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Boiler Cqlculotions
A. Whnt is equivalent evaporation?
Ans It is the quantity of water evaporated from and
at 100'C to produce dry saturated steam at 100"C by
absorbing the same anount of heat as used in the
boiler under actual operating conditions.
M"q= Mn (H - Hwt)/539
where M"n - equivalent evaporation
Mact = actual mass of steam generated per unit
mass of fuel burnt
I/ - total specifrc enthalpy of steam under operat-
ing conditions, kcaUkg
H*, = specific enthalpy of feedwater, kcaVkg
Latent heat of dry, saturated steam at 100"c is 539
kcaVkg.
A. What is factor of evaporation?
Ans, It is the factor to be multiplied with the quan-
tity of steam generated under working conditions to
get the equivalent evaporation.
Equivalent evaporation - Actual evaporation
x (f)
or M"n= M*.(f )
or M$(H - Hwt)/539 = M*rf
f=(H_H*t)/539
a. Wtut is boiler eftcienq?
Ans. It is lhe ratio of the heat load of he generated
steam to the heat supplied by the fuel over the same
period.
Heat load of generated steam
= G"(H - H*,) kcaVs
where G, = rate of steam generation, kg/s
Rate of heat supplied by fuel - Gf X (CV)lkcaVs
where Gr - rate of fuel buming, kgls
n. .. - 
G' (H - H*t)
'rnoiler Gf (Cv),
= Mnr(H-Hnr)/(U)1
where G"/G1- actual evaporation - M*t
O. What is ecornmizer eficiency?
Ans. It is dcfined as the ratio of the heat absorbed
by the BFW in the economizer to the heat supplied by
the flue gases in the economizcr, the temperature of
flue gases being reckoned above the temperature of
the air supplied to the boiler
6 - 
Mac tA@
rleon = MrrCn (@1- @6)
where AO - rise in BFW temperature in the
economizer
M1, = r[&SS of flue gases per unit mass of fuel
Cp - sPecific heat of flue gases
@f - flue gas temperature at inlet to economizer
O"ir - temperature of air delivered to the boiler
Ptoblem 6.1 A boiler generates 4.5 t of super-
heated steam (500'C,9}kgflcmz abs.) per ton of coal
feed.
The BFW temperature - 45'C
What is the equivalent evaporation from and at
100'C pcr ton ofcoal?
Solution
Specific Enthalpy
= 809 kcaVkg
82 Boller Operolion Englneedng
Sensible heat of feedwater at 45"C - 45 kcal/kg ^
Heat required to produce 4.5 t steam (90 kgflcm'
abs.,500"C)
= 4.5 x f03 x (809 - 45)
= 3 438 x 103 kcal
Ijtent heat of dry, saturated steam at 100'C
- 539 kcaVkg - 539 x ld kcaUt
Therefore, equivalent evaporation from and at
100'c
3 438 x td tcat
= __:_ _ 6.379 t per ton of coal
539 x 105 kcat/t
Ans.
Problem 5.2 A steam boiler generates 7.5 tons of
steam per ton of coal burned. Calculate the equivalent
evaporation ftom and at 100'C per ton ofcoal from
the following data
Steampdssure - 10 kgflcm2. abs.
Dryness fraction - 0.95
Feedwater temperature - 50"C
fulution Working formula
Mg(H - H*,)
M.s=-B-
Mnr- 7'5 Ut of coal
"*,'S kcaUkg - 50 x 103 kcaVt
Hr" = 181.3 x 103 kcal /t
.r - 0.95
L-483x 103kcaVt
H = H* + x L =t181.3 + 0.95 (483)l x 103
= 640 x lG kcal/t
M.n=1.5 (6<O.tS - 50) x rc3/fi9
= 8.211 t of steam/t of coal
Ans.
Problem 6.3 A boiler is working at 14 bar and
evaporates 8.5 kg of water per kg of coal fired from
BFW entering at 39"C. Determine the equivalent
evaporation fromand at 100"C if the steamis 0.96 dry
at the stop valve.
Solution The equivalent evaporation from and at
100"C is
M"q=Mn (H-Hwt)/Lrco
Now
Mrct- 8.5 kg steam Per kg of coal
H*r'163'4kl&;g
H,r, - 830 kJ/kg (at 14 bar)
-r - 0.96
L - 1957.7 kJ&g (at 14 bar)
H = H w + x L
= 830 + 0.96 (1957.7)
=27W.39kJ/kg
Itoo - 2257 klkg
M"q = 8.5 Q7w.39-163'4)/2257
= 9.588 kg stearn/kg of coal
= 9.59 kg steam,zkg of coal
Ans.
Prcblem 6.4 A boiler produces 220 t o^f dry
saturated steam pcr hour at a pressure 60 kgflcm', abs.
from feedwater at a temperature of 120"c.
Coal consumption - 1200 VdaY
Calorific value of coal - 42O0 kcaVkg
1% of coal escapes unbumt.
Determine
(a) the equivalent evaporation per ton of coal
fred
(b) the eff,rciencY of the boiler
(c) the overall efficiency of the boiler
Solution
Step (I) Heat l-oad of Steam/Ton
inrhalpy of dry, saturated steam at 60 kgflcm2 abs.
:665.4 kcaVkg
Enthalpy of water at 120"C = 120 kcaVkg
Therefore, heat required to raise I ton of steam
= td (66s.+- l2o) = 545.4x 103 kcal
Step (II) Equivalent EvaPoration
Coal consumption - 1200 VdaY = 59 Y6
Steam gencrated per ton ofcoal fred
-22U50 - 4.4t
Therefore, equivalent evaporation
= 
4.4 (545.4 x to3)
539 x 103
- 4.452 ton of steam/t of coal
Ans,
Step (IID Boiler Efficiency
Energy output =220 (545.4 x 103) kcaVh
Coal charged to the boiler = 50 t/h
Actual coal bumt - 50 (l - l/lm) - 49.5tJh
Therefore, energy input - 49.5 x I 03 (4200) kcaUh
_ 
220 (s4s.4x rd)
'rboiler - 
49.5 x lo3 x42oo
= 0.577 i.e. 57.7Vo
Ans.
Step (IV) Overall Efficiency of the Boiler
r- 1 220 (545.4x t03)r - -
L'rboilerl- 5ox lo3 x42cn
= 0.5713
= 57.137o
Ans.
Problem 6.5 A boilerconsurrc,s224tons of coal
to produce 1864 tons ofsteamperday. The steam is
dry, saturated at 90 afrrr abs. Calculate the boiler
thermal efficiency, and the equivalent evaporation
per ton of coal if the calorific value of coal is 5400
kcal/kg of coal, the specihc enthalpy of feedwater
being a25.036 U/kg of water.
Solution
Step (I) Rate of Evaporation
Mass of steam produced - 1864 ton
Mass of coal consumed -224ton
Actual evaporation capacity = 1864/224
= 8.321 t /t of coal
Step (II) Equivalent Evaporation
Evaporation capacity, M^= 8.321tlt of coal
Sp. enthalpy of dry, satd. steam (90 atm. abs.),
H =2705kJ/kg
=2705 x t03 kJtt
Sp. enthalpy of BFW, Hw = 425.036W1k1
= 425.036 x l0r U/t
Boller Colculoflont 83
Equivalentevaporation, M"= M^(H - H)/L
= 8.321 (2705 - 425.036) x r03 /Q257 x t}l)
- 8.405 ton stearn/ton of coal
Ans.
Step (III) Boiler Thermal Efficiency
Working Formula: Boiler ttrermal efficiency
- energy to stearn/energy from fuel
Energy to steam
= 8.32r (2705 - 425.03Ox t03 U
Energy from fuel
= 5400 x t03 kcavt ofcoal
= 5400 x 4.1868 x 103 kJ/t of coal
B oiler thermal effi ciency
_8.32r (2705 - 425.036)x rO3
5 4 0 0 x 4 . 1 8 6 8 x 1 0 3
= 0.8391
=83.91%
- 8 4 %
Ans.
Problem 6.6 A boilersenerates 7.5 tons of steam
per hour at 18 bar ( 1 b; - td ttlm2. The steam
temperature is 598K and the feedwatertemperature is
328K.
When fired with oil of calorific value 47 250kJkg,
the boiler plant achicves an efficiency of 85%.
The generated steam is fed to drive a turbine which
develops 0.75 MW and exhausts at 1.8 bar, the dry-
ness fraction of the steam being 0.97.
Determine the rate of fuel consumption and the
fraction of enthalpy drop, through turbine, converted
to useful work.
If the turbine exhaust is directed for process heat-
ing, estimate the heat transfer available per ton of
exhaust steam above 322.4K.
Solution
Step (I) Energy to Raise Steam
Specific enthafy of generated steam
= 3 106 - 0.84(3 105 - 3083)
= 3 086.6 kJAg (by interpolation)
Specific enthalpy of BFW at 328 K
&1 Boller Operotlon Englneedng
-230.TlakJkE
Specific energy to nrise steam
= 3086.68 -230.274
=2 856.4kJ&g.
Step (II) Rate of Fuel Oil Consumption
Rale of steam generation -7.5 tlh - 7500 kg/h
sp. energy to raise steilrl - 2856.a kJ&g
Energy input to steanr/h - 7500 (2 856.4) kJ
Boilerefficien"y=:ry4Energy from fuel./h
=85* �
= 0.85
Rate of firel consumption (Energy from oiVh)
- 7500 Q856.4Y (0.85)(47250)
- 533.408 kg 
Ans.
Step (III) Rate of Sp. Enthatpy Drop in Turbine
Sp. enthalPY of exhaust steam'
H 2 = H n + x ' L
= 49O.7 + 0.97 Qzrl)
=2635.37 kJ/kg
Sp. enthalpy of inlet steam, Hr - 3086.68 kJlkg
Sp. enthalPY droP in turbine,.
A , H = H t - H z
= 3086.68 -2635.37
= 451.31 kJAg of steam
Steam feed - 7 .5 tlh - 7500/3600 kds
Rate of sp. enthalpy drop in turbine
- 451.31 (7 500/3 600) kJ/s
-940.229kJ|s
Step (IV) Fractionof Enthalpy Converted to Use'
fulWork
Energy output ftom turbine
-0.75 MW
- 0.75 x 103 kw
- 0.75 x 103 U.zs
Rate of enthalpy drop in turbine '94O'229 kJls
Fraction of enthalpy drop converted to useful work
- 0.75 xro3/90.229
- 0.7976 - 0.8
Step (V) Heat Transfer Available in Exhaust
Steam Above 322.4K
Sp. enthalpy of exhaust steam - 2635.37 kJftg
Sp. enthalpy of water at322.4 K = 207 kJ/kB
Heat transfer available in exhaust steam above
322.4K
-2635.37 -207
-242E.37 kl/Kg Ans.
Problem 6.7 The following observations were
made in the case of a boiler fitted with an economizen
Rate of steam generation - 5 Vt of coal
Equivalent evaporation from and at 100'C
= 5.5 Vt of coal
Boiler feedwater temp. inlet to economizer
= 100"C
Temperature of BFW inlet to boiler - 180"C
Temlrrature of air supplied to the boiler - 30'C
Temperature of flue gases entering the economizer
- 4O0"C
Weight of flue gases produced per ton of dry coal
= 1 5 t
Mean specific heat of flue glses - 0.20 kcaVkg "C
Calorific value of coal - 5400 kcal/kg
Determine
(a) the boiler efficiencY
(b) the economizer efficiencY
(c) lhe combined efficiency of the whole plant
Solution
Step (I) Heat OutPut
Steamgeneratedfromandat 100"C- 5.5 Vtof coal
burnt
Therefore, heat output = 5.5 x 10' (539) kcaVt of
coal burnt
Step (II) Heat InPut
Calorific value of coal * 5400 kcaVkg
Therefore, heat input - 5400 x l0r kcaVt of coal
burnt
Step (lII) Boiler EfficiencY
Heat Otttotrt 5.5 x l0' (539)
I . . . =l lboiler- Heatllput 
- 
54O0x 103
= 0.5489 i.e. 557o (approx)
Ans.
Step 0V) Heat of Flue Gases
Heat of the flue gases entering the economizer
= 15 x 103 (0.20) (400 - 30)kcayt
= 111 x t04kcaVtofcoal
Step (V) Heat Absorbed by BFW in the
Economizer
Heat absorbed by BFW in the economizer
= 5 x ld (180 - 100) kcaVt of coal
=40x t04fcaUtofcoal
Step (VI) Economizer Efficiency
I"-o = (e0 x t04 )/(ttl x 104)-0.3603 i.e.36%
step (vII) combined Efficiency 
Ans'
Heat absorbed in the boiler- 5.5 x ld (539)kcaVt
ofcoal
Heat absorbed in the @onomizer - 40 x 104 kcaVt
ofcoal
Total heat absorbed in boiler and economizercom-
bined
- 5.5 x 103(539) + 40 x loa
- 336.45x 104 kcaVt of coal
Energy released by burning co3l - 5a00 x ld
kcaVt ofcoal
n^^-. = 
336'45 x lOa 
-0.623 i.e.62.30%'rcornD 
5aoo x ld
Ans.
Problem 6.6 A boiler produces steam at 90
kgttcn? abs. at the rate 150 t/h from the feedwater at
120"C. The steam is dry, saturated. What is the boiler
horse power?
Solution
Steanr,90kgflcm2
abs. Dry,sanrrated Total heat - 655.7 kcal/kg
Sensible heat of BFlv at 120'C - 120 kcaVkg
Equivalent evaporation from and at 100'C
- 150 x 103 (655.7 - t20)/539
- 149.08 x td fgn
Therefore, boiler horse power
= 149.08 xrc3/$.653
=9524.15
Ans.
Boller Cslculollonr 85
Boiler Horse Power is a very commonly used unit
for measuring the capacity of a boiler. ASME
(American Society for Mechanical Engineers)
defines a unitboiler horse power as the boilercapacity
to evaporate 15.653 kg of BFW per hour from and at
373 K into dry, satunted steam or equivalent in
heating effect
Boiler h.p. - Equivalent evaporation from and at
373"K per hour/l5.653
Problem 6.9 Aboilergenerates 6.5 tof steamper
ton of coal fired.
The steam is at l8 kgflcm2 gauge
The boiler feedwater temperature - 110"C
downstream of deaerator
Boilerefficiency -75%
Factor of evaporation - l.l5
Co of steam - 0.55 kcaVkg "C
Determine
(a) the tempcratxre of the steam
(b) thedegree of superheat, if any
(c) the equivalent evaporation per ton of coal
burnt
(d) thecalorific value ofcoal
Solution
Step P (I) Steam Parameters
Pressure - 19 kgf/cm2. abs.
Sensible heat, I/* - 213.1 kcakg
l,atent heat, L-455.1 kcaVkg
Satrrration temp., @" - 20E.E'C
Step (II) Degree ofSuperheat
Total heat of the steam - Hw + L + Co AO
where, AO - degree of superheat
Sp. enthalpy of feedwater - Hrw
Therefore, the factor of evaporation
_ H * + L + C o L / g . - H r *
or 1.15 =
539
213.1 + 455.1 + 0.55 (AO) - l l0
539
AO= 112.09"C
Ans.
Step (III) Superheated Steam Temperature
AO - lrz.Ogrc
86 Boller Operollon Englneerlng
or , @-@"-112.09"C
or O=208.8+ 112.09= 320.89'C -321"C
Ans.
Step (IV) Heat Output
Heat required to generate stearn
= 6.5 x ld (213.1 + 455.r + 0.55 AO - 110)
kcaVt of coal
= 6.5 x 103 (558.2 + 0.55 x 112.09) kcaVt of coal
= 4029.021x 103 kcavt of coal
Step (V) Heat Input
Calorific value of coal - CV kcaVkg
Energy released per ton of coal burnt
- 103 x CV kcal
Step (VI) Boiler Efficiency
11. .. HeatoutDut -4029'o2lxlo'''borrer=E;ffi 
ld x cv
or 0.75 - 4029.021 ICY;
CY - 5372 kcaVkg ofcoal
Ans.
Step (VII) Equivalent Evaporation
6.5 (2r3.r + 455.1 + 0.55 x 112.09 - 110)
,r"q =
= 7 .474 t/t of coal bumt
Ans.
Prcblem 6.10 fbe following observations were
made during the trial run of a boiler.
Steam generation rate = 16 Uh
Feedwater temperature = 30"C
Steam qualitY = 0.9 dry
Steam pressure - 15 kgflcm' abs.
Coal consumption = 2.5 t/h
Calorific value ofcoal = 6540 kcaVkg
Ash + unburnt coal collected from beneath the
grrtes = 0.2llh (Calorific value - 700 kcal/kg)
Weight of flue gts€s = 15 Vt of coal fred
Flue gas temperature - 350"C
Average specific heat of flue gases = 0.25
kcaVkg'C
Ambient air temperature - 25"C
Calculate
(a) the boilerefficiency
(b) the percentage of heat loss to the flue gases
(c) the percentage of heat loss to the ash
(d) the percentage of heat loss unaccounted for
Solution
Step (I) Steam Parameters
Sensible Heat,If* - ?frO.1 kcaVkg
l---Latent Heat, L - 466 kcakg
Step (II) Heat Output Rate
Rate of stearn generation - 16 t/h
Heatoutputrate= 16x t03 (H, +xL-Hp)
= 16 x 103 (200.7 + 0.g x 466 -30)
=9441.6x 103 kcaUh
Step (III) Heat Input Rate
Coal consumption - 2.5 Uh
Calorific value ofcoal = 6540 kcaVkg
Heat input rate =2.5 x 103 x 6540 kcaVh
Step (IY) Boiler Efficiency
Iboiler = Heat output rate,/Heat input rate
= 9441.6x 103/Q.5 x 103 x 6540)
=0.5774 i.e. 57.74V0
Ans.
Step (V) Heat Load of Flue Gases
Flue gases generated - 15 Ut ofcoal
Heat load of flue gases
= 15 x 103 x 0.25 (350 - 25)
= 1218.75 x ld kcaVt of coal
Step (VI) Heat Generated By l Ton of Coal
Heatproduced by ltofcoal - 103 x 65210 kcal
Step (VID Percentage of Heat llss to Flue Gases
= Ir2t8.7 5 x r03 / 651CIx l03l (100)
= 18.63%
Ans.
Step (VIID Percentage of Heat Loss to Ash
Ash + unbuntt coal collected - 0.2 Vh
Heat loss due to ash + unburnt coal
=0.2x ld x 700 kcaVh
Heat gencratcd in the fumace
=2.5x ld x 654O kcaVh
Therefore, percentage of heat loss to ash
0.2 x 103 x7ffi ,
__________ 
. 
_ rl00)=0.85%
2.5 x 10'x 6540
Ans.
Step (IX) Percentage of Unrccounted Heat
Useful beat- 57.74%
Heat lost to flue gases - 18.63%
Heat lost to ash - 0.85%
Total rccounted heat = 57.74+18.63+0.85
--77.22%.
.'. Unaccounted heat - 100 - 77.22 - 22.77 %
Ans.
Ptoblem 6.ll A boiler genentes 75 t of steam
per hour at pressure 1.8 MN/h' and temperature
325'C ftom feedwater at 49.4"C. When fircd with oil
of caiorific value 45 MJftg, the boiler attains an
efficiency of 78%. The steam (325'C, is fed to a
hubine that develops 650 kW and exhausts at 0.18
I\,N/m2, the dryness fraction of steam being 0.95.
Determine
(a) the mass of oil fired Per hour(b) the fraction of the enthalpy drop through the
turbine which is converted to useful work
Also determine the heat transfer available per kg
of exhaust steam above 49.4"C, if the turbine exhaust
is used for process heating.
blution
Step (I) Specific Enthalpy of Generated Steam
H =3rM - 0.84 (3106 - 3083)
= 3086.45 kJ/ke by interpolation)
Step (II) Specific Enthalpy of BFW (49A"C)
Hr*=2A6.9tJ�/lrg
Step (IID Heat OutPut
Energy required to generate steam
= 3086.45 -206.9
=2879.55kJftg
The rate of steam generation- 7.5 Uh
.'. Heat ouput =7.5 x 103 x2 879.55 U/h
Step (IV) Rate of Oil Burning
I.et the mass of oil fired be fi vgn
Heat input - Ifr x 45 x 103 U/h
Boilerefficierc! =78%
0.78 = Heat ouput/Heat input
Boller Cdculctlonr 87
=7.5 x rG x2879.55/tfr x 45 x ld
rt =firlkelh
Step (V) Specific Enthalpy Drop in Turbine
Specific enthalpy ofelhaust steam
- H* + x L - 490.7 + 0.95 (2210.8)
- 2590.96 U/kg
.'. Specific enthalpy drop in turbine
= 3086.45 -2590.96
= 495.45kl/lrg
Step (VI) Rate of Enthalpy Drop in Turblne
Rate of steam fed to turbine
-7 .5 l lh
- 7.5 x 1d/3600 kgs
Specific enthalpy drop in h[bine - 495.45kJlkg
.'. Rate of enthalpy drop in turbine
= 495.45 (7.5 x 1dl3600) kJls
= 1032.187 U/s
Step (YID Fraction of Enthalpy Drop Converted
to Useful l{ork
Energy outprt from trbine
- 650 kw
- 650 kJ/s
Energy inputto turbine - 1032.187 U/s
.'. Fraction of enthalpy drop converted to useful
work
- 65Cl/1032.187
-0.629
Ans.
Step (VIID Heat Transfer from Exhaust Steam
The net heat available, for process heating, from
exhaust steam above 49.4"C - ?59O.96-?n6.9 -
2384.06kI/lr9
Ans.
Pr<iblem 6.12 A steam generation plant supplies
8500 kg of steam per hour at pressure 0.75 MN/m".
The steam is 0.95 dry.
Feedwater temlrnture = 41.5"C
Coal consumption - 900 kg/h
88 Bolbt Operollon Englneedng
Calorific value of coal -32450kJilrg
Determine
(a) theboilerefficiency
(b) the equivalent evaporation from and at 100"C
(c) the saving in fuel consurnption, if by installing
an economizer it is estimated that the feed-
water ternperatue could be raised to 100"C,
assuming that other coditions rcmained un-
changed and the efficiercy of the boiler in-
creases by 6%.
blwion
Step (I) Steam Generation Fer Ton of Coal
Rate of steam generation- 8500 kgh
Coal consumption - 900 kg/h
Therefore, steam generation per kg of coal
- 8500/900
- 9.44 kg
.'. St€am generation/ton of coal
- 9440 kg
-9.44t
Step (II) Specific Enthalpy of Steam Rais€d
H -- H* + x L -7@.3 + 0.95 (2055.5)
=2662.025kl/kg
Step (III) Energr Required to Generate Steam
Specific enthalpy of steamraised
-262.0?5kr/ltg
Specific enfhalpy of BFW - l139kJ/lKE
Hear ouput = 26tr2025 - 173.9
=2488.125 kJAg of steam
Step (IV) Boiler Efficiency
_ 
Heat outDut/ks of coal 2488.125 ,^tl*r",=ffi#ff=ffi(g.u't
=0.7238 i.e. 72.38%
Ans.
Step (V) Equivalent Evaporatbn
Steam raised per kg of coal - 9.4k9
Energy required to gerrcmte this steam
-9.44 (2488.125) U&g coal
Specific enthalpy of evaporation frorn and at
100"C - 2?56.9kJft:g.
Equivalent evaporation = 9.44Q488.125>/2?56.9
= 10.40 ke/ke of coal
Step (YI) Energy Required to Generate Sm
Under New Conditions
Specific enthalpy of BFW at 100'C - 419.1 U/kg
Energy required to generate steam when
econqnizer is incorporated
-2662.U25 - 419.1
-2242.9?5Wfrg
Energy to steury'h
=2242.925 x 8 500kJ
Step (VII) Rate of Coal Consumption when
Economizer is Fitted
Erergy output -2242.98 x 8500 kJ/h
Energy input- Ifr x32a50Hth
Boiler efficiercy -72.38 + 6 -78.38%
2242.Y25 x 8500
0.7838 - Nl x32450
rt =749.57 kglh
Step (VIII) Saving in Fuel Consumption
Initial fuel consumption rate - 900 kg/h
Modified fuel consumption rate when economizer
is fiued -749.57 kglh
Saving in frcl consumption = 900 - 749.57
= 150.43 kg coal,zh
Ans.
PtoHem 6.13 The following obsenations were
made during tbe trial run of a boilen
Rate of steam generation = 5 Uh
Steam quality: dry, saturated
Steam pressure = l0 kgf/cm2 gauge
Average specific beat of steam - 0.55 kcaUkg. "C
Redwater ternperature - 85'C
Room ternperature - 25"C
Atnospheric pressure - 1 kgflcm2
Fuel consumption - 650 kg coaVh
Calorific value ofcoal - 7500 kcaVkg ofcoal
Moisture content of coal -2.5%
Fuel contains: C - 86%: H - 5%: Ash - 9%
Flue gas temperature - 300"C
Mean sp. heat of flue gines - 0.25 kcaVkg "C
Analysis of dry flue gases:
C O z - l 0 % ; O 2 - 8 % ; N 2 - 8 2 %
Produce a complete heat balance sheet taking I kg
dry coal as the basis.
Solution
Step (I) Energr to Steam Per kg of Dry Coal
Steampressure = l0 kgf/cn? garye
= l1 kgflcm2 abs.
Sensible heat of steam at 1l kgflcm2 abs.
- 185.7 kcaUkg
Latent beat ofevaporation at 1 I kgflcm2 abs.
- 478.4 kcaVkg
Specific enthalpy of dry, saturated steam generaEd
- 185.7 + 478.4
- 664.1kcal/kg
Specific enthalpy of feedwater = 85 kcal./kg
Coal consumption
- 650kdh
Dry, coal consumption = 650 (100 -2.r/lm
= 650 x 0.975kg/h
Energy to steam/kg coal
- (664.r _ 85) (5000)/(650 x 0.975)
- 4568.836 kcal
Step (II) Flue Gas Analysis
Basis: 100 m3 of dry flue gas
Boiler Cqlculqlions 89
Step (III) Coal Analysis
Basis: lkg dry coal
Constitu{hemicalReaction Remorks
nt during combustion
C + O r - + C O ,
(r2) (44)
ur+)or-+ Hro
(2) (18 )
WL of moisture fired
=0.025/0.975
= 0.02564 kg/kg of dry coal
Total wt. of water vapour in flue gases
- 0.45 + 0.02564
= 0.4756 kg/kg of dry coal.
Step (IV) Heat L,oad of Water Vapour
= 0.4756 [638.8 + 0.55(300 - 90) - 251
=344.5 kcaVkg of dry coal
where 638.8 kcaVkg = total heat of water vapour
at I kgflcm2 abs. to which flue gases arc discharged
Step (V) Heat Load of Dry Flue Gases
Heat loss to flue gases
=24.741(0.25) (300 - 25)
= 17N.944 kcaVkg of dry coal
Step (VI) Heat Balance
Basis: I kg of dry coal
Wt. of dry flue gas
produced
- (2992n04)(86trm)
- 24.741 kglkg of coal
Wt. of water vapour
produced
- (18/2)(5/ loo)
-0.a5kglkg of dry coal
Con-
stituent
Volume
-3
Mol.v,t. Proportional
Mass
Remarks
coz
o2
10
8
82
44
32
28
44(lO)= 440
32(8\ -2s6
28(82) -229<
Carbon
content
= 410(12/44)
= l2O
100 t04 2992
Heat Input % Heat Expenditure 7o
Total hcat
supplied
- 7500 kcal
t00 Ijcat consumed in stearn
fonnation
- 4568.836 kcal
Ileat lost to flue gas
- 1700.944kcal
Heat lost to vapour
- 344kcal
Heat unaccounted for
- 886.220 kcal
@.92
22.70
4.ffi
I 1 .80
7 500 kcal 100 7 500 kcal 100.00
90 Bolter OPerollon Englneedng
Prcbtem 6.t4 During the tial run of aboilerthe
following data were recorded
(ii) suPerbeater
(iii) air heater
(iv) economizer
(e) heat lost in the flue gas
Summarize the overall result on the basis of I kg
coal burnt.
Sohttion
Step (I) Theoretical Air Requirements
Basis: 100kg coal
Coal consumPtion
Steam produced
Boiler:
Steam Pressure
Steam temPeraturc
Superheater:
Superheated steam tempcrature
Economizer:
Water inlet temPerature
Watelr outlct tcmPerature
Air heater:
Air inlet temPerature
Air outlet temPerature
FIue gas inlet temPerature
Flue gas outlet temPerature
83.1 t
606 t
1.461MN/m2
-14.42^tn
470 K
610 K
353 K
400 K
320K
380 K
503 K
405 K
Constituenl % W
Element Weight
Molecular klmol
Weight
kmol of Ot
requircd for
complete
combustion
62.5
4.25
5 . 1 I
t .2
c
H
o
N
t2
2
32
28
32
IE
5.208
2.r25
0.159
o.o42
o.o27
5.208
2.12512
- 1.0625
(-) 0.lse
o.027
E - 6.1385c
H
o
N
s
Ash
Moisture
62.5%
4.25%
5. t t%
t2%
o.85%
9.85%
16.24%
Coal analysis (bY weight) Fluc gas analYsis (drY basis)
13.2% (by volume)
4.85% ( -do- )
81.95% ( do- )
Total rco%
S 0.85
Ash 9.85
Moisture 16.24
There, theoretical air requirement
= 6.1385 (100/21)
49.23 kmoV100kg coal
=29.23 (28.9) k9100 kg coal
=844.744 kglm kg coal
=8.447 kgftg of coal 
AnS.
(c/ The average molecular weight of air is 28'9)
Step (II) Actual Air SirPPlied
100 kg coal contains 5.208 kmol of C
100 kmol of dry flue gas contains 13.2 kmol of C
Therefore, the amount of flue gas produced
- 5.208 (100/13.2)
=39.45 kmoV100 kg coal
Let r mol of air be supplied per 100 kg of coal
burnL
therefore by nitogen balance we get'
79 x 4ffi + 0.042- H (39.45)
coz
o2
N2
Gross calorific value - 30550 kJftg (dry coal)
Boiler housetemPerature = 298 K
Enthalpy of dry, saturated steam at 1.451 MN/m2
=2791L<IkE.
Substances S p e c if c H e ats (kJltg'K)
Dry flue gas
Water vapor in flue gas
Water
1.005
2.095
4.t81
Determine
(a) theoretical air requirements per kg of coal
(b) actual air supplied per kg of coal fred
(c) weight of flue gas per kg of coal bumed
(d) thermal efficiencies of
(i) boiler
coz
o2
N2
.'. .r - 4O.87 kmoU100 kg coal
Therefure, the weight of air supplied
- 40.87 (28.9)
- l l8 l . l4kg/ l00kgcoal
- 11.81 kgfu coal
Nore % excess xi1= (ll.8l -8.47)(tU.,/8.#
=39.81%
Step (III) Weight of Flue Gas
Basis: 100 kmol
Fluc Gas koiol Mohca- Wcight
Conslitaen ls
t Weight
Wcight in39.45
kaolof Fluc Gas
r3.2 4
4.85 32
81.95 28
13.2(44) 13.2(,14X39.45/100)
-229.t25
4.85(32) 4.8s(32)(3e.4sll m)
- 61.226
8 1.95(2E) 8 l.9s(28x39.4sl IOO)
= 905.219
E= il95.57 k9
Water produced due to combustion of hydrogen
content ofcoal
-2.18 kmol - 2.125 (18) - 3E.25 kg
Free moisture - l6.24kg
Therefore, the total weight of wet flue gas
- 1195.57 + 38.25 + 16.24
- I 250.06 kglm kg coal
- 12.50 kgltg coal
Step (IV) Thermal Efliciencies 
Ans'
l. Boiler
Total heat content of steam at 1.461 MN/m2
-279lkItkg
Total heat content of water charged to boiler
=4.187 (4N-273)
= 531.749 U/kg
Therefore, the net heat tansfened to steam
=2791- 531.749
Boller Colculollonc 9l
-2E9.25lU.kg
Rate of steam generation/t of coal
=66/83.1r/t
-7 
.292 Vt of coal (or kJlkg coal)
Therefore, the heat transferred to steany'kg of coal
burned
=7.2V) (2259.251)
= 16474.458 kJ./kg coal
Gross calorific value of coal as fired
= 30550 (100 - 16.24)/100
= 25588.68 kI/kg coal
Therefore, thermal efficiency of the boiler
= 16474.458/25588.68
= 0.6438 i.e. 64.38%
2. Superheater
Ans,
Net heat tansferred to steam in the superheater
=7.2T2 (2.095) (610 _ 470)
=2138.74 kJ/kg coal
Therefore, thermal efficiency of lhe superheater
= 2138.7 4/25588.68
= 0.0835 i.e. 8.35%
Ans.
3. Air Heater
Weight of air charged to the boiler
- 11.81 kJ/kg coal
Heat absorbed by air in the air heater
= 11.81 (1.005X380 - 320)
=712.143 kJ/kg coal
Therefore, the efficiency of the air heater
=712.143/25588.68
= 0.0278 i.e. 2.78%
Ans.
4. Economizer
Heattransfcncd to BFW
=7.292 (4.187X400 - 353)
: 1434.985 kJ/kg coal
92 Boller Operotlon Englneedng
Therefore, thermal efficiency of the economizpr
- r434.985t25 588.68
:0.0560 i.e. 5.6Vo
Ans.
Step (V) Heat Lost to the Flue Gas
Weight of the dry flue gas
= 11.95 kJ/kg coal
Enthalpy of the dry flue gas
= 11.95 (1.005) (405 - 298)
= 1285 U/kg coal
Water content in flue gases when 100 kg coal burnt
-38.25 + 16.24
= 54.49k9
Therefore, the weight of water vapour/kg of coal
burnt
- 0.5,149 kg
Therefore, the enthalpy of water in the flue gases
=0.5M9 t2.095 (405 - 311*) +24t1.2*
+ 4.187 (405 - 298)
= 1665.29kJ/kgcoal
* fDew Poittt ofWet Flue Gasl
kmol
- (29 50.29 t?55 88.68)( 1 00)
= 1L.52%
Tabulation of Result (Basis: I kg coal bumed)
Heatrecovered %GCV % Elficiency
(kJ) t^,ct coal
Boiler
Air Heater
Superheater
Economizer
Heat to flue gas
Heat
unaccounted
r5762
712
2t39
r435
29fl
259r
61 .591
2J8 |8.36 r
s.6 lJ
I l.53\
1o. t3 i
78.34
21.66
>- 100.00
Flue Gas
Conslituent
%
Composition
ProUem 5.15 A boiler generates 6000 kg steam
per hour at 10 kgf/cnt2 from BFW at 4O"C. The steam
is 0.97 dry. The boiler is fired with coal at the rate of
700 ke/h using 16 kg ofair (at 15'C) per kg ofcoal
fired.
Assuming tlre boiler efficiency to be 70%, deter-
mine
(a) excess air coefficient
(b) flue gas ternperaure leaving the boiler
Given
The coal is composed of carbon and hydrogen
besides its ash contcnt12%.
Combustion Heat Of Combustion
coz
o2
N2
Hzo
(13.2/100x39.4s) - s.2
(4.82/lm)(3e.4s) - l.e
(8 1.95/ 1 00)(3e .4s) - 32.33
54.49t18 -3.02
=242.46
C + O r - r C O ,
Hr+ Or-+ H, O
8075 kcaVkg ofcarbon
34500 kcaVkg of hydrogen
12.24
4.47
76.13
1 . t l
> - 99.95
Therefore, the vapour pressure of water vapour
- (7.ru99.9s) (101.3)
=7.z}kN/m2
which corresponds to the dew point 311'K and
latent beat of evaporation 241 1.2 kJ kg
Therefore, the total heat lost to flue gas
- 1285 + 1665.29
=2950.29 kJ/kg coa.
Hence, the percentage of heat lost to the flue gas
Specific heat of flue gas - 0.25 kcaVkg 'C
18% of total heat generated by coal is lost to
substances other than coal.
Solution
Step (I) Heat Content of Generated Steam
H = H w + x L - H r n
= 181 + 0.97 (482) - { = 608.54 kcal/kg
Step (II) Calorific Value of Coal
It can be determined from the boiler efficiencv
relationship.
Iboil", = Heat outpuvHeat input
Now Heat Output = 6000 (608.54) kcaVh
Heat input - 700 x CV where CV - calorif,rc value
ofcoal
or 0.70 - 6000 (608.54)/(700 x CV)
CV - 7 4l?kcallkg
Step (IID Carbon and Hydrogen Contents of'Coal
Basis: I kg coal
Therefore, the coal contains (l - 0.12), i.e.
0.88 kg ofC + H per kg ofcoal fired.
If .r be the part of CYkg of coal, then
.t (8075) + (0.88 - -rX3a500) =7452
.r = 0.8669
lc-"'b* l*lHyd'os;l
I kg Coal 0.8669 kg 0.0131 kg
Step (IV) Theoretical Air Requirement for Com-
plete Combustion
Basis: I kg coal
Elzme
nt
Boller Colculqllonr 93
Heat Rcceivcd by
(kcaVh)
by coal: 70{l,Q452)
- 521640,0
Flue gas: 16.88(700X0.25XA@)
- 29s4(A@)
Steam: 6000(608.54) - 3651240
Sirbstances other than flue gas:
7 OO(7 4s2)(O.r8) - 9389s2
By heat balancing
2954 (A@) + 365Qa0 + 938952 - 52164A0
. ' . AO=211.98"C *212C
Hence the ternperature of the flue gas atthe biler
outlet
-212 + L5 -227'C
Ans.
PtoUem 6.16 A boiler is fired with coal having
following percentage composition by mass:
C45Vo; H-54o; S-lVo; O-2.59o: Incom-
bustible-6.5%.
Determine the boiler efficiency from the given
data:
Excess air supplied = 407o
Flue gas temperature at boiler exit = 170"C
Ambient air temperature - 25"C
Specific heat of flue gas - 0.25 kcaVkg 'C
Specific heat of steam = 0.48 kcaVkg"C
Heat Generated
(kcaVh)
Combustion
Reaction
Weight OrRequirement
C + O, -+ CO, 0.8669 kg
(r2) (32)
2H, + O, + 2HrO 0.0131 kg
(4) (32'�)
o.E669QAn)
-2.3rr7 kg
o.or31(32t4)
- 0.1048 kg
E - 2.4165 kg
Combustion Heat of CombustionSince air contains 23% Oz by mass, the
stoichiometric (theoretical) air requirenrnt for com-
plete combustion
-2.4165 (100/23)
= t0.506 kg
Step (V) Excess Air Coefficient
Excess air coefficient = Actual air,/minimum air
= 16/10.506
= I.523
AIlr.
Step (VI) Enthalpy of Flue Gas
Mass of coal + mass of air = Mass of flue gas
(1 - 0.12) kg + 16 kg = Mass of flue gas
or Mass of flue gas - 16.88 kg.
Unaccounted heat loss - lSVo
Solution
Step (I) Calorific Value of Coal
Cv - 8075(c) + 2220(5) + 3a500(H- O/8)
where, C, S, H & O stand for carbon, sulphur,
hydrogen and oxygen percentage.
- 8075(0.85) + 2220(0.0r)
+ 3a500(0.05 - 0.025/8)
- 8503 kcaUkg
C + Or--+ CO,
S + Or---r SO,
H, + Or----+ HrO
8075 kcaVkg
222Okcdlkg
34500 kcaVkg
94 Boiler Operolion Englneerlng
Step (II) Stoichiometric Oxygen
Basis: I kg fuel
Element Oxygen Rcquircd Per
Kg of Fuel
C + Or --r COt(r2) (32)
S + O, ----+ SOt
(32) (32)
ZHr+ Or-+ZHrO
(32t 12)(0.8 s) - 2.2666 ks
(3u32)(o.ot) - 0.01 kg
(324) (0.05) - 0.4 kg
2-2.6766k9
Since the fuel contains 0.025 kg oxygen/kg of fuel'
the actualmass of 02 requirementperkg of coal burnt
-2.6766 - 0.025
=2.6516k9
Step (III) Air Supplied
Theoretical mass of air requtement
=2.5516 (1m/23)
- 11.5289 kg
40% excess air suPPlied.
Hence the actual air suPPlied
= 1.4 (11.5289)
= L6.l4kg/kg of coal
Step (IV) Mass of Flue Gas
Mass of combustibles Per kg of coal
- 1 - 0.065
= 0.935 kg
Fuel + Air = Flue Gas
0.935 kg 16.14 kg = Flue Gas
Hence the total mass of flue gas(inclusive of waler
vapour) produced per kg of coal burnt
= 0.935 + 16.14
- t7.075 kg
Step (V) Mass of Dry Flue Gas
2H2 + O, ------+ 2HrO
(4\ (2 x l8)
= 2 (18\14
- 9 k g
Mass of water produced/kg of coal (H-content :
0.5%) burned
= 9 (0.05) kg
= 0.45 kg
Mass ofdry flue gas produced/kg ofcoal burnt
= 17.075 - 0.45
= 16.625k9
Step (VI) Heat Balance
Basis: I kg coal
Heat Evolved Heal Lost To
8 503 kcal Flue gas (dry)
- r6.62s (0.25) (l7O-2s)
- 602.65 kcal
Stcam (l atm pr€s$.) generated from
fuel burning
= 0.45 [/J + Cp (A@) -@.ir]
= 0.45 [639 + 0.48 (170 - 100) - 25]
=297.42kcal
Unaccountcd sourceg
- (18/100) (E 503)
- I 530.54 kcal
Total - 2 424.61kca|
Heat utilized = 8503 -2424.61
= 6078.39 kcalAg coal
Step (VID Boiler Efficiency
Heat utilizedTlboiLr = H*t ga"*aLd
6078.39
= 
8503
= 0.7 I 48 i.e. 7 1.487o
Ans.
fuoblem 6.17 A water tube boiler operates
8400 h/year al 80o/o efficiency. The unit rated at
272L5kgh operates at7.82 atm.
It burns natural gas for six months of the year and
No.2 fuel oil for the rest.
Average aurual boiler loading is 6O% with an
input of 11347303 kcaVh.
Combustion
Reoction
Mass of water produced/kg of H2 burned
lYlthout Economlzer 
^ \
Natural g:rs consumption = 1274,25 Nmr/h latwo
Fuel oil (No.2)consumption = t.+S9 m3n,J bd
Afier Addlng An Economl,zer
BFW flowate (including blowdown) at60%lofi
- 17145 kg/h
Feedwater tenperature at the economizer inlet
= 105"C
Feedwater temperature at the economizer outlet
- 136'C
Fluegas t€mperature ateconomizer inlet = 260'C
Fluegas temperatur€ at economizer outlet - 149"C
Determine
(a) the fuel saving using the economizer
(b) totat annual fuel cost without insralling the
economizer
(c) total annual saving of fuel alter installing the
economizer
(d) the payback months, if the economizer cost is
Rs. 500,000 installed
Given: Nanral gas cost = Rs. 1.06 per Nm3 of
gas
Fuel oil (No.2) cost - Rs I 255 perm3 of F.O.
Solwion The addition of an economizqr to awater-
tube boiler system reduces fuel cosl
The fuel saving using the economizer is
- 
Ilx 100
"=___
where S- fuel saving in perceht
H= heatrecovere4 kcavh =ry
F - BFW flowrate, kg/h
AO = 6z - 8l = temperaturc difference of BFW
before and after the economizer
Or - BFW temperatue at economizer inlet, 'C
Oz = BFW temperaturc at the economizer outlet,
'c
B = boiler efTiciency
(a) H= (17t45X136- 105)/0.8
= 664368.78kca1/h
Boiler Colculotionr
Flg. 6.1 Figure to the Problem 6.17
s -664368.78 x 100:5.85%
rr347303
(b) Total operating perid = 8400 hlyear 
Ans'
Natural gas burned for 4200 h and F.O. bumed for
rest 4200 h over the year.
Annual cost of natural gas
/ ^ - r \
=,zzo.zs[*l' lx +200 rj-), r.06 r-R'-)
\ n / 1v .n / 1N* , . , ;
= Rs.5672961.
Annual cost of fuel oil
/ r \
= r.45e f+ l,<4200 r-!-)x tzss [4)
\o / \Year ; l - ' J
- Rs. 7690389
Total annual cost of fuel prior to installation
of economizer
= Rs.5672961 + Rs. 7690389
F l u o g o ! o x h o u r t
96 BollEr OPerollon Englneedng
- Rs. 13,363,350/-
(c) After the installation of the
5.85% saving in fuel results'
Annual saving in natural gas
= Rs. 5572961 x 5'85/100
= Rs. 331868/-
Annual saving in fuel oil cost
= Rs.76903g9 x 5.95.2100
- Rs. 449888/-
Number of operating days per year - 270
Cost of waste heat boiler installed
= Rs 3,000 000/-
Maintenance atrd overhead expenses
- 12% of the cAst of WHB
Rate of interest payable - 20% on the principal
arnount
DG Set Load StcanGeneration Rate
4O VdaY
80 Uday
Ans.
economizer, a
@%
too%
.'. Total annual saving of fuel cost after instalation
of the economizer
= Rs. 331868 + Rs' ut49888 = Rs' 781756/-Ans.
(d) The PaYback is
o - E x 1 2r - A
whereP-paybackmonths
E - installed economizer cost' Rs
A - annual fuel savings with economizer' Rs
. p -!''9SS ,. rz
' - Rs.781756
= 9.21 month, 
on .
Problem 6-t8 A waste heat boiler is hooked up
with a diesel generator set to produce steam from
waste heat.
ATQVoDG' setload' saturated steamof l0kg/c#
g is produced in the waste heat boiler at the rate of 40
tor/daY.--- 
nn.tug" ebcric energy generated per day varies
from65 to 70MWH.
Esttnate
tui-,n" oonomics of incorporating the waste heat
boiler
(b) th; payback period of the waste heat boiler
Given
ih" "o.t of purchased electrical energy from the
prid - Rs l.IZKWH
'- 
- 
cor, of generated electricity after the installation
of WHB - Rs0.95/KWH'- 
(t* includes overheads and depreciation char-
ges)
1 ton ofcoal generates 4'5 t of saturated steam at
10 kg/cm2 g
Cost of coal - Rs 750 Per ton
blution
ia) Bco"o*ics of lncorporating y-HB,.'-'rn" 
"ott of ebcricity puphased from the grid
-Rs l.l?KWH
The cost of generated electricity after the installa-
tion of WHB '0.95/KWH-- 
frfonetary savings per unit electricity generated
= Rs (1.12 - 0.95)/KWH = Rs 0'17/KWH
Average electricity generatior/day
=g#*MwH =67.5MwH
Number of working days per lear -270
Therefore, monetary savings with respectto power
purchased frorn the grid
- o/tl [nffr)' 67.5 x'.* [#, )
, zzo (,days')= Rs 3,098 250/- per year
[vear ''1
(b) Payback Period Of WHB
Cost of waste heat boiler
= Rs 3,000'000/-
Maintenarpe ard overheads
= l2Vo of caPital cost of WHB
= Rs 360'000/-
Interest ol PrinciPal arnount
= 20% of Rs.3,000,000/-
= Rs 600,000./-
Total steam generated on 100% load - 80 Vday
4.5t ofcoal generat€ I ton of steam
Amount of coal saved - 8C/4.5 - 17.777 tlday
Monetary savings, on the basis of coal, per year
- 750 f-n,-)x n.t77 (tro')" zzo [-@-)(ton/ (oayJ [vear.1
- Rs 359984?-
WHB is an energy saving equipment. So it
qualifies for 100% depreciation in the lst year.
Approximate savings in corporate taxes (@ 55%)
per ye:u
- Rs 3000000 x 0.55
- Rs 16500004
Net Savings per year
- Rs (3599842 - 360000 - 600000 + 1650000)
- Rs 4289842-
Ps 3000000Payback period = ffiffi x 12 months
= 8.39 months
Ans,
Ptoilem d.l9 lretermine the
(a) rate of fuel consumption in kg/h
(b) efficiency of WHB of problem 6.18
Given
I lt. of fuel generates 4.025 KWH of electrical
energy
Specific gravity of liquid fuel - 0.90
Exhaust gas flowrate and temperature at 68% load
are 7.55 kg/s and 325'C respectively.
Flue gas temperature at WHB inlet - 320'C
Flue gas tempenture at WHB outlet = 170'C
Average feedwater temperature to the boiler
- 75'C
Specific heat of flue gas - 0.26 kcaVkg 'C
Assume 59o radiation loss suffered by the flue gas
in the WHB.
Solution The determination of efficiency of the
waste heat boiler is to be made on the basis of heat
balance.
Boller Colculollons
(a) Rate of Fuel Consumption
Average elcclric energy generated per day
:67500 KWH
Average fuel consumption per day
-6750014.025
- 16770 lL
Mass rate of fuel consumption
---^ / rt ) (. r<e) I I= 16770 [dj,. o.e [sf,j' ;i $6
- 628.875 ke/h
Ans.
(b) Flowrate of Flue Gas at 60% l-oad
-7.55 x 3600 - 27180 kgh
(c) Useful Heat of Flue Gas
Total heat rejected by hot flue gas in the WHB
= 27180 f+l'0.26 f'9)x ezl -r70) ("c)(n J [K8-u '
= 1060020 kcaVh
Heat lost to radiation
- 106(n20 x 5/100 kcaVh
= 53001 kcaVlt
Useful heat available for steam generation
= 1060020-53001
= 1007019 kcaVh
- Heat input rate
(d) Heat Output
Average steanr (10 kg/cm2 g and saturated)
generation rate
- 40ttday
= 40x l0ffil24kg/b
= 1666.66 kg/h
Average feedwater temperature = 75'C
Heat required to generate 1666.66 kg steam
(10 kg/cmz g and saturated)
- 1666.66 x (183 - 75) + 1666.66 x 478.4 kcal
=977329 kcal
Heat output rate = 917329 kcat/h
Heat input rate = 1007019 kcaVh
98 Boller OPerotlon Englneerlng
.'. Efficiency of waste heat boiler
-mxLoo%
-97'05% 
Ans.
BOILER HEAT BALANCE CALCULATIONS
Basis: lkg fuel
Heat Input
(A) Ilr- Gross calorific value of fuel' kcal
(B) Hz= Heat input of fuel
= c ' (@r- @r) 'kcal
c - specific heat of fuel, kcaVkg "C
Ot - temperature of fuel, 'C
@r - r€ference temperature, "C
(c) tt = Heat input of air
= M"cr(@" - @r), kcal
M" - mass of input dry airlkg fuel, kg airlkg fuel
ca - sP€ciltc heat of humid air
-0.24+ 0.46H, kcaVkg dry air "C
I/ - humidig of air, kg moisture&g dry air
@a - air temperature, 'C
Total heat input,I{ = Hr* Hr+ H3,kcal
Heat Output
(A) Heat consumed in generating steam
l. Economizer Ho= Mw (he,, - h6r), kcal
Mw - fllass of feedwater per unit mass of fuel,
kg/kg tuel
iew - eothalpy of water at economizer outlet'
kcaVkg
ftfw = enthalpY of feedwater, kcaVkg
2. Evaporator (Boiler) Hs= M"(/t, - i"*),kcal
M. - mass of steam generation per unit mass of
fuel, kg steam/kg fuel
h, = enthalpy of steam generated, kcaVkg
3. SuPerheater Hu= M"(Hu- hr)' kcal
/rr. = enthalpy of superheated steam, kcal/kg
(B) Heat lost in flue gases
Ht =2 n; ?0, (@s, - 25)' kcal
ni - number of moles of i-th component present in
the flue gas produced due to combustion of 1 kg fuel.
En, - the mean specific heat of i-th component at
ofc
@1, = flue gas tcmPerature' "C
(C) Heat loss due to evaPoration
l. Moisture is formed due to combustion of
hydrogen in the fuel. Loss of heat to evaporate this
moisture
H8= M^L' kcal
Mm = tnilss of moisture formed by burning of
hydrogen per kg of t'uel, kg H2Olkg fuel
L = latcnt hcat of evaporalion of the moisture at the
dew point of the flue gases, kcaVkg
2. Heat loss due to evaporation of moisture
present in thc luel
Hs= M*f L,kcal
Mnrf = mass of moisture present in the fuel' kg/kg
fuel
(D) Heat Loss due to incomplete combustion of
carbon as carbon monoxide.
I co l .a,o = 
fEofrE.] x c x 5 636.7 kcal
CO - Vo (by volume) of carbon monoxide in the
flue gas
COt- qo (by volume) of carbon dioxide in the flue
gas
C = c:ubon bumt per kg fuel burnt, kglkg fuel
(E) Heat loss due to unburnt carbon
Htr= M"(7 837'5) ' kcal
M" = 62ss of unconsunred carbon in refuse, kg/kg
fuel
(F) Heat loss due to blowdorvn
H tz= Mt t (hu* - /t1.."), kcal
Mbl = mass of blowdown water, kg/kg fuel
ftuw = enthalpy of boilerwater, kcaVkg
(G) Unaccounted heat loss
Hrt= Hi- (Hc+ H5 + Hu+ H, + Hg
+ Hn+ Hrc+ H1.+ Hp)
Prodem 6.fr A stoker-fired waterhrbe boiler
burns coal-at the rate of 4 Uh to generate steam of 30
kglcfl:P abs and 430"C at the rate of 30 Vhour.
Evaluate the boiler performance from the following
data
(a) Component Proximate analysis of coal
Ash 12.7% (by weight)
Moisture 7.9% (by weight)
(b) Gross calorific value of coal - 6 250kcaUkg
(c) Component Flue Gas Arnlysis
coz t2.85%
02 6.580
N2 rest
(d) Carbon present in the cinder as unburnt com-
bustible -2.75%
(e) The feedwater tempenture - 90"C
(f) Flue gas temperaturc at economizer outlet
- 150\C
Flue gas pressure at economizer outlet
- 755 mmHg
(g) Air tempqratures at burner inlet 30"C DB and
22"C WB. Ignore the presence of sulphur and
oxygen in coal.
blwion The boiler performance, i.e. the overall
thermal efficiency of the boiler is to be evaluated on
the basis of heat balance.
Basis: 100 kmol of dry flue gas.
1. Oxygen Supplied with Combustion Air
N, in the flue gases
- 100 - (12.85 + 6.5)
- 80.65 kmol
l0Okmol aircontains 79kmol N2 and 21 kmol 02
.'. 02 supplied for combustion
-(21179) x 80.65 - 21.438 kmol
2. Water Vapour Produced During
Combustion
C + 02 -----+ CO2
I kmol I kmol I kmol
I kmol of CO2 requires I kmol of O, forcombus-
tion.
12.85 kmol of CO2 require 12.85 kmol of 02
for combustion.
Boller Colculotlons 99
Therefore, the oxygen utilized for hydrogen burn-
ing of fuel
=21.438- (12 .85+6.5)
- 2.088 kmol
2Hz + 02 -----s 2H2O
2 kmols I kmol 2 kmols
Hydrogen bumt = 2 (2.088) = 4.176 kmol
Water produced - 4.176 kmol
3. Unburnt Carbon for 100 kmol of
Dry Flue Gas
Carbon retained in the cinder
) 1 \
-ff ix 4x 1000kg/h
- l l 0 k 9 h
(FC+ VM) in coal
= 100 - (asb% + moistureTo) in coal
= 100- (12.7 +7.9)=79.4Vo
. 
unbumt carbon
" (FC + VM - unburnt carbon)
) 7 \
===-? =0.0358't9.4 
-2.75
Carbon in the flue gas
- carbon in the CO2 in the flue gas
- 12.85 kmol
= 12.85 x 12
= 154.2 kg
Hydrogen in the flue gas
- 4.176 kmol
= 4 .176x2
= 8.352 kg
Total burnt combustible
- 154.2 + 8.352
- r62.552kg
Carbon unburnt for 100 kmol ofdry flue gas
= 0.0358 x 162552
= 5.8193 kg
= 0.4849 kmol
100 Boller Operotlon Englneedng
4. Excess Air
C + O 2 + C O 2
l kmol l kmol l kmol
I kmol of C requires I kmol . 3r for combustion
.'. 0.4899 kmol of C requires 0.4899 kmol of 02
for combustion
.'. Oxygen required to bum that unbumt carbon -
0.4899 kmol
.'. Excess 02 supplied
= 6.5 - 0.4899 = 6.0151 kmol
Total stoichiometric 02 required
= 12.85 + 2.088 + 0.4899
= 15.4229lnnol
Therefore, excess air supplied
_ 
6.0151 Y rf l )
15.4229 " '""
=39Vo.
5. Moisture Content of the Flue Gas
100 kg coal contains 7.9 kg free moisture and
ll00 - (12.7 + 7.9)l i.e. T9.4kgcombustibles
Free moisture appearing with the combustion of
162.552 kg combustibles
? o
t9 .4
- 16.173kg= t6.tl3t18 kmol - 0.8985 kmol
From psychrometric chart, humidity of air at 30'C
DB and 22'C WB
- 13.4 grr/kg dry air
1 ? 4 I
= 
lg . 1000 kmol HrO$ kmol dry air
0.02158 kmol water
= -- 
ktnol dry uit
Therefore, the total moisture entering the combus-
tion zone
= 0.02158 x kmol of air containing 21.438 kmol
o2
= (0.02158) r [+ x 2r.+38-]
l L t I
- 2.203 kmol
Therefore, total moisture in the flue gases
- (4.176 + 0.8985 + 2.203) kmol
=7.277Sklrrol
6. Composition of Flue Gas
Component kmol mol%
12.85 rr.97
6.5 6.05
10
*x21.$8-80.&'7 75.179
z l
7.275 6.781
lm.272 D.980
7. Heat Input Rate
Rate of fuel buming - 4 ton coaVh
Gross cloritic value of fuel - 6250 kcaVkg
Heat input rate = 4 x 1000 x 6250
= 25000000 kcaVh
Ignoring the heat input of air (at 30'C), the net heat
input rate - 25000000 kcavh
8. Heat Output Rate
[A] Heat absorbed in generating superheat,ed
steam
Hn+ Hr+ Hu
- 30,000 x (787.8 - 90.04) kcaUh
= 20932800kcaI/h
where 787.8 kcal/kg - enthalpy of steam at
30 kgf/cm2 abs and at 4t0"C = h*
90.04 kcaVkg = enthalPY of water at 90'C
lBl Combustibles left in the cinder (as C)
- I l0 kgih
Calorihc value of carbon (GCV - NCV)
= 94.05 kcaVmol
q4 ()5
=# x 1000kcal/kgt z
Heat lost in the combustibles
Hr r=Wx 1000x l l0kcaVh
=862125 kcal,/h
lCl Total free moisture evaporated from coal
=l#x4ooo"%E
= 305.055 kg/h
coz
o2
N2
Hzo
Now, partial pressure of water vapour in the flue
gas
= 755 x rnol frrction of water vap. in the flue gas
- 7 55 (7 .I7 5l W7 .2:12) mmltg
-51.20mmH9
Dew point of flue gases - 38"C
Latent heat of water at 38"C - 575.83 kcavkg
Heat lost due to evaporation of free mois[re
- 305.055 x 575.83 - 175660 kcaYh
[D] Heat loss due to evaporation of moisture
fonned due to cornbustion of hydrogen in the coal
burnt
,, =l#r)G.*) [4#] (, sx57s.s3)rca,/h
I tmot urO lI 1g coarlf ks cornbustiblesl
' ' - - - - - - - - - - - - - - - - '
fkgcorrbustibleslf h lL kgcoal I
I tgHro lf r""rt| _-- ll -:- l- 815408 kcarl/h
I kmol H2oll tg J
lEl Heat lost in flue gases (I/r) is evaluated m the
basis of rnean specific heat data of flue gas com-
ponents:
Componcfi Mcan Sp. Hcat intlu Rangc25"-l50"C
9.5 kcal/tmol'C
8.12kcalrtmol "C
7.l2kcal/kmol "C
7.00 kcal,/kmol'C
Thercfore, for 1f/.272krnol of flue gas
r\c^ = 12.E5 (95) +7.275 (8.12) + 6.5 (7.12\
' r t
+8O.&7 (7) kcaV 'C
=79l.957kcal/"C
Boiler Colculoliom l0l
Heatrost in nue 83s€s ='lrt:lrT"i!;"'
Therefore, the rate of heat loss of flue gases
- ee,ee 4.62( 4,00cy r d2. s rr> (#EJ *"ro
= t,Ag6,06aY
9. Heat Balarrce
Hcd Input
Rate
(kcd/h)
Heat Output Rate
(kcaYh)25,000,000 $1eam generation
Heat loss due to
unburnt combustibles
Heat loss due to
evaporation offree
moisture
Heat loss due to
-2W3280083.73%
-862125 3.4%
- 175660 0.70%
- El64$ 3.26%
coz
Hzo
o2
N2
evapcation of moishre
formed due to
combustion of
hydrogen h thc fuel
Heat lost to flue gases - 1886064 7.54%
Unaccountedheatloss -326943 Ln%
(by differerce)
Total - 25 000 000
(10) Overall Thermal Efficiency of the Boiler
_ 
Heat oubut rate (steam generation)
Heat input rate (fuel combustion)
=#ffffix roo= 83.73%
Steam tables
Pressu re
bar kPa
Temperature
Specif ic EnthalPY I Dpecrrrc
Water (hf) | Evaporation (hfs ) lSteam (hn ) l Steam
kJ/kq I kJ /ks I kJ /kg lmJ/kg
abso lu te
0.30 30.0 69.10 289.23 2336.1 2625.3 5.229
0.50 50.0 81.33 340.49 2305.4 2645.9 3.24Q
o.7s 75 .O 91.78 384.39 2278.6 2663.0 2 .217
0.95 95.0 98.20 411.43 2261 .8 2673.2 1.777
gauge
0 0 100 .00 419 .04 2257 .0 2676 .0 1 .673
0.10 10.0 102.66 430.2 2250.2 2680.2 1.533
0 .20 20 .0 105 .10 440 .8 2243 .4 2644 .2 1 .414
0 .30 30 .0 107 .39 450 .4 2237 .2 2647 .6 ' t . 312
0 .40 40 .0 109 .55 459 .7 2231 . 3 2691 . 0 1 .225
0 .50 50 .0 1 1 1 . 61 468 .3 2225 .6 2693 .9 1 .149
0 .60 60 .0 1 13 .56 476 .4 2220 .4 2696 .8 1 .083
0.70 70 .o 115.40 444.1 2215.4 2699.s 1 .024
0 .80 80 .0 117 .14 491 .6 2210 .5 2702 .1 0 .971
0 .90 90 .0 1 18 .80 498 .9 2205 .6 2704 .5 0 .923
1.00 100.0 120.42 505.6 2201 .1 2706.7 0.881
1 .10 110 .0 121 . 96 512 .2 2197 .0 2709 .2 0 .841
1 . 2 0 120.0 123.46 518.7 9 2 . 8 2 7 1 1 . 5 0 . 8 0 6
1 . 3 0 130.0 124.90 524.6 8 8 . 7 2 7 1 3 . 3 0 . 7 7 3
'1 
.40 140.0 126.28 530.5 z 84 .8 2715 .3 0 .743
1 . 5 0 150.0 1 2 7 . 6 2 5 3 6 . 1 z 8 1 . 0 2 7 1 7 . 1 0 . 7 1 4
1 . 6 0 160.0 128 .89 541 . 6 7 7 . 3 2 7 1 8 . 9 0 . 6 8 9
1 . 7 0 170.0 1 3 0 . 1 3 5 4 7 . 1 73.7 2720 .8 0 .665
1 . 8 0 180.0 131 .37 552.3 z 70 .1 2722 .4 0 .643
1 . 9 0 190.0 132.54 557.3 z 66.7 2724.0 0.622
2.00 200.0 '1 33.69 562.2 63 .3 2725 .5 0 .603
2.20 220.0 135 .88 571 . 7 56 .9 2724 .6 0 .568
2.40 240.O 1 3 8 . 0 1 580.7 2 50 .7 2731 . 4 0 .536
2.60 260.0 140.00 589.2 2 44 .7 2733 .9 0 .509
2.80 280.0 141 .92 597.4 39.0 2736.4 0 .483
3.00 300.0 143.75 605.3 33 .4 2738 .7 0 .461
3.20 320.0 145.46 612.9 24.1 2741 .O 0.440
3.40 340.0 147 .20 620.0 22.9 2742.9 0 .422
3.60 360.0 148.84 627.1 I 17 .8 2744 .9 0 .405
3.80 380.0 150.44 634.0 z 12 .9 27 46 .9 0 .389
4.00 400.0 151 . 96 640 .7 08 .1 2748 .4 0 .374
4.50 450.0 155.55 656.3 2096.7 2753.0 0.342
5.00 500.0 158.92 670.9 2086.0 2756.9 0 .315
5.50 550.0 162.08 684.6 2075.7 2760.3 0.292
6.00 600.0 |65 .04 697.5 2066.0 2763.5 0 .272
6.50 650.0 167.83 709.7 2056.8 2766.5 0.255
7 .00 700 .0 170 .50 721 .4 2047 .7 2769 .1 0 .240
7.50 750.0 173.02 732.5 2039.2 2771.7 0.227
8 .00 800 .0 175 .43 743 .1 2030 .9 2774 .0 0 .215
8.50 850.0 177.75 753.3 2022.9 2776.2 0 .204
9.00 900.0 179.97 763.0 2015.1 2778.1 0 .194
9.50 950.0 182.10 772.5 2007 .5 2780.0 0 .185
1 0 . 0 0 1 0 0 0 . 0 ' ' | 8 4 . 1 3 7 8 1 . 6 2 0 0 0 . 1 2 7 8 1 . 7 0 . 1 7 7
1 0 . 5 0 1 0 5 0 . 0 1 8 6 . 0 5 7 9 0 . 1 1 9 9 3 . 0 2 7 8 3 . 3 0 . 1 7 1
1 . 00 1 1 00 .0 1 88 .02 798 .8 1986 .0 2744 .4 0 .163
EDB/1
splrax' tsarco 1 . 0 1
tables
Pressu re
bar gauge kPa
Temperature
Speci f ic Enthalpy Specif ic
Vo lume
Steam
ms/kg
Water (h1 )
kJ/kg
Evaporation (h1n)
kJ/kg
Steam (hn )
kJ/kg
I t . 5 U
1 2 . o 0
1 2 5 0
ft-oo
I J . 3 U
t4-oo
- 1450
I C . U U
.-
I 3 . J U
16-00
1 7 .o0
t s J o
19-OO
20^oo
E.oo
22nO
23^00
24l �0
25-00
26-OO
27 SO
Z8-oo
2 9 . o 0
30.00
3 1 . 0 0
32 '00
33.00
34^OO
35-OO
36-OO
37-OO
38-00
39-OO
40-oo
?loo
43-OO
44nO
45-OO
46-OO
nno
Zs.oo
49nO
50s0
5 1 i 0
52-00
53.0O
54.oO
C C . U U
56-OO
1 1 5 0 . 0 1 8 9 . 8 2 8 0 7 . 1 f f i
-
io_
ias
T
iei#r
irffi
ffi
ffi
ffi
ffi
1.02
splrax' 
.lsarco EDB/1
Steam tables
Pressu re
bar qauoe kPa
Temperature
"C
Speci f ic Enthalpy
@kJ/ks I kJ/ks
l D p e o l i l c ;
Vo lume
Steam (ho) | Steam
kJ/ko 
- | m3/ko
57.00 s700.0 273.45 2 0 2 . 1 584.5 2786.6 0 .0337
5 8 . 0 0 5 8 0 0 . 0 2 7 4 . 5 5 2 0 7 . 8 577.7 2785.5 0 .0331
59.00 5900.0 275.65 213.4 5 7 1 . 0 2 7 8 4 . 4 0 . 0 3 2 5
6 0 . 0 0 6 0 0 0 . 0 2 7 6 . 7 3 2 1 8 . 9 564.4 2783.3 0 .0319
6 1 . 0 0 6 1 0 0 . 0 2 7 7 . 8 0 2 2 4 . 5 5 5 7 . 6 2 7 8 2 . 1 0 . 0 3 1 4
62.00 6200.0 278.85 2 3 0 . 0 550.9 2780.9 0 .0308
6 3 . 0 0 6 3 0 0 . 0 2 7 9 . 8 9 235.4 5 4 4 . 3 2 7 7 9 . 7 0 . 0 3 0 3
64.00 6400.0 280.92 240.8 537.3 2778.5 0 .0298
6 5 . 0 0 6 5 0 0 . 0 2 8 1 . 9 5 2 4 6 . 1 531 .2 2777.3 0 .0293
66.00 6600.0 282.95 251 . 4 5 2 4 . 7 2 7 7 6 . 1 0 . 0 2 8 8
6 7 . 0 0 6 7 0 0 . 0 2 8 3 . 9 5 256.7 5 1 8 . 1 2 7 7 4 . 8 0 . 0 2 8 3
68.00 6800.0 284.93 2 6 1 . 9 5 1 1 . 6 2 7 7 3 . 5 0 . 0 2 7 8
6 9 . 0 0 6 9 0 0 . 0 2 8 5 . 9 0 2 6 7 . 0 5 0 1 . 1 2 7 7 2 . 1 0 . 0 2 7 4
70 .00 7000 .0 28 t i . 85 272 . ' l 498 .7 2770 .4 O .A '274
7 1 . 0 0 7 1 0 0 . u 2 4 7 . 4 o 2 7 7 3 4 9 2 . 2 2 7 b 9 . 5 0 . 0 2 6 6
72.OO 7200.0 244.7 5 2 8 2 . 3 485 .8 2768 . ' t 0 . 0262
73 .00 7300 .0 289 . t i 9 2 4 7 . 3 4 7 9 . 4 2 7 6 b . 7 0 . 0 2 5 8
74 .OO 7400 .0 290 . t t 0 2 9 2 . 3 473 .0 2765 .3 0 .0254
7 5 . 0 0 7 5 0 0 . 0 2 9 1 . 5 1 2 9 7 . 2 4 t t 6 . 6 2 7 6 3 . A 0 . 0 2 5 0
7 6 . 0 0 7 6 0 0 . 0 2 9 2 . 4 1 302.3 460 .2 2762 .5 0 .0246
77 .OO 7700 .0 293 .91 3 0 7 . 0 453 .9 2760 .9 Q .O242
7 8 . 0 0 7 8 0 0 . 0 2 9 4 . 2 0 3 1 1 . 9 4 4 7 . 6 2 7 5 9 . 5 0 . 0 2 3 9
2 9 . 0 0 7 9 0 0 . 0 2 9 5 . 1 0 J t o 7 4 4 ' t . 3 2 7 5 8 . 0 0 . 0 2 3 6
8 0 . 0 0 8 0 0 0 . 0 2 9 5 . 9 6 321 .5 435 .0 2756 .5 0 .0233
8 1 . 0 0 8 1 0 0 . 0 2 9 6 . 8 1 326.2 424 .7 2754 .9 0 "0229
8 2 . 0 0 8 2 0 0 . 0 2 9 7 . 6 6 3 3 0 . 9 422 .5 2753 .4 0 .0226
8 3 . 0 0 8 3 0 0 . 0 2 9 8 . 5 0 3 3 5 . 7 416 .2 2751 . 9 0 .0223
8 4 . 0 0 8 4 0 0 . 0 2 9 9 . 3 5 3 4 0 . 3 4 1 0 . 0 2 7 5 0 . 3 0 . 0 2 2 0
8 5 . 0 0 8 5 0 0 . 0 3 0 0 . 2 0 3 4 5 . 0 4 0 3 . 8 2 7 4 8 . 8 0 . 0 2 1 7
8 6 . 0 0 8 6 0 0 . 0 3 0 1 . 0 0 3 4 9 . 6 3 9 7 . 6 2 7 4 7 . 2 0 . 0 2 1 4
8 7 . 0 0 8 7 0 0 . 0 3 0 1 . 8 1 354.2 3 9 1 . 3 2 7 4 5 . 5 0 . 0 2 1 1
8 8 . 0 0 8 8 0 0 . 0 3 0 2 . 6 1 3 5 8 . 8 3 8 5 . 2 2 7 4 4 . O 0 . 0 2 0 8
8 9 . 0 0 8 9 0 0 . 0 3 0 3 . 4 1 J O J . J 3 7 9 . 0 2 7 4 2 . 3 0 . 0 2 0 5
90.00 9000.0 304.20 3 6 7 . 8 372.7 27 40 .5 0 .0202
92.00 9200.0 305.77 3 7 6 . 8 3 6 0 . 3 2 7 3 7 . 1 0 . 0 9 7
94.00 9400.0 307.24 3 8 5 . 7 3 4 8 . 0 2 7 3 3 . 7 0 . 0 9 2
9 6 . 0 0 9 6 0 0 . 0 3 0 8 . 8 3 3 9 4 . 5 3 3 5 . 7 2 7 3 0 . 2 0 . 0 87
9 8 . 0 0 9 8 0 0 . 0 3 1 0 . 3 2 403.2 3 2 3 . 3 2 7 2 6 . 5 0 . 0 83
1 0 0 . 0 0 0 0 0 0 . 0 3 1 1 . 7 9 4 1 1 I 3 1 0 . 9 2 7 2 2 . 8 0 . 0 7 8
1 0 2 . 0 0 0200.0 313.24 420.5 2 9 8 . 7 2 7 1 9 . 2 0 . 0 74
1 0 4 . 0 0 0400.0 314.67 429.0 2 8 6 . 3 2 7 1 5 . 3 0 . 0 70
0 6 . 0 0 1 0 6 0 0 . 0 3 1 6 . 0 8 4 3 7 . s 2 7 4 . 0 2 7 1 1 . 5 0 . 0 1 6 6
0 8 . 0 0 1 0 8 0 0 . 0 3 1 7 . 4 6 445.9 2 6 1 . 7 2 7 0 7 . 6 0 . 0 1 6 2
1 0 . 0 0 1 0 0 0 . 0 3 1 8 . 8 3 454.3 249 .3 27 03.6 0 .0 5 8
1 2 . 0 0 1 2 0 0 . 0 3 2 0 . 1 7 4 6 2 . 6 2 3 7 . 0 2 6 9 9 . 6 0 . 0 5 4
1 4 . 0 0 1 4 0 0 . 0 3 2 1 . 5 0 4 7 0 . 8 224 .6 2655 .4 0 .0
1 6 . 0 0 1 6 0 0 . 0 3 2 2 . 8 1 4 7 9 . O 2 1 2 . 2 2 6 9 1 . 2 0 . 0 47
1 8 . 0 0 ' 1 8 0 0 . 0 3 2 4 . 1 0 487.2 199.8 2687.0 0 .0 44
2 0 . 0 0 2 0 0 0 . 0 3 2 s . 3 8 495.4 187.3 2682.7 0 .0 4 1
EDB/1
sprrax' tsarco 1.03