/(m2⋅K) (d) ΔPi = 53.159kPa; ΔPo = 19.702kPa (e) Over-surface = 41.5% 39 
 40 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 129 
7 Heat Exchanger Design 1 
When dealing with heat exchanger design, the engineer’s most common tasks are twofold: 2 
1. Sizing a new heat exchanger. 3 
2. Rating the performance of an existing heat exchanger. 4 
The sizing procedure will be detailed in Section 7.4, while the rating task is discussed in 5 
Section 7.5. In principle, sizing or rating a heat exchanger is a simple matter of following the 6 
calculations in a rather straightforward routine; however, the engineer need to make important 7 
decisions in key steps, which, in the end, may determine the effectiveness of the design. Among these 8 
decisions are the fluid allocation and the acceptable margin embedded in the final design. 9 
7.1 Fluid Allocation 10 
Commonly, one of the first decisions the engineer needs to make relates to the fluid allocation, 11 
i.e., which fluid will be channeled through the inner tube or the annulus (shell) of the double-pipe heat 12 
exchanger. The fluids allocation has a direct impact on the equipment capital and operating costs. 13 
There is no strict rule for choosing if a particular fluid stream should flow internally or 14 
externally, therefore each heat transfer service has to be carefully considered under its own 15 
particularities. As a general guide, a more effective design can be achieved if the stream allocated in 16 
the inner tube has the following characteristics ([95], [96], [94]): 17 
1. More corrosive, erosive or hazardous 18 
The larger allowances for corrosion or erosion make the exchanger more expensive, once the 19 
required tube thickness may increase significantly. Placing the more aggressive fluid inside the tube 20 
prevents the selection of heavier schedules for both inner and outer tubes, since only the inner tube 21 
will be exposed to an accelerated damage. 22 
As might seem obvious, the reason for allocating the hazardous (i.e harmful or toxic) stream 23 
inside the tube is safety. If the leakage of one fluid involves great risk for people or to the 24 
environment, it is strongly recommended the allocation in the inner tube. In this case, if the fluid leaks, 25 
it remains contained by the outer pipe. 26 
2. More fouling 27 
In general, the inner tube passage is more accessible for mechanical cleaning. Therefore, the 28 
allocation of the stream with an increased propensity for fouling in the tube results in lower 29 
maintenance costs and shorter downtime. 30 
While the fouling formation may not be critical for fluid allocation in double-pipe exchangers, 31 
it might be a key factor to make this decision when designing multi-tube and finned heat exchangers, 32 
or even other types such as shell and tube exchangers. 33 
3. Excessively high temperature and operating pressure 34 
Tubular heat exchangers operating with streams in extreme conditions of temperature and 35 
pressure require thick tubes to sustain the physical integrity even under high mechanical stresses. 36 
Allocating the high temperature or high-pressure stream in the inner tube usually leads to the most cost 37 
effective design, since a thinner outer tube (shell) may be applied. 38 
4. Lower viscosity 39 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
130 7 Heat Exchanger Design 
Usually, the fluid of lower viscosity produces the higher heat transfer coefficient. It is easier 1 
and less expensive to install external fins; therefore, having the controlling fluid in the annulus (shell) 2 
allows the placement of external fins, which may increase significantly the overall heat transfer 3 
coefficient of the exchanger. 4 
5. Hotter fluid 5 
Typically, the methods for designing heat exchangers presume that all energy lost by the hot 6 
fluid goes to the cold fluid, i.e. no energy is lost to the environment. Allocating the hot fluid in the 7 
annulus undermines this assumption, especially if the stream operates at relatively high temperature. 8 
To minimize the problem, the heat exchanger may be thermally insulated, implying in an additional 9 
cost. 10 
7.2 Design Margin 11 
In broad terms, “design margin” in an engineering project refers to a performance excess 12 
attainable by the equipment, beyond what was primarily requested to the designer. The same concept 13 
appears with several denominations in the technical literature, such as “over−area”, “excess area”, 14 
“excess surface”, “over-sizing”, “safety factor”, among others. The engineer must be aware of the 15 
exact definition stated in these sources. 16 
For a heat exchanger, performance is usually related to its heat duty or heat transfer area. A 17 
deep discussion of design margins for heat exchangers is done in Refs. [6], [94], [97], which are 18 
recommended for the interested student; however the key aspects are the following: 19 
1. Design margin can be good and can be bad, depending on its magnitude. 20 
2. An appropriate design margin provides flexibility and operational robustness to the 21 
equipment against unadvised process disturbances, feedstock changes, environmental 22 
variations, etc. 23 
3. The design margin provides room for plant revamps or increases of production without 24 
significant equipment changes. 25 
4. While a design margin may increase the equipment response time, and consequently less 26 
sensitive to sudden variations, an excessive margin can make the control less manageable. 27 
5. An excessive design margin increases the capital costs, produces larger and heavier 28 
exchangers (very undesirable in offshore applications) and decreases streams velocities, 29 
promoting fouling. This acceleration of the deposit formation is particularly a serious 30 
drawback. 31 
When designing heat exchangers, there are three “kinds” of heat transfer areas considered for 32 
the equipment: the clean, design and actual (final) heat transfer areas, denoted by Ac, Ad and A, 33 
respectively. The clean heat transfer area is obtained from the design equation (Eq. (2.1)) using the 34 
clean overall heat transfer coefficient Uc, as: 35 
𝐴𝑐 =
𝑞
𝑈𝑐𝛥𝑇𝑚
 (7.1) 
After combining the fouling factors, the design overall heat transfer coefficient Ud is obtained, 36 
allowing the evaluation of the design heat transfer area as: 37𝐴𝑑 =
𝑞
𝑈𝑑𝛥𝑇𝑚
 (7.2) 
Commonly, the design heat transfer area Ad is rounded up to adhere to some standardization or 38 
requirement of the proposed design (e.g. available pipe length to enforce an even number of 39 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 131 
exchangers in a bank or by adding safety factors). After that, the actual heat transfer area A for the 1 
exchanger is obtained using the actual (final) overall heat transfer coefficient U. 2 
The relation among these areas is such that: A > Ad >Ac. Therefore, for a heat exchanger, the 3 
“design margin” can be defined as the relative difference between the actual (final) heat transfer area 4 
(A) and a reference heat transfer area required to perform properly the specified service (Ad or Ac). 5 
These quantities are named over-design and over-surface, respectively. 6 
7.2.1 Over-design 7 
The excess area between the actual (A) and design (Ad) areas is usually named over-design, 8 
evaluated in the form: 9 
𝑂𝑣𝑒𝑟 − 𝑑𝑒𝑠𝑖𝑔𝑛(%) = 100 �
𝐴
𝐴𝑑
− 1� = 100 �
𝑈𝑑
𝑈
− 1� (7.3) 
7.2.2 Over-surface 10 
The over-surface is a measure of the total excess area taking as reference the heat transfer area 11 
(Ac) required by a clean heat exchanger to meet the duty: 12 
𝑂𝑣𝑒𝑟 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(%) = 100 �
𝐴
𝐴𝑐
− 1� = 100 �
𝑈𝑐
𝑈
− 1� (7.4) 
7.2.3 Fouling over-surface 13 
The impact of the provision of fouling resistances upfront in the design can be quantified using 14 
the definition of “fouling over-surface”, which is expressed in the form of Eq. (7.5). 15 
𝐹𝑜𝑢𝑙𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(%) = 100 �
𝐴𝑑
𝐴𝑐
− 1� = 100 �
𝑈𝑐
𝑈𝑑
− 1� (7.5) 
Fouling factors should be prescribed with care, since they can introduce an excessive 16 
overdesign, not necessary to meet the actual process requirements. In fact, the application of high 17 
fouling factors can be the very cause of the deposit formation, as a result of the lower fluid velocities. 18 
Even though the indiscriminate use of fouling factors is not recommended, if a process stream is 19 
known to be prone to solids deposition, they must be applied by the designer. As a general (but not 20 
rigid) rule, in typical situations, the excess area introduced by the fouling resistances should outcome a 21 
fouling over-surface of around 30% or less. 22 
7.2.4 Design Margins Ranges 23 
There is no absolute rule for defining the margin built in a heat exchanger design. After all, this 24 
is a designer/engineer call, based on their specific knowledge about the process in question and the 25 
accuracy of available tools (e.g. methods, software, etc.) and physical property data. As a general 26 
reference, an over-design of 5 – 10% may be considered acceptable. 27 
For designs involving dirty streams, the over-surface can be appreciably higher, since it 28 
includes the used fouling factors. In such cases, typical values may easily approach 40 − 50%. 29 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
132 7 Heat Exchanger Design 
7.3 Physical Properties of Chemical Components 1 
Physical properties of substances are a cornerstone of any chemical process design. They are 2 
reasonably susceptible to theoretical and/or experimental uncertainties and must be used with much 3 
care. The designer should be always suspicious about physical property data and crosscheck it twice 4 
against multiple sources, if possible. On the other hand, physical properties of chemical compounds 5 
are not abundantly available and having multiples data sources to compare with is usually a luxury. 6 
The situation is worse for mixtures, since the variation of compositions and concentrations create 7 
unlimited possibilities. 8 
Nowadays, the day-to-day routine of chemical process designers and analysts involve the 9 
evaluation of physical properties directly from process simulation computer programs. These 10 
simulators allow the estimation of physical properties for pure components and mixtures for a given 11 
pressure and temperature, by the application of the some appropriate mixing rule. There are paid (e.g. 12 
Aspen [98] , Hysys [99], Pro/II [100], Chemcad [101], etc.) and free (e.g. COCO [102], EMSO [103] 13 
etc.) chemical process simulators, and a few of them have the invaluable bonus of being free and open 14 
source (e,g, DWSim [104], ASCEND [105], etc.). A comprehensive list of commercial simulators may 15 
be found is Refs. [106] and [107]. 16 
A chemical process simulator that I personally recommend is DWSim [108]. Its open source 17 
nature makes it an ideal tool for teaching purposes, since the students may see in bare eyes all the 18 
models and methods that go on under the hood when performing a chemical process simulation. 19 
Besides that, this great piece of software is continuously becoming more reliable and powerful for 20 
practical industrial applications. 21 
Although the daily engineering practice comprises the use of property packages integrated in 22 
the process simulator software, we need “in paper” data to learn the subject matter presented herein. 23 
To offer support in the solution of the problems proposed in this text, Appendix C brings a complete 24 
set of the required physical properties for a number of chemical components. The provided plots of 25 
dynamic viscosity, specific heat, specific mass and thermal conductivity are based on the same 26 
methods available in the DWSim simulator, which means that the student or chemical process 27 
professional interested in learning or validating by themselves the calculations underneath can do that 28 
by direct inspection of the simulator source code. 29 
It is worth to note that the physical properties given in Appendix C have to sole purpose of 30 
being instruction support material. Even though this data may be considered with sufficient accuracy 31 
for some practical applications, the reader is encouraged to use experimental data from manuals and 32 
handbooks available in the technical literature wherever possible. 33 
7.4 Sizing 34 
Sizing a heat exchanger comprises essentially the thermal and hydraulic specification of a new 35 
equipment that should be built in order to accomplish a given heat transfer service. This includes the 36 
selection of a heat exchanger type and configuration, determine its dimensions, flow arrangement and, 37 
in some cases, the inlet and outlet temperatures or flow rate of a utility fluid. 38 
For a heat exchanger that contacts two process streams, their operating temperatures and flow 39 
rates are dictated by the process as a whole, i.e. by the operations upstream and downstream the heat 40 
exchanger. If the equipment transfers heat from/to a process stream and a utility fluid, such as cooling 41 
water, steam or thermal fluids, there is more freedom for the specification of the outlet conditions and 42 
flow rates of the utility, since the impact on the operation of the process is mostly limited to the 43 
equipment producing or recycling such fluids. 44 
Once defined the flow arrangement and heat exchanger type/model, the sizing procedure aims 45 
to answer two questions: 46 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 133 
1. What is the required heat transfer area (size) to match the service? 1 
2. Does the exchanger with the designed size exceed the available pressure drop in the process 2 
line? 3 
Notably, the most important question is (1): the required heat transfer area. Question (2) is a 4 
verification step; however its importance should not be underestimated, since the installation of a heat 5 
exchanger with exceeded pressure drop may cause the flow shutdown of the whole equipment train in 6 
a process line. 7 
7.4.1 The Design Variables 8 
The designer may specify the surface necessary to perform a heat transfer duty in several ways. 9 
Provided the type/model of the heat exchanger is selected,some design variable is tuned in order to 10 
yield the necessary heat transfer area. 11 
Typically, the design variable is the dimension or the number of the equipment surface 12 
element, which varies according to the heat exchanger class or geometry. For example, in the case of 13 
planar heat exchangers, the surface elements are “plates”, therefore, with complete freedom, the 14 
designer is able to select the plate size and shape, and calculate the number of these plates needed to 15 
attain the required duty. In some cases, the plate size and shape are restricted, for whatever reason, 16 
such as available stock, space limitations, existent supplier contracts, standardization/code, 17 
recommended practice, etc. Therefore, the designer cannot choose the plate characteristics, and the 18 
only remaining design variable is the quantity of plates to be assembled. 19 
For tubular heat exchangers, e.g. double pipe, multitubular hairpins or shell and tube, the 20 
surface elements are, as expected, tubes, and the design variable may be the diameter, length or 21 
number of tubes. Sometimes, all of them are to be specified by the designer, however, it is 22 
commonplace to have the tube set selected by some heuristic rule and the length of the tubes to be the 23 
design variable used to define the heat exchanger area. If there are space restrictions limiting the 24 
length of the exchanger, the diameter or number of tubes becomes the design variable adjusted to 25 
fulfill the heat transfer area. 26 
Usually, several double pipe exchangers (hairpins) should be associated in banks to accomplish 27 
an ordinary energy duty. This fact introduces the number of hairpins as a design variable for tuning the 28 
heat transfer surface. Hence, for this type of heat exchanger, depending on the previously imposed 29 
design restrictions, the engineer should possibly deal with three design variables: 30 
1. Inner and outer pipe diameters. 31 
2. Pipe length. 32 
3. Number of hairpins (associated to form a bank). 33 
If all three variables above are unrestricted, there are literally infinite designs to choose from, 34 
and some optimization method upon an objective function is the only way to find the best solution. For 35 
“manual” designs, some heuristics are necessary to select a pipe set, with specified diameters and 36 
length. After the hairpin pipe diameters and length are defined, the sizing task reduces to the 37 
determination of the number of hairpins connected in the bank. 38 
7.4.2 Sizing Outline 39 
The problem of designing the double pipe heat exchanger may be summarized in the following 40 
steps: 41 
1. Thermal Design (sizing) 42 
1.1. Collect process data parameters. 43 
1.1.1. Fluid temperatures. 44 
1.1.2. Physical properties. 45 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
134 7 Heat Exchanger Design 
1.2. Energy balance using mean bulk temperatures of the fluids. 1 
1.2.1. Determine any unknown fluid temperature or flow rate. 2 
1.2.2. Or just check the energy balance between fluids for correctness. 3 
1.3. Select a heat exchanger type: double pipe, multi-tube or finned heat exchanger in the present 4 
case. 5 
1.4. Select the design variable: 6 
1.4.1. Pipe length, if diameter is specified. 7 
1.4.2. Pipe diameter, if length is specified. 8 
1.4.3. Number of hairpins, if the service should be performed by a set of standardized hairpin 9 
exchangers. 10 
1.5. Evaluate the actual mean temperature difference to be used in the heat exchanger design 11 
equation. 12 
1.6. Overall heat transfer coefficient. 13 
1.6.1. Calculate the inner pipe heat transfer coefficient. 14 
1.6.2. Calculate the annulus heat transfer coefficient. 15 
1.6.3. Calculate wall temperature. 16 
1.6.4. Iterate steps 1.6.1 to 1.6.3 until convergence of the wall temperature is achieved. 17 
1.6.5. Calculate the clean overall heat transfer coefficient Uc. 18 
1.6.6. Calculate the design (required) overall heat transfer coefficient Ud by introducing the 19 
fouling factors Rdi and Rdo. 20 
1.7. Heat transfer areas 21 
1.7.1. Calculate the design (required) heat transfer area Ad. 22 
1.7.2. Specify the heat exchanger geometry to match the required heat transfer surface. 23 
1.7.2.1. Number of hairpins, 24 
1.7.2.2. Pipe diameter and/or number of internal tubes for multi-tube hairpins. 25 
1.7.3. Calculate the actual (final) heat transfer area A. 26 
1.7.4. Calculate the actual (final) overall heat transfer coefficient U for the exchanger. 27 
1.8. Calculate the over-surface and over-design. 28 
2. Hydraulic Design (verify if allowable pressure drop is not exceeded). 29 
2.1. Evaluate the inner pipe pressure drop. 30 
2.2. Evaluate the annulus pressure drop. 31 
Worked Example 7–1 32 
Determine the required number of hairpins to heat a benzene stream of 0.7 kg/s from 20 °C to 33 
55 °C using 2-butanol cooling from 60 °C to 40 °C. There are available double pipe heat exchangers 34 
with 5 m length, built with NPS 32 × 50 mm, sch 40, stainless steel grade 316 pipes. For this service, 35 
the hot fluid should be allocated in the inner pipe. 36 
Solution 37 
1) Fluids physical properties 38 
a) Mean temperatures of the fluids 39 
The four process temperatures are specified; therefore, we are able to estimate the necessary 40 
fluid properties using the arithmetic mean of the inlet and outlet temperatures for each stream: 41 
𝑇1𝑚 =
𝑇1,𝑖𝑛
2
+
𝑇1,𝑜𝑢𝑡
2
=
293.15
2
+
328.15
2
= 310.65 𝐾 
𝑇2𝑚 =
𝑇2,𝑖𝑛
2
+
𝑇2,𝑜𝑢𝑡
2
=
313.15
2
+
333.15
2
= 323.15 𝐾 
b) Fluid properties at these mean stream temperatures are evaluated using Appendix C.9 and 42 
Appendix C.10 for 2-butanol and benzene respectively: 43 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 135 
Fluid 1 (cold) → Annulus Fluid 2 (hot) → Tube 
𝑩𝒆𝒏𝒛𝒆𝒏𝒆 
𝟑𝟏𝟎.𝟔𝟓 𝑲− 𝟏𝟎𝟏𝟑𝟐𝟓 𝑷𝒂 
𝟐 − 𝒃𝒖𝒕𝒂𝒏𝒐𝒍 
𝟑𝟐𝟑.𝟏𝟓 𝑲− 𝟏𝟎𝟏𝟑𝟐𝟓 𝑷𝒂 
𝜇 = 0.000511s ⋅ Pa 
𝜇 = 0.511cP 
𝜌 = 860.0
kg
m3
 
𝐶𝑝 = 1530.0
J
(kg ⋅ K)
 
𝑘 = 0.14
W
(m ⋅ K)
 
𝑃𝑟 = 5.58 
𝜇 = 0.00134s ⋅ Pa 
𝜇 = 1.34cP 
𝜌 = 865.0
kg
m3
 
𝐶𝑝 = 2370.0
J
(kg ⋅ K)
 
𝑘 = 0.13
W
(m ⋅ K)
 
𝑃𝑟 = 24.5 
c) The energy balance is used to determine the heat load and the 2-butanol flow rate: 1 
𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� + 𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 0 
Solving for W2, we have the 2-butanol flow rate: 2 
𝑊2 =
𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛�
𝐶𝑝2�𝑇2,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡�
=
0.7 ⋅ 1530.0(328.15 − 293.15)
2370.0(−313.15 + 333.15)
= 0.791
𝑘𝑔
𝑠
 
Checking the enthalpy change for both fluids, we confirm they are about the same within 3 
round-off approximations. Therefore, the energy balance is correct. 4 
𝑞1 = 𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 0.7 ⋅ 1530.0(−293.15 + 328.15) = 37485.0 𝑊 
𝑞2 = 𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� = 0.791 ⋅ 2370.0(313.15 − 333.15) = −37493.0 𝑊 
2) Logarithmic mean temperature difference 5 
For counterflow arrangement, the fluids temperatures at the heat exchanger terminals are: 6 
𝑇1,𝑖𝑛 = 293.15𝐾 → 𝑇1,𝑜𝑢𝑡 = 328.15𝐾 
𝑇2,𝑜𝑢𝑡 = 313.15𝐾 ← 𝑇2,𝑖𝑛 = 333.15𝐾 
 
And the logarithmic mean temperature difference evaluated from: 7 
𝛥𝑇𝑙𝑚,𝑐 =
𝑇1,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡 + 𝑇2,𝑖𝑛 − 𝑇1,𝑜𝑢𝑡
𝑙𝑛 �
𝑇1,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡
𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛
�
=
293.15 − 313.15 − 328.15 + 333.15
𝑙𝑛 �293.15 − 313.15328.15 − 333.15�
= −10.82 𝐾 
Notice that the order for taking the fluid temperature differences at the terminals was chosen 8 
purposefully to illustrate the fact that the negative sign of the ΔTlm is irrelevant in this context. It 9 
simply indicates that the terminal temperature differences were calculated from the cold to the hot 10 
fluid, in this order, and that the heat transfer rate should be used with negative sign when calculating 11 
the heat transfer area from the design equation q = UA ΔTlm. 12 
3) Pipe dimensions 13 
The inner pipe is NPS 32 mm − sch 40 and the external pipe is specifiedas NPS 50 mm/sch 40, 14 
from Table 1 in Appendix A.1 the inner and outer diameters for the inner pipe are: 15 
𝐼𝐷 = 𝐷𝑖 = 35.08 𝑚𝑚 
𝑂𝐷 = 𝐷𝑜 = 42.0 𝑚𝑚 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
136 7 Heat Exchanger Design 
For the external pipe, only the inner diameter is necessary in the heat transfer calculations, 1 
which is: 2 
𝐼𝐷 = 𝐷𝑠 = 52.48 𝑚𝑚 
4) Tube heat transfer coefficient 3 
Cross section of the inner pipe, mass flux and Reynolds number are: 4 
𝑆𝑖 =
𝜋𝐷𝑖2
4
=
𝜋0.035082
4
= 0.00096652 𝑚2 
𝐺𝑖 =
𝑊𝑖
𝑆𝑖
=
0.791
0.00096652
= 818.4
𝑘𝑔
(𝑚2 ⋅ 𝑠)
 
𝑅𝑒𝑖 =
𝐷𝑖𝐺𝑖
𝜇𝑖
=
0.03508
0.00134
818.4 = 21425.0 → 𝑓𝑢𝑙𝑙𝑦 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 
With this Reynolds number in common flow conditions, the regime is turbulent. The Sieder-5 
Tate equation (Eq. (4.10)) may be used to estimate the heat transfer coefficient. As the first 6 
approximation, let us disregard the viscosity correction by assuming ϕi = (µi/µw)0,14 = 1 as below: 7 
𝑁𝑢 = 0.027𝑃𝑟0.33𝑅𝑒0.8 �
𝜇𝑖
𝜇𝑤
�
0.14
 
ℎ𝑖 =
0.027𝑘
𝐷𝑖
√𝑃𝑟3 𝑅𝑒
4
5 =
0.027√24.53
0.03508
0.13 ⋅ 21425.0
4
5 = 847.3
𝑊
(𝑚2 ⋅ 𝐾)
 
Notice that the application of Eq. (4.10) assumes fully developed flow. Although the whole 8 
pipe length is not known at this time, for the minimum number of only one hairpin in the bank, the 9 
correspondent number of diameters at the entrance surpasses by far the needed to achieve the full 10 
development: 11 
#𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 =
𝐿ℎ𝑝
𝐷𝑖
=
5.0
0.03508
= 142.53 
5) Annulus heat transfer coefficient 12 
The annulus equivalent diameter used in the Reynolds number is taken here as equal to the 13 
hydraulic diameter: 14 
𝐷𝑒 = 𝐷𝑠 − 𝐷𝑜 = 0.05248 − 0.042 = 0.01048 𝑚 
𝑆𝑜 =
𝜋
4
(𝐷𝑠2 − 𝐷𝑜2) =
𝜋
4
(0.052482 − 0.0422) = 0.00077766 𝑚2 
𝐺𝑜 =
𝑊𝑜
𝑆𝑜
=
0.7
0.00077766
= 900.14
𝑘𝑔
(𝑚2 ⋅ 𝑠)
 
𝑅𝑒𝑜 =
𝐷𝑒𝐺𝑜
𝜇𝑜
=
0.01048
0.000511
900.14 = 18461.0 → 𝑓𝑢𝑙𝑙𝑦 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 
For fully developed turbulent flow, and ignoring the viscosity correction as ϕo = (µo/µw)0.14 = 1, 15 
the annulus film coefficient is estimated from the Sieder-Tate equation: 16 
𝑁𝑢 = 0.027𝑃𝑟0.33𝑅𝑒0.8 �
𝜇𝑜
𝜇𝑤
�
0.14
 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 137 
ℎ𝑜 =
0.027𝑘
𝐷𝑒
√𝑃𝑟3 𝑅𝑒
4
5 =
0.027√5.583
0.01048
0.14 ⋅ 18461.0
4
5 = 1655.8
𝑊
(𝑚2 ⋅ 𝐾)
 
6) Variable properties correction factor 1 
a) Wall temperature and correction factors 2 
To further improve the accuracy of the heat transfer coefficient, the wall temperature is 3 
calculated using the first approximations of the heat transfer coefficients. 4 
𝑡𝑤 =
𝐷𝑖ℎ𝑖𝑡𝑖 + 𝐷𝑜ℎ𝑜𝑡𝑜
𝐷𝑖ℎ𝑖 + 𝐷𝑜ℎ𝑜
=
0.03508 ⋅ 323.15 ⋅ 847.3 + 0.042 ⋅ 1655.8 ⋅ 310.65
0.03508 ⋅ 847.3 + 0.042 ⋅ 1655.8
= 314.39 𝐾 
The fluid properties at the wall (from Appendix C.9 and Appendix C.10 ) are applied to correct 5 
the pipe and annulus film coefficients as: 6 
Fluid 1 (cold) Fluid 2 (hot) 
𝑩𝒆𝒏𝒛𝒆𝒏𝒆 
𝟑𝟏𝟒.𝟑𝟗 𝑲− 𝟏𝟎𝟏𝟑𝟐𝟓 𝑷𝒂 
𝟐 − 𝒃𝒖𝒕𝒂𝒏𝒐𝒍 
𝟑𝟏𝟒.𝟑𝟗 𝑲− 𝟏𝟎𝟏𝟑𝟐𝟓 𝑷𝒂 
𝜇 = 0.000488s ⋅ Pa 𝜇 = 0.00177s ⋅ Pa 
𝜙𝑖 = �
𝜇𝑖
𝜇𝑤
�
0.14
= �
0.00134
0.00177
�
0.14
= 0.96179 
𝜙𝑜 = �
𝜇𝑜
𝜇𝑤
�
0.14
= �
0.000511
0.000488
�
0.14
= 1.0065 
The corrected heat transfer coefficients are: 7 
ℎ𝑖,𝑐𝑜𝑟𝑟 = ℎ𝑖𝜙𝑖 = 0.96179 ⋅ 847.3 = 814.92
𝑊
(𝑚2 ⋅ 𝐾)
 
ℎ𝑜,𝑐𝑜𝑟𝑟 = ℎ𝑜𝜙𝑜 = 1.0065 ⋅ 1655.8 = 1666.6
𝑊
(𝑚2 ⋅ 𝐾)
 
7) Overall heat transfer coefficient 8 
The clean overall heat transfer coefficient, including the wall conductive resistance, the internal 9 
and external convective resistances is: 10 
𝑈𝑐 =
1
𝐷𝑜
2𝑘 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 1ℎ𝑜
+ 𝐷𝑜𝐷𝑖ℎ𝑖
=
1
0.042 𝑙𝑛 � 0.0420.03508�
2 ⋅ 16.269 +
1
1666.6 +
0.042
0.03508 ⋅ 814.92
= 434.48
𝑊
(𝑚2 ⋅ 𝐾)
 
The provision for fouling for both fluids is given by the required fouling resistances Rdi and 11 
Rdo. 12 
𝑅𝑑𝑖 = 0.0002
𝑚2 ⋅ 𝐾
𝑊
 
𝑅𝑑𝑜 = 0.0002
𝑚2 ⋅ 𝐾
𝑊
 
Therefore, the design overall heat transfer coefficient used to size the exchanger area is given 13 
by: 14 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
138 7 Heat Exchanger Design 
𝑈𝑑 =
1
𝑅𝑑𝑜 +
1
𝑈𝑐
+ 𝐷𝑜𝑅𝑑𝑖𝐷𝑖
=
1
0.0002 + 0.00020.03508 0.042 +
1
434.48
= 364.82
𝑊
(𝑚2 ⋅ 𝐾)
 
8) Number of hairpins 1 
a) Heat transfer area 2 
Solving the design equation for the required heat transfer area Ad for performing the service, 3 
using the calculated design overall heat transfer coefficient, we get: 4 
𝑞 = 𝐴𝑑𝑈𝑑𝛥𝑇𝑙𝑚 
𝐴𝑑 =
𝑞
𝑈𝑑𝛥𝑇𝑙𝑚
=
37485.0
10.82 ⋅ 364.82
= 9.4962 𝑚2 
b) Total pipe length and number of exchangers 5 
The hairpin linear heat transfer area (based on the external surface of the inner pipe) is: 6 
𝐴𝑙𝑖𝑛𝑒𝑎𝑟 = 𝜋𝐷𝑜 = 𝜋0.042 = 0.13195 �
𝑚2
𝑚
� 
Therefore, given the total required area of 9.6685 m2, the necessary pipe length and number of 7 
exchangers in the bank are: 8 
𝐿 =
𝐴𝑑
𝐴𝑙𝑖𝑛𝑒𝑎𝑟
=
9.4962
0.13195
= 71.968 𝑚 
𝑁ℎ𝑝 =
𝐿
2𝐿ℎ𝑝
=
71.968
2 ⋅ 5.0
= 7.1968 → 8 ℎ𝑎𝑖𝑟𝑝𝑖𝑛𝑠 
Worked Example 7–2 9 
Consider a service where 10000 lb/h of benzene is heated from 60 °F to 120 °F in a bank of 10 
double pipe heat exchangers using a stream of aniline cooled from 150 °F to 100 °F. There are 11 
available hairpins NPS 1.25×2 in-Sch 40, 16 ft long, manufactured in 316 stainless steel with thermal 12 
conductivity k = 9.4 BTU/(h ft °F) and absolute roughness ε = 0.000895 in. The annulus fluid return is 13 
of bonnet type. Due to pumping restrictions, the allowable pressure drop for each line is 20 psi. 14 
Propose a feasible design, specifying the number and configuration of an exchanger bank for 15 
performing this duty. A fouling factor of Rd = 0.001 (ft2 h °F)/BTU should be provisioned for each 16 
stream. 17 
Solution 18 
This same problem was proposed in Ref. [28] as a solved example. Let us take that as a 19 
benchmark and approaching the same design, but relying on a distinct source of physical properties 20 
and some alternative equations for heat transfer coefficients and friction factors. For easier 21 
comparison, if the reader is interested in doing so, the solution will be developed in US customary 22 
units. 23 
Design 1 24 
For the first configuration, we assume that benzene (fluid 1) is allocated in the inner tube and 25 
aniline (fluid 2) flowing inside the annulus. 26 
1) Fluids physical properties 27 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 139 
a) Mean temperatures of the fluids 1 
The mean temperatures for each stream will be used for initial estimation of the physical 2 
properties: 3 
𝑇1𝑚 =
𝑇1,𝑖𝑛
2
+
𝑇1,𝑜𝑢𝑡
2
=
120.00
2
+
60.000
2
= 90.000 °𝐹 
𝑇2𝑚 =
𝑇2,𝑖𝑛
2
+
𝑇2,𝑜𝑢𝑡
2
=
100.00
2
+
150.00
2
= 125.00 °𝐹 
b) Fluid properties at these mean stream temperatures are evaluated using Appendix C.10 and 4 
Appendix C.12 for benzene and aniline, respectively: 5 
Fluid 1 (cold) Fluid 2 (hot) 
𝑩𝒆𝒏𝒛𝒆𝒏𝒆 
𝟗𝟎.𝟎 °𝑭 − 𝟏𝟒.𝟕 𝒑𝒔𝒊 
𝑨𝒏𝒊𝒍𝒊𝒏𝒆 
𝟏𝟐𝟓.𝟎 °𝑭 − 𝟏𝟒.𝟕𝒑𝒔𝒊 
𝜇 = 1.32
lb
(ft ⋅ h)
 
𝜇 = 0.546 cP 
𝜌 = 54.0
lb
ft3
 
𝐶𝑝 = 0.36
BTU
(lb ⋅ °F)
 
𝑘 = 0.082
BTU
(ft ⋅ h ⋅ °F)
 
𝑃𝑟 = 5.8 
𝜇 = 4.31
lb
(ft ⋅ h)
 
𝜇 = 1.78 cP 
𝜌 = 64.3
lb
ft3
 
𝐶𝑝 = 0.414
BTU
(lb ⋅ °F)
 
𝑘 = 0.0985
BTU
(ft ⋅ h ⋅ °F)
 
𝑃𝑟 = 18.1 
c) The energy balance is used to determine the heat load and the flow rate of aniline: 6 
𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� + 𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 0 
Isolating W2 and evaluating, we have: 7 
𝑊2 =
𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛�
𝐶𝑝2�𝑇2,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡�
=
0.36000 ⋅ 10000. (120.00 − 60.000)
0.41400(−100.00 + 150.00)
= 10435.
𝑙𝑏
ℎ
 
The heat load can be confirmed to be the same for both streams as: 8 
𝑞1 = 𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 0.36000 ⋅ 10000. (120.00 − 60.000) = (2.1600𝑒 + 5)
𝐵𝑇𝑈
ℎ
 
𝑞2 = 𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡− 𝑇2,𝑖𝑛� = 0.41400 ⋅ 10435. (100.00 − 150.00) = (−2.1600𝑒 + 5)
𝐵𝑇𝑈
ℎ
 
2) Logarithmic mean temperature difference 9 
For counterflow arrangement, the fluids temperatures at the heat exchanger terminals are: 10 
Terminal 1 Terminal 2 
𝑇1,𝑖𝑛 = 60.0 °F → 𝑇1,𝑜𝑢𝑡 = 120.0 °F 
𝑇2,𝑜𝑢𝑡 = 100.0 °F ← 𝑇2,𝑖𝑛 = 150.0 °F 
 
Δ𝑇1 = −40.000 °F Δ𝑇2 = −30.000 °F 
With the temperature differences at the terminals, the log mean temperature difference is: 11 
𝛥𝑇1 = 𝑇1,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡 = 60.000 − 100.00 = −40.000 °𝐹 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
140 7 Heat Exchanger Design 
𝛥𝑇2 = 𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛 = 120.00 − 150.00 = −30.000 °𝐹 
𝛥𝑇𝑙𝑚,𝑐 =
𝛥𝑇1 − 𝛥𝑇2
𝑙𝑛 �𝛥𝑇1𝛥𝑇2
�
=
(−40.000) − (−30.000)
𝑙𝑛 �−40.000−30.000�
= −34.761 °𝐹 
Note that the negative sign in this result is irrelevant, since it is just a consequence of taking the 1 
temperature differences from the cold fluid (1) to the hot fluid (2). The negative sign can be dropped 2 
when appropriate. Also note that the heat transfer rates q1 and q2 calculated previously have opposite 3 
signs, since the hot fluid loses and the cold fluid gains enthalpy. 4 
3) Pipe dimensions 5 
The detailed size specifications for both pipes are obtained from the Appendix A.1 as: 6 
NPS 1(1/4) in - Schedule 40 
ID = Di = 1.38 in (inner diameter) 
OD = Do = 1.66 in (outer diameter) 
NPS 2 in - Schedule 40 
ID = Ds = 2.067 in (inner diameter) 
OD = 2.375 in (outer diameter) 
4) Inner pipe heat transfer coefficient 7 
Cross section of the inner pipe, mass flux and Reynolds number are: 8 
𝑆𝑖 =
𝜋𝐷𝑖2
4
=
𝜋0.115002
4
= 0.010387 𝑓𝑡2 
𝐺𝑖 =
𝑊𝑖
𝑆𝑖
=
10000.
0.010387
= (9.6274𝑒 + 5)
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 
𝑅𝑒𝑖 =
𝐷𝑖𝐺𝑖
𝜇𝑖
=
(9.6274𝑒 + 5)0.11500
1.3200
= 83875. 
Therefore, the regime is turbulent and the Sieder-Tate equation (Eq. (4.10)) may be used to 9 
estimate the heat transfer coefficient. As the first approximation, let us disregard the viscosity 10 
correction by assuming ϕi = (µi/µw)0,14 = 1 as below: 11 
𝑁𝑢 = 0.027𝑃𝑟0.33𝑅𝑒0.8 �
𝜇𝑖
𝜇𝑤
�
0.14
 
ℎ𝑖 =
0.027𝑘
𝐷𝑖
√𝑃𝑟3 𝑅𝑒
4
5 =
0.027√5.80003
0.11500
0.082000 ⋅ 83875.
4
5 = 300.51
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
Notice that the application of Eq. (4.10) assumes fully developed flow. Although the whole 12 
pipe length is not known at this time, for the minimum number of only one hairpin in the bank, the 13 
correspondent number of diameters at the entrance surpasses by far the needed to achieve the full 14 
development in turbulent flow, which is about 10 to 20 diameters. The number of diameters for just 15 
one hairpin leg is: 16 
#𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 =
𝐿ℎ𝑝
𝐷𝑖
=
16.000
0.11500
= 139.13 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 141 
5) Annulus heat transfer coefficient 1 
The annulus equivalent diameter used in the Reynolds number is taken here as equal to the 2 
hydraulic diameter: 3 
𝐷𝑒 = 𝐷𝑠 − 𝐷𝑜 = 0.17225 − 0.13833 = 0.033920 𝑓𝑡 
𝑆𝑜 =
𝜋
4
(𝐷𝑠2 − 𝐷𝑜2) =
𝜋
4
(0.172252 − 0.138332) = 0.0082741 𝑓𝑡2 
𝐺𝑜 =
𝑊𝑜
𝑆𝑜
=
10435.
0.0082741
= (1.2612𝑒 + 6)
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 
𝑅𝑒𝑜 =
𝐷𝑒𝐺𝑜
𝜇𝑜
=
(1.2612𝑒 + 6)0.033920
4.3100
= 9925.7 
The annulus Reynolds number is still in the transition region; although quite near from the 4 
turbulent region. For developing transition flow, the convective heat transfer may be estimated by the 5 
Hausen’s equation (Eq. (4.30)). The developing flow term will be ignored, since for just one 6 
exchanger leg the number of diameters is significant, given by: 7 
#𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 =
𝐿ℎ𝑝
𝐷𝑒
=
16.000
0.033917
= 471.74 
Also, ignoring the viscosity correction as ϕo = (µo/µw)0.14 = 1, the annulus film coefficient is 8 
estimated as: 9 
𝑁𝑢 = 𝑃𝑟0.33(0.116𝑅𝑒0.67 − 14.5) ��
𝐷𝑒
𝐿
�
0.67
+ 1� �
𝜇
𝜇𝑤
�
0.14
 
ℎ𝑜 =
0.116√𝑃𝑟3
𝐷𝑒
𝑘 �𝑅𝑒
2
3 − 125.0� =
0.116√18.1003
0.033920
0.098500 �9925.7
2
3 − 125.0�
= 297.93
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
6) Variable properties correction factor 10 
a) Wall temperature and correction factors 11 
To improve the accuracy of the heat transfer coefficient further, the wall temperature is 12 
calculated using the first approximations of the heat transfer coefficients. 13 
𝑡𝑤 =
𝐷𝑖ℎ𝑖𝑡𝑖 + 𝐷𝑜ℎ𝑜𝑡𝑜
𝐷𝑖ℎ𝑖 + 𝐷𝑜ℎ𝑜
=
0.11500 ⋅ 300.51 ⋅ 90.000 + 0.13833 ⋅ 125.00 ⋅ 297.93
0.11500 ⋅ 300.51 + 0.13833 ⋅ 297.93
= 109.04 °𝐹 
The fluid properties at the wall (from Appendix C.9 and Appendix C.10) are applied to correct 14 
the pipe and annulus film coefficients as: 15 
Fluid 1 (cold) Fluid 2 (hot) 
𝑩𝒆𝒏𝒛𝒆𝒏𝒆 
𝟏𝟎𝟗.𝟎 °𝑭 − 𝟏𝟒.𝟕𝒑𝒔𝒊 
𝑨𝒏𝒊𝒍𝒊𝒏𝒆 
𝟏𝟎𝟗.𝟎 °𝑭 − 𝟏𝟒.𝟕𝒑𝒔𝒊 
𝜇 = 1.16
lb
(ft ⋅ h)
 𝜇 = 5.31
lb
(ft ⋅ h)
 
𝜙𝑖 = �
𝜇𝑖
𝜇𝑖,𝑤
�
0.14
= �
1.3200
1.1600
�
0.14
= 1.0183 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
142 7 Heat Exchanger Design 
𝜙𝑜 = �
𝜇𝑜
𝜇𝑜,𝑤
�
0.14
= �
4.3100
5.3100
�
0.14
= 0.97121 
Then, we have the corrected heat transfer coefficients as: 1 
ℎ𝑖,𝑐𝑜𝑟𝑟 = ℎ𝑖𝜙𝑖 = 1.0183 ⋅ 300.51 = 306.01
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
ℎ𝑜,𝑐𝑜𝑟𝑟 = ℎ𝑜𝜙𝑜 = 0.97121 ⋅ 297.93 = 289.35
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
7) Overall heat transfer coefficient 2 
The clean overall heat transfer coefficient, including the wall conductive resistance, the internal 3 
and external convective resistances is: 4 
𝑈𝑐 = �
𝐷𝑜
2𝑘
𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� +
1
ℎ𝑜
+
𝐷𝑜
𝐷𝑖ℎ𝑖
�
−1
= �
0.13833 𝑙𝑛 �0.138330.11500�
2 ⋅ 9.4000
+
1
289.35
+
0.13833
0.11500 ⋅ 306.01
�
−1
= 114.34
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
The fouling resistances required for both fluids are Rdi and Rdo: 5 
𝑅𝑑𝑖 = 0.001
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
𝑅𝑑𝑜 = 0.001
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
Therefore, the design overall heat transfer coefficient used to size the exchanger area is given 6 
by: 7 
𝑈𝑑 = �𝑅𝑑𝑜 +
1
𝑈𝑐
+
𝐷𝑜𝑅𝑑𝑖
𝐷𝑖
�
−1
= �0.0010000 +
1
114.34
+
0.0010000
0.11500
0.13833�
−1
= 91.335
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
8) Number of hairpins 8 
a) Heat transfer area 9 
From the heat exchanger design equation, the clean heat transfer area is: 10 
𝐴𝑐 =
𝑞
𝑈𝑐𝛥𝑇𝑙𝑚
=
(−2.1600𝑒 + 5)
−34.761 ⋅ 114.34
= 54.345 𝑓𝑡2 
And the design required heat transfer surface is: 11 
𝐴𝑑 =
𝑞
𝑈𝑑𝛥𝑇𝑙𝑚
=
(−2.1600𝑒 + 5)
−34.761 ⋅ 91.335
= 68.034 𝑓𝑡2 
b) Total pipe length and number of exchangers 12 
The heat transfer area per unit length, based on the external surface of the inner pipe, is: 13 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 143 
𝐴𝑙𝑖𝑛𝑒𝑎𝑟 = 𝜋𝐷𝑜 = 𝜋0.13833 = 0.43458 �
𝑓𝑡2
𝑓𝑡
� 
Therefore, the necessary pipe length and number of exchangers in the bank and the actual 1 
specified surface are: 2 
𝐿 =
𝐴𝑑
𝐴𝑙𝑖𝑛𝑒𝑎𝑟
=
68.034
0.43458
= 156.55 𝑓𝑡 
𝑁ℎ𝑝 =
𝐿
2𝐿ℎ𝑝
=
156.55
2 ⋅ 16.000
= 4.8922 → 5 ℎ𝑎𝑖𝑟𝑝𝑖𝑛𝑠 
𝐴 = 2𝐴𝑙𝑖𝑛𝑒𝑎𝑟𝐿ℎ𝑝𝑁ℎ𝑝 = 2 ⋅ 0.43458 ⋅ 16.000 ⋅ 5.0000 = 69.533 𝑓𝑡2 
The actual overall heat transfer and fouling factor for the exchanger are: 3 
𝑈 =
𝑞
𝐴𝛥𝑇𝑙𝑚
=
(−2.1600𝑒 + 5)
−34.761 ⋅ 69.533
= 89.366
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
𝑅𝑑 =
𝑈𝑐 − 𝑈
𝑈𝑐𝑈
=
114.34 − 89.366
114.34 ⋅ 89.366
= 0.0024441
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
Taking the required heat transfer surface as reference, the over-design is only about 2%, as 4 
calculated below: 5 
𝑂𝑣𝑒𝑟 − 𝑑𝑒𝑠𝑖𝑔𝑛(%) = 100 �
𝐴
𝐴𝑑
− 1� = 100 �
69.533
68.034
− 1� = 2.2033% 
𝑂𝑣𝑒𝑟 − 𝑑𝑒𝑠𝑖𝑔𝑛(%) = 100 �
𝑈𝑑
𝑈
− 1� = 100 �
91.335
89.366
− 1� = 2.2033% 
Because of the fouling factors, the over-surface is significantly higher: 6 
𝑂𝑣𝑒𝑟 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(%) = 100 �
𝐴
𝐴𝑐
− 1� = 100 �
69.533
54.345
− 1� = 27.947% 
𝑂𝑣𝑒𝑟 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(%) = 100 �
𝑈𝑐
𝑈
− 1� = 100 �
114.34
89.366
− 1� = 27.946% 
The fouling over-surface is evaluated as: 7 
𝐹𝑜𝑢𝑙𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(%) = 100 �
𝐴𝑑
𝐴𝑐
− 1� = 100 �
68.034
54.345
− 1� = 25.189%𝐹𝑜𝑢𝑙𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(%) = 100 �
𝑈𝑐
𝑈𝑑
− 1� = 100 �
114.34
91.335
− 1� = 25.187% 
9) Pressure drop 8 
a) Pressure drop in the tube 9 
𝑓𝐹𝑖 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑖
�
10
9 + 6.9𝑅𝑒𝑖
�
=
0.41
𝑙𝑛2 �0.23 �(7.4583𝑒 − 5)0.11500 �
10
9
+ 6.983875.�
= 0.0052745 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
144 7 Heat Exchanger Design 
𝜙𝑓,𝑖 =
1.02
� 𝜇𝑖𝜇𝑖,𝑤
�
0.14 =
1.02
�1.32001.1600�
0.14 = 1.0017 
𝑓𝐹𝑖,𝑐𝑜𝑟𝑟 = 𝑓𝐹𝑖𝜙𝑓,𝑖 = 0.0052745 ⋅ 1.0017 = 0.0052835 
(1) Distributed pressure drop − tube 1 
𝐺𝑖 = 962740.0
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 
𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 =
4𝐺𝑖2𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑖
𝐷𝑖𝜌𝑖
=
4(9.6274𝑒 + 5)2 ⋅ 16.000 ⋅ 5.0000 ⋅ 0.0052835
0.11500 ⋅ 54.000
= (2.5235𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 4.2027 𝑝𝑠𝑖 
(2) Localized pressure drop − tube 2 
𝐷𝑖(𝑖𝑛) = 1.38 𝑖𝑛 
𝐾𝑖 = 0.7 +
2000
𝑅𝑒𝑖
+
0.7
𝐷𝑖(𝑖𝑛)
= 0.7 +
2000
83875.
+
0.7
1.3800
= 1.2311 
𝛥𝑃𝑖,𝑙𝑜𝑐 =
𝐺𝑖2𝐾𝑖
2𝜌𝑖
�2𝑁ℎ𝑝 − 1� =
(9.6274𝑒 + 5)2 ⋅ 1.2311
2 ⋅ 54.000
(2 ⋅ 5.0000 − 1) = (9.5089𝑒 + 10)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 1.5836 𝑝𝑠𝑖 
𝛥𝑃𝑖 = 𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑖,𝑙𝑜𝑐 = (2.5235𝑒 + 11) + (9.5089𝑒 + 10) = (3.4744𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 5.7864 𝑝𝑠𝑖 
b) Pressure drop in the annulus 3 
i) Friction factor 4 
𝑓𝐹𝑜 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑒
�
10
9 + 6.9𝑅𝑒𝑜
�
=
0.41
𝑙𝑛2 �0.23 �(7.4583𝑒 − 5)0.033920 �
10
9
+ 6.99925.7�
= 0.0084696 
 5 
𝜙𝑓,𝑜 =
1.02
� 𝜇𝑜𝜇𝑜,𝑤
�
0.14 =
1.02
�4.31005.3100�
0.14 = 1.0502 
𝑓𝐹𝑜,𝑐𝑜𝑟𝑟 = 𝑓𝐹𝑜𝜙𝑓,𝑜 = 0.0084696 ⋅ 1.0502 = 0.0088948 
ii) Distributed pressure drop − annulus 6 
𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 =
4𝐺𝑜2𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑜
𝐷ℎ𝜌𝑜
=
4(1.2612𝑒 + 6)2 ⋅ 16.000 ⋅ 5.0000 ⋅ 0.0088948
0.033920 ⋅ 64.300
= (2.0758𝑒 + 12)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 34.571 𝑝𝑠𝑖 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 145 
iii) Localized pressure drop − annulus 1 
𝐷ℎ(𝑖𝑛) = 0.40704 𝑖𝑛 
𝐾𝑜 = 1.75 +
2000
𝑅𝑒𝑜
+
1.75
𝐷ℎ(𝑖𝑛)
= 1.75 +
2000
9925.7
+
1.75
0.40704
= 6.2508 
𝛥𝑃𝑜,𝑙𝑜𝑐 =
𝐺𝑜2𝐾𝑜𝑁ℎ𝑝
2𝜌𝑜
=
(1.2612𝑒 + 6)2 ⋅ 6.2508 ⋅ 5.0000
2 ⋅ 64.300
= (3.8657𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 6.4381 𝑝𝑠𝑖 
iv) Total pressure drop 2 
𝛥𝑃𝑜 = 𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑜,𝑙𝑜𝑐 = (2.0758𝑒 + 12) + (3.8657𝑒 + 11) = (2.4624𝑒 + 12)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 41.01 𝑝𝑠𝑖 
From the last result, we can see that the previous design is not practicable; once the aniline 3 
pressure drop in the annulus exceeds by far the allowable exchanger pressure drop. A secondary, but 4 
also important, issue is the barely turbulent Reynolds number (Re ~ 10000) for the annulus fluid. 5 
Generally, it is not recommended to design an exchanger in the transition regime, because of the 6 
instabilities on the heat transfer coefficients and friction factor, which may cause sudden variations on 7 
the heat exchanger performance and vibration due to significant oscillations on the pressure drop. 8 
Design 2 9 
Our second attempt for sizing this exchanger should approach the previously identified 10 
problems. In order to reduce the hot fluid pressure drop, let us try a series-parallel bank by dividing the 11 
annulus fluid (aniline) in two separate streams with half flow rate. The drawback from halving the 12 
aniline stream is to reduce the Reynolds number further into the transition region, as we can calculate 13 
below. 14 
- Two parallel streams of aniline through the annulus: 15 
𝐺𝑜 =
𝑊𝑜
𝑆𝑜
=
(10435./2)
0.0082741
= (6.3058𝑒 + 5)
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 
𝑅𝑒𝑜 =
𝐷𝑒𝐺𝑜
𝜇𝑜
=
0.033920(6.3058𝑒 + 5)
4.3100
= 4962.7 → 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 
We are going to refuse this alternative, in order to avoid the transition regime difficulties. 16 
Noticing the benzene is less viscous than the aniline and that the flow rates are similar, one possibility 17 
is to switch the fluid allocation by placing the benzene in the annulus divided in two parallel streams, 18 
and the aniline passing through the inner tube. 19 
- Two parallel streams of benzene through the annulus: 20 
𝐺𝑜 =
𝑊𝑜
𝐴𝑜
=
(10000.0/2)
0.0082741
= (6.0430𝑒 + 5)
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 
𝑅𝑒𝑜 =
𝐷𝑒𝐺𝑜
𝜇𝑜
=
0.033920(6.0430𝑒 + 5)
1.3200
= 15529.→ 𝑓𝑢𝑙𝑙𝑦 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 
Therefore, the change of fluid allocation turns the annulus stream into fully turbulent flow. At 21 
this point, the analysis leads us to the following configuration: 22 
1) Aniline allocated in the inner tube. 23 
2) Benzene allocated in the annulus. 24 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
146 7 Heat Exchanger Design 
3) Divide the annulus (hot fluid) in two parallel streams. 1 
The division of the benzene stream (fluid 1) in two parallel streams through the annulus 2 
channel and the allocation of aniline (fluid 2) inside the internal pipe results in the design flow rates 3 
of: 4 
𝑊𝑜 = 𝑊1 = 5000.0
𝑙𝑏
ℎ
 (𝑎𝑛𝑛𝑢𝑙𝑢𝑠 𝑖𝑛 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙) 
𝑊𝑖 = 𝑊2 = 10435.0
𝑙𝑏
ℎ
 (𝑡𝑢𝑏𝑒 𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠) 
1) Mean temperature difference for series-parallel arrangement 5 
The mean temperature for a hairpin bank with the hot fluid passing in series and the cold fluid 6 
partitioned in nc = 2 parallel streams is given from Eqs. (6.5) – (6.7), as: 7 
𝑃1 =
𝑇2 − 𝑡1
𝑇1 − 𝑡1
=
100.00 − 60.000
150.00 − 60.000
= 0.44444 
𝑅1 =
𝑇1 − 𝑇2
𝑛𝑐(𝑡2 − 𝑡1)
=
−100.00 + 150.00
2.0000(120.00 − 60.000)
= 0.41667 
𝐹𝑠𝑝 =
𝑃1 + 𝑅1(−𝑃1 + 1) − 1
𝑅1𝑛𝑐 𝑙𝑛 �
1
𝑅1
�𝑅1 �
1
𝑃1
�
1
𝑛𝑐 − � 1𝑃1
�
1
𝑛𝑐 + 1��
=
0.41667(−0.44444 + 1) + 0.44444 − 1
0.41667 ⋅ 2.0000 𝑙𝑛 � 10.41667�0.41667 �
1
0.44444�
1
2.0000
− � 10.44444�
1
2.0000
+ 1��
= 0.32300 
Hence, with Eq. (6.4): 8 
𝛥𝑇𝑚 = 𝐹𝑠𝑝(𝑇1 − 𝑡1) = 0.32300(150.00 − 60.000) = 29.070 °𝐹 
2) Inner pipe heat transfer coefficient 9 
𝐺𝑖 =
𝑊𝑖
𝑆𝑖
=
10435.
0.010387
= (1.0046𝑒 + 6)
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 
𝑅𝑒𝑖 =
𝐷𝑖𝐺𝑖
𝜇𝑖
=
0.11500(1.0046𝑒 + 6)
4.3100
= 26805. 
Using the turbulent Sieder and Tate equation (Eq. (4.10)) assuming initially ϕi = (µi/µw)0,14 = 1: 10 
𝑁𝑢 = 0.027𝑃𝑟0.33𝑅𝑒0.8 �
𝜇
𝜇𝑤
�
0.14
 
ℎ𝑖 =
0.027𝑘√𝑃𝑟3
𝐷𝑖
𝑅𝑒
4
5 =
0.027 ⋅ 0.098500 ⋅ √18.1003
0.11500
26805.
4
5 = 211.79
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
3) Annulus heat transfer coefficient 11 
𝐺𝑜 =
𝑊𝑜
𝑆𝑜
=
5000.0
0.0082741
= (6.0430𝑒 + 5)
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 147 
𝑅𝑒𝑜 =
𝐷𝑒𝐺𝑜
𝜇𝑜
=
0.033920(6.0430𝑒 + 5)
1.3200
= 15529. 
From Eq. (4.10) with ϕo = (µo/µw)0,14 = 1: 1 
ℎ𝑜 =
0.027𝑘√𝑃𝑟3
𝐷𝑒
𝑅𝑒
4
5 =
0.027 ⋅ 0.082000 ⋅ √5.80003
0.033920
15529.
4
5 = 264.31
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
4) Variable properties correction factor 2 
a) Wall temperature and correction factors 3 
𝑡𝑤 =
𝐷𝑖ℎ𝑖𝑡𝑖 + 𝐷𝑜ℎ𝑜𝑡𝑜
𝐷𝑖ℎ𝑖 + 𝐷𝑜ℎ𝑜
=
0.11500 ⋅ 125.00 ⋅ 211.79 + 0.13833 ⋅ 264.31 ⋅ 90.000
0.11500 ⋅ 211.79 + 0.13833 ⋅ 264.31
= 103.99 °𝐹 
Fluid 1 (cold) Fluid 2 (hot) 
𝑩𝒆𝒏𝒛𝒆𝒏𝒆 
𝟏𝟎𝟒.𝟎 °𝑭 − 𝟏𝟒.𝟕𝒑𝒔𝒊 
𝑨𝒏𝒊𝒍𝒊𝒏𝒆 
𝟏𝟎𝟒.𝟎 °𝑭 − 𝟏𝟒.𝟕𝒑𝒔𝒊 
𝜇 = 1.2
lb
(ft ⋅ h)
 𝜇 = 5.71
lb
(ft ⋅ h)
 
The correction factor for non-isothermal flow and the new heat transfer coefficients are: 4 
𝜙𝑖 = �
𝜇𝑖
𝜇𝑖,𝑤
�
0.14
= �
4.3100
5.7100
�
0.14
= 0.96139 
𝜙𝑜 = �
𝜇𝑜
𝜇𝑜,𝑤
�
0.14
= �
1.3200
1.2000
�
0.14
= 1.0134 
Using the viscosity correction factors, the heat transfer coefficients are: 5 
ℎ𝑖,𝑐𝑜𝑟𝑟 = ℎ𝑖𝜙𝑖 = 0.96139 ⋅ 211.79 = 203.61
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
ℎ𝑜,𝑐𝑜𝑟𝑟 = ℎ𝑜𝜙𝑜 = 1.0134 ⋅ 264.31 = 267.85
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
5) Overall heat transfer coefficient 6 
𝑈𝑐 = �
𝐷𝑜
2𝑘
𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� +
1
ℎ𝑜
+
𝐷𝑜
𝐷𝑖ℎ𝑖
�
−1
= �
0.13833 𝑙𝑛 �0.138330.11500�
2 ⋅ 9.4000
+
1
267.85
+
0.13833
0.11500 ⋅ 203.61
�
−1
= 90.907
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
Considering the required fouling factors for each stream: 7 
𝑅𝑑𝑖 = 0.001
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
𝑅𝑑𝑜 = 0.001
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
The design overall heat transfer coefficientis: 8 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
148 7 Heat Exchanger Design 
𝑈𝑑 = �𝑅𝑑𝑜 +
1
𝑈𝑐
+
𝐷𝑜𝑅𝑑𝑖
𝐷𝑖
�
−1
= �0.0010000 +
0.0010000
0.11500
0.13833 +
1
90.907
�
−1
= 75.740
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
10) Number of hairpins 1 
a) Heat transfer areas 2 
𝐴𝑐 =
𝑞
𝑈𝑐𝛥𝑇𝑚
=
(2.1600𝑒 + 5)
29.070 ⋅ 90.907
= 81.736 𝑓𝑡2 
𝐴𝑑 =
𝑞
𝑈𝑑𝛥𝑇𝑚
=
(2.1600𝑒 + 5)
29.070 ⋅ 75.740
= 98.103 𝑓𝑡2 
b) Total pipe length and number of exchangers 3 
𝐿 =
𝐴𝑑
𝐴𝑙𝑖𝑛𝑒𝑎𝑟
=
98.103
0.43458
= 225.74 𝑓𝑡 
𝑁ℎ𝑝 =
𝐿
2𝐿ℎ𝑝
=
225.74
2 ⋅ 16.000
= 7.0544 
Therefore, we must use two identical banks in parallel with four exchangers each. The total 4 
number of hairpins for the service is: 5 
𝑁ℎ𝑝,𝑡𝑜𝑡𝑎𝑙 = 8.0 
The total heat transfer area available from all exchangers is: 6 
𝐴 = 2𝐴𝑙𝑖𝑛𝑒𝑎𝑟𝐿ℎ𝑝𝑁ℎ𝑝 = 2 ⋅ 0.43458 ⋅ 16.000 ⋅ 8.0000 = 111.25 𝑓𝑡2 
The actual overall heat transfer coefficient and fouling factor are: 7 
𝑈 =
𝑞
𝐴𝛥𝑇𝑚
=
(2.1600𝑒 + 5)
111.25 ⋅ 29.070
= 66.790
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
𝑅𝑑 =
𝑈𝑐 − 𝑈
𝑈𝑐𝑈
=
90.907 − 66.788
66.788 ⋅ 90.907
= 0.0039725
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
Since the required surface is 98.103 ft2, the over-design is evaluated as: 8 
𝑂𝑣𝑒𝑟 − 𝑑𝑒𝑠𝑖𝑔𝑛(%) = 100 �
𝐴
𝐴𝑑
− 1� = 100 �
111.25
98.103
− 1� = 13.401 % 
𝑂𝑣𝑒𝑟 − 𝑑𝑒𝑠𝑖𝑔𝑛(%) = 100 �
𝑈𝑑
𝑈
− 1� = 100 �
75.740
66.788
− 1� = 13.404 % 
And the over-surface is given by: 9 
𝑂𝑣𝑒𝑟 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(%) = 100 �
𝐴
𝐴𝑐
− 1� = 100 �
111.25
81.736
− 1� = 36.109 % 
𝑂𝑣𝑒𝑟 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(%) = 100 �
𝑈𝑐
𝑈
− 1� = 100 �
90.907
66.788
− 1� = 36.113 % 
Over-surface for fouling: 10 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 149 
𝐹𝑜𝑢𝑙𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(%) = 100 �
𝐴𝑑
𝐴𝑐
− 1� = 100 �
98.103
81.736
− 1� = 20.024% 
𝐹𝑜𝑢𝑙𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(%) = 100 �
𝑈𝑐
𝑈𝑑
− 1� = �
90.907
75.740
− 1� = 20.025% 
11) Pressure drop 1 
a) Pressure drop in the tube 2 
i) Friction factor 3 
Using the Haaland equation, the isothermal tube friction factor is: 4 
𝑓𝐹𝑖 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑖
�
10
9 + 6.9𝑅𝑒𝑖
�
=
0.41
𝑙𝑛2 �0.23 �(7.4583𝑒 − 5)0.11500 �
10
9
+ 6.926805.�
= 0.0063480 
For turbulent flow (Re > 10000), the corrected friction factor is evaluated as: 5 
𝜙𝑓,𝑖 =
1.02
� 𝜇𝑖𝜇𝑖,𝑤
�
0.14 =
1.02
�4.31005.7100�
0.14 = 1.0610 
𝑓𝐹𝑖,𝑐𝑜𝑟𝑟 = 𝑓𝐹𝑖𝜙𝑓,𝑖 = 0.0063480 ⋅ 1.0610 = 0.0067352 
ii) Distributed pressure drop − tube 6 
𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 =
4𝐺𝑖2𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑖
𝐷𝑖𝜌𝑖
=
4(1.0046𝑒 + 6)2 ⋅ 16.000 ⋅ 8.0000 ⋅ 0.0067352
0.11500 ⋅ 64.300
= (4.7065𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 7.8384 𝑝𝑠𝑖 
iii) Localized pressure drop – tube 7 
(1) Pressure loss coefficient 8 
𝐷𝑖(𝑖𝑛) = 1.38 𝑖𝑛 
𝐾𝑖 = 0.7 +
2000
𝑅𝑒𝑖
+
0.7
𝐷𝑖(𝑖𝑛)
= 0.7 +
2000
26805.
+
0.7
1.3800
= 1.2819 
(2) Localized pressure drop 9 
𝛥𝑃𝑖,𝑙𝑜𝑐 =
𝐺𝑖2𝐾𝑖
2𝜌𝑖
�2𝑁ℎ𝑝 − 1� =
(1.0046𝑒 + 6)2 ⋅ 1.2819
2 ⋅ 64.300
(2 ⋅ 8.0000 − 1) = (1.5090𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 2.5131 𝑝𝑠𝑖 
iv) Total pressure drop − tube 10 
𝛥𝑃𝑖 = 𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑖,𝑙𝑜𝑐 = (1.5090𝑒 + 11) + (4.7065𝑒 + 11) = (6.2155𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 10.351 𝑝𝑠𝑖 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
150 7 Heat Exchanger Design 
b) Pressure drop in the annulus 1 
i) Friction factor 2 
𝑓𝐹𝑜 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑒
�
10
9 + 6.9𝑅𝑒𝑜
�
=
0.41
𝑙𝑛2 �0.23 �(7.4583𝑒 − 5)0.033920 �
10
9
+ 6.915529.�
= 0.0077709 
𝜙𝑓,𝑜 =
1.02
� 𝜇𝑜𝜇𝑜,𝑤
�
0.14 =
1.02
�1.32001.2000�
0.14 = 1.0065 
𝑓𝐹𝑜,𝑐𝑜𝑟𝑟 = 𝑓𝐹𝑜𝜙𝑓,𝑜 = 0.0077709 ⋅ 1.0065 = 0.0078214 
ii) Distributed pressure drop − annulus 3 
𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 =
4𝐺𝑜2𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑜
𝐷ℎ𝜌𝑜
=
4(6.0430𝑒 + 5)2 ⋅ 16.000 ⋅ 4.0000 ⋅ 0.0078214
0.033920 ⋅ 54.000
= (3.9919𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 6.6482 𝑝𝑠𝑖 
iii) Localized pressure drop – annulus 4 
(1) Pressure loss coefficient 5 
𝐷ℎ(𝑖𝑛) = 0.40704 𝑖𝑛 
𝐾𝑜 = 1.75 +
2000
𝑅𝑒𝑜
+
1.75
𝐷ℎ(𝑖𝑛)
= 1.75 +
2000
15529.
+
1.75
0.40704
= 6.1781 
(2) Localized pressure drop 6 
𝛥𝑃𝑜,𝑙𝑜𝑐 =
𝐺𝑜2𝐾𝑜𝑁ℎ𝑝
2𝜌𝑜
=
(6.0430𝑒 + 5)2 ⋅ 4.0000 ⋅ 6.1780
2 ⋅ 54.000
= (8.3558𝑒 + 10)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 1.3916 𝑝𝑠𝑖 
iv) Total pressure drop − annulus 7 
𝛥𝑃𝑜 = 𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑜,𝑙𝑜𝑐 = (3.9919𝑒 + 11) + (8.3558𝑒 + 10) = (4.8275𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 8.0399 𝑝𝑠𝑖 
Conclusion 8 
The last configuration of 8 hairpins with the annulus connected in two parallel banks of 4 9 
exchangers, and the inner pipe connected in series, matches all the design requirements of heat transfer 10 
area and pressure drop, with the additional benefit that neither stream is operating in the transition 11 
flow regime. Also, the over-design of about 13% is a reasonable value. 12 
7.5 Rating 13 
The quantitative evaluation of how an existing heat exchanger equipment – or a proposed 14 
design – performs against a given heat transfer duty is usually named “analysis” or “rating”. The heat 15 
exchanger specification should be available with sufficient detail to allow the evaluation of the several 16 
heat transfer parameters necessary to estimate the overall heat transfer coefficient for the equipment, 17 
such as the pertinent nondimensional numbers for each process stream, e.g. Prandtl, Reynolds, 18 
Nusselt, Graetz, friction factors, etc. Additionally, the pressure drop developed should be calculated, 19 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 151 
in order to check its accordance within the allowable pressure drop in the process line. Therefore, in 1 
summary, the designer rates a heat exchanger to answer the following questions: 2 
1. Is the heat transfer surface (size) enough to transfer the required heat between the fluids? 3 
2. Is the added pressure drop for both streams acceptable for the line in which the heat 4 
exchanger will be installed? 5 
To answer the first question, it is necessary to determine how the analyzed heat exchanger 6 
would perform under the specified process conditions. From this calculation, we get the required heat 7 
transfer surface to meet the proposed service. The heat exchanger (or bank) is considered “thermally 8 
satisfactory” if the available heat transfer area is greater than the required one. In other words, the 9 
equipment must have some excess area, quantified as over-design, as calculated from Eq. (7.3), and 10 
repeated below for convenience: 11 
𝑂𝑣𝑒𝑟 − 𝑑𝑒𝑠𝑖𝑔𝑛(%) = 100 �
𝐴
𝐴𝑑
− 1� = 100 �
𝑈𝑑
𝑈
− 1� (7.6) 
For the purpose of rating, A is the effective heat transfer surface of the tested exchanger, and 12 
Ad is the required heat transfer area, including the provisioned fouling resistances. The same result can 13 
be obtained using the fouled (Ud) and actual (U) overall heat transfer coefficients. 14 
Along with being thermally satisfactory, the heat exchanger must not block the process line, 15 
nor impare the product flow rate when installed, therefore we have to answer the second question, in 16 
order to make sure the developed pressure drop does not exceed the available pressure drop in the line. 17 
7.5.1 Rating Outline 18 
1. Thermal Design 19 
1.1. Collect process data parameters. 20 
1.1.1. Fluid temperatures. 21 
1.1.2. Physical properties. 22 
1.2. Energy balance using mean bulk temperatures of the fluids. 23 
1.2.1. Determine any unknown fluid temperature or flow rate. 24 
1.2.2. Or just check the energy balance between fluids for correctness. 25 
1.3. Select a heat exchanger type: double pipe heat exchanger in the present case. 26 
1.4. Select the design variable: 27 
1.4.1. Pipe length, if the diameter is specified. 28 
1.4.2. Pipe diameter, if length is specified. 29 
1.4.3. Number of hairpins, if the service should be performed by a set of standardized hairpin 30 
exchangers. 31 
1.5. Evaluate the actual mean temperaturedifference to be used in the heat exchanger design 32 
equation. 33 
1.6. Overall heat transfer coefficient. 34 
1.6.1. Calculate the inner pipe heat transfer coefficient. 35 
1.6.2. Calculate the annulus heat transfer coefficient. 36 
1.6.3. Calculate wall temperature. 37 
1.6.4. Iterate steps 1.6.1 to 1.6.3 until convergence of the wall temperature is achieved. 38 
1.6.5. Calculate the clean overall heat transfer coefficient Uc. 39 
1.6.6. Calculate the fouled overall heat transfer coefficient Ud by introducing the fouling 40 
factors Rdi and Rdo. 41 
1.7. Heat transfer areas 42 
1.7.1. Calculate the required clean heat transfer area Ac, using Uc. 43 
1.7.2. Calculate the required fouled heat transfer area Ad, using Ud. 44 
1.7.3. Calculate the available heat transfer area A, from the equipment geometric 45 
specifications. 46 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
152 7 Heat Exchanger Design 
1.7.4. Calculate the required overall heat transfer coefficient U, using the available area A, the 1 
heat duty q and the appropriate mean temperature difference (e.g. ΔTlm) 2 
1.8. Calculate the over-design excess area, from Eq. (7.6). 3 
2. Hydraulic Design (verify if the allowable pressure drop is not exceeded). 4 
2.1. Evaluate the inner pipe pressure drop. 5 
2.2. Evaluate the annulus pressure drop. 6 
7.6 The Data Sheet 7 
Once the design procedure is done, all the data necessary to undertake the detailed mechanical 8 
design and building the heat exchanger must be provided to the manufacturer. Accordingly, 9 
fundamental information such as exchanger type, piping specs, operating conditions (flow rates, 10 
temperatures, stream pressure, pressure drop, fouling factors, etc.), performance requirements, and any 11 
special requests are packed in a formulary generally called “data sheet”, “specification sheet” or 12 
“spec-sheet”. 13 
The data sheet is filled in progressively in steps. In some cases, if the heat exchanger type is 14 
proprietary or patented the process engineer just specify the fluids and operating conditions, and let the 15 
contractor do the thermal and mechanical designs. A common situation is to have the engineering staff 16 
of a chemical plant performing the thermal and hydraulic calculations, based on their better knowledge 17 
about the process. Then, the data sheet with the derived design information filled in goes to the heat 18 
exchanger vendor for accomplishing the complete mechanical specification. 19 
There is no rigid format for the fields included in a heat exchanger data sheet, and, in fact, 20 
many manufacturers set their own standard, however typically they consist in some modification of 21 
the format defined by the TEMA standard [109]. The illustration of a typical TEMA data sheet is 22 
presented in Figure 31. Though this data sheet is originally targeted to shell and tube heat exchangers, 23 
it is suitable for double-pipe, multi-tube and finned heat exchangers as well. The lines in the data sheet 24 
are numbered for precise and easy reference of the information therein. 25 
Therefore, the data sheet is a quite important document, and the designer must insure its 26 
correctness. Commonly, it is the sole document forwarded to the equipment supplier. Although the 27 
detailed calculation log of the design procedure may comprise dozens of pages, the exchanger 28 
construction will be based on the few pages in the data sheet! 29 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 153 
1 CUSTOMER NO. SPEC. NO
2 LOCATION PAGE OF
3 PLANT SECTION
4 SERVICE
5 Size TEMA Type Connected in (series/parallel)
6 Surface per Unit Shells per Unit Surface per Shell m²
7
8 Shell Side Tube Side
9 Fluid Name
10 Flow Total kg/h
11 Vapor kg/h (in/out)
12 Liquid kg/h (in/out)
13 Steam kg/h (in/out)
14 Water kg/h (in/out)
15 Noncondensable kg/h (in/out)
16 Temperature (In/Out) °C (in/out)
17 Density kg/m3
18 Viscosity cP
19 Molecular Weight, vapor
20 Specif ic Heat J/(kg °C)
21 Thermal Conductivity W/(m °C)
22 Latent Heat J/kg
23 Inlet Pressure kPa(a) (inlet)
24 Velocity m/s
25 Press Drop Allow /Calc kPa(g)
26 Fouling Factor (m² °C)/W
27 Heat Exchanged W MTD (corrected) °C
28 Heat Transfer Coeff icient W/(m² °C)
29 Service Coeff. W/(m² °C) Dirty Clean
30
31 Shell Side Tube Side
32 Design/Test Press kPa(g)
33 Design Temperature °C
34 No. Passes per Shell
35 Corrosion Allow ance mm
36 In
37 Out
38 Intermediate
39 Tubes No. OD, mm Gauge Length, m. Pitch layout, deg.
40 Type Material Pitch ratio
41 Shell OD, mm ID, mm Material
42 Channel or Bonnet OD, mm Thick Channel Cover
43 Tubesheet Type
44 Floating Heat Cover Impingement Protection
45 Baffles Cross (number) % Cut (d) Spacing C/C, mm
46 Baffles Long Seal Type No
47 Supports Tube U-Bend Type
48 Bypass Seal Arrangement Tube-Tubesheet Joint
49 Expansion Joint No. Type
50 ρV2-Inlet Nozzle Bundle Entrance Bundle Exit
51 Gaskets - Shell Side Tube Side
52 Floating Heat Cover Supports
53 Code Requirements TEMA Class
54 Weight per shell kg Filled w /w ater Bundle
55
56
57
58
59
60
61
62
63
64
HEAT EXCHANGER SPECIFICATION SHEET
REVISION NO.CUSTOMER
PROJECT NO.
TAG
DATE
Notes
Sketch
PROCESS SPEC. VER.
EQUIP. NO
CONSTRUCTION DATA FOR ONE SHELL
REVISION NO.
QUALITY LEVEL
PREPARED BY
CHECKED BY
APPROVED BY
Connections Size & 
Rating
PERFORMANCE OF ONE UNIT
 1 
Figure 31: Typical heat exchanger data sheet. [Adapted from Ref. 2 
[109]] 3 
Unless stated otherwise, the manufacturer’s guarantee will cover the thermal performance and 4 
the mechanical integrity when the heat exchanger operates strictly in the conditions given in the data 5 
sheet. 6 
The information comprising the data sheet is usually grouped in the following sections: 7 
Identification and control 8 
This section records customer and management information about the design task, e.g. the 9 
process service, the chemical plant and its location, equipment id, etc. An overview of the equipment 10 
is included, such as the total heat transfer area (surface per unit, line 6), number of connected 11 
exchangers (shells per unit, line 6), flow arrangement (connected in, line 5), etc. 12 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
154 7 Heat Exchanger Design 
1 CUSTOMER NO. SPEC. NO.
2 LOCATION PAGE OF
3
4 SERVICE
5 Size TEMA Type Connected in (series/parallel)
6 Surface per Unit Shells per Unit Surface per Shell m²
REVISION NO.CUSTOMER
PROJECT NO.
TAG
EQUIP. NO.PLANT SECTION
 1 
Streams and physical properties 2 
The flow rates, inlet/outlet temperatures, and physical properties of the fluids are listed here. 3 
The fluid allocation is also specified. The physical properties are evaluated at the average bulk 4 
temperatures (design temperature, line 33). Notice that intermediary temperatures (e.g. wall 5 
temperatures) used in the calculations do not appear in the data sheet. 6 
8 Shell Side Tube Side
9 Fluid Name
10 Flow Total kg/h
11 Vapor kg/h (in/out)
12 Liquid kg/h (in/out)
13 Steam kg/h (in/out)
14 Water kg/h (in/out)
15 Noncondensable kg/h (in/out)
16 Temperature (In/Out) °C (in/out)
17 Density kg/m3
18 Viscosity cP
19 Molecular Weight, vapor
20 Specif ic Heat J/(kg °C)
21 Thermal Conductivity W/(m °C)
22 Latent Heat J/kg 7 
Thermal and hydraulic performance 8 
A summary of the thermal and hydraulic calculations is compiled in this section. Although the 9 
heat transfer coefficient for each stream is not part of the standard TEMA data sheet, it is useful 10 
information during the design development and revision steps. 11 
23 Inlet Pressure kPa(a) (inlet)
24 Velocity m/s
25 Press Drop Allow /Calc kPa(g)
26 Fouling Factor (m² °C)/W
27 Heat Exchanged W MTD (corrected) °C
28 Heat Transfer Coeff icient W/(m² °C)
29 Service Coeff. W/(m² °C) Dirty Clean 12 
Inner tubes (bundle) and outer tube (shell) configuration 13 
Basic constructioninformation for the inner tube − or tube bundle, in case of multi-tube 14 
exchangers − is presented in lines 31-54. This is not the mechanical design, which is considerably 15 
more detailed in the specification of each assembled component in the heat exchanger, such as 16 
gaskets, flanges, valves, bolts, welding, etc. 17 
31 Shell Side Tube Side
32 Design/Test Press kPa(g)
33 Design Temperature °C
34 No. Passes per Shell
35 Corrosion Allow ance mm
36 In
37 Out
38 Intermediate
39 Tubes No. OD, mm Gauge Length, m. Pitch layout, deg.
40 Type Material Pitch ratio
41 Shell OD, mm ID, mm Material
42 Channel or Bonnet OD, mm Thick Channel Cover
43 Tubesheet Type
44 Floating Heat Cover Impingement Protection
45 Baffles Cross (number) % Cut (d) Spacing C/C, mm
46 Baffles Long Seal Type No
47 Supports Tube U-Bend Type
48 Bypass Seal Arrangement Tube-Tubesheet Joint
49 Expansion Joint No. Type
50 ρV2-Inlet Nozzle Bundle Entrance Bundle Exit
51 Gaskets - Shell Side Tube Side
52 Floating Heat Cover Supports
53 Code Requirements TEMA Class
54 Weight per shell kg Filled w /w ater Bundle
Sketch
Connections Size & 
Rating
 18 
Annotations 19 
Any important assumption and special requirement must be recorded in the annotation section. 20 
The area for annotation can grow considerably to several pages in a practical design; therefore, the 21 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 155 
notes need to be numbered to facilitate the identification. It is usual to fill some fields in the data sheet 1 
with a reference to a given note. 2 
55
56
57
Notes
 3 
Revision control 4 
The specification and acquisition of heat exchangers typically involves a considerable cost. For 5 
safety, the design process is subjected to overseeing and cooperation of several engineers. This section 6 
serves to the purpose of tracking the revisions during the design phase. 7 
58
59
60
61
62
63
64 DATE
PROCESS SPEC. VER.
REVISION NO.
QUALITY LEVEL
PREPARED BY
CHECKED BY
APPROVED BY
 8 
Worked Example 7–3 9 
A residual hot water stream of 19842 lb/h must be cooled down from 158°F to 140°F before 10 
treatment. The energy should be recovered to a product stream of benzoic acid being heated from 95°F 11 
to 122°F. There is available a mounted bank of 12 carbon steel double-pipe heat exchangers of length 12 
16 ft, built with NPS 2 in × 3-1/2 in - Sch 40 pipes. The hairpins have the tube and annulus both 13 
connected in series and the annulus return is of bonnet-type. Fouling factors of 0.002 ft2 h °F/BTU and 14 
0.001 ft2 h °F/BTU are required for water and benzoic acid, respectively. Consider that water is 15 
allocated in the tube and an allowable pressure drop for the tube and annulus is 20 psi. Determine if 16 
the available equipment is able to perform this service. 17 
Solution 18 
1) Fluid physical properties 19 
a) Mean temperature of the fluids 20 
Since the benzoic acid flow rate is not specified, we need to determine its value from the 21 
energy balance using the average bulk temperatures of the fluids. The process temperatures are given 22 
as: 23 
𝑊𝑎𝑡𝑒𝑟: 𝑇1,𝑖𝑛 = 158°𝐹 → 𝑇1,𝑜𝑢𝑡 = 140°𝐹 
𝐵𝑒𝑛𝑧𝑜𝑖𝑐 𝑎𝑐𝑖𝑑: 𝑇2,𝑜𝑢𝑡 = 122°𝐹 ← 𝑇2,𝑖𝑛 = 95°𝐹
 
The average bulk temperatures and the heat capacities for both fluids are: 24 
𝑇1𝑚 =
𝑇1,𝑖𝑛
2
+
𝑇1,𝑜𝑢𝑡
2
=
140.00
2
+
158.00
2
= 149.00°𝐹 
𝑇2𝑚 =
𝑇2,𝑖𝑛
2
+
𝑇2,𝑜𝑢𝑡
2
=
122.00
2
+
95.000
2
= 108.50°𝐹 
Fluid 1 (hot) − Tube Fluid 2 (cold) − Annulus 
Water 
609.0 °R = 149.0°F 14.7psi 
Benzoic acid 
568.0 °R = 108.0°F 14.7psi 
Cp = 0.983
BTU
(lb ⋅ °F)
 𝐶𝑝 = 0.324
BTU
(lb ⋅ °F)
 
b) Benzoic acid flow rate 25 
From the enthalpy balance, the mass flow rate of the benzoic acid can be evaluated as: 26 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
156 7 Heat Exchanger Design 
𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� + 𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 0 
𝑊2 =
𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛�
𝐶𝑝2�𝑇2,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡�
=
0.98300 ⋅ 19842. (140.00 − 158.00)
0.32400(−122.00 + 95.000)
= 40133.
𝑙𝑏
ℎ
 
Let’s check the heat load for both fluids using the four process temperatures and the respective 1 
flow rates: 2 
𝑞1 = 𝑊1𝐶𝑝1�−𝑇1,𝑖𝑛 + 𝑇1,𝑜𝑢𝑡� = 0.98300 ⋅ 19842. (140.00 − 158.00) = (−3.5108𝑒 + 5)
𝐵𝑇𝑈
ℎ
 
𝑞2 = 𝐶𝑝2𝑊2�−𝑇2,𝑖𝑛 + 𝑇2,𝑜𝑢𝑡� = 0.32400 ⋅ 40133. (122.00 − 95.000) = (3.5108𝑒 + 5)
𝐵𝑇𝑈
ℎ
 
Within the accuracy of five significant figures used in the calculations, the energy exchange is 3 
balanced. 4 
Using Appendix C.12 and C.13 , the physical properties at the bulk temperatures are estimated 5 
as: 6 
Fluid 1 (hot) Fluid 2 (cold) 
Water 
609.0 °R = 149.0°F 14.7psi 
Benzoic acid 
568.0 °R = 108.0°F 14.7psi 
μ = 1.04 lb/(ft⋅h) 
μ = 0.43 cP 
ρ = 64.0 lb/ft3 
Cp = 0.983 BTU/(lb⋅°F) 
k = 0.38 BTU/(ft⋅h⋅°F) 
Pr = 2.68 
μ = 34.6 lb/(ft⋅h) 
μ = 14.3 cP 
ρ = 73.7 lb/ft3 
Cp = 0.324 BTU/(lb⋅°F) 
k = 0.0949 BTU/(ft⋅h⋅°F) 
Pr=118.0 
2) Logarithmic mean temperature difference. 7 
For counterflow arrangement, the fluid temperatures at the heat exchanger terminals are: 8 
Terminal 1 Terminal 2 
𝑇1,𝑖𝑛 = 158°𝐹 → 𝑇1,𝑜𝑢𝑡 = 140°𝐹 
𝑇2,𝑜𝑢𝑡 = 122°𝐹 ← 𝑇2,𝑖𝑛 = 95°𝐹
 
Δ𝑇1 = 36.000°F Δ𝑇2 = 45.000°F 
With the temperature differences at the terminals, the log mean temperature difference is: 9 
𝛥𝑇1 = 𝑇1,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡 = −122.00 + 158.00 = 36.000°𝐹 
𝛥𝑇2 = 𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛 = 140.00 − 95.000 = 45.000°𝐹 
𝛥𝑇𝑙𝑚,𝑐 =
𝛥𝑇1 − 𝛥𝑇2
𝑙𝑛 �𝛥𝑇1𝛥𝑇2
�
=
36.000 − 45.000
𝑙𝑛 �36.00045.000�
= 40.333°𝐹 
Notice that, considering its bulk viscosity of 14.3 cP, the benzoic acid is a relatively viscous 10 
stream, what indicates a possible increase of the overall heat transfer coefficient if a parallel flow 11 
arrangement is employed. In such case, as a later analysis refinement, it is worth to verify the 12 
performance of this exchanger in parallel flow. 13 
3) Pipe dimensions 14 
The detailed size specifications for both pipes are obtained from the Appendix A.1 as: 15 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 157 
Inner tube: NPS 2 in – Sch STD/40/40S 
ID = Di = 2.067 in (inner diameter) 
OD = Do = 2.375 in (outer diameter) 
Outer tube: NPS 3−1/2 in - Sch STD/40/40S 
ID = Ds = 3.548 in (inner diameter) 
OD = 4.0 in (outer diameter) 
 1 
Carbon steel 
Absolute roughness: ϵ = 0.001378 in = 0.00011483 ft 
Heat conductivity: k = 29.467 BTU/(ft⋅h⋅°F) 
4) Inner pipe heat transfer coefficient 2 
The cross section of the inner pipe, mass flux and Reynolds number are: 3 
𝑆𝑖 =
𝜋𝐷𝑖2
4
=
𝜋0.172252
4
= 0.023303𝑓𝑡2 
𝐺𝑖 =
𝑊𝑖
𝑆𝑖
=
19842.
0.023303
= (8.5148𝑒 + 5)
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 
𝑅𝑒𝑖 =
𝐷𝑖𝐺𝑖
𝜇𝑖
=
(8.5148𝑒 + 5)0.17225
1.0400
= (1.4103𝑒 + 5) 
Therefore, the flow regime is turbulent. Considering the carbon steel absolute roughness, the 4 
isothermal friction factor coefficient can be estimated with the Haaland’s equation as: 5 
𝑓𝐹𝑖 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑖
�
10
9 + 6.9𝑅𝑒𝑖
�
=
0.41
𝑙𝑛2 �0.23 �0.000114830.17225 �
10
9
+ 6.9(1.4103𝑒 + 5)�
= 0.0050019 
The convective heat transfer coefficient can be evaluated from the Petukhov-Popov equation 6 
(Eq. (4.11)): 7 
𝐾1 = 13.6𝑓𝐹𝑖 + 1 = 13.6 ⋅ 0.0050019 + 1 = 1.0680 
𝐾2 = 11.7 +
1.8
�𝑃𝑟𝑖
3 = 11.7 +
1.8
√2.68003
= 12.996 
𝑁𝑢𝑖 =
𝑃𝑟𝑖𝑅𝑒𝑖𝑓𝐹𝑖
2𝐾1 + √2𝐾2�𝑓𝐹𝑖 �𝑃𝑟𝑖
2
3 − 1�
=
(1.4103𝑒 + 5)0.0050019 ⋅ 2.6800
√2√0.0050019 ⋅ 12.996 �2.6800
2
3 − 1� + 2 ⋅ 1.0680
= 565.33 
ℎ𝑖 =
𝑘𝑁𝑢𝑖
𝐷𝑖
=
0.38000
0.17225
565.33 = 1247.2
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
5) Annulus heat transfer coefficient 8 
𝐷𝑒 = 𝐷𝑠 − 𝐷𝑜 = 0.29567 − 0.19792 = 0.097750 𝑓𝑡 
𝑆𝑜 =
𝜋
4
(−𝐷𝑜2 + 𝐷𝑠2) =
𝜋
4
(−0.197922 + 0.295672) = 0.037894𝑓𝑡2 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
158 7 Heat Exchanger Design 
𝐺𝑜 =
𝑊𝑜
𝑆𝑜
=
40133.
0.037894
= (1.0591𝑒+ 6)
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 
𝑅𝑒𝑜 =
𝐷𝑒𝐺𝑜
𝜇𝑜
=
(1.0591𝑒 + 6)0.097750
34.600
= 2992.1 
𝑁𝑢𝑜 = 𝑃𝑟𝑜0.33(0.116𝑅𝑒𝑜0.67 − 14.5) = 118.000.33(0.116 ⋅ 2992.10.67 − 14.5) = 47.366 
ℎ𝑜 =
𝑘𝑁𝑢𝑜
𝐷𝑒
=
0.094900
0.097750
47.366 = 45.985
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
6) Variable physical property factor 1 
a) Wall temperature 2 
Disregarding the wall conductive and fouling resistances, the wall temperature may be 3 
estimated as: 4 
𝑡𝑤 =
𝐷𝑖ℎ𝑖𝑡𝑖 + 𝐷𝑜ℎ𝑜𝑡𝑜
𝐷𝑖ℎ𝑖 + 𝐷𝑜ℎ𝑜
=
0.17225 ⋅ 1247.2 ⋅ 149.00 + 0.19792 ⋅ 108.50 ⋅ 45.985
0.17225 ⋅ 1247.2 + 0.19792 ⋅ 45.985
= 147.35°𝐹 
Fluid 1 (hot) Fluid 2 (cold) 
Water 
607.0 °R = 147.0 °F 14.7psi 
Benzoic acid 
607.0 °R = 147.0 °F 14.7psi 
μ = 1.05 lb/(ft⋅h) 
μ = 0.434 cP 
ρ = 64.0 lb/ft3 
Cp = 0.983 BTU/(lb⋅°F) 
k = 0.38 BTU/(ft⋅h⋅°F) 
Pr = 2.72 
μ = 15.9 lb/(ft⋅h) 
μ = 6.57 cP 
ρ = 72.4 lb/ft3 
Cp = 0.337 BTU/(lb⋅°F) 
k = 0.0922 BTU/(ft⋅h⋅°F) 
Pr = 58.1 
Due to the strong change in the benzoic acid viscosity, it is recommended the use of the 5 
Petukhov or Gnielinski factors to account for the effect of the varying physical properties on the heat 6 
transfer coefficients. Then, from Eq. (4.25), we have: 7 
𝜙𝑖 = �
𝑃𝑟𝑖
𝑃𝑟𝑖,𝑤
�
0.11
= �
2.6800
2.7200
�
0.11
= 0.99837 
𝜙𝑜 = �
𝑃𝑟𝑜
𝑃𝑟𝑜,𝑤
�
0.11
= �
118.00
58.100
�
0.11
= 1.0811 
And the corrected heat transfer coefficients are20: 8 
ℎ𝑖,𝑐𝑜𝑟𝑟 = ℎ𝑖𝜙𝑖 = 0.99837 ⋅ 1247.2 = 1245.2
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
ℎ𝑜,𝑐𝑜𝑟𝑟 = ℎ𝑜𝜙𝑜 = 1.0811 ⋅ 45.985 = 49.714
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
7) Overall heat transfer coefficient 9 
20 Even in this case, where the cold stream is highly viscous, the divergence from the Sieder and Tate correction 
factor, based on the viscosity on the wall, is not quite significant: ϕi = (μi/μi,w)0.14 = (1.0400/1.0500)0.14 = 0.99866 and ϕo = 
(μo/μo,w)0.14 = (34.600/15.900)0.14 = 1.1150. 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
 
7 Heat Exchanger Design 159 
Using the corrected convective coefficients, the clean overall heat transfer coefficient is given 1 
by: 2 
𝑈𝑐 =
1
𝐷𝑜
2𝑘 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 1ℎ𝑜
+ 𝐷𝑜𝐷𝑖ℎ𝑖
=
1
0.19792 𝑙𝑛 �0.197920.17225�
2 ⋅ 29.467 +
1
49.714 +
0.19792
0.17225 ⋅ 1245.2
= 46.502
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
The required fouling factors for both streams are: 3 
𝑅𝑑𝑖 = 0.002
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
𝑅𝑑𝑜 = 0.001
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
Therefore, the fouled overall heat transfer coefficient is: 4 
𝑈𝑑 =
1
𝑅𝑑𝑜 +
1
𝑈𝑐
+ 𝐷𝑜𝑅𝑑𝑖𝐷𝑖
=
1
0.0010000 + 0.00200000.17225 0.19792 +
1
46.502
= 40.319
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
8) Heat transfer areas 5 
a) Required heat transfer areas 6 
The required heat transfer surface for performing the service is: 7 
𝐴𝑐 =
𝑞
𝑈𝑐𝛥𝑇𝑙𝑚
=
(3.5108𝑒 + 5)
40.333 ⋅ 46.502
= 187.19𝑓𝑡2 
Including the fouling resistances, the required surface increases to: 8 
𝐴𝑑 =
𝑞
𝑈𝑑𝛥𝑇𝑙𝑚
=
(3.5108𝑒 + 5)
40.319 ⋅ 40.333
= 215.89𝑓𝑡2 
9) Heat exchanger performance 9 
a) Available heat transfer area 10 
The bank is composed of Nhp = 12 hairpins associated in series-series, hence the actual heat 11 
transfer surface is given by: 12 
𝐴 = 2𝐴𝑙𝑖𝑛𝑒𝑎𝑟𝐿ℎ𝑝𝑁ℎ𝑝 = 2 ⋅ 0.62178 ⋅ 12.000 ⋅ 16.000 = 238.76 𝑓𝑡2 
For the required heat duty and mean temperature difference, the actual overall heat transfer 13 
coefficient and fouling factor are: 14 
𝑈 =
𝑞
𝐴𝛥𝑇𝑙𝑚
=
(3.5108𝑒 + 5)
238.76 ⋅ 40.333
= 36.457
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
𝑅𝑑 =
𝑈𝑐 − 𝑈
𝑈𝑐𝑈
=
−36.457 + 46.502
36.457 ⋅ 46.502
= 0.0059251
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
As we see, the actual fouling factor provided by the whole area of the bank surpasses the value 15 
specified for the service. Additionally, using the required surface in the fouled condition, the over-16 
design margin is calculated as: 17 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
160 7 Heat Exchanger Design 
𝑂𝑣𝑒𝑟 − 𝑑𝑒𝑠𝑖𝑔𝑛(%) =
100𝐴
𝐴𝑑
− 100 = −100 +
100
215.89
238.76 = 10.593% 
These results state that the available hairpins bank is enough to perform the requested heat 1 
duty, with some operational margin. Subsequently, it is necessary to confirm that the available 2 
pressure drop for both streams is not exceeded. 3 
10) Pressure drop 4 
a) Inner tube pressure drop 5 
The previously calculated Reynolds number points to turbulent flow: 6 
𝑅𝑒𝑖 = 141030.0 → 𝑓𝑢𝑙𝑙𝑦 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 
Then we can evaluate the friction factor from the Haaland’s equation: 7 
𝑓𝐹𝑖 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑖
�
10
9 + 6.9𝑅𝑒𝑖
�
=
0.41
𝑙𝑛2 �0.23 �0.000114830.17225 �
10
9
+ 6.9(1.4103𝑒 + 5)�
= 0.0050019 
To include the effect of variable physical properties, the correction factor and adjusted friction 8 
factor are: 9 
𝜙𝑓,𝑖 =
1.02
� 𝜇𝑖𝜇𝑖,𝑤
�
0.14 =
1.02
�1.04001.0500�
0.14 = 1.0214 
𝑓𝐹,𝑐𝑜𝑟𝑟 = 𝑓𝐹0𝜙𝑓,𝑖 = 0.0050019 ⋅ 1.0214 = 0.0051089 
i) Distributed pressure drop 10 
𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 =
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹
𝐷𝑖𝜌𝑖
𝐺𝑖2 =
4(8.5148𝑒 + 5)2 ⋅ 0.0051089 ⋅ 12.000
0.17225 ⋅ 64.000
16.000
= (2.5805𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 4.2976 𝑝𝑠𝑖 
ii) Localized pressure drop 11 
With the inner tube diameter expressed in inches, the friction loss coefficient for the internal 12 
pipe and the localized pressure drop are calculated as follows: 13 
𝐷𝑖(𝑖𝑛) = 2.067𝑖𝑛 
𝐾𝑖 = 0.7 +
2000
𝑅𝑒𝑖
+
0.7
𝐷𝑖(𝑖𝑛)
= 0.7 +
0.7
2.0670
+
2000
(1.4103𝑒 + 5)
= 1.0528 
𝛥𝑃𝑖,𝑙𝑜𝑐 =
𝐺𝑖2𝐾𝑖
2𝜌𝑖
�2𝑁ℎ𝑝 − 1� =
(8.5148𝑒 + 5)2 ⋅ 1.0528
2 ⋅ 64.000
(2 ⋅ 12.000 − 1) = (1.3716𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 2.2843𝑝𝑠𝑖 
𝛥𝑃𝑖 = 𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑖,𝑙𝑜𝑐 = (1.3716𝑒 + 11) + (2.5805𝑒 + 11) = (3.9521𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 6.582𝑝𝑠𝑖 
b) Annulus pressure drop 14 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 161 
The annulus fluid operates in the transition range, as indicated the previously calculated 1 
Reynolds number: 2 
𝑅𝑒𝑜 = 2992.1 → 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 
The friction factor accounting for variable physical properties is evaluated as: 3 
𝑓𝐹𝑜 = 2.3 ⋅ 10−8𝑅𝑒𝑜1.5 + 0.0054 = 2.3 ⋅ 10−8 ⋅ 2992.11.5 + 0.0054 = 0.0091644 
𝜙𝑓,𝑜 =
1.02
� 𝜇𝑜𝜇𝑜,𝑤
�
0.14 =
1.02
�34.60015.900�
0.14 = 0.91480 
𝑓𝐹,𝑐𝑜𝑟𝑟 = 𝑓𝐹0𝜙𝑓,𝑜 = 0.0091644 ⋅ 0.91480 = 0.0083836 
i) Distributed pressure drop 4 
And the distribute pressure drop comes from: 5 
𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 =
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹
𝐷ℎ𝜌𝑜
𝐺𝑜2 =
4(1.0591𝑒 + 6)2 ⋅ 0.0083836 ⋅ 12.000
0.097750 ⋅ 73.700
16.000
= (1.0025𝑒 + 12)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 16.696 𝑝𝑠𝑖 
ii) Localized pressure drop 6 
Taking the hydraulic diameter as the annulus equivalent diameter, the localized pressure drop 7 
for an annulus with a straight pipe return can be evaluated as: 8 
𝐷ℎ(𝑖𝑛) = 1.173𝑖𝑛 
𝐾𝑜 = 2.8 +
2000
𝑅𝑒𝑜
+
2.8
𝐷ℎ(𝑖𝑛)
= 2.8 +
2000
2992.1
+
2.8
1.1730
= 5.8555 
𝛥𝑃𝑜,𝑙𝑜𝑐 =
𝐺𝑜2𝐾𝑜𝑁ℎ𝑝
2𝜌𝑜
=
(1.0591𝑒 + 6)2 ⋅ 12.000 ⋅ 5.8555
2 ⋅ 73.700
= (5.3471𝑒 + 11)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 8.9052𝑝𝑠𝑖 
𝛥𝑃𝑜 = 𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑜,𝑙𝑜𝑐 = (1.0025𝑒 + 12) + (5.3471𝑒 + 11) = (1.5372𝑒 + 12)
𝑙𝑏
(𝑓𝑡 ⋅ ℎ2)
= 25.601𝑝𝑠𝑖 
Conclusion 9 
The battery of heat exchangers has enough surface to undergo the requested service, however, 10 
for the given process conditions, the developed pressure drop in the annulus cannot be accepted, 11 
implying that the available equipment is not suitable, at least in the current configuration. 12 
Worked Example 7–4 13 
In an attempt to fulfill the service specified in the Worked Example 7–3, a series-parallel 14 
configuration with the annulus fluid divided in two parallel streams was proposed to overcome the 15 
problem of exceeding benzoic acid pressure drop. Rate this design and determine if it can accomplish 16 
the heat duty adequately. 17 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
162 7 Heat Exchanger Design 
Solution 1 
We can anticipatethat splitting the benzoic acid in two halved parallel streams should solve the 2 
annulus pressure drop excess, since a decrease to about 1/8 of the original pressure loss is expected. 3 
Nevertheless, considering that the division of the stream generally impairs the heat exchanger 4 
performance, we need to test if the available heat transfer surface is still sufficient to achieve the 5 
demanded heat duty. 6 
Much of the calculation done in the former problem is valid; hence let us just adapt the 7 
necessary steps. 8 
1) Benzoic acid flow rate 9 
The cold fluid will pass through the annulus in two parallel streams, therefore, the actual flow 10 
rate used in the calculations is: 11 
𝑊2 =
𝑊2,𝑡𝑜𝑡𝑎𝑙
𝑛
=
40133.
2.0000
= 20066.
𝑙𝑏
ℎ
 
2) Mean temperature difference 12 
For the hot fluid in series and the cold fluid split in nc parallel streams, the effective mean 13 
temperature difference is given by Eq. (6.5), hence, evaluating the terms: 14 
𝑃1 =
𝑇2 − 𝑡1
𝑇1 − 𝑡1
=
140.00 − 95.000
158.00 − 95.000
= 0.71429 
𝑅1 =
𝑇1 − 𝑇2
𝑛𝑐(−𝑡1 + 𝑡2)
=
−140.00 + 158.00
2.0000(122.00 − 95.000)
= 0.33333 
The series-parallel temperature difference factor Fsp is: 15 
𝐹𝑠𝑝 =
𝑃1 + 𝑅1(−𝑃1 + 1) − 1
𝑅1𝑛𝑐 𝑙𝑛 �
1
𝑅1
�𝑅1 �
1
𝑃1
�
1
𝑛𝑐 − � 1𝑃1
�
1
𝑛𝑐 + 1��
=
0.33333(−0.71429 + 1) + 0.71429 − 1
0.33333 ⋅ 2.0000 𝑙𝑛 � 10.33333�0.33333 �
1
0.71429�
1
2.0000
− � 10.71429�
1
2.0000
+ 1��
= 0.62604 
Finally, the effective temperature difference in this configuration is: 16 
𝛥𝑇𝑚 = 𝐹𝑠𝑝(𝑇1 − 𝑡1) = 0.62604(158.00 − 95.000) = 39.441°𝐹 
3) Tube heat transfer coefficient 17 
The isothermal heat transfer coefficient for the inner fluid remains unaltered, then: 18 
ℎ𝑖 = 1247.2
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
4) Annulus heat transfer coefficient 19 
The Reynolds number for the halved benzoic acid flow rate is: 20 
𝐺𝑜 =
𝑊𝑜
𝑆𝑜
=
20066.
0.037894
= (5.2953𝑒 + 5)
𝑙𝑏
(𝑓𝑡2 ⋅ ℎ)
 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 163 
𝑅𝑒𝑜 =
𝐷𝑒𝐺𝑜
𝜇𝑜
=
(5.2953𝑒 + 5)0.097750
34.600
= 1496.0 → 𝑙𝑎𝑚𝑖𝑛𝑎𝑟 
Using the Sieder-Tate equation for the laminar regime, the Nusselt number and convective 1 
coefficient are: 2 
𝑁𝑢𝑜 = 1.86 �
𝐷𝑒𝑃𝑟𝑜
𝐿
𝑅𝑒𝑜�
0.33
= 1.86 �
0.097750
16.000
118.00 ⋅ 1496.0�
0.33
= 19.096 
ℎ𝑜 =
𝑘𝑁𝑢𝑜
𝐷𝑒
=
0.094900
0.097750
19.096 = 18.539
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
5) Variable properties correction factor 3 
a) Wall temperature 4 
𝑡𝑤 =
𝐷𝑖ℎ𝑖𝑡𝑖 + 𝐷𝑜ℎ𝑜𝑡𝑜
𝐷𝑖ℎ𝑖 + 𝐷𝑜ℎ𝑜
=
0.17225 ⋅ 1247.2 ⋅ 149.00 + 0.19792 ⋅ 108.50 ⋅ 18.539
0.17225 ⋅ 1247.2 + 0.19792 ⋅ 18.539
= 148.32°𝐹 
The difference between this wall temperature and the one obtained with the previous 5 
arrangement is neglectable, therefore we are going to use the isothermal heat transfer coefficients to 6 
calculate the overall heat transfer coefficients. 7 
6) Overall heat transfer coefficient 8 
The clean overall heat transfer coefficient is given by: 9 
𝑈𝑐 =
1
𝐷𝑜
2𝑘 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 1ℎ𝑜
+ 𝐷𝑜𝐷𝑖ℎ𝑖
=
1
0.19792 𝑙𝑛 �0.197920.17225�
2 ⋅ 29.467 +
1
18.539 +
0.19792
0.17225 ⋅ 1247.2
= 18.074
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
With the fouling factors: 10 
𝑅𝑑𝑖 = 0.002
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
𝑅𝑑𝑜 = 0.001
𝑓𝑡2 ⋅ ℎ ⋅ °𝐹
𝐵𝑇𝑈
 
The fouled overall heat transfer coefficient can be evaluated as: 11 
𝑈𝑑 =
1
𝑅𝑑𝑜 +
1
𝑈𝑐
+ 𝐷𝑜𝑅𝑑𝑖𝐷𝑖
=
1
0.0010000 + 0.00200000.17225 0.19792 +
1
18.074
= 17.057
𝐵𝑇𝑈
(𝑓𝑡2 ⋅ ℎ ⋅ °𝐹)
 
From the effective mean temperature difference and overall heat transfer coefficients, the clean 12 
and fouled heat transfer surfaces are: 13 
𝐴𝑐 =
𝑞
𝑈𝑐𝛥𝑇𝑚
=
(1.1392𝑒 + 17)
(1.0556𝑒 + 13)21.912
= 492.51𝑓𝑡2 
𝐴𝑑 =
𝑞
𝑈𝑑𝛥𝑇𝑚
=
(3.5108𝑒 + 5)
17.057 ⋅ 39.441
= 521.86𝑓𝑡2 
The total available area on the exchanger battery is: 14 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
164 7 Heat Exchanger Design 
𝐴 = 2𝐴𝑙𝑖𝑛𝑒𝑎𝑟𝐿ℎ𝑝𝑁ℎ𝑝 = 2 ⋅ 0.62178 ⋅ 12.000 ⋅ 16.000 = 238.76𝑓𝑡2 
Therefore the excess area given by the over-design margin is: 1 
𝑂𝑣𝑒𝑟 − 𝑑𝑒𝑠𝑖𝑔𝑛(%) =
100𝐴
𝐴𝑑
− 100 =
100
521.86
238.76 − 100 = −54.248% 
Conclusion 2 
The proposed series-parallel association cannot match the required heat duty. As indicated by 3 
the negative design margin of about −54%, the splitting of the cold fluid in two parallel streams 4 
lowered severely the overall heat transfer coefficient to the point that the whole heat transfer surface of 5 
the bank is not enough to achieve the specified heat transfer rate. 6 
Problems 7 
7.1) A stream of 1.4 kg/s of hot water must be cooled from 73°C to 35°C before discharge, 8 
using cooling water at 20°C. The exit temperature of the cooling water must not exceed 45°C. There is 9 
available 4.5 m long stainless steel double-pipe hairpins (bonnet type), with tube diameters 10 
DN 50×80 mm/40S. The fouling factor is 0.00021 m2K/W and 0.00035 m2K/W for the hot water and 11 
cooling water, respectively. A pressure loss of 70 kPa is allowed for both streams. Determine (a) the 12 
number of exchangers with their tube and annulus connected in series required to perform this service 13 
and answer: (b) the tube and annulus heat transfer coefficient, (c) the clean overall heat transfer 14 
coefficient, (d) fouling overall heat transfer coefficient, (e) tube and annulus pressure drop and (f) the 15 
resulting over-surface of this project. 16 
Answer: (a) Nhp = 9 exchangers (b) hi = 4312.4 W/(m2⋅K); ho = 4775.0 W/(m2⋅K) 17 
(c) Uc = 1212.3 W/(m2⋅K) (d) Ud = 695.88 W/(m2⋅K) (e) ΔPi = 23.387 kPa; ΔPo = 42.065 kPa 18 
(f) Over-surface = 77.6%. 19 
7.2) Examine an alternative design by changing the fluid allocation in the service presented in 20 
Problem 7.1. Determine (a) the number of exchangers with their tube and annulus connected in series 21 
required to perform this service and answer: (b) the tube and annulus heat transfer coefficient, (c) the 22 
clean overall heat transfer coefficient, (d) fouling overall heat transfer coefficient, (e) tube and annulus 23 
pressure drop and (f) the resulting over-surface of this project. (g) Do you think this new design is 24 
feasible? Why? 25 
Answer: (a) Nhp = 9 exchangers (b) hi = 3564.7 W/(m2⋅K); ho = 5708.5 W/(m2⋅K) 26 
(c) Uc = 1181.3 W/(m2⋅K) (d) Ud = 695.5 W/(m2⋅K) (e) ΔPi = 10.187 kPa; ΔPo = 96.533 kPa 27 
(f) Over-surface = 73.1%(g) No. Although the design margin is about the same of the previous 28 
configuration, the permitted pressure drop is exceeded in the annulus. 29 
7.3) Design a bank of double-pipe exchangers to recover heat from a 360 kg/h flue gas stream, 30 
using water at an initial temperature of 20°C. The needed water exit temperature is 90°C. The flue gas 31 
is allocated in the inner pipe and must be cooled from 350°C to 175°C. A number of 2 m long 32 
DN 40×65 mm/40S stainless steel double-pipe (bonnet type return) hairpins are available. A fouling 33 
factor of 0.00018 m2K/W is provided for the water stream. It was verified that to avoid excessive 34 
noise, the velocity of the gas stream should not exceed 20 m/s. Assume that the flue gas has the same 35 
physical properties of the air and determine: 36 
a) Number of gas parallel streams in the tube. 37 
b) Required number of heat exchangers in the bank. 38 
c) Final overall heat transfer coefficient. 39 
d) Final fouling factor. 40 
e) Percentage of heat transfer over-surface. 41 
f) Pressure drop in the tube and annulus. 42 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
7 Heat Exchanger Design 165 
Answer: (a) 6 parallel streams (b) 6 heat exchangers (c) U = 25.645 W/(m2⋅K) 1 
(d) Rd = 0.010301 (m2⋅K)/W (e) Over-surface = 35.904% (f) ΔPi = 0.47740 kPa; 2 
ΔPo = 0.13603 kPa. 3 
7.4) During the winter, a 468 kg/h flow rate of ambient air at −10°C must be heated to 20°C 4 
before feeding the ventilation system of a chemical plant facility. To reduce energy costs, it is used a 5 
137kg/h stream of exhausted combustion gases (mostly CO2) at 250°C. There is in standby a number 6 
of 3 m long DN 40×65 mm/40S carbon steel double-pipe hairpins. The annulus has a straight-pipe 7 
type return. Because of noise generation restrictions, the maximum velocity for both streams is set to 8 
50 m/s. The fouling factors are 0.00176 m2⋅K/W and 0.00035 m2⋅K/W for the air and combustion 9 
gases, respectively. The allowable pressure drop is 80 kPa for both streams. Design a battery of 10 
exchangers to accomplish this service. (a) Choose the fluid allocation (justify). (b) Define the 11 
configuration of the heat exchangers battery (justify). Determine (c) the heat transfer coefficients for 12 
the tube and annulus, (d) the final overall heat transfer coefficient, (e) percentage of heat transfer over-13 
surface, (f) the pressure drop in the tube and annulus. 14 
Answer: (a) Combustion gases in the tube and air in the annulus (justify!) (b) tube in series 15 
and annulus divided in two streams (justify!) (c) hi = 81.976 W/(m2⋅K), ho = 155.08 W/(m2⋅K) 16 
(d) U = 11.295 W/(m2⋅K) (e) Over-surface = 323.04% (f) ΔPi = 3.7927 kPa; ΔPo = 20.469 kPa. 17 
7.5) Assume that the exchanger required in the Problem 7.4 will be installed in a housed 18 
remote place, in a way that no restrictions on noise production holds anymore. Redesign the heat 19 
exchanger bank to accomplish this service and determine (a) the heat transfer coefficients for the tube 20 
and annulus, (b) the final overall heat transfer coefficient, (c) percentage of heat transfer over-surface, 21 
(d) the pressure drop in the tube and annulus. 22 
Answer: (a) hi = 81.976 W/(m2⋅K), ho = 265.17 W/(m2⋅K) (b) U = 22.462 W/(m2⋅K) (c) Over-23 
surface = 143.92% (d) ΔPi = 1.6812 kPa; ΔPo = 79.311 kPa. 24 
 25 
 26 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
166 8 Multi-tube Heat Exchangers 
8 Multi-tube Heat Exchangers 1 
A method to attenuate the disadvantage of low compactness of double-pipe exchangers is to 2 
use an internal bundle of tubes instead of a single tube-inside-tube design. Such type of heat exchanger 3 
is commonly referred as “multi-tube” or “multi-pipe” exchanger and Figure 32 is representative of its 4 
structure. The installation of multiple internal tubes may increase the heat transfer area per unit length 5 
of the exchanger a few times, in comparison with a regular double-pipe exchanger. 6 
 7 
Figure 32: Multi-tube hairpin heat exchanger. [Adapted from Ref. 8 
[110]] 9 
When the process requires a temperature cross between the cold and hot fluids, usually the 10 
multi-tube exchanger is the most effective option for transferring greater heat loads. Standard shell 11 
pipes with diameters up to 900 mm (36 in), and surface areas as large as 930 m2 (10,000 ft2) per 12 
exchanger section [110], are found in the market, but the most common range is about 80 to 400 mm 13 
(3 to 16 in) [28]. 14 
8.1 Tube count 15 
In a multi-tube heat exchanger, the quantity of tubes comprising the tube bundle can differ 16 
significantly, ranging from only two to as many as a few hundred, depending on the outer tube (shell) 17 
size. Furthermore, the number of tubes installed inside a shell with a given internal diameter may also 18 
vary from a manufacturer to another, because of particularities in the mechanical construction. 19 
Typically, manufacturers offer standardized tables for tube counting, which must be consulted for 20 
selecting up front a valid number of tubes when designing a multi-tube heat exchanger. With the sole 21 
purpose of convenient reference, Table 6 in Appendix A.4 shows a tube counting table, compiled 22 
from a few heat exchanger suppliers. 23 
8.2 Design Method 24 
Within practical accuracy, the method outlined in Section 7 for designing and rating regular 25 
double-pipe heat exchangers may be adapted using the hydraulic diameter as an equivalent diameter 26 
for the channel delimited by the tube bundle and the external tube shell. 27 
8.2.1 Cross-flow Area 28 
Given the mass flow rates for the internal and external fluids in a multi-tube exchanger, the 29 
cross sectional area used in the calculation of the mass fluxes must be evaluated accordingly. 30 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
8 Multi-tube Heat Exchangers 167 
For the internal fluid, the total cross section Si is the summation of the individual cross sections 1 
over all Nt tubes in the bundle, with the internal diameter Di: 2 
𝑆𝑖 =
𝜋𝑁𝑡
4
𝐷𝑖2 (8.1) 
The cross-flow area So for the external channel is delimited by the external diameter of the 3 
internal tubes Do and the internal diameter of the outer tube Ds: 4 
𝑆𝑜 =
𝜋
4
(𝐷𝑠2 − 𝐷𝑜2𝑁𝑡) (8.2) 
8.2.2 Shell Equivalent Diameter 5 
From the definition of hydraulic diameter, we have: 6 
𝐷ℎ = 4 ×
𝑓𝑙𝑜𝑤 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
 
Defining as Do the outer diameter of a single tube in the bundle containing Nt tubes, and the 7 
inner diameter of the outer tube (shell) as Ds, the flow cross section and wetted perimeter are given by: 8 
𝑓𝑙𝑜𝑤 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 =
𝜋
4
(𝐷𝑠2 − 𝐷𝑜2𝑁𝑡) 
𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 𝜋(𝐷𝑠 + 𝐷𝑜𝑁𝑡) 
Substituting in the hydraulic diameter expression, the equivalent diameter De to be used on the 9 
multi-tube exchanger calculations is: 10 
𝐷𝑒 = 𝐷ℎ =
𝐷𝑠2 − 𝐷𝑜2𝑁𝑡
𝐷𝑠 + 𝐷𝑜𝑁𝑡
 (8.3) 
Worked Example 8–1 11 
Consider a service where 1.8 kg/s of acrylic acid is cooled from 70 °C to 45 °C before storage. 12 
Nearby the acrylic acid storage tank, there is a line of 1 kg/s of isopropanol with 32 °C that can be 13 
used as the cooling fluid. There are available multi-tube hairpins 5 m long built in carbon steel with a 14 
bundle of 9 tubes DN 10 mm-sch 40 inside a shell DN 80 mm-sch 40. The carbon steel absolute 15 
roughness is estimated as ε = 0.035 mm and its thermal conductivity is k = 51 W/mK. Design a bank 16 
of multi-tube heat exchangers to perform this service assuming a fouling factor of Rd = 0.0001 m2K/W 17 
and the allowable pressure drop of 90 kPa for each stream. The shell fluid return is of “straight pipe” 18 
type. 19 
Solution 20 
1) Fluids physical properties 21 
a) Mean temperatures of the fluids 22 
Within the problem specification, we do not have the outlet temperature of the cooling fluid, 23 
isopropanol; therefore, the bulk mean temperature cannot be determined. Let us use the energy balance 24 
to calculate the exit temperature of the isopropanol. 25 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
168 8 Multi-tube Heat Exchangers 
Assuming a counter current flow, the given operating temperatures are: 1 
𝐴𝑐𝑟𝑦𝑙𝑖𝑐 𝑎𝑐𝑖𝑑: 𝑇1,𝑖𝑛 = 343.15𝐾 → 𝑇1,𝑜𝑢𝑡 = 318.15𝐾 
𝐼𝑠𝑜𝑝𝑟𝑜𝑝𝑎𝑛𝑜𝑙: 𝑇2,𝑜𝑢𝑡 = ? ← 𝑇2,𝑖𝑛 = 305.15𝐾
 
As a first approximation, we are not going to guess the exit temperature of the cold fluid, 2 
instead, let us evaluate the specific heats using the entrance temperature of the cold fluid and the mean 3 
bulk temperature of the hot fluid, given by: 4 
𝑇1𝑚 =
𝑇1,𝑖𝑛
2
+
𝑇1,𝑜𝑢𝑡
2
=
318.15
2
+
343.15
2
= 330.65 𝐾 
The specific heat for both fluids can be obtained from Appendix C.11 and Appendix C.12 : 5 
Fluid 1 (hot) Fluid 2 (cold) 
𝑨𝒄𝒓𝒚𝒍𝒊𝒄 𝒂𝒄𝒊𝒅 
𝟑𝟑𝟎.𝟔𝟓𝑲− 𝟏𝟎𝟏𝟑𝟐𝟓𝑷𝒂 
𝑰𝒔𝒐𝒑𝒓𝒐𝒑𝒂𝒏𝒐𝒍 
𝟑𝟎𝟓.𝟏𝟓𝑲− 𝟏𝟎𝟏𝟑𝟐𝟓𝑷𝒂 
𝐶𝑝 = 1870.0
J
(kg ⋅ K)
 𝐶𝑝 = 2520.0
J
(kg ⋅ K)
 
Assuming no heat lost for the environment, from the energy balance between both streams, we 6 
have: 7 
𝐶𝑝2𝑊2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� + 𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 0 
Solving for the isopropanol exit temperature, we get: 8 
𝑇2,𝑜𝑢𝑡 =
1
𝐶𝑝2𝑊2
�𝐶𝑝2𝑇2,𝑖𝑛𝑊2 + 𝑇1,𝑖𝑛𝑊1𝐶𝑝1 − 𝑇1,𝑜𝑢𝑡𝑊1𝐶𝑝1�
=
1
1.0000 ⋅ 2520.0
(1.0000 ⋅ 2520.0 ⋅ 305.15 − 1.8000 ⋅ 1870.0 ⋅ 318.15 + 1.8000
⋅ 1870.0 ⋅ 343.15) = 338.54𝐾 
In addition, the mean bulk temperature of the cold fluid is now evaluatedas: 9 
𝑇2𝑚 =
𝑇2,𝑖𝑛
2
+
𝑇2,𝑜𝑢𝑡
2
=
305.15
2
+
338.54
2
= 321.85𝐾 
Finally, let us check the energy balance, using the calculated outlet temperature: 10 
𝑞1 = 𝑊1𝐶𝑝1�𝑇1,𝑜𝑢𝑡 − 𝑇1,𝑖𝑛� = 1.8000 ⋅ 1870.0(318.15 − 343.15) = −84150.𝑊 
𝑞2 = 𝑊2𝐶𝑝2�𝑇2,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛� = 1.0000 ⋅ 2520.0(−305.15 + 338.54) = 84143.𝑊 
As we see, the enthalpy changes of both streams are in reasonable agreement within the 11 
accuracy of five significant figures used in the calculations. 12 
Using the obtained mean bulk temperatures, from Appendix C.12 and Appendix C.11, the 13 
physical properties are estimated as: 14 
Fluid 1 (hot) Fluid 2 (cold) 
𝑨𝒄𝒓𝒚𝒍𝒊𝒄 𝒂𝒄𝒊𝒅 
𝟑𝟑𝟎.𝟔𝟓𝑲− 𝟏𝟎𝟏𝟑𝟐𝟓𝑷𝒂 
𝑰𝒔𝒐𝒑𝒓𝒐𝒑𝒂𝒏𝒐𝒍 
𝟑𝟐𝟏.𝟖𝟓𝑲− 𝟏𝟎𝟏𝟑𝟐𝟓𝑷𝒂 
𝜇 = 0.000651 s ⋅ Pa 
𝜇 = 0.651cP 
𝜇 = 0.00108s ⋅ Pa 
𝜇 = 1.08cP 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
8 Multi-tube Heat Exchangers 169 
𝜌 = 1090.0
kg
m3
 
𝐶𝑝 = 1870.0
J
(kg ⋅ K)
 
𝑘 = 0.15
W
(m ⋅ K)
 
𝑃𝑟 = 8.15 
𝜌 = 771.0
kg
m3
 
𝐶𝑝 = 2610.0
J
(kg ⋅ K)
 
𝑘 = 0.133
W
(m ⋅ K)
 
𝑃𝑟 = 21.3 
2) Logarithmic mean temperature difference 1 
For counterflow arrangement, the fluids temperatures at the heat exchanger terminals are: 2 
Terminal 1 Terminal 2 
𝑇1,𝑖𝑛 = 343.15K → 𝑇1,𝑜𝑢𝑡 = 318.15K 
𝑇𝑐,𝑜𝑢𝑡 = 338.54K ← 𝑇2,𝑖𝑛 = 305.15K
 
Δ𝑇1 = 4.6100 K Δ𝑇2 = 13.000 K 
With the temperature differences at the terminals, the log mean temperature difference is: 3 
𝛥𝑇1 = 𝑇1,𝑖𝑛 − 𝑇2,𝑜𝑢𝑡 = −338.54 + 343.15 = 4.6100 𝐾 
𝛥𝑇2 = 𝑇1,𝑜𝑢𝑡 − 𝑇2,𝑖𝑛 = −305.15 + 318.15 = 13.000 𝐾 
𝛥𝑇𝑙𝑚,𝑐 =
𝛥𝑇1 − 𝛥𝑇2
𝑙𝑛 �𝛥𝑇1𝛥𝑇2
�
=
4.6100 − 13.000
𝑙𝑛 �4.610013.000�
= 8.0928 𝐾 
3) Pipe dimensions 4 
The detailed size specifications for both pipes are obtained from the Appendix A.1 as: 5 
DN 10 mm - Schedule 40 
Di = ID = 12.48 mm (inner diameter) 
Do = OD = 17.1 mm (outer diameter) 
DN 80 mm - Schedule 40 
Ds = ID = 77.92 mm (inner diameter) 
OD = 88.9 mm (outer diameter) 
4) Inner pipes heat transfer coefficient 6 
The internal bundle is composed of Nt = 9 tubes, therefore the total flow cross section, mass 7 
flux and Reynolds number for the internal fluid are: 8 
𝑆𝑖 =
𝜋𝑁𝑡
4
𝐷𝑖2 =
𝜋9.0000
4
0.0124802 = 0.0011009 𝑚2 
𝐺𝑖 =
𝑊𝑖
𝑆𝑖
=
1.0000
0.0011009
= 908.35
𝑘𝑔
(𝑚2 ⋅ 𝑠)
 
𝑅𝑒𝑖 =
𝐷𝑖𝐺𝑖
𝜇𝑖
=
0.012480
0.0010800
908.35 = 10496. (𝑓𝑢𝑙𝑙𝑦 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤) 
For this Reynolds number, the flow regime is considered fully turbulent, and then we may 9 
apply the Petukhov−Popov equation (Eq. (4.11)) to estimate the heat transfer coefficient. Since the 10 
pipe absolute roughness is known, for increased accuracy, we can use a friction equation that takes 11 
into account the effect of the surface roughness on the flow, such as the Haaland’s equation: 12 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
170 8 Multi-tube Heat Exchangers 
𝑓𝐹 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑖
�
10
9 + 6.9𝑅𝑒𝑖
�
=
0.41
𝑙𝑛2 �0.23 �(3.5000𝑒 − 5)0.012480 �
10
9
+ 6.910496.�
= 0.0085752 
The Petukhov-Popov’s parameters are evaluated as: 1 
𝐾1 = 13.6𝑓𝐹 + 1 = 13.6 ⋅ 0.0085752 + 1 = 1.1166 
𝐾2 = 11.7 +
1.8
√𝑃𝑟3
= 11.7 +
1.8
√21.3003
= 12.349 
Solving the Petukhov-Popov equation for hi, and applying the previously calculated 2 
parameters, we have: 3 
ℎ𝑖 =
𝑃𝑟𝑅𝑒𝑖𝑓𝐹𝑘
𝐷𝑖 �2𝐾1 + √2𝐾2𝑃𝑟
2
3�𝑓𝐹 − √2𝐾2�𝑓𝐹�
=
0.0085752 ⋅ 0.13300 ⋅ 10496.21.300
0.012480�√2√0.0085752 ⋅ 12.349 ⋅ 21.300
2
3 − √2√0.0085752 ⋅ 12.349 + 2 ⋅ 1.1166�
= 1566.5
𝑊
(𝑚2 ⋅ 𝐾)
 
This heat transfer coefficient assumes fully developed flow, condition that is physically valid 4 
in the present case, because for a single hairpin leg the number of tube diameters is already high: 5 
#𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 =
𝐿ℎ𝑝
𝐷𝑖
=
5.0000
0.012480
= 400.64 ≫ 10 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 (𝑓𝑢𝑙𝑙𝑦 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑓𝑙𝑜𝑤) 
Considering the relatively low viscosity of both fluids, the viscosity correction factor for 6 
variable physical properties will be ignored, i.e. ϕi = (µi/µw)0.14 = 1. 7 
5) Shell heat transfer coefficient 8 
The annulus equivalent diameter used in the Reynolds number is taken here as equal to the 9 
hydraulic diameter considering the 9 tubes in the internal bundle: 10 
𝐷𝑒 = 𝐷ℎ =
𝐷𝑠2 − 𝐷𝑜2𝑁𝑡
𝐷𝑠 + 𝐷𝑜𝑁𝑡
=
0.0779202 − 0.0171002 ⋅ 9.0000
0.017100 ⋅ 9.0000 + 0.077920
= 0.014838 𝑚 
𝑆𝑜 =
𝜋
4
(𝐷𝑠2 − 𝐷𝑜2𝑁𝑡) =
𝜋
4
(0.0779202 − 0.0171002 ⋅ 9.0000) = 0.0027016 𝑚2 
𝐺𝑜 =
𝑊𝑜
𝑆𝑜
=
1.8000
0.0027016
= 666.27
𝑘𝑔
(𝑚2 ⋅ 𝑠)
 
𝑅𝑒𝑜 =
𝐷𝑒𝐺𝑜
𝜇𝑜
=
0.014838
0.00065100
666.27 = 15186. (𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤) 
This Reynolds number indicates turbulent flow, and the number of equivalent diameters for the 11 
shell in a single exchanger is: 12 
#𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 =
𝐿ℎ𝑝
𝐷𝑒
=
5.0000
0.014838
= 336.97 ≫ 10 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 (𝑓𝑢𝑙𝑙𝑦 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑓𝑙𝑜𝑤) 
Therefore, for further accuracy, we can use Petukhov-Popov equation combined with the 13 
Haaland’s friction factor as: 14 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
8 Multi-tube Heat Exchangers 171 
𝑓𝐹 =
0.41
𝑙𝑛2 �0.23 � 𝜖𝐷𝑒
�
10
9 + 6.9𝑅𝑒𝑜
�
=
0.41
𝑙𝑛2 �0.23 �(3.5000𝑒 − 5)0.014838 �
10
9
+ 6.915186.�
= 0.0078639 
𝐾1 = 13.6𝑓𝐹 + 1 = 13.6 ⋅ 0.0078639 + 1 = 1.1069 
𝐾2 = 11.7 +
1.8
√𝑃𝑟3
= 11.7 +
1.8
√8.15003
= 12.594 
Solving for ho, and substituting the calculated parameters, we have: 1 
ℎ𝑜 =
𝑃𝑟𝑅𝑒𝑜𝑓𝐹𝑘
𝐷𝑒 �2𝐾1 + √2𝐾2𝑃𝑟
2
3�𝑓𝐹 − √2𝐾2�𝑓𝐹�
=
0.0078639 ⋅ 0.15000 ⋅ 15186.8.1500
0.014838 �√2√0.0078639 ⋅ 12.594 ⋅ 8.1500
2
3 − √2√0.0078639 ⋅ 12.594 + 2 ⋅ 1.0954�
= 1404.0
𝑊
(𝑚2 ⋅ 𝐾)
 
The factor for correction of variable physical properties will be ignored also for the shell fluid, 2 
due to the low viscosity, then ϕo = (µo/µw)0.14 = 1. 3 
6) Overall heat transfer coefficient 4 
The overall heat transfer coefficient, including the wall conductive resistance, the internal and 5 
external convective resistances is: 6 
𝑈𝑐 =
1
𝐷𝑜
2𝑘 𝑙𝑛 �
𝐷𝑜
𝐷𝑖
� + 1ℎ𝑜
+ 𝐷𝑜𝐷𝑖ℎ𝑖
=
1
0.017100 𝑙𝑛 �0.0171000.012480�
2 ⋅ 51.000 +
1
1404.0 +
0.017100
0.012480 ⋅ 1566.5
= 609.85
𝑊
(𝑚2 ⋅ 𝐾)
 
The fouling resistances required for both fluids are Rdi and Rdo: 7 
𝑅𝑑𝑖 = 0.0001
𝑚2 ⋅ 𝐾
𝑊
 
𝑅𝑑𝑜 = 0.0001
𝑚2 ⋅ 𝐾
𝑊
 
Therefore the final design overall heat transfer coefficient used to size the exchanger area is 8 
given by: 9 
𝑈𝑑 =
1
𝑅𝑑𝑜 +
1
𝑈𝑐
+ 𝐷𝑜𝑅𝑑𝑖𝐷𝑖
=
1
0.00010000 + 0.000100000.012480 0.017100 +
1
609.85
= 532.83
𝑊
(𝑚2 ⋅ 𝐾)
 
7) Number of hairpins 10 
a) Heat transfer area 11 
From the heat exchanger design equation, the required clean heat transfer surface is: 12 
𝐴𝑐 =
𝑞
𝑈𝑐𝛥𝑇𝑙𝑚
=
84143.
609.85 ⋅ 8.0928
= 17.049𝑚2 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
172 8 Multi-tube Heat Exchangers 
And the fouled heat transfer area is calculated as: 1 
𝐴𝑑 =
𝑞
𝑈𝑑𝛥𝑇𝑙𝑚
=
84143.
532.83 ⋅ 8.0928
= 19.513𝑚2 
b) Total pipe length and number of exchangers 2 
The heat transfer area per unit length, based on the external surface of the internal bundle, is: 3 
𝐴𝑙𝑖𝑛𝑒𝑎𝑟 = 𝜋𝐷𝑜𝑁𝑡 = 𝜋0.017100 ⋅ 9.0000 = 0.48349 �
𝑚2
𝑚
� 
Therefore, the necessary pipe length L and number of exchangers Nhp in the bank and the final 4 
specified surface are: 5 
𝐿 =
𝐴𝑑
𝐴𝑙𝑖𝑛𝑒𝑎𝑟
=
19.513
0.48349
= 40.359 𝑚 
𝑁ℎ𝑝 =
𝐿
2𝐿ℎ𝑝
=
40.359
2 ⋅ 5.0000
= 4.0359 → 5 ℎ𝑎𝑖𝑟𝑝𝑖𝑛𝑠 
The final design heat transfer area, overall heat transfer and fouling factor for the exchanger 6 
are: 7 
𝐴 = 2𝐴𝑙𝑖𝑛𝑒𝑎𝑟𝐿ℎ𝑝𝑁ℎ𝑝 = 2 ⋅ 0.48349 ⋅ 5.00002 = 24.174 𝑚2 
𝑈 =
𝑞
𝐴𝛥𝑇𝑙𝑚
=
84143.
24.174 ⋅ 8.0928
= 430.10
𝑊
(𝑚2 ⋅ 𝐾)
 
𝑅𝑑 =
𝑈𝑐 − 𝑈
𝑈𝑐𝑈
=
609.85 − 430.09
430.09 ⋅ 609.85
= 0.00068535
𝑚2 ⋅ 𝐾
𝑊
 
8) Pressure drop 8 
a) Pressure drop in the internal tubes 9 
The friction factor evaluated previously from Haaland’s equationis: 10 
𝑓𝐹𝑖 = 0.0085752 
(1) Distributed pressure drop − tube 11 
For a bank with 5 heat exchangers, the distributed pressure drop along the internal tube bundle 12 
is: 13 
𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 =
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑖
𝐷𝑖𝜌𝑖
𝐺𝑖2 =
4 ⋅ 0.0085752 ⋅ 5.00002
0.012480 ⋅ 771.00
908.352 = 73533.
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 73.533 𝑘𝑃𝑎 
(2) Localized pressure drop − tube 14 
𝐷𝑖(𝑚𝑚) = 12.48 𝑚𝑚 
𝐾𝑖 = 0.7 +
2000
𝑅𝑒𝑖
+
17.78
𝐷𝑖(𝑚𝑚)
= 0.7 +
17.78
12.480
+
2000
10496.
= 2.3152 
𝛥𝑃𝑖,𝑙𝑜𝑐 =
𝐺𝑖2𝐾𝑖
2𝜌𝑖
�2𝑁ℎ𝑝 − 1� =
2.3152 ⋅ 908.352
2 ⋅ 771.00
(2 ⋅ 5.0000 − 1) = 11149.
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 11.149 𝑘𝑃𝑎 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
8 Multi-tube Heat Exchangers 173 
𝛥𝑃𝑖 = 𝛥𝑃𝑖,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑖,𝑙𝑜𝑐 = 11149. +73533. = 84682.
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 84.682 𝑘𝑃𝑎 
b) Pressure drop in the shell 1 
i) Friction factor 2 
The shell friction factor was calculated previously as: 3 
𝑓𝐹𝑜 = 0.0078639 
ii) Distributed pressure drop − shell 4 
𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 =
4𝐿ℎ𝑝𝑁ℎ𝑝𝑓𝐹𝑜
𝐷ℎ𝜌𝑜
𝐺𝑜2 =
4 ⋅ 0.0078639 ⋅ 5.00002
0.014838 ⋅ 1090.0
666.272 = 21584.
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 21.584 𝑘𝑃𝑎 
iii) Localized pressure drop – shell 5 
Considering a straight pipe return connecting the shells of the two legs of the exchanger, the 6 
localized pressure drop is evaluated from Eqs. (5.40) and (5.32) as: 7 
𝐷ℎ(𝑚𝑚) = 14.838 𝑚𝑚 
𝐾𝑜 = 2.8 +
2000
𝑅𝑒𝑜
+
71.12
𝐷ℎ(𝑚𝑚)
= 2.8 +
2000
15186.
+
71.12
14.838
= 7.7248 
𝛥𝑃𝑜,𝑙𝑜𝑐 =
𝐺𝑜2𝐾𝑜𝑁ℎ𝑝
2𝜌𝑜
=
5.0000 ⋅ 666.272 ⋅ 7.7248
2 ⋅ 1090.0
= 7865.0
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 7.865 𝑘𝑃𝑎 
iv) Total pressure drop 8 
𝛥𝑃𝑜 = 𝛥𝑃𝑜,𝑑𝑖𝑠𝑡 + 𝛥𝑃𝑜,𝑙𝑜𝑐 = 21584. +7865.0 = 29449.
𝑘𝑔
(𝑚 ⋅ 𝑠2)
= 29.449 𝑘𝑃𝑎 
Conclusion 9 
The proposed service can be performed by a battery of 5 multi-tube hairpin exchangers with 10 
both tube and shell associated in series. The pressure loss for both fluids is below the allowable value 11 
of 90 kPa. 12 
Problems 13 
8.1) A stream of isopropanol with a flow rate of 19842.0 lb/h is cooled in the annulus (shell) of 14 
a multi-tube heat exchanger from 122°F to 77°F, using treated cooling water at 73.4°F and flow rate of 15 
23810.0 lb/h. The available exchangers are carbon steel hairpins 16.4 ft long with 31 internal tubes of 16 
OD = 0.75 in and ID = 0.584. The external pipe is NPS 6 in schedule 40S, also made of carbon steel. 17 
The annulus return is a rod through straight pipe. Evaluate (a) the annulus flow area, (b) equivalent 18 
diameter, (c) the internal and (d) external heat transfer coefficients and (e) the overall heat transfer 19 
coefficient. (f) How many exchangers are enough to meet this service? 20 
Answer: (a) So = 0.10559 ft2 (b) De = 0.055033 ft (c) hi = 570.43 BTU/(ft2h°F) 21 
(d) ho = 40.055 BTU/(ft2h°F) (e) Uc = 36.339 BTU/(ft2h°F) (f) 7 hairpins connected with tube 22 
and annulus in series. 23 
8.2) The configuration used in the Problem 8.1 produced a tube heat transfer coefficient one 24 
order of magnitude greater than the heat transfer coefficient of the annulus. Such substantial difference 25 
may indicate that an inverted fluid allocation can possibly improve the design, requiring less heat 26 
transfer area to perform the same heat load. Verify the feasibility of this alternative design allocating 27 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
174 8 Multi-tube Heat Exchangers 
the isopropanol in the tubes and the water in the annulus. Calculate (a) the internal and (b) external 1 
heat transfer coefficients, (c) the overall heat transfer coefficient, (d) the new number of exchangers 2 
necessary to perform the service, the pressure loss (e) in the tube and (f) in the annulus. 3 
Answer: (a) hi = 89.126 BTU/(ft2h°F) (b) ho = 285.65 BTU/(ft2h°F) 4 
(c) Uc = 54.901 BTU/(ft2h°F) (d) 5 hairpins connected with tube and annulus in series 5 
(e) ΔPi = 3.1721 psi (f) ΔPo = 0.94645 psi. 6 
8.3) Cooling water is used to cool a benzene stream in a bank of multi-tube hairpin exchangers, 7 
where the water is the internal fluid. The average temperatures are 99.5°F and 86.1°F for benzene and 8 
water, respectively. The heat transfer coefficients are hi = 1027.5 BTU/(ft2h°F) and 9 
ho = 289.3 BTU/(ft2h°F). The inner pipes are of carbon steel (k = 26.001 BTU/(ft h°F)), 10 
ID = 0.505 in and OD = 0.625 in. The fouling factors for the fluids are Rdi = 0.0019874 ft2h°F/BTU 11 
and Rdo = 0.0010221 ft2h°F/BTU. Determine the clean overall heat transfer coefficient (a) disregarding 12 
the tube wall resistance and (b) including the pipe wall conductive resistance. Estimate the (c) internal 13 
and (d) external wall temperatures. 14 
Answer: (a) Uc = 214.54 BTU/(ft2h°F) (b) Uc = 205.13 BTU/(ft2h°F) (c) twi = 89.4°F 15 
(d) two = 90.0°F. 16 
8.4) A battery of 3 multi-tube carbon steel hairpins is used to cool down 547.2 kg/h of benzene 17 
from 60.0°C to 23.0°C using 792 kg/h of cooling water at 20.0°C, allocated inside the tubes. The heat 18 
exchangers are 5 m long, with outer pipe DN 100 schedule 40 and 19 inner tubes 19 
ID = 12.83 mm/OD = 15.88 mm. The annulus fluid return is a straight pipe and the fouling factor is 20 
0.00018 m2K/W and 0.00035 m2K/W for the benzene and cooling water, respectively. A pressure drop 21 
of 70 kPa is allowed in both streams. (a) Do you think the pipe set of these exchangers is adequate for 22 
this service? Determine (b) the heat load, (c) inner and outer convective heat transfer coefficients, 23 
(d) tubes and annulus pressure loss and (e) the over-surface. 24 
Answer: (a) No (justify!) (b) q = 8721.2 W (c) hi = 238.82 W/(m2⋅K); ho = 45.462 W/(m2⋅K) 25 
(d) ΔPi = 511.43 Pa; ΔPo = 115.3 Pa (e) Over-surface = 41.3%. 26 
8.5) Reassessing the service presented in the Problem 8.4, an engineer suggested the division 27 
of the cooling water in two parallel streams allocated inside the tubes as a presumably more cost 28 
effective design. (a) Does the new proposed design make sense, in your judgment? Determine (b) the 29 
effective mean temperature difference, (c) the heat transfer coefficient of the tube and annulus, (d) the 30 
clean overall heat transfer coefficient, (e) the resulting over-surface. (f) How many hairpins are 31 
necessary to match the desired heat load using this configuration? 32 
Answer: (a) No (justify!) (b) 10.77°C (c) hi = 189.53 W/(m2⋅K); ho = 45.462 W/(m2⋅K) 33 
(d) Uc = 35.008 W/(m2⋅K) (e) Over-surface = 63.9 % (f) Nhp = 4 exchangers. 34 
© 2015 Samuel Jorge Marques Cartaxo Hairpin Heat Exchangers Explained 
9 Finned Tube Heat Exchangers 175 
9 Finned Tube Heat Exchangers 1 
When specifying an exchanger with a relatively low heat transfer coefficient for one or both 2 
fluids, the extension of the heat transfer surface using fins may be an effective solution. If only one 3 
fluid is a gas, a very viscous liquid or develops a low Reynolds number (small flow rate), possibly it 4 
will control the heat transfer process, and ultimately determine the size of the equipment. The low 5 
convective heat transfer coefficient can be balanced by the additional area provided by the fins 6 
installed over the base pipe surface. In these cases, finned double-pipe heat exchangers often surpass 7 
the performance of shell-and-tube exchangers, being more cost effective [44]. An in depth 8 
presentation of general finned surfaces and methods of heat transfer rate calculation is found in Ref. 9 
[30]. 10 
Fins can be attached to the pipe external surface by extrusion, soldering or welding. In 11 
common low-pressure applications, the typical thickness of extruded or soldered fins up to 12.7 mm 12 
high is 0.5 mm, while for higher fins the ordinary thickness is 0.8 mm. Welded fins are 0.89 mm thick 13 
for heights up to 25.4 mm [48]. 14 
For services with very viscous or condensing fluids, it is recommended the use of longitudinal 15 
fins, as exemplified in Figure 33 [18]. The reason is that longitudinal fins, aligned with the fluid flow, 16 
produce