Solution Koretsky ch02

Prévia do material em texto

Chapter 2 Solutions 
Engineering and Chemical Thermodynamics 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Wyatt Tenhaeff 
Milo Koretsky 
 
Department of Chemical Engineering 
Oregon State University 
 
koretsm@engr.orst.edu 
 
 
 
 
 2
2.1 
There are many possible solutions to this problem. Assumptions must be made to solve the 
problem. One solution is as follows. First, assume that half of a kilogram is absorbed by the 
towel when you dry yourself. In other words, let 
 
 [ ]kg 5.0
2
=OHm 
 
Assume that the pressure is constant at 1.01 bar during the drying process. Performing an energy 
balance and neglecting potential and kinetic energy effects reveals 
 
 hq ˆˆ ∆= 
 
Refer to the development of Equation 2.57 in the text to see how this result is achieved. To find 
the minimum energy required for drying the towel, assume that the temperature of the towel 
remains constant at K 298.15Cº 25 ==T . In the drying process, the absorbed water is 
vaporized into steam. Therefore, the expression for heat is 
 
 l OH
v
OH hhq 22
ˆˆˆ −= 
 
where is v OHh 2
ˆ is the specific enthalpy of water vapor at bar 01.1=P and K 15.298=T and 
l
OHh 2
ˆ is the specific enthalpy of liquid water at bar 01.1=P and K 15.298=T . A hypothetical 
path must be used to calculate the change in enthalpy. Refer to the diagram below 
 
P = 1 [atm]
liquid vapor
liquid vapor
 ∆ ˆ h 1
P
 ∆ ˆ h 2
 ∆ ˆ h 3
∆ ˆ h 
 Psat
1 atm
3.17 kPa
 
 
By adding up each step of the hypothetical path, the expression for heat is 
 
( ) ( )[ ] ( ) ( )[ ]
( ) ( )[ ]Cº 25ˆbar 01.1 C,º 25ˆ Cº 25
ˆCº 25ˆbar 1.01 C,º 25 ˆCº 25ˆ
ˆ
,
,,,
321
22
2222
satv
OH
v
OH
satl
OH
satv
OH
l
OH
satl
OH
hh
hhhh
hhhq
−+
−+−=
∆+∆+∆=
 
 
 3
However, the calculation of heat can be simplified by treating the water vapor as an ideal gas, 
which is a reasonable assumption at low pressure. The enthalpies of ideal gases depend on 
temperature only. Therefore, the enthalpy of the vapor change due to the pressure change is 
zero. Furthermore, enthalpy is weakly dependent on pressure in liquids. The leg of the 
hypothetical path containing the pressure change of the liquid can be neglected. This leaves 
 
 ( ) ( )kPa 3.17 C,º 25ˆ kPa .173 C,º 25ˆˆ
22
l
OH
v
OH hhq −= 
 
From the steam tables: 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 2.2547ˆ ,
2
satv
OHh (sat. H2O vapor at 25 ºC) 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 87.104ˆ ,
2
satl
OHh (sat. H2O liquid at 25 ºC) 
 
which upon substitution gives 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 3.2442qˆ 
 
Therefore, 
 
 [ ]( ) [ ]kJ 2.1221
kg
kJ 442.32kg 5.0 =⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡=Q 
 
To find the efficiency of the drying process, assume the dryer draws 30 A at 208 V and takes 20 
minutes (1200 s) to dry the towel. From the definition of electrical work, 
 
 [ ]( ) [ ]( ) [ ]( ) [ ]kJ 7488s 1200V 208A 30 === IVtW 
 
Therefore, the efficiency is 
 
 [ ][ ] %3.16%100kJ 7488
kJ 221.21%100 =⎟⎟⎠
⎞⎜⎜⎝
⎛ ×=⎟⎠
⎞⎜⎝
⎛ ×=
W
Qη 
 
There are a number of ways to improve the drying process. A few are listed below. 
• Dry the towel outside in the sun. 
• Use a smaller volume dryer so that less air needs to be heated. 
• Dry more than one towel at a time since one towel can’t absorb all of the available 
heat. With more towels, more of the heat will be utilized. 
 
 4
2.3 
In answering this question, we must distinguish between potential energy and internal energy. 
The potential energy of a system is the energy the macroscopic system, as a whole, contains 
relative to position. The internal energy represents the energy of the individual atoms and 
molecules in the system, which can have contributions from both molecular kinetic energy and 
molecular potential energy. Consider the compression of a spring from an initial uncompressed 
state as shown below. 
 
M
State 1 State 2
∆x
uncompressed compressed
 
 
Since it requires energy to compress the spring, we know that some kind of energy must be 
stored within the spring. Since this change in energy can be attributed to a change of the 
macroscopic position of the system and is not related to changes on the molecular scale, we 
determine the form of energy to be potential energy. In this case, the spring’s tendency to restore 
its original shape is the driving force that is analogous to the gravity for gravitational potential 
energy. 
 
This argument can be enhanced by the form of the expression that the increased energy takes. If 
we consider the spring as the system, the energy it acquires in a reversible, compression from its 
initial uncompressed state may be obtained from an energy balance. Assuming the process is 
adiabatic, we obtain: 
 
 WWQE =+=∆ 
 
We have left the energy in terms of the total energy, E. The work can be obtained by integrating 
the force over the distance of the compression: 
 
 2
2
1 kxkxdxdxFW ==⋅−= ∫∫ 
Hence: 
 
 2
2
1 kxE =∆ 
 
We see that the increase in energy depends on macroscopic position through the term x. 
 
It should be noted that there is a school of thought that assigns this increased energy to internal 
energy. This approach is all right as long as it is consistently done throughout the energy 
balances on systems containing springs.
 5
2.4 
For the first situation, let the rubber band represent the system. In the second situation, the gas is 
the system. If heat transfer, potential and kinetic energy effects are assumed negligible, the 
energy balance becomes 
 
 WU =∆ 
 
Since work must be done on the rubber band to stretch it, the value of the work is positive. From 
the energy balance, the change in internal energy is positive, which means that the temperature 
of the system rises. 
 
When a gas expands in a piston-cylinder assembly, the system must do work to expand against 
the piston and atmosphere. Therefore, the value of work is negative, so the change in internal 
energy is negative. Hence, the temperature decreases. 
 
In analogy to the spring in Problem 2.3, it can be argued that some of the work imparted into the 
rubber band goes to increase its potential energy; however, a part of it goes into stretching the 
polymer molecules which make up the rubber band, and the qualitative argument given above 
still is valid. 
 6
2.5 
To explain this phenomenon, you must realize that the water droplet is heated from the bottom. 
At sufficiently high temperatures, a portion of the water droplet is instantly vaporized. The 
water vapor forms an insulation layer between the skillet and the water droplet. At low 
temperatures, the insulating layer of water vapor does not form. The transfer of heat is slower 
through a gas than a liquid, so it takes longer for the water to evaporate at higher temperatures. 
 
 7
2.6 
 
Apartment
System
Fr
id
ge
+
-W
Q
HOT
Surr.
 
 
If the entire apartment is treated as the system, then only the energy flowing across the apartment 
boundaries (apartment walls) is of concern. In other words, the energy flowing into or out of the 
refrigerator is not explicitly accounted for in the energy balance because it is within the system. 
By neglecting kinetic and potential energy effects, the energy balance becomes 
 
 WQU +=∆ 
 
The Q term represents the heat from outside passing through the apartment’s walls. The W term 
represents the electrical energy that must be supplied to operate the refrigerator. 
 
To determine whether opening the refrigerator door is a good idea, the energy balance with the 
door open should be compared to the energy balance with the door closed. In both situations, Q 
is approximately the same. However,the values of W will be different. With the door open, 
more electrical energy must be supplied to the refrigerator to compensate for heat loss to the 
apartment interior. Therefore, 
 
 shutajar WW > 
 
where the subscript “ajar” refers the situation where the door is open and the subscript “shut” 
refers to the situation where the door is closed. Since, 
 
 shutajar QQ = 
 shutshutshutajarajarajar WQUWQU +=∆>+=∆ 
 shutajar TT ∆>∆∴ 
 
The refrigerator door should remain closed. 
 8
2.7 
The two cases are depicted below. 
 
 
System
H
ea
te
r
Q
Surr.
W
System
H
ea
te
r
Q
Surr.
Heater off Heater on
 
 
Let’s consider the property changes in your house between the following states. State 1, when 
you leave in the morning, and state, the state of your home after you have returned home and 
heated it to the same temperature as when you left. Since P and T are identical for states 1 and 2, 
the state of the system is the same and ∆U must be zero, so 
 
 0=+=∆ WQU 
 
or 
 
 WQ =− 
 
where -Q is the total heat that escaped between state 1 and state 2 and W is the total work that 
must be delivered to the heater. The case where more heat escapes will require more work and 
result in higher energy bills. When the heater is on during the day, the temperature in the system 
is greater than when it is left off. Since heat transfer is driven by difference in temperature, the 
heat transfer rate is greater, and W will be greater. Hence, it is cheaper to leave the heater off 
when you are gone. 
 9
2.8 
The amount of work done at constant pressure can be calculated by applying Equation 2.57 
 
 QH =∆ 
 
Hence, 
 
 hmQH ˆ∆==∆ 
 
where the specific internal energy is used in anticipation of obtaining data from the steam tables. 
The mass can be found from the known volume, as follows: 
 
 
[ ]( )
[ ]kg 0.1
kg
m0.0010
L
m001.0L1
ˆ 3
3
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡
==
v
Vm 
 
As in Example 2.2, we use values from the saturated steam tables at the same temperature for 
subcooled water at 1 atm. The specific enthalpy is found from values in Appendix B.1: [ ]( ) [ ]( ) ⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡−⎥⎦⎤⎢⎣⎡=−=∆ kgkJ 15.314kgkJ 87.104kgkJ 02.419C25at ˆC100at ˆˆ o1,o2, ll uuu 
Solve for heat: 
 [ ]( ) [ ]kJ 15.314
kg
kJ 05.314kg 0.1ˆ =⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡=∆= umQ 
and heat rate: 
 
 [ ][ ]( ) [ ] [ ]kW 52.0
min
s 60min. 01
kJ 15.314 =
⎟⎠
⎞⎜⎝
⎛== t
QQ& 
 
This value is the equivalent of five strong light bulbs. 
 
 10
2.9 
 
(a) 
From Steam Tables: 
 
⎥⎦
⎤⎢⎣
⎡=
kg
kJ 8.2967ˆ1u (100 kPa, 400 ºC) 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 8.2659ˆ2u (50 kPa, 200 ºC) 
 
⎥⎦
⎤⎢⎣
⎡−=−=∆
kg
kJ 0.308ˆˆˆ 12 uuu 
 
(b) 
From Equations 2.53 and 2.63 
 
( )∫∫ −==−=∆ 2
1
2
1
12
T
T
P
T
T
v dTRcdTcuuu 
 
 From Appendix A.2 
 
)( 322 ETDTCTBTARcP ++++= − 
[ ]∫ −++++=∆ −2
1
 1322
T
T
dTETDTCTBTARu 
 
Integrating 
 
⎥⎦
⎤⎢⎣
⎡ −+−−−+−+−−=∆ )(
4
)11()(
3
)(
2
))(1( 41
4
2
12
3
1
3
2
2
1
2
212 TT
E
TT
DTTCTTBTTARu 
 
The following values were found in Table A.2.1 
 
0
1021.1
0
1045.1
470.3
4
3
=
×=
=
×=
=
−
E
D
C
B
A
 
 
Substituting these values and using 
 11
 
K 473.15K) 15.273200(
K 673.15K) 15.273400(
Kmol
J 314.8
2
1
=+=
=+=
⎥⎦
⎤⎢⎣
⎡
⋅=
T
T
R
 
 
provides 
 
⎥⎦
⎤⎢⎣
⎡−=⎟⎠
⎞⎜⎝
⎛⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−=∆
⎥⎦
⎤⎢⎣
⎡−=∆
kg
kJ 1.308
J 1000
kJ 1
kg 1
g 1000
O]H g[ 0148.18
O]H [mol 1
mol
J 5551ˆ
mol
J 5551
2
2u
u
 
 
The values in parts (a) and (b) agree very well. The answer from part (a) will serve as the basis 
for calculating the percent difference since steam table data should be more accurate. 
 
 ( ) %03.0%100
0.308
1.308308% =×−
−−−=Difference 
 
 
 
 12
2.10 
(a) 
Referring to the energy balance for closed systems where kinetic and potential energy are 
neglected, Equation 2.30 states 
 
WQU +=∆ 
 
(b) 
Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the 
process is isothermal 
 
0=∆U 
 
According to Equation 2.77 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛=⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
2
11
1
2 lnln
P
PRTn
P
PnRTW 
 
From the ideal gas law: 
 
1111 VPRTn = 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
2
11 ln P
PVPW 
 
Substitution of the values from the problem statement yields 
 
 
( )( )
[ ]J 940
bar 8
bar 5lnm 105.2Pa 108 335
−=
⎟⎠
⎞⎜⎝
⎛××= −
W
W
 
 
The energy balance is 
 
 [ ]J 940
J 0
=∴
+=
Q
WQ
 
 
(c) 
Since the process is adiabatic 
 
0=Q 
 
The energy balance reduces to 
 
WU =∆ 
 13
 
The system must do work on the surroundings to expand. Therefore, the work will be negative 
and 
 
12
0
0
2
1
TT
TcnU
U
v
T
T
<∴
<∆=∆
<∆
∫ 
T2 will be less than 30 ºC 
 
 
 14
2.11 
 
(a) 
(i). 
2
P
 [b
ar
]
v [m3/mol]
1
2
3
0.01 0.030.02
1
Path A
Path B
 
 
(ii). 
Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the 
process is isothermal 
 
 0=∆u 
 
Equation 2.48 states that enthalpy is a function of temperature only for an ideal gas. Therefore, 
 
 0=∆h 
 
Performing an energy balance and neglecting potential and kinetic energy produces 
 
0=+=∆ wqu 
 
For an isothermal, adiabatic process, Equation 2.77 states 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
2ln
P
PnRTW 
 
or 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛==
1
2ln
P
PRT
n
Ww 
 
Substituting the values from the problem statement gives 
 
 15
( ) ⎟⎠
⎞⎜⎝
⎛+⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅= bar 3
bar 1lnK )15.27388(
Kmol
J 8.314w 
⎥⎦
⎤⎢⎣
⎡−=
mol
J 3299w 
 
Using the energy balance above 
 
⎥⎦
⎤⎢⎣
⎡=−=
mol
J 3299wq 
 
(b) 
(i). See path on diagram in part (a) 
 
(ii). 
Since the overall process is isothermal and u and h are state functions 
 
 0=∆u 
 0=∆h 
 
The definition of work is 
 
∫−= dvPw E 
 
During the constant volume part of the process, no work is done. The work must be solved for 
the constant pressure step. Since it is constant pressure, the above equation simplifies to 
 
)( 12 vvPdvPw EE −−=−= ∫ 
 
The ideal gas law can be used to solve for 2v and 1v 
 
( )
( )
⎥⎦
⎤⎢⎣
⎡=×
+⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡ ⋅
==
⎥⎦
⎤⎢⎣
⎡=×
+⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡ ⋅
==
mol
m 010.0
Pa 103
K )15.27388(
K
molJ 314.8
mol
m 030.0
Pa 101
K )15.27388(
K
molJ 314.8
3
5
1
1
1
3
5
2
2
2
P
RTv
P
RTv
 
 
Substituting in these values and realizing that 1PPE = since the process is isobaric produces 
 
 16
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡−
⎥⎥⎦
⎤
⎢⎢⎣
⎡×−=
mol
m 010.0
mol
m 030.0Pa) 103(
33
5w 
⎥⎦
⎤⎢⎣
⎡−=
mol
J 6000w 
 
Performing an energy balance and neglecting potential and kinetic energy results in 
 
0=+=∆ wqu 
⎥⎦
⎤⎢⎣
⎡=−=∴
mol
J 6000wq 
 
 
 17
2.12 
First, perform an energy balance. No work is done, and the kinetic and potential energies can be 
neglected. The energy balance reduces to 
 
 QU =∆ 
 
We can use Equation 2.53 to get 
 
 ∫= 2
1
T
T
vdTcnQ 
 
which can be rewritten as 
 
 ∫= 2
1
T
T
PdTcnQ 
 
since the aluminum is a solid. Using the atomic mass of aluminum we find 
 
 mol 3.185
mol
kg 0.02698
kg5 =
⎥⎦
⎤⎢⎣
⎡=n 
 
Upon substitution of known values and heat capacity data from Table A.2.3, we get 
 
 ( ) ( )∫ −×+⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅=
K 15.323
K 15.294
3 1049.1486.2
Kmol
J 314.8mol 3.185 dTTQ 
 [ ]kJ 61.131=Q 
 
 18
2.13 
First, start with the energy balance. Potential and kinetic energy effects can be neglected. 
Therefore, the energy balance becomes 
 
 WQU +=∆ 
 
The value of the work will be used to obtain the final temperature. The definition of work 
(Equation 2.7) is 
 
 ∫−= 2
1
V
V
EdVPW 
 
Since the piston expands at constant pressure, the above relationship becomes 
 
 ( )12 VVPW E −−= 
 
From the steam tables 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 02641.0ˆ
3
1v (10 MPa, 400 ºC) 
 [ ]33111 m 07923.0kgm 02641.0kg) 3(ˆ =⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡== vmV 
 
Now 2V and 2v are found as follows 
 
 [ ]36312 m 4536.0Pa 100.2 J 748740m 07923.0 =×−−=−= EPWVV 
 [ ][ ] ⎥⎥⎦⎤⎢⎢⎣⎡=== kgm 1512.0kg 3 m 4536.0ˆ
33
2
2
2 m
Vv 
 
Since 2vˆ and 2P are known, state 2 is constrained. From the steam tables: 
 
 [ ]Cº 4002 =T ⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡
kg
m 0.1512 bar, 20
3
 
 
Now U∆ will be evaluated, which is necessary for calculating Q . From the steam tables: 
 
 19
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 2.2945ˆ2u ⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡
kg
m 0.1512 bar, 20
3
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 4.2832ˆ1u ( )Cº 400 bar, 001 
 
( ) [ ]( ) [ ]kJ 4.338
kg
kJ 4.2832
kg
kJ 2.2945kg 3ˆˆ 121 =⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡=−=∆ uumU 
 
Substituting the values of U∆ and W into the energy equation allows calculation of Q 
 
 WUQ −∆= 
 [ ]( ) [ ]J 1009.1J 748740 [J] 338400 6×=−−=Q 
 
 20
2.14 
In a reversible process, the system is never out of equilibrium by more than an infinitesimal 
amount. In this process the gas is initially at 2 bar, and it expands against a constant pressure of 
1 bar. Therefore, a finite mechanical driving force exists, and the process is irreversible. 
 
To solve for the final temperature of the system, the energy balance will be written. The piston-
cylinder assembly is well-insulated, so the process can be assumed adiabatic. Furthermore, 
potential and kinetic energy effects can be neglected. The energy balance simplifies to 
 
 WU =∆ 
 
Conservation of mass requires 
 
 21 nn = 
 Let 21 nnn == 
 
The above energy balance can be rewritten as 
 
 ∫∫ −= 2
1
2
1
V
V
E
T
T
v dVPdTcn 
 
Since vc and EP are constant: 
 
 ( ) ( )1212 VVPTTnc Ev −−=− 
 
2V and 1T can be rewritten using the ideal gas law 
 
 
2
2
2 P
nRTV = 
 
nR
VPT 111 = 
 
Substituting these expressions into the energy balance, realizing that 2PPE = , and simplifying 
the equation gives 
 
 
nR
VPP
T
2
7
2
5
112
2
⎟⎠
⎞⎜⎝
⎛ +
= 
 
Using the following values 
 
 21
 
[ ]
[ ]
[ ]
[ ]
⎥⎦
⎤⎢⎣
⎡
⋅
⋅=
=
=
=
=
Kmol
barL 08314.0
mol 0.1
L 10
bar 1
bar 2
1
2
1
R
n
V
P
P
 
 
results in 
 
 [ ]K 2062 =T 
 
To find the value for work, the energy balance can be used 
 
 ( )12 TTncUW v −=∆= 
 
Before the work can be calculated, 1T must be calculated 
 
 [ ]( ) [ ]( )
[ ]( )
[ ]K 241
Kmol
barL 08314.0mol 1
L 10bar 211
1 =
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅
⋅== nR
VPT 
 
Using the values shown above 
 
 [ ]J 727−=W 
 
 
 22
2.15 
The maximum work can be obtained through a reversible expansion of the gas in the piston. 
Refer to Section 2.3 for a discussion of reversible processes. The problem states that the piston 
assembly is well-insulated, so the heat transfer contribution to the energy balance can be 
neglected, in addition to potential and kinetic energy effects. The energy balance reduces to 
 
 WU =∆ 
 
In this problem, the process is a reversible, adiabatic expansion. For this type of process, 
Equation 2.90 states 
 
 [ ]11221
1 VPVP
k
W −−= 
 
From the problem statement (refer to problem 2.13), 
 
 
[ ]
[ ]
[ ]bar 1
L 10
bar 2
2
1
1
=
=
=
P
V
P
 
 
To calculate W, 2V must be found. For adiabatic, reversible processes, the following 
relationship (Equation 2.89) holds: 
 
 constPV k = 
 
where k is defined in the text. Therefore, 
 
 kkV
P
PV
1
1
2
1
2 ⎟⎟⎠
⎞
⎜⎜⎝
⎛= 
 
Noting that 
5
7==
v
P
c
ck and substituting the proper values provides 
 
 [ ]L 4.162 =V 
 
Now all of the needed values are available for calculating the work. 
 
 [ ] [ ]J 900barL 9 −=⋅−=W 
 
From the above energy balance, 
 
 [ ]J 900−=∆U 
 23
 
The change in internal energy can also be written according to Equation 2.53: 
 
 ∫=∆ 2
1
T
T
vdTcnU 
 
Since vc is constant, the integrated form of the above expression is 
 
 ( )122
5 TTRnU −⎟⎠
⎞⎜⎝
⎛=∆ 
 
Using the ideal gas law and knowledge of 1P and 1V , 
 
 [ ]K 6.2401 =T 
 
and 
 
 [ ]K 3.1972 =T 
 
The temperature is lower because more work is performed during the reversible expansion. 
Review the energy balance. As more work is performed, the cooler the gas will become. 
 
 24
2.16 
Since the vessel is insulated, the rate of heat transfer can be assumed to be negligible. 
Furthermore, no work is done on the system and potential and kinetic energy effects can be 
neglected. Therefore, the energy balance becomes 
 
 0ˆ =∆u 
 
or 
 12 ˆˆ uu = 
 
 
From the steam tables 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 2.2619ˆ1u (200 bar, 400 ºC) 
 ⎥⎦
⎤⎢⎣
⎡=∴
kg
kJ 2.2619ˆ2u 
 
The values of 2uˆ and 2P constrain the system. The temperature can be found from the steam 
tables using linear interpolation: 
 
 Cº 5.3272 =T [ ] ⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡=
kg
kJ 2.2619ˆ ,bar 100 2u 
 
Also at this state, 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 02012.0ˆ
3
2v 
 
Therefore, 
 
 [ ]( ) [ ]332 m 020.0kgm 02012.0kg 0.1 =⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡== mvVvessel 
 
 
 
 
 
 
 
 
 
 25
2.17 
 
Let the entire tank represent the system. Since no heat or work crosses the system boundaries, 
and potential and kinetic energies effects are neglected, the energy balance is 
 
 0=∆u 
 
Since the tank contains an ideal gas 
 
 
K 300
0
12
12
==
=−
TT
TT
 
 
The final pressure can be found using a combination of the ideal gas law and conservation of 
mass. 
 
 
22
2
11
1
VP
T
VP
T = 
 
We also know 
 
 12 2VV = 
 
Therefore, 
 
 bar 5
2
1
2 == PP 
 
 26
2.18 
(a) 
First, as always, simplify the energy balance. Potential and kinetic energy effects can be 
neglected. Therefore, the energy balance is 
 
 WQU +=∆ 
 
Since, this system contains water, we can the use the steam tables. Enough thermodynamic 
properties are known to constrain the initial state, but only one thermodynamic property is 
known for the final state: the pressure. Therefore, the pressure-volume relationship will be used 
to find the specific volume of the final state. Since the specific volume is equal to the molar 
volume multiplied by the molecular weight and the molecular weight is constant, the given 
expression can be written 
 
 constvP =5.1ˆ 
 
This equation can be used to solve for 2vˆ . 
 
 
5.1
1
5.1
1
2
1
2 ˆˆ ⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛= v
P
Pv 
 
Using 
 
 
[ ]
[ ]
[ ]
[ ]bar 100
kg
m 1.0
kg 10
m 1.0 ˆ
bar 20
2
33
1
1
=
⎥⎦
⎤⎢⎣
⎡==
=
P
v
P
 
 
gives 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 0.0342 ˆ
3
2v 
 
Now that the final state is constrained, the steam tables can be used to find the specificinternal 
energy and temperature. 
 
 
[ ]
⎥⎦
⎤⎢⎣
⎡=
=
kg
kJ 6.3094ˆ
K 7.524
2
2
u
T
 
 
To solve for the work, refer to the definition (Equation 2.7). 
 27
 
 ∫−= 2
1
V
V
EdVPW 
or 
 ∫−= 2
1
ˆ
ˆ
ˆˆ
v
v
E vdPw 
 
Since the process is reversible, the external pressure must never differ from the internal pressure 
by more than an infinitesimal amount. Therefore, an expression for the pressure must be 
developed. From the relationship in the problem statement, 
 
 constvPvP == 5.1115.1 ˆˆ 
 
Therefore, the expression for work becomes 
 
 ∫∫ −=−= 2
1
2
1
ˆ
ˆ
5.1
5.1
11
ˆ
ˆ
5.1
5.1
11 ˆ
ˆ
1ˆˆ
ˆ
ˆˆ
v
v
v
v
vd
v
vPvd
v
vPw 
 
Integration and substitution of proper values provides 
 
 ⎥⎦
⎤⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⋅=
kg
kJ 284
kg
mbar 840.2ˆ
3
w 
 [ ]( ) [ ]kJ 2840
kg
kJ 284kg 10 =⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡=∴W 
 
A graphical solution is given below: 
 
2
P
 [b
ar
]
v [m3/kg]
20
60
100
0.02 0.100.06
1
Work
 
 28
 
To solve for Q , U∆ must first be found, then the energy balance can be used. 
 
 ( ) [ ]( ) [ ]kJ 4918 
kg
kJ 8.2602
kg
kJ 6.3094kg 10ˆˆ 12 =⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡=−=∆ uumU 
 
Now Q can be found, 
 
 [ ] [ ] [ ]kJ 2078kJ 2840kJ 4918 =−=−∆= WUQ 
 
(b) 
Since the final state is the same as in Part (a), U∆ remains the same because it is a state function. 
The energy balance is also the same, but the calculation of work changes. The pressure from the 
weight of the large block and the piston must equal the final pressure of the system since 
mechanical equilibrium is reached. The calculation of work becomes: 
 
 ∫−= 2
1
ˆ
ˆ
ˆ
v
v
E vdmPW 
 
All of the values are known since they are the same as in Part (a), but the following relationship 
should be noted 
 
 2PPE = 
 
Substituting the appropriate values results in 
 
 [ ]kJ 6580=W 
 
Again we can represent this process graphically: 
 
2
P
 [b
ar
]
v [m3/kg]
20
60
100
0.02 0.100.06
1
Work
 
 29
 
Now Q can be solved. 
 
 [ ] [ ] [ ]kJ 1662kJ 6580kJ 4918 −=−=−∆= WUQ 
 
 
(c) 
This part asks us to design a process based on what we learned in Parts (a) and (b). Indeed, as is 
characteristic of design problems there are many possible alternative solutions. We first refer to 
the energy balance. The value of heat transfer will be zero when 
 
 WU =∆ 
 
For the same initial and final states as in Parts (a) and (b), 
 
 [ ]kJ 4918=∆= UW 
 
There are many processed we can construct that give this value of work. We show two 
alternatives which we could use: 
 
Design 1: 
If the answers to Part (a) and Part (b) are referred to, one can see that two steps can be used: a 
reversible compression followed by an irreversible compression. Let the subscript “i" represent 
the intermediate state where the process switches from a reversible process to an irreversible 
process. The equation for the work then becomes 
 
[ ]J 4918000ˆ
ˆ
1ˆ)ˆˆ(
ˆ
ˆ
5.1
5.1
1122
1
=⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−−= ∫i
v
v
i vd
v
vPvvPmW 
 
Substituting in known values (be sure to use consistent units) allows calculation of ivˆ : 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 0781.0ˆ
3
iv 
 
The pressure can be calculated for this state using the expression from part (a) and substituting 
the necessary values. 
 
 
[ ]bar 0.29
5.1
1
1
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
i
i
i
P
v
vPP 
 
 30
Now that both iP and ivˆ are known, the process can be plotted on a P-v graph, as follows: 
 
 
2
P
 [b
ar
]
v [m3/kg]
20
60
100
0.02 0.100.06
1
i
Work
 
 
 
Design 2: 
In an alternative design, we can use two irreversible processes. First we drop an intermediate 
weight on the piston to compress it to an intermediate state. This step is followed by a step 
similar to Part (b) where we drop the remaining mass to lead to 100 bar external pressure. In this 
case, we again must find the intermediate state. Writing the equation for work: 
 
 [ ] [ ]J 4918000)ˆˆ()ˆˆ( 221 =−−−−= iii vvPvvPmW 
 
However, we again have the relationship: 
 
 
5.1
1
1 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
i
i v
vPP 
 
Substitution gives one equation with one unknown vi: 
 
[ ]J 4918000)ˆˆ()ˆˆ( 221
5.1
1
1 =⎥⎥⎦
⎤
⎢⎢⎣
⎡ −−−⎟⎟⎠
⎞
⎜⎜⎝
⎛−= ii
i
vvPvv
v
vPmW 
 
There are two possible values vi to the above equation. 
 
Solution A: 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
m 043.0ˆ
3
iv 
 
 31
which gives 
 
 [ ]bar 8.70=iP 
 
This solution is graphically shown below: 
 
2
P
 [b
ar
]
v [m3/kg]
20
60
100
0.02 0.100.06
1
i
Work
 
 
Solution B 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
m 0762.0ˆ
3
iv 
 
which gives 
 
 [ ]bar 0.30=iP 
 
This solution is graphically shown below: 
 
2
P
 [b
ar
]
v [m3/kg]
20
60
100
0.02 0.100.06
1
i
Work
 
 32
2.19 
 
(a) 
Force balance to find k: 
 
Piston
Fspring=kx Fatm=PatmA
Fgas=PgasA
Fmass=mg
 
 
 Pgas = Patm + mgA − kxA 
since ∆V=Ax 
 Pgas = Patm + mgA + k∆VA2 
 
since ∆V is negative. Now solve: 
 
 ( )( ) ( )( )22
3
2
2
55
m 1.0
m 02.0
m 1.0
m/s 81.9kg 2040Pa 101Pa 102 k−+×=× 
 ⎥⎦
⎤⎢⎣
⎡×=∴
m
N1001.5 4k 
 
Work can be found graphically (see P-V plot) or analytically as follows: Substituting the 
expression in the force balance above: 
 
WA = Patm + mgA + k∆VA2
⎛ ⎝ ⎞ ⎠ ∫ dV 
( )∫ ∫
∆
∆
∆∆+⎟⎠
⎞⎜⎝
⎛ +=
f
i
f
i
V
V
V
V
atmA Vd
A
VkdV
A
mgPW 2 
( )( ) [ ]( ) ( ) ⎥⎥⎦
⎤
⎢⎢⎣
⎡ −×+−×=
2
0m02.0
m 1.0
N/m 1001.5m 05.003.0Pa1000.3
223
22
4
25
AW 
 33
[ ]J 105 3×−=AW 
 
 V [m3]
1
2
3
0.01 0.03 0.05
Patm
mg
A
k∆V
A2
-W 10squares 0.5kJsquare = 5kJ
Work
P
 [b
ar
]
2
1
=
 
 
(b) 
You need to find how far the spring extends in the intermediate (int) position. Assume 
PVn=const (other assumptions are o.k, such as an isothermal process, and will change the answer 
slightly). Since you know P and V for each state in Part (a), you can calculate n. 
 
 PiPf =
Vi
V f
⎛ 
⎝ ⎞ ⎠ 
n
 or 10
5 Pa
2×105 Pa =
0.03 m3
0.05 m3
⎛ ⎝ ⎞ ⎠ 
n
 ⇒ n = 1.35 
 
Now using the force balance 
 
Pgas,int = Patm + mgA + k∆VA2 
 
with the above equation yields: 
 
 Pint = Pi ViV int( )n = Patm + mgA + k Vint −Vi( )A2 
 
This last equality represents 1 equation. and 1 unknown (we know k), which gives 
 
 3int m 0385.0=V 
Work can be found graphically (see P-V plot) or analytically using: 
 34
 
WB = Pgas
Vi
V int∫ dV + Pgas
Vint
V f
∫ dV 
 
expanding as in Part (a) 
 
WB = 2 ×105 Nm2
⎛ ⎝ ⎞ ⎠ 
Vi
V int∫ dV + 5 ×106 N
m5
∆V
∆Vi
∆Vint∫ d ∆V( )+ 3 ×105 Nm2⎛ ⎝ ⎞ ⎠ Vint
Vf
∫ dV + 5 ×106 N
m 5
∆V
∆Vint
∆Vif
∫ d ∆V( )
 
Therefore, 
 
 [ ]kJ85.3−=BW 
 
just like we got graphically. 
 
V [m3]
1
2
3
0.01 0.03 0.05
Patm
mg
A
P
 [b
ar
]
2
1
Work
mg
A
 
 
(c) 
The least amount of work is required by adding differential amounts of mass to the piston. This 
is a reversible compression. For our assumption that PVn = const, we have the following 
expression: 
 
 ∫∫ ==
f
i
f
i
V
V
V
V
gasrevC dV
V
constdVPW 35.1, 
 
Calculate the constant from the initial state 
 35
 
 ( )( ) 335.13 5 1075.1m05.0Pa101. ×=×=const 
 
Therefore, 
 
WC,rev = 1.75 ×103
.05
.03
∫ dV
V1.35
 = − 1.75×1030.35 V −0.35 .05
.03
 [ ]J2800−=V [m3]
1
2
3
0.01 0.03 0.05
Patm
P
 [b
ar
]
2
1
Workrev
 
 
 36
2.20 
Before this problem is solved, a few words must be said about the notation used. The system 
was initially broken up into two parts: the constant volume container and the constant pressure 
piston-cylinder assembly. The subscript “1” refers to the constant volume container, “2” refers 
the piston-cylinder assembly. “i" denotes the initial state before the valve is opened, and “f” 
denotes the final state. 
 
To begin the solution, the mass of water present in each part of the system will be calculated. 
The mass will be conserved during the expansion process. Since the water in the rigid tank is 
saturated and is in equilibrium with the constant temperature surroundings (200 ºC), the water is 
constrained to a specific state. From the steam tables, 
 
 
[ ] kPa 8.1553
kg
kJ 3.2595ˆ
kg
kJ 64.850ˆ
kg
kJ 12736.0ˆ
kg
kJ 001156.0ˆ
,1
,1
,1
,1
=
⎥⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡=
sat
v
i
l
i
v
i
l
i
P
u
u
v
v
 (Sat. water at 200 ºC) 
 
Knowledge of the quality of the water and the overall volume of the rigid container can be used 
to calculate the mass present in the container. 
 
 ( ) ( )vil i vmvmV ,11,111 ˆ95.0ˆ05.0 += 
 
Using the values from the steam table and [ ]31 m 5.0=V provides 
 
 [ ]kg 13.41 =m 
 
Using the water quality specification, 
 
 
[ ]
[ ]kg 207.005.0
kg 92.395.0
11
11
==
==
mm
mm
l
v
 
 
For the piston-cylinder assembly, both P and T are known. From the steam tables 
 
 37
 
⎥⎦
⎤⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
kJ 9.2638ˆ
kg
m 35202.0ˆ
,2
3
,2
i
i
u
v
 (600 kPa, 200 ºC) 
 
Enough information is available to calculate the mass of water in the piston assembly. 
 
 [ ]kg 284.0
ˆ2
2
2 == v
Vm 
 
Now that the initial state has been characterized, the final state of the system must be determined. 
It helps to consider what physically happens when the valve is opened. The initial pressure of 
the rigid tank is 1553.8 kPa. When the valve is opened, the water will rush out of the rigid tank 
and into the cylinder until equilibrium is reached. Since the pressure of the surroundings is 
constant at 600 kPa and the surroundings represent a large temperature bath at 200 ºC, the final 
temperature and pressure of the entire system will match the surroundings’. In other words, 
 
 ⎥⎦
⎤⎢⎣
⎡==
kg
kJ 9.2638ˆˆ ,2 if uu (600 kPa, 200 ºC) 
 
Thus, the change in internal energy is given by 
 
 l i
l
i
v
i
v
iif
l
i
v
i umumumummmU ,1,1,1,1,22,1,12 ˆˆˆˆ)( −−−++=∆ 
 
Substituting the appropriate values reveals 
 
 [ ]kJ 0.541=∆U 
 
To calculate the work, we realize the gas is expanding against a constant pressure of 600 kPa 
(weight of the piston was assumed negligible). From Equation 2.7, 
 
 )( ifE
V
V
E VVPdVPW
f
i
−−=−= ∫ 
 
where 
 
 
[ ] [ ][ ] [ ] [ ]333
3
,2,1,12
m 6.0m 5.0m 0.1
m 55.1ˆ)(
Pa 600000
=+=
=++=
=
i
i
l
i
v
if
E
V
vmmmV
P
 
 38
Note: iv ,2ˆ was used to calculate fV because the temperature and pressure are the same 
for the final state of the entire system and the initial state of the piston-cylinder assembly. 
 
The value of W can now be evaluated. 
 
 [ ]kJ 570−=W 
 
The energy balance is used to obtain Q. 
 
 [ ] [ ]( ) [ ]kJ 1111kJ 570kJ 0.541 =−−=−∆= WUQ 
 
 
 39
2.21 
A sketch of the process follows: 
10 cm50 cm
A B
p1,B = 20 bar
T1,B = 250 oC
p1,A = 10 bar
T1,A = 700 oC
H2O H2O
system
boundary
Process
A B
p2 
T2
H2O
system
boundary p2 T2
H2O
 
The initial states are constrained. Using the steam tables, we get the following: 
 
 State 1,A State 1,B 
p 10 [bar] 20 [bar] 
T 700 [oC] 250 [oC] 
v 0.44779 [m3/kg] 0.11144 [m3/kg] 
u 3475.35 [kJ/kg] 2679.58 [kJ/kg] 
V 0.01 m3 0.05 m3 
 
m = V
v
 0.11 [kg] 0.090 [kg] 
 
All the properties in the final state are equal. We need two properties to constrain the system: 
We can find the specific volume since we know the total volume and the mass: 
 
 [ ][ ] ⎥⎥⎦⎤⎢⎢⎣⎡==++= kgm 30.0kg 0.20 m 06.0
33
,1,1
,1,1
2
BA
BA
mm
VV
v 
 
We can also find the internal energy of state 2. Since the tank is well insulated, Q=0. Since it is 
rigid, W=0. An energy balance gives: 
 
 ∆U = Q −W = 0 
Thus, 
 
 U2 =U1 = m1,Au1, A + m1,Bu1,B 
or 
 
 ⎥⎦
⎤⎢⎣
⎡=+
+==
kg
kJ 3121
,1,1
,1,1,1,1
2
2
2
BA
BBAA
mm
umum
m
Uu 
 
We have constrained the system with u2 and v2, and can find the other properties from the steam 
Tables. Very close to 
 
 T2 = 500 [oC] and P2 = 1200 [kPa] 
 
Thus, 
 40
 [ ]( ) [ ]m 267.0kg 0.09
kg
m 30.0
3
,2,2,2 =⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡== AAA mvV and 
 
 ( ) [ ]m 167.01.0267.0 =−=∆x 
 
 
 41
2.22 
We start by defining the system as a bubble of vapor rising through the can. We assume the 
initial temperature of the soda is 5 oC. Soda is usually consumed cold; did you use a reasonable 
estimate for T1? A schematic of the process gives: 
 
State 1: 
T1 = 278 K 
P1 = 3 bar
State 2: 
T2= ? 
P2= 1.01 bar
 
 
where the initial state is labeled state 1, and the final state is labeled state 2. To find the final 
temperature, we perform an energy balance on the system, where the mass of the system (CO2 in 
the bubble) remains constant. Assuming the process is adiabatic and potential and kinetic energy 
effects are negligible, the energy balance is 
 
 wu =∆ 
 
Expressions for work and internal energy can be substituted to provide 
 
 ( ) ( )1212 vvPdvPTTc EEv −−=−=− ∫ 
 
where cv = cP – R. Since CO2 is assumed an ideal gas, the expression can be rewritten as 
 
 ( ) ⎥⎦
⎤⎢⎣
⎡ −−=⎥⎦
⎤⎢⎣
⎡ −−=−
1
12
2
1
1
2
2
12 P
TPTR
P
T
P
TRPTTc Ev 
 
where the equation was simplified since the final pressure, P2, is equal to the external pressure, 
PE. Simplifying, we get: 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
1
2
12 11 Pc
RPT
c
RT
vv
 
or 
 
 K 2371
1
2
12 =⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
P
v
v c
c
Pc
RPTT 
 
 42
2.23 
The required amount of work is calculated as follows: 
 
 VPW ∆−= 
 
The initial volume is zero, and the final volume is calculated as follows: 
 
 ( ) 3333 m0436.0ft 54.1ft 5.0
3
4
3
4 ==== ππrV 
 
Assuming that the pressure is 1 atm, we calculate that 
 
 ( )( ) J4417m 0m 0436.0Pa1001325.1 33 5 =−×=W 
 
This doesn’t account for all of the work because work is required to stretch the rubber that the 
balloon is made of. 
 
 
 
 
 43
2.24 
 
(a) 
Since the water is at its critical point, the system is constrained to a specific temperature, 
pressure, and molar volume. From Appendix B.1 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 003155.0ˆ
3
cv 
 
Therefore, 
 
 [ ] [ ]kg 17.3
kg
m 003155.0
m 01.0
ˆ 3
3
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡== cv
Vm 
 
(b) 
The quality of the water is defined as the percentage of the water that is vapor. The total volume 
of the vessel can be found using specific volumes as follows 
 
( )[ ] ( ) vlvvll vxmvmxvmvmV ˆˆ1ˆˆ +−=+= 
 
where x is the quality of the water. To solve for the quality, realize that starting with saturated 
water at a pressure of 1 bar constrains the water. From the steam tables, 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 001043.0ˆ
kg
m 6940.1ˆ
3
3
l
v
v
v
 (sat. H2O at P = 1 bar) 
 
Now the quality can be found 
 
 00125.0=x 
 
Thus, the quality of the water is 0.125%. 
 
(c) 
To determine the required heatinput, perform an energy balance. Potential and kinetic energy 
effects can be neglected, and no work is done. Therefore, 
 
 QU =∆ 
 
 44
where 
 
 ( )[ ] ( )[ ]vl uxmumxumU 112 ˆˆ1ˆ +−−=∆ 
 
From the steam tables 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 58.2029ˆ2u (H2O at its critical point) 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 1.2506ˆ
kg
m 33.417ˆ
3
1
3
1
v
l
u
u
 (sat. H2O at P=1 bar) 
 
Evaluation of the expression reveals 
 
 [ ] [ ]J 1010.5kJ 6.5102 6×==∆U 
 
 
 
 45
2.25 
 
(a) 
Consider the air in ChE Hall to be the system. The system is constant volume, and potential and 
kinetic energy effects can be neglected. Furthermore, disregard the work. The energy balance is 
 
 q
dt
du &= 
 
since the temperature of the system changes over time. Using the given expression for heat 
transfer and the definition of dU , the expression becomes 
 
 ( )surrv TThdt
dTc −−= 
 
We used a negative sign since heat transfer occurs from the system to the surroundings. If vc is 
assumed constant, integration provides 
 
 ( ) ChtTTc surrv +−=−ln 
 
where C is the integration constant. Therefore, 
 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+= tc
h
surr
veCTT 1 
 
where C1 is a constant. Examining this equation reveals that the temperature is an exponential 
function of time. Since the temperature is decreasing, we know that the plot of temperature vs. 
time shows exponential decay. 
 
time
Te
m
pe
ra
tu
re
T0
Tsurr
 
 
 46
(b) 
Let time equal zero at 6 PM, when the steam is shut off. At 6 PM, the temperature of the hall is 
22 ºC. Therefore, 
 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+= tc
h
surr
veCTT 1 
 )0(1Cº 2 Cº 22 eC+= 
 Cº 201 =∴C 
 
After 10 PM, ( hr 4=t ), the temperature is 12 ºC. 
 
 ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+= hr) 4(Cº 20Cº 2 Cº 21 vc
h
e 
 1-hr 173.0=∴
vc
h 
 
At 6 AM, hr 12=t . Substitution of this value into the expression for temperature results in 
 
 Cº 5.4=T 
 
 
 47
2.26 
The gas leaving the tank does flow work as it exits the valve. This work decreases the internal 
energy of the gas – lowering the temperature. During this process, water from the atmosphere 
will become supersaturated and condense. When the temperature drops below the freezing point 
of water, the water forms a solid. 
 
Attractive interactions between the compressed gas molecules can also contribute to this 
phenomena, i.e., it takes energy to pull the molecules apart as they escape; we will learn more of 
these interactions in Chapter 4. 
 
 
 
 
 48
2.27 
Mass balance 
 
 inoutin mmmdt
dm &&& =−= 
 
Separating variables and integrating: 
 
 ∫∫ = t in
m
m
dtmdm
0
2
1
& 
 
or 
 ∫=− t indtmmm
0
12 & 
 
Energy balance 
Since the potential and kinetic energy effects can be neglected, the open system, unsteady state 
energy balance is 
 
 ∑ ∑ ++−=⎟⎠⎞⎜⎝⎛ out sin ininoutoutsys WQhmhmdt
dU &&&& 
 
The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream 
and not outlet stream. Therefore, the energy balance simplifies to 
 
 inin
sys
hm
dt
dU & =⎟⎠
⎞⎜⎝
⎛ 
 
The following math is performed 
 
 
( ) in
t
inin
t
inin
U
U
hmmumumUU
dtmhdthmdU
ˆˆˆ 12112212
00
2
1
−=−=−
== ∫∫∫ && 
 
where the results of the mass balance were used. Both 2m and 1m can be calculated by dividing 
the tank volume by the specific volume 
 
 49
 
1
1
2
2
ˆ
ˆ
v
Vm
v
Vm
=
=
 
 
Substitution of these relationships and simplification results in 
 
 
( ) ( )
0
ˆ
ˆ
ˆ
ˆ
1
1
2
2 =−−−
v
hu
v
hu inin 
 
From the steam tables: 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡=
kg
m 19444.0ˆ
kg
kJ 6.2583ˆ
3
1
1
v
u
 (sat. H2O vapor at 1 MPa) 
 
⎥⎦
⎤⎢⎣
⎡=
kg
kJ 2.3177iˆnh (6 MPa, 400 ºC) 
 
There are still two unknowns for this one equation, but the specific volume and internal energy 
are coupled to each other. To solve this problem, guess a temperature and then find the 
corresponding volume and internal energy values in the steam tables at 6 MPa. The correct 
temperature is the one where the above relationship holds. 
 
Cº 600=T : Expression = 4427.6 
Cº 500=T : Expression = 1375.9 
Cº 450=T : Expression = -558.6 
 
Interpolation between 500 ºC and 450 ºC reveals that the final temperature is 
 
 Cº 4.4642 =T 
 
 50
2.28 
We can pick room temperature to be 295 K 
 
 Tin = T1 = 295 K[ ] 
 
Mass balance 
 
 inoutin nnndt
dn &&& =−= 
 
Separating variables and integrating: 
 
 ∫∫ = t in
n
n
dtndn
0
2
1
& 
 
or 
 ∫=− t indtnnn
0
12 & 
 
Energy balance 
Neglecting ke and pe, he unsteady energy balance, written in molar units is written as: 
 
WQhnhn
dt
dU
outoutinin
sys
&&&& −+−=⎟⎠
⎞⎜⎝
⎛ 
The terms associated with flow out, heat and work are zero. 
 
 inin
sys
hn
dt
dU &=⎟⎠
⎞⎜⎝
⎛ 
 
Integrating both sides with respect to time from the initial state where the pressure is 10 bar to 
the final state when the tank is at a pressure of 50 bar gives: 
 
 dtnhdthndU
t
inin
t
inin
U
U
∫∫∫ ==
00
2
1
&& 
since the enthalpy of the inlet stream remains constant throughout the process. Integrating and 
using the mass balance above: 
 
 ( ) inhnnunun 121122 −=− 
 51
 
 
Now we do some math: 
 
 ( ) inhnnunun 121122 −=− 
 
 ( ) ( )inin hunhun −=− 1122 
 
By the definition of h 
 
 11 RTuRTuvPuh inininininin +=+=+= 
 
so 
 ( ) ( ) 1111112122 TRnuunRTnuun −−=−− 
 
 ( ) 1112122 RTnRTnTTcn v −=−− 
 
Since RRcc Pv 2
3=−= 
 
 
n2
3
2
T2 − T1( )− n2T1 = −n1T1 
 
or 
111222 253 TnTnTn −=− 
 
dividing by n1: 
 
 11
1
2
2
1
2 253 TT
n
nT
n
n −=− 
 
Using the ideal gas law: 
 
 
21
12
1
2
TP
TP
n
n = 
 
so 
 11
21
12
2
21
12 253 TT
TP
TPT
TP
TP −=⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡
 
 
or 
 52
 [K] 434
23
5
1
2
1
12
2 =
⎥⎦
⎤⎢⎣
⎡ +⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
=
P
P
P
TP
T 
 
(b) Closed system 
 
 ∆u = q − w = q 
 
( ) ( ) ⎥⎦
⎤⎢⎣
⎡=−=−=∆
kg
kJ 9.28
2
5ˆ 1212 TTMW
RTT
MW
cu v 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 9.28q 
 
(c) P2T2 = P3T3 
 [ ]bar 34
3
22
3 == T
TPP 
 
 53
2.29 
Mass balance 
 
 inoutin nnndt
dn &&& =−= 
 
Separating variables and integrating: 
 
 ∫∫ =
=
t
in
n
n
dtndn
00
2
1
& 
 
or 
 ∫= t indtnn
0
2 & 
 
Energy balance 
Neglecting ke and pe, the unsteady energy balance, in molar units, is written as: 
 
WQhnhn
dt
dU
outoutinin
sys
&&&& ++−=⎟⎠
⎞⎜⎝
⎛ 
 
The terms associated with flow out and heat are zero. 
 
 Whn
dt
dU
inin
sys
&& +=⎟⎠
⎞⎜⎝
⎛ 
 
Integrating both sides with respect to time from the empty initial state to the final state gives: 
 
 WnhWdtnhdtWdthndU in
t
inin
tt
inin
U
U
+=+=+= ∫∫∫∫ 2
000
2
1
&&& 
 
since the enthalpy of the inlet stream remains constant throughout the process. The work is 
given by: 
 
22122 )( vPnvvPnW extext −=−−= 
 
 
 54
 [ ]2222 vPhnun extin −= 
 
Rearranging, 
 
 222 vPvPuvPhu extinininextin −+=−= 
 
22 vPvPuu extininin −=− 
 
 ( )
2
2
2 P
TPRRTTTc extininv −=− 
so 
 
 ( ) [ ]K 333
2
2 =
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
+= in
ext
v
v T
R
P
Pc
RcT 
 
 55
2.30 
valve maintains 
pressure in system 
constant
T1 = 200 oC
x1 = 0.4
V = 0.01 m3
v
lMass balance 
 
 outoutin mmmdt
dm &&& −=−= 
 
Separating variables and integrating: 
 
 ∫∫ −= t out
m
m
dtmdm
0
2
1
& 
 
or 
 ∫−=− t outdtmmm
0
12 & 
 
Energy balance 
 
Qhm
dt
dU
outout
sys
&& +−=⎟⎠
⎞⎜⎝
⎛ ˆ 
 
Integrating 
 
 [ ]∫ ∫∫∫ +−=+−= t tt outoutoutoutum
um
dtQdtmhdtQhmdU
0 00
ˆ
ˆ
ˆˆ
22
11
&&&& 
 
Substituting in the mass balance and solving for Q 
 
( ) outhmmumumQ ˆˆˆ 121122 −−−= 
 
 56
 
We can look up property data for state 1 and state 2 from the steam tables: 
 
⎥⎦
⎤⎢⎣
⎡=×+×=+−=
kg
m 0.0511274.04.0001.6.0ˆˆ)1(ˆ
3
1 gf vxvxv 
⎥⎦
⎤⎢⎣
⎡=
kg
m 1274.0ˆ
3
2v 
 
So the mass in each state is: 
 
 [ ] [ ]kg 196.0
kg
m 0.051
m 01.0
ˆ 3
3
1
1
1 =
⎥⎦
⎤⎢⎣
⎡== v
Vm 
 
 [ ] [ ]kg 0785.0
kg
m 0.1274
m 01.0
ˆ 3
3
2
2
2 =
⎥⎦
⎤⎢⎣
⎡== v
Vm 
 
 [ ]kg 1175.012 −=− mm 
 
And for energy and enthalpy 
 
 ⎥⎦
⎤⎢⎣
⎡=×+×=+−=
kg
kJ 54915.25974.064.8506.0ˆˆ)1(ˆ1 gf uxuxu 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 3.2595ˆ2u 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 2.2793oˆuth 
 
Solving for heat, we get 
 
( ) [ ]kJ 228ˆˆˆ 121122 =−−−= outhmmumumQ 
 
 
 57
2.31 
Consider the tank as the system. 
 
Mass balance 
 
 inoutin mmmdt
dm &&& =−= 
 
Separating variables and integrating: 
 
 ∫∫ = t in
m
m
dtmdm
0
2
1
& 
 
or 
 ∫=− t indtmmm
0
12 & 
 
Energy balance 
Since the potential and kinetic energy effects can be neglected, the open system, unsteady state 
energy balance is 
 
 ∑ ∑ ++−=⎟⎠⎞⎜⎝⎛ out sin ininoutoutsys WQhmhmdt
dU &&&& 
 
The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream 
and not outlet stream. Therefore, the energy balance simplifies to 
 
 inin
sys
hm
dt
dU & =⎟⎠
⎞⎜⎝
⎛ 
 
The following math is performed 
 
 
in
t
inin
t
inin
U
U
hmumU
dtmhdthmdU
ˆˆ 2222
000
2
1
==
== ∫∫∫
=
&&
 
 
where the results of the mass balance were used. Thus, 
 
inhu ˆˆ2 = 
 
From the steam tables 
 58
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 5.3632ˆ2u (9 MPa, 800 ºC) 
 
so 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 5.3632iˆnh 
 
We can use the value of inh and the fact that the steam in the pipe is at 9 MPa to find the 
temperature. 
 
 Cº 600=inT 
 
 
 59
2.32 
 
(a) 
First, the energy balance must be developed. Since the problem asks how much energy is stored 
in the battery after 10 hours of operation, the process is not steady-state. Let the battery be the 
system. Potential and kinetic energy effects can be neglected. Furthermore, heating of the 
battery as it is charged can be ignored. The energy balance is 
 
 s
sys
WQ
dt
dU && +=⎟⎠
⎞⎜⎝
⎛ 
 
No shaft work is performed, but electrical is supplied to the battery, which must be accounted for 
in sW& . The value of Q& is given explicitly in the problem statement. Both of these values remain 
constant over time, so integration provides 
 
 ( )tWQU s&& +=∆ 
 
From the problem statement 
 
 
[ ]
[ ]
[ ]s 36000
kW 1
kW 5
=
−=
=
t
Q
Ws
&
&
 
 
Substituting these values allows the calculation of the amount of energy stored: 
 
 [ ] [ ]MJ 144kJ 000,144 ==∆U 
 
(b) 
To calculate the velocity of the falling water, an energy balance must be developed with the 
water passing through the electricity generator (probably a turbine) as the system, where the 
water enters with a velocity 1V
r
 and leaves with a negligible velocity, which will be approximated 
as 0. Assume that potential energy changes can be neglected. Furthermore, assume that the 
temperature of the water does not change in the process, so the change in internal energy is zero. 
Also, view the process as adiabatic. The energy balance reduces to 
 
 riverK WE && =∆ 
 
where riverW& is the power of the flowing water. The actual power being provided by the stream 
can be calculated using the efficiency information. Let η represent the efficiency. 
 
 60
 [ ] [ ]kW 10
5.0
kW 5 ===∴
=
η
η
s
river
river
s
WW
W
W
&&
&
&
 
 
The value of riverW& should be negative since the water is supplying work that is stored electrical 
energy. Therefore, the energy balance becomes 
 
 [ ]W 10000−=∆ KE& 
 
This expression can be rewritten as 
 
 ( ) [ ]W 10000
2
1 2
1
2
2 −=−VVm
rr
& 
 
From the problem statement and the assumptions made, 
 
 
⎥⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡=
s
m 0
s
kg 200
2V
m
r
&
 
 
Therefore, 
 
 ⎥⎦
⎤⎢⎣
⎡=
s
m 101V
r
 
 
There are a number of reasons for the low conversion efficiency. A possible potential energy 
loss inherent in the design of the energy conversion apparatus decreases the efficiency. Heat is 
lost to the surroundings during conversion. Some of the energy is also lost due to friction (drag) 
effects. 
 
 61
2.33 
Considering the turbine to be the system, rearrangement of the steady-state, open system energy 
balance provides 
 
∑ ∑ +=++−++
out
s
in
inpkinoutpkout WQeehneehn &&&& )()( 
 
Performing a mass balance reveals 
 
 21 nnnn outin &&&& === 
 
Assuming the rate of heat transfer and potential energy effects are negligible and realizing that 
there is one inlet and one outlet allows the simplification of the above equation to 
 
 ( ) ( )[ ]1,2,122 KKs eehhnW −+−= && 
 ( )12 hh − can be rewritten using Equations 2.58 and Appendix A.2 
 
( ) ( )∫∫ ++++==− −2
1
322
12
T
T
p dTETDTCTBTARdTchh 
 
Since the quantity ( )1,2, kk ee − is multiplied by n& , it is rewritten as follows for dimensional 
homogeneity 
 
( ) ( )21221,2, )(21 VVMWee airkk rr −=− 
 
To solve for n& , the ideal gas law is used 
 
2
22
2
21222
RT
VPn
RTnVP
&
&
&&
=
=
 
 
To solve for the volumetric flow rate, the fluid velocity must be multiplied by the cross-sectional 
area 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
4
2
2
2
2
VDV
r
& π 
 
The energy balance is now 
 
 62
( ) ( )
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−+⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
++++⎟⎟⎠
⎞
⎜⎜⎝
⎛= ∫ − 2122322222
2
2 )(
2
1
4
2
1
VVMWdTETDTCTBTARVD
RT
PW air
T
T
s
rrr& π 
Substituting values from Table A.2.1 and the problem statement results in 
 
 [MW] -4.84[W] 1084.4 6 =×−=W 
 
 63
2.34 
First, a sketch of the process is useful: 
 
q
30 bar
100 oC
20 bar
150 oC
 
 
To find the heat in we will apply the 1st law. Assuming steady state, the open system energy 
balance with one stream in and one stream out can be written: 
 
 ( ) Qhhn && +−= 210 
 
which upon rearranging is: 
 
 12 hhn
Q −=&
&
 
 
Thus this problem reduces to finding the change in the thermodynamic property, enthalpy from 
the inlet to the outlet. We know 2 intensive properties at both the inlet and outlet so the values 
for the other properties (like enthalpy!) are already constrained. From Table A.2.1, we have an 
expression for the ideal gas heat capacity: 
 
263 10392.410394.14424.1 TT
R
c p −− ×−×+= 
 
with T in (K). Since this expression is limited to ideal gases any change in temperature must be 
under ideal conditions. From the definition of heat capacity: 
 
 ( )∫ −− ×−×+=−= 2
1
263
12 10392.410394.14424.1
T
T
dTTThh
n
Q
&
&
 
 
By integrating and substituting the temperatures, we obtain: 
 
 ⎥⎦
⎤⎢⎣
⎡=
mol
J5590
n
Q
&
&
 
 
 64
2.35 
A schematic of the process follows: 
 
T1 = 350 oC
v1 = 600 cm3/mol P2= 1 atm
Ý W s
T1 = 308 oC
 
 
To solve for nWs && /we need a first law balance. With negligible eK and eP, the 1st law for a 
steady state process becomes: 
 
 ( ) sWQhhn &&& ++−= 210 
 
If heat transfer is negligible, 
 
 h
n
Ws ∆=&
& 
 
We can calculate the change in enthalpy from ideal gas heat capacity data provided in the 
Appendix. 
 
[ ]∫∫ −− ×−×+==∆= 2
1
2
1
263 10824.810785.28213.1
T
T
T
T
p
s dTTTRdTch
n
W
&
&
 
 
Integrate and evaluate: 
 
 ⎥⎦
⎤⎢⎣
⎡−=
mol
J 5358
n
Ws
&
&
 
 
 65
2.36 
 
(a) 
First start with the energy balance. Nothing is mentioned about shaft work, so the term can be 
eliminated from the energy balance. The potential and kinetic energy effects can also be 
neglected. Since there is one inlet and one outlet, the energy balance reduces to 
 
 1122 hnhnQ &&& −= 
 
A mass balance shows 
 
 12 nn && = 
 
so the energy balance reduces to 
 
 ( )121 hhnQ −= && 
 
Using the expressions from Appendix A.1, the energy balance becomes 
 
 ( )∫ ++++= −2
1
322
1
T
T
dTETDTCTBTARnQ && 
 
Using 
 
 
0
10031.0
0
10557.0
376.3
5
3
=
×−=
=
×=
=
−
E
D
C
B
A
 
 ⎥⎦
⎤⎢⎣
⎡
⋅= Kmol
J 314.8R 
 ⎥⎦
⎤⎢⎣
⎡=
s
mol 201n& 
 
[ ]
[ ]K 15.773
K 15.373
2
1
=
=
T
T
 
 
gives 
 
 [ ] [ ]kW 1.245W 245063 ==Q& 
 
 
 66
(b) 
To answer this question, think about the structure of n-hexane and carbon monoxide. N-hexane 
is composed of 20 atoms, but carbon monoxide has two. One would expect the heat capacity to 
be greater for n-hexane since there are more modes for molecular kinetic energy (translational, 
kinetic, and vibrational). Because the heat capacity is greater and the rate of heat transfer is the 
same, the final temperature will be less. 
 
 67
2.37 
First start with the energy balance around the nozzle. Assume that heat transfer and potential 
energy effects are negligible. The shaft work term is also zero. Therefore, the energy balance 
reduces to 
 
 0)()( 1122 =+−+ KK ehnehn && 
 
A mass balance shows 
 
 21 nn && = 
 
On a mass basis, the energy balance is 
 
 ( )22212,1,12 21ˆˆˆˆ VVeehh KK rr −=−=− 
 
Since the steam outlet velocity is much greater than the velocity of the inlet, the above 
expression is approximately equal to 
 
 ( )2212 21ˆˆ Vhh r−=− 
 
The change in enthalpy can be calculated using the steam tables. 
 
 ⎥⎦
⎤⎢⎣
⎡×=
kg
J 109.2827 31h (10 bar, 200 ºC) 
 ⎥⎦
⎤⎢⎣
⎡×=
kg
J 105.2675 32h (sat. H2O(v) at 100 kPa) 
 
Therefore, 
 
 ⎥⎦
⎤⎢⎣
⎡=
s
m 5522V
r
 
 
To solve for the area, the following relationship is used 
 
 
2
2
vˆ
VAm
r
& = 
 
From the steam tables 
 
 68
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 6940.1ˆ
3
2v 
 
Now all but one variable is known. 
 
 [ ]23 m 1007.3 −×=A 
 
 
 
 
 69
2.38 
First start with the energy balance around the nozzle. Assume that heat transfer and potential 
energy effects are negligible. The shaft work term is also zero. Therefore, the energy balance 
reduces to 
 
 0)()( 1122 =+−+ kk ehnehn && 
 
The molar flow rates can be eliminated from the expression since they are equal. Realizing that 
1,2, KK ee >> since the velocity of the exit stream is much larger than the velocity of the inlet 
stream simplifies the energy balance to 
 
 2,12 kehh −=− 
 
Using Appendix A.2 and the definition of kinetic energy 
 
 ( ) 2232212 832
1
)(
2
1 VMWdTETDTCTBTARhh HC
T
T
r−=++++=− ∫ − 
 
From Table A.2.1 
 
 
0
0
10824.8
10785.28
213.1
6
3
=
=
×−=
×=
=
−
−
E
D
C
B
A
 
 
It is also important that the units for the molecular weight and universal gas constant are 
consistent. The following values were used 
 
 ⎥⎦
⎤⎢⎣
⎡
⋅= Kmol
J 314.8R 
 ⎥⎦
⎤⎢⎣
⎡=
mol
kg 0441.0)(
83HC
MW 
 
Integration of the above expression and then solving for 2T provides 
 
 [ ]K 2.4192 =T 
 
 
 70
2.39 
First start an energy balance around the diffuser. Assume that heat transfer and potential energy 
effects are negligible. The shaft work term is also zero. The energy balance reduces to 
 
 0)()( 1122 =+−+ kk ehnehn && 
 
A mass balance reveals 
 
 21 nn && = 
 
The molar flow rates can be eliminated from the expression. Using the definitions of enthalpy 
and the kinetic energy, the equation can be rewritten as 
 
 ( )2122)(212
1
VVMWdTc air
T
T
P
rr −−=∫ 
 
The temperature and velocity of the outlet stream are unknown, so another equation is needed to 
solve this problem. From the conservation of mass, 
 
 
( ) ( )
2
222
1
111
1
11
T
VAP
T
VAP
T
VP
rr& == 
 
where A2, the cross-sectional area of the diffuser outlet, is twice the area of the inlet. Therefore, 
 
 1
1
2
2
1
2 2
1 V
T
T
P
PV
rr
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛= 
 
Using Appendix A.2 and the above expression, the energy balance becomes 
 
 ( ) ⎥⎥⎦
⎤
⎢⎢⎣
⎡
−⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛−=++++∫ − 21
2
1
1
2
2
1322
2
1)(
2
12
1
VV
T
T
P
PMWdTETDTCTBTAR air
T
T
rr
 
 
Substituting values from the problem statement provides an equation with one unknown: 
 
 K 3812 =T 
 
Therefore, 
 
 ( ) ⎥⎦
⎤⎢⎣
⎡=⎟⎟⎠
⎞
⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
s
m 111m/s300
K15.343
K 381
bar 5.1
bar 1
2
1
2V
r
 
 71
2.40 
To find the minimum power required for the compressor, one must look at a situation where all 
of the power is used to raise the internal energy of the air. None of the power is lost to the 
surroundings and the potential and kinetic energy effects must be neglected. Therefore, the 
energy balance becomes 
 
 sWhnhn &&& +−= 22110 
 
Performing a mass balance reveals 
 
 21 nn && = 
 
The energy balance reduces to 
 
 ( )121 hhnWs −= && 
 
Using Equation 2.58 and Appendix A.2, the equation becomes 
 
 ( )∫ ++++= −2
1
322
1
T
T
s dTETDTCTBTARnW && 
 
Table A.2.1 and the problem statement provide the following values 
 
0
1600
0
10575.0
355.3
3
=
−=
=
×=
=
−
E
D
C
B
A
 
 
K 300
s
mol 50
1
1
=
⎥⎦
⎤⎢⎣
⎡=
T
n&
 
 
To find the work, we still need T2. We need to pick a reasonable process to estimate T2. Since 
the heat flow is zero for this open system problem, we choose an adiabatic, reversible piston 
situation. For this situation, 
 
.constPV k = 
 
Since we are assuming the air behaves ideally, we can rewrite the equation as 
 
 72
 
kk
P
RTnP
P
RTnP ⎟⎟⎠
⎞
⎜⎜⎝
⎛=⎟⎟⎠
⎞
⎜⎜⎝
⎛
2
22
2
1
11
1 
kkkk TPTP 2
1
21
1
1
−− = 
 
Substituting values from the problem statement, we obtain 
 
 
( ) ( )( )( )( ) K 579bar 10
bar1K300
7/5
5/71
5/71
5/7
2 =⎥⎥⎦
⎤
⎢⎢⎣
⎡= −
−
T 
 
 
Substitute this value into the expression for the work and evaluate: 
 
 [ ]kW 4.417=sW& 
 
 73
2.41 
 
(a) 
Perform a mass balance: 
 
 outnnn &&& =+ 21 
 
Apply the ideal gas law: 
 
 
 outnRT
VP
RT
VP &
&& =+
2
22
1
11 
 
Substitute values from the problem statement: 
 
 ( )( )( )( ) ( )( )( )( ) outn&=××+×× −− 15.293314.8 105.210215.373314.8 105101 3535 
 [ ]mol/s 366.0=outn& 
 
(b) 
No work is done on the system, and we can neglect potential and kinetic energy effects. We will 
assume the process is also adiabatic. The energy balance reduces to 
 
 out
out
outin
in
in hnhn ∑∑ −= &&0 
 ( ) ( ) ( ) 022112211 =−+−=+− hhnhhnhnhnhn outoutoutout &&&&& 
 
We can calculate the enthalpy difference from the given ideal heat capacity: 
 
 ( ) ( ) 010324.5267.310324.5267.315.293
3
2
15.373
3
1 =×++×+ ∫∫ −− outout
TT
dTTRndTTRn && 
 
Again, we must calculate the molar flow rates from the ideal gas law. Upon substitution and 
evaluation, we obtain 
 
 [ ]K 329=outT 
 
 74
2.42 
 
(a) 
Since the temperature, pressure, and volumetric flow rate are given, the molar flow rate is 
constrained by the ideal gas law. 
 
Note: [ ] [ ]min/cm 1/sm 1067.1 338 =× − 
 ( ) [ ]( )
[ ]( )( ) [ ]mol/s 1045.7K 15.273KJ/mol 314.8
/sm 1067.1Pa100135.1 7385 −− ×=⋅
××==
RT
VPn
&
& 
 
To recap, we have shown 
 
 [ ] [ ]mol/s 1045.7SCCM 1 7−×= 
 
(b) 
Assumptions: N2 is an ideal gas 
 All power supplied by the power supply is transferred to the N2 
 Uniform temperature radially throughout sensor tube 
 Kinetic and potential energy effects negligible in energy balance 
 
Let x represent the fraction of N2 diverted to the sensor tube, and sn& represent the molar flow 
rate through the sensor tube. Therefore, the total molar flow rate, totaln& , is 
 
x
nn stotal
&& = 
 
We can use temperature and heat load information from the sensor tube to find the molar flow 
rate through the sensor tube. First, perform an energy balance for the sensor tube: 
 
 ( ) QhhnH inouts && =−=∆ 
 
The enthalpy can be calculated with heat capacity data. Therefore, 
 
 
∫
=
2
1
T
T
P
s
dTc
Qn
&
& 
 
Now, we can calculate the total molar flow rate. 
 
 75
 
∫
=
2
1
T
T
P
total
dTcx
Qn
&
& 
 
To find the flow rate in standard cubic centimeters per minute, apply the conversion factor found 
in Part (a) 
 
 [ ][ ]⎟⎟⎠
⎞
⎜⎜⎝
⎛
×= −∫ mol/s 1045.7
SCCM 1)SCCM( 72
1
T
T
P
total
dTcx
Qv
&
& 
 
(c) 
To find the correction factor for SiH4, re-derive the expression for flow rate for SiH4 and then 
divide it by the expression for N2 for the same power input, temperatures, and fraction of gas 
diverted to the sensor tube. 
 
 
[ ]
[ ]
[ ]
[ ] ∫
∫
∫
∫
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
×
⎟⎟⎠
⎞
⎜⎜⎝
⎛
×
==
−
−
2
1
4
2
1
2
2
1
2
2
1
4
2
4
,
,
7
,
7
,
,
,
mol/s 1045.7
SCCM 1
mol/s 1045.7
SCCM 1
T
T
SiHP
T
T
NP
T
T
NP
T
T
SiHP
Ntotal
SiHtotal
dTc
dTc
dTcx
Q
dTcx
Q
v
v
Factor &
&
&
&
 
 
If we assume that heat capacities are constant, the conversion factor simplifies: 
 
 
4
2
,
,
SiHP
NP
c
c
Factor = 
 
Using the values in Appendix A.2.2 at 298 K, we get 
 
 67.0=Factor 
 
 76
2.43 
 
(a) 
It takes more energy to raise the temperature of a gas in a constant pressure cylinder. In both 
cases the internal energy of the gas must be increased. In the constant pressure cylinder work, 
Pv work must also be supplied to expand the volume against the surrounding’s pressure. This is 
not required with a constant volume. 
 
(b) 
As you perspire, sweat evaporates from your body. This process requires latent heat which cools 
you. When the water content of the environment is greater, there is less evaporation; therefore, 
this effect is diminished and you do not feel as comfortable. 
 
 77
2.44 
From the steam tables at 10 kPa: 
 
 
 
 
 
 
 
 
( ) ⎥⎦
⎤⎢⎣
⎡
⋅+=⎥⎦
⎤⎢⎣
⎡
⋅+=⎟⎠
⎞⎜⎝
⎛=
Kmol
J 001434.0516.3
Kkg
kJ 0006624.06241.1 RTT
dT
dhc
P
P 
 
Now compare the above values to those in Appendix A.2. 
 
 
 
 
 
 
 
 
T(K) h 
323.15 2592.6 
373.15 2687.5 
423.15 2783 
473.15 2879.5 
523.15 2977.3 
573.15 3076.5 
673.15 3279.5 
773.15 3489 
873.15 3705.4 
973.15 3928.7 
1073.15 4159.1 
1173.15 4396.4 
1273.15 4640.6 
1373.15 4891.2 
1473.15 5147.8 
1573.15 5409.7 
 h vs. T
y = 0.0003x 2 + 1.6241x + 2035.7
R2 = 1
0
1000
2000
3000
4000
5000
6000
0 500 1000 1500 2000
T
Series1
Poly. (Series1)
 A B 
Steam Tables 3.516 0.001434 
Appendix A 3.470 0.001450 
% difference 1.3 1.4 
 78
2.45 
For throttling devices, potential and kinetic energy effects can be neglected. Furthermore, the 
process is adiabatic and no shaft work is performed. Therefore, the energy balance for one inlet 
and one outlet is simplified to 
 
2211 hnhn && = 
 
which is equivalent to 
 
 2211 ˆˆ hmhm && = 
 
Since mass is conserved 
 
 221 ˆˆ hmh &= 
 
From the steam tables: 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 3.33981ˆh (8 MPa, 500 ºC) 
 ⎥⎦
⎤⎢⎣
⎡=∴
kg
kJ 3.3398ˆ2h 
 
Now that we know 2uˆ and 2P , 2T is constrained. Linear interpolation of steam table data gives 
 
 [ ]Cº 4572 =T 
 
 
 
 
 79
2.46 
 
(a) 
An expression for work in a reversible, isothermal process was developed in Section 2.7. 
Equation 2.77 is 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
2ln
P
PnRTW 
 
Therefore, 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
2ln
P
PRTw 
 
Evaluating the expression with 
 
 [ ]
[ ]
[ ]kPa 500
kPa 100
K 300
Kmol
J 314.8
1
2
=
=
=
⎥⎦
⎤⎢⎣
⎡
⋅=
P
P
T
R
 
 
gives 
 
 ⎥⎦
⎤⎢⎣
⎡−=
kg
J 4014w 
 
(b) 
Equation 2.90 states 
 
 [ ]121 TTk
Rw −−= 
 
Since the gas is monatomic 
 
 
Rc
Rc
v
P
⎟⎠
⎞⎜⎝
⎛=
⎟⎠
⎞⎜⎝
⎛=
2
3
2
5
 
 
and 
 
 80
 
3
5=k 
 
2T can be calculated by applying the polytropic relation derived for adiabatic expansions. From 
Equation 2.89 
 
 
constPv
constPV
k
k
=∴
= 
 
By application of the ideal gas law 
 
P
RTv = 
Since R is a constant, substitution of the expression for P into the polytropic relation results in 
 
 
( )
( ) ( ) kkkk
kk
TPTP
constTP
2
1
2
1
1
1
1
−−
−
=∴
=
 
 
This relation can be used to solve for 2T . 
 
 [ ]K 6.1572 =T 
 
Now that 2T is known, value of work can be solved. 
 
 ⎥⎦
⎤⎢⎣
⎡−=
kg
J 9.1775w 
 
 81
2.47 
 
(a) 
The change in internal energy and enthalpy can be calculated using 
 
 
( ) ( )
( ) ( )Cº 0 ,atm 1ˆCº 100 ,atm 1ˆˆ
Cº 0 ,atm 1ˆCº 100 ,atm 1ˆˆ
ll
ll
uuu
hhh
−=∆
−=∆
 
 
We would like to calculate these values using the steam tables; however, the appendices don’t 
contain steam table data for liquid water at 0.0 ºC and 1 atm. However, information is provided 
for water at 0.01 ºC and 0.6113 kPa. Since the enthalpy and internal energy of liquid water is 
essentially independent of pressure in this pressure and temperature range, we use the steam 
table in the following way 
 
( ) ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 02.419Cº 100 ,atm 1ˆlh 
( ) ( ) ⎥⎦
⎤⎢⎣
⎡==
kg
kJ 0Cº 01.0 ,kPa .61130ˆCº 0 ,atm 1ˆ ll hh 
( ) ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 91.418Cº 100 ,atm 1ˆ lu 
 ( ) ( ) ⎥⎦
⎤⎢⎣
⎡==
kg
kJ 0Cº 01.0 ,kPa .61130ˆCº 0 ,atm 1ˆ ll uu 
 
Therefore, 
 
 ⎥⎦
⎤⎢⎣
⎡=∆
kg
kJ 02.419hˆ 
 ⎥⎦
⎤⎢⎣
⎡=∆
kg
kJ 91.418uˆ 
 
 
(b) 
The change in internal energy and enthalpy can be calculated using 
 
 
( ) ( )
( ) ( )Cº 100 ,atm 1ˆCº 100 ,atm 1ˆˆ
Cº 100 ,atm 1ˆCº 100 ,atm 1ˆˆ
lv
lv
uuu
hhh
−=∆
−=∆
 
 
From the steam tables 
 
 82
 
⎥⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡=
kg
kJ 5.2506ˆ
kg
kJ 02.419ˆ
kg
kJ 0.2676ˆ
v
l
v
u
h
h
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 91.418ˆ lu 
 
Therefore, 
 
 
⎥⎦
⎤⎢⎣
⎡=∆
⎥⎦
⎤⎢⎣
⎡=∆
kg
kJ 59.2087ˆ
kg
kJ 99.2256ˆ
u
h
 
 
The change in internal energy for the process in Part (b) is 5.11 times greater than the change in 
internal energy calculated in Part (a). The change in enthalpy in Part (b) is 5.39 times greater 
than the change in enthalpy calculated in Part (a). 
 
 
 
 
 832.48 
To calculate the heat capacity of Ar, O2, and NH3 the following expression, with tabulated values 
in Table A.2.1, will be used, 
 
 322 ETDTCTBTA
R
cP ++++= − 
 
where T is in Kelvin. From the problem statement 
 
 [ ]K 300=T 
 
and from Table 1.1 
 
 ⎥⎦
⎤⎢⎣
⎡
⋅= Kmol
J 314.8R 
 
To find the A-E values, Table A.2.1 must be referred to. 
 
Formula A 310×B 610×C 510−×D 910×E 
Ar - - - - - 
O2 3.639 0.506 0 -0.227 0 
NH3 3.5778 3.02 0 -0.186 0 
 
The values are not listed for Ar since argon can be treated as a monatomic ideal gas with a heat 
capacity independent of temperature. The expression for the heat capacity is 
 
 Rc ArP ⎟⎠
⎞⎜⎝
⎛=
2
5
, 
 
Now that expressions exist for each heat capacity, evaluate the expressions for [ ]K 300=T . 
 
 
⎥⎦
⎤⎢⎣
⎡
⋅=
⎥⎦
⎤⎢⎣
⎡
⋅=
⎥⎦
⎤⎢⎣
⎡
⋅=
Kmol
J 560.35
Kmol
J 420.29
Kmol
J 785.20
3
2
,
,
,
NHP
OP
ArP
c
c
c
 
 
By examining the heat capacity for each molecule, it should be clear that the magnitude of the 
heat capacity is directly related to the structure of the molecule. 
 
 
 
 84
Ar 
• Since argon is monatomic, translation is the only mode through which the atoms can 
exhibit kinetic energy. 
O2 
• Translation, rotation, and vibration modes are present. Since oxygen molecules are 
linear, the rotational mode of kinetic energy contributes RT per mol to the heat capacity. 
NH3 
• Translation, rotation, and vibration modes are present. Ammonia molecules are non-
linear, so the rotation mode contributes 3RT/2 per mole to the heat capacity. 
 
The vibration contributions can also be analyzed for oxygen and ammonia, which reveals that the 
vibration contribution is greatest for ammonia. This is due to ammonia’s non-linearity. 
 
 85
2.49 
For a constant pressure process where potential and kinetic energy effects are neglected, the 
energy balance is given by Equation 2.57: 
 
 HQ ∆= 
 
The change in enthalpy can be written as follows 
 
 ( )12 hhmH −=∆ 
 
From the steam tables: 
 
 ( ) ⎥⎦
⎤⎢⎣
⎡==
kg
kJ 6.2584kPa 10at vapor sat.ˆ2 hh 
 ( ) ⎥⎦
⎤⎢⎣
⎡==
kg
kJ 81.191kPa 10at liquid sat.1ˆ hh 
 
Therefore, 
 
 ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡=
kg
kJ 81.191
kg
kJ 6.2584kg 2Q 
 [ ]kJ 6.4785=Q 
 
We can find the work from its definition: 
 
 ∫−=
f
i
V
V
EdVPW 
 
The pressure is constant, and the above equation can be rewritten as follows 
 
 ( )12 ˆˆ vvmPW E −−= 
 
From the steam tables: 
 
 ( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡==
kg
m 674.14kPa 10at vapor sat.ˆˆ
3
2 vv 
 ( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡==
kg
m 00101.0kPa 10at liquid sat.ˆˆ
3
1 vv 
 
Therefore, 
 86
 
 ( )( ) kJ 5.293
kg
m 00101.0
kg
 674.14kg2Pa10000
3
−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡−=W 
 
 87
2.50 
First, perform an energy balance on the system. Potential and kinetic energy effects can be 
neglected. Since nothing is mentioned about work in the problem statement, W can be set to 
zero. Therefore, the energy balance is 
 
 UQ ∆= 
 
Performing a mass balance reveals 
 
 12 mm = 
 
where 
 
 vl mmm 111 += 
 
 
Now the energy balance can be written as 
 
 ( )vvll umumumuumQ 111121121 ˆˆˆ)ˆˆ( +−=−= 
 
Since two phases coexist initially (water is saturated) and 1P is known, state 1 is constrained. 
From the saturated steam tables 
 
 
⎥⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡=
kg
kJ 9.2437ˆ
kg
kJ 79.191ˆ
1
1
v
l
u
u
 (sat. H2O at 10 kPa) 
 
As heat is added to the system, the pressure does not remain constant, but saturation still exists. 
One thermodynamic property is required to constrain the system. Enough information is known 
about the initial state to find the volume of the container, which remains constant during heating, 
and this can be used to calculate the specific volume of state 2. 
 
 vl
vvll
mm
vmvm
m
V
m
Vv
11
1111
1
1
2
2
2
ˆˆ
+
+=== 
 
From the saturated steam tables 
 
 88
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 674.14ˆ
kg
m 001010.0ˆ
3
1
3
1
l
l
v
v
 (sat. H2O at 10 kPa) 
 
Therefore, 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 335.1ˆ
3
2v 
 
The water vapor is now constrained. Interpolation of steam table data reveals 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 6.2514ˆ2u 
 
Now that all of the required variables are known, evaluation of the expression for Q is possible. 
 
 [ ]kJ 11652=Q 
 
 89
2.51 
Let the mixture of ice and water immediately after the ice has been added represent the system. 
Since the glass is adiabatic, no work is performed, and the potential and kinetic energies are 
neglected, the energy balance reduces 
 
 0=∆H 
 
We can split the system into two subsystems: the ice (subscript i) and the water (subscript w). 
Therefore, 
 
 0=∆+∆=∆ wwii hmhmH 
and 
 wwii hmhm ∆−=∆ 
 
We can get the moles of water and ice. 
 
 [ ] [ ]kg 399.0
kg
m 001003.0
m .00040
ˆ 3
3
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡== w
w
w v
Vm 
 ( )
[ ]( ) [ ]mol 15.22
mol
kg 0180148.0
kg 399.0
2
=
⎥⎦
⎤⎢⎣
⎡== OH
w
w MW
mn 
 ( ) [ ]mol 55.5
2
==
OH
i
i MW
mn 
 
Now, let’s assume that all of the ice melts in the process. (If the final answer is greater than 0 
ºC, the assumption is correct.) The following expression mathematically represents the change 
in enthalpy. 
 
 ( )( ) ( )[ ] ( )Cº 25º 0Cº 100 ,,, −−=−+∆−−− fwPwfwPfusiPi TcnCTchcn 
Note: Assumed the heat capacities are independent of temperature to obtain this 
expression. 
 
From Appendix A.2.3 
 
 Rc iP 196.4, = 
 Rc wP 069.9, = 
 
and 
 
⎥⎦
⎤⎢⎣
⎡−=∆
mol
kJ0.6fush 
 90
 
Substitution of values into the above energy balance allows calculation of Tf. 
 
 Cº 12.3=fT 
 
(Our assumption that all the ice melts is correct.) 
 
(b) 
To obtain the percentage of cooling achieved by latent heat, perform the following calculation 
 
 
( )( )iwfwPw fusilatent TTcn
hn
Fraction
,, −−
∆−= 
[ ]( )
[ ]( ) ( ) 911.0Cº 25Cº 12.3
Kmol
J 314.8069.9mol 15.22
mol
J 6000mol 55.5
=
−⎥⎦
⎤⎢⎣
⎡
⋅⋅−
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
=latentFraction 
%1.91=latentPercent 
 
 
 91
2.52 
A mass balance shows 
 
vl nnn 112 += 
 
To develop the energy balance, neglect kinetic and potential energy. Also, no shaft work is 
performed, so the energy balance becomes 
 
 QU =∆ 
 
The energy balance can be expanded to 
 
 ( ) vvllvl unununnQ 1111211 −−+= 
 
If the reference state is set to be liquid propane at 0 ºC and 4.68 bar, the internal energies become 
 
 ( )
( ) ( )dTRcuu
uu
u
Pvap
vap
v
l
∫ −+∆=
∆=
=
2T
K 273
2
1
1
Cº 0
Cº 0
0
 
 
Once the change in internal energy for vaporization and temperature of state 2 is determined, Q 
can be solved. As the liquid evaporates, the pressure increases. At state 2, where saturated 
propane vapor is present, the ideal gas law states 
 
 
2
2
2 v
RTP = 
 
To find 2v , assume that 
lv vv 11 >> . The volume of the rigid container is 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡===
mol
m 00485.0
3
1
1
111 P
RTnvnV vvv 
 
Therefore, 
 
 ( ) ⎥⎥⎦
⎤
⎢⎢⎣
⎡=+= mol
m 00243.0
3
11
2 vl nn
Vv 
 
 92
Also, since the propane is saturated, 2P and 2T are not independent of each other. They are 
related through the Antoine Equation, 
 
 
 ( )
CT
BAP sat
sat
+−=ln 
 
where 
 
 2PP
sat = and 2TT sat = 
 
Substitution provides, 
 
 
CT
BA
v
RT
+−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
22
2ln 
 
Using values from Table A.1.1 and 
⎥⎥⎦
⎤⎢⎢⎣
⎡
⋅
⋅×= −
Kmol
mbar 10314.8
3
5R 
 
 K 7.3012 =T 
 
To find vapu∆ , refer to the definition of enthalpy. 
 
 ( )lvlvvap PvuPvuhhh +−+=−=∆ )( 
 
Since lv vv >> , the above the change in internal energy of vaporization can be written as 
 
 ( ) RThPvhu vapvvapvap −∆=−∆=∆ 
 
Therefore, 
 
 ( ) ⎥⎦
⎤⎢⎣
⎡=∆
mol
kJ 39.14Cº 0vapu 
 
Evaluation of the following equation after the proper values have been substituted from Table 
A.2.1 
 
 ( ) ( ) ( ) ( ) ( )Cº 00Cº 0 11K 301.7
K 273
11 vap
vl
Pvap
vl unndTRcunnQ ∆−−⎥⎥⎦
⎤
⎢⎢⎣
⎡
−+∆+= ∫ 
 93
gives 
 
 [ ]kJ 1.18=Q 
 
 
 94
2.53 
The equation used for calculating the heat of reaction is given in Equation 2.72. It states 
 
 ( )∑ °° =∆ ifirxn hvh 
 
This equation will be used for parts (a)-(e). Since the heat of reaction at 298 K is desired, values 
from Appendix A.3 can be used. 
 
(a) 
First the stoichiometric coefficient must be determined for each species in the reaction. 
 
 
1
1
1
1
)(
)(
)(
)(
2
2
2
4
=
=
−=
−=
gOH
gCO
gO
gCH
v
v
v
v
 
 
From Tables A.3.1 and A.3.2 
 
 
( )
( )
( )
( ) ⎥⎦⎤⎢⎣⎡−=
⎥⎦
⎤⎢⎣
⎡−=
⎥⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡−=
°
°
°
°
mol
kJ 82.241
mol
kJ 51.393
mol
kJ 0
mol
kJ 81.74
)(298,
)(298,
)(298,
)(298,
2
2
2
4
gOHf
gCOf
gOf
gCHf
h
h
h
h
 
 
From Equation 2.72, the equation for the heat of reaction is 
 ( ) ( ) ( ) ( )
)(298,)()(298,)()(298,)()(298,)(298, 22222244 gOHfgOHgCOfgCOgOfgOgCHfgCHrxn
hvhvhvhvh °°°°° +++=∆
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−+⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−+⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−−=∆ °
mol
kJ 82.241
mol
kJ 51.393
mol
kJ 0
mol
kJ 81.74298,rxnh 
⎥⎦
⎤⎢⎣
⎡−=∆ °
mol
kJ 52.560298,rxnh 
 
Now that a sample calculation has been performed, only the answers will be given for the remaining 
parts since the calculation process is the same. 
 
 
 
 95
(b) 
⎥⎦
⎤⎢⎣
⎡−=∆ °
mol
kJ 53.604298,rxnh 
 
(c) 
⎥⎦
⎤⎢⎣
⎡−=∆ °
mol
kJ 12.206298,rxnh 
 
(d) 
⎥⎦
⎤⎢⎣
⎡−=∆ °
mol
kJ 15.41298,rxnh 
 
(e) 
⎥⎦
⎤⎢⎣
⎡−=∆ °
mol
kJ 38.905298,rxnh 
 
 96
2.54 
The acetylene reacts according to the following equation 
 
 C2H2(g) + (5/2)O2(g) Æ 2CO2(g) + H2O(g) 
 
(a) 
First, choose a basis for the calculations. 
 
 mol 1
22
=HCn 
 
Calculate the heat of reaction at 298 K using Equation 2.72 and Appendix A.3 
 
 ( )∑ °° =∆ ifirxn hvh 
 ( ) ( ) ( ) ( ) OHfCOfOfHCfrxn hhhhh 22222 25.2 °°°°° ++−−=∆ 
 ( ) [ ]J 10255.1
mol
J 10255.1
6
298,
6
22
×−=∆=∆
⎥⎦
⎤⎢⎣
⎡×−=∆
°
°
rxnHCrxn
rxn
hnH
h
 
 
The required amount of oxygen is calculated as follows 
 ( ) mol 5.2)(5.2 11 222 == HCO nn 
 
The compositions for both streams are 
 
Streams 
22 HCn 2On 2Nn 2COn OHn 2 
1 (Inlet) 1 2.5 0 0 0 
2 (Outlet) 0 0 0 2 1 
 
From Table A.2.2 
 
Species A 310×B 610×C 510−×D 910×E 
C2H2 6.132 1.952 0 -1.299 0 
O2 3.639 0.506 0 -0.227 0 
CO2 5.457 1.045 0 -1.157 0 
H2O 3.470 1.45 0 0.121 0 
 
Integration of the following equation provides an algebraic expression where only 2T is 
unknown. 
 
 97
 ( ) ( ) 02
298
2298, =+∆ ∫ ∑
T
i
iPirxn dTcnH 
 
Substituting the proper values into the expression gives 
 
 K 61692 =T 
 
(b) 
The calculations follow the procedure used in Part (a), but now nitrogen is present. The basis is 
 
 mol 1
22
=HCn 
 
The heat of reaction is the same as in Part (a), but the gas composition is different. Since 
stoichiometric amount of air is used, 
 
 ( ) mol 5.2
12
=On 
 ( ) ( ) mol 40.9
2
2
22 11
=⎟⎟⎠
⎞
⎜⎜⎝
⎛=∴
airO
N
ON y
y
nn 
 
The composition of the streams are summarized below 
 
Streams 
22 HCn 2On 2Nn 2COn OHn 2 
1 1 2.5 9.40 0 0 
2 0 0 9.40 2 1 
 
From Appendix A.2 
 
Species A 310×B 610×C 510−×D 910×E 
C2H2 6.132 1.952 0 -1.299 0 
O2 3.639 0.506 0 -0.227 0 
CO2 5.457 1.045 0 -1.157 0 
H2O 3.470 1.45 0 0.121 0 
N2 3.280 0.593 0 0.04 0 
 
Therefore, 
 
 K 27922 =T 
 
 
 
 
 98
(c) 
Now excess air is present, so not all of the oxygen reacts. The heat of reaction remains the same 
because only 1 mole of acetylene reacts. Since the amount of air is twice the stoichiometric 
amount 
 
 ( ) mol 5
12
=On 
 ( ) ( ) mol 80.18
2
2
22 11
=⎟⎟⎠
⎞
⎜⎜⎝
⎛=∴
airO
N
ON y
y
nn 
 
The compositions of the streams are summarized below 
 
Streams 
22 HCn 2On 2Nn 2COn OHn 2 
1 1 5 18.8 0 0 
2 0 2.5 18.8 2 1 
 
The table of heat capacity data in Part (b) will be used for this calculation. Using the expression 
shown in Part (a) 
 
 K 17872 =T 
 
 
 99
2.55 
(a) 
The combustion reaction for propane is 
 
 C3H8(g) + 5O2(g) Æ 3CO2(g) + 4H2O(g) 
 
For all subsequent calculations, the basis is one mole of propane. The heat of reaction is 
calculated as follows 
 ( ) ( ) ( ) ( ) OHfCOfOfHCfrxn hhhhh 22283 435 °°°°° ++−−=∆ 
 ( ) [ ]J 10044.2
mol
J 10044.2
6
298,
6
83
×−=∆=∆
⎥⎦
⎤⎢⎣
⎡×−=∆
°
°
rxnHCrxn
rxn
hnH
h
 
 
The required amount of oxygen for complete combustion of propane is 
 
 ( ) mol 5)(5 11 832 == HCO nn 
 ( ) ( ) mol 81.18
2
2
22 11
=⎟⎟⎠
⎞
⎜⎜⎝
⎛=∴
airO
N
ON y
y
nn 
The stream compositions are listed below 
 
Streams 
83 HCn 2On 2Nn 2COn OHn 2 
1 (Inlet) 1 5 18.8 0 0 
2 (Outlet) 0 0 18.8 3 4 
 
From Table (a)2.2 
 
Species A 310×B 610×C 510−×D 910×E 
N2 3.280 0.593 0 0.04 0 
CO2 5.457 1.045 0 -1.157 0 
H2O 3.470 1.45 0 0.121 0 
 
Now all of the necessary variables for the following equation are known, except 2T . 
 
 ( ) ( ) 02
298
2298, =+∆ ∫ ∑
T
i
iPirxn dTcnH 
 
Solving the resulting expression provides 
 
 K 23742 =T 
 100
(b) 
The combustion reaction for butane is 
 
 C4H10(g) + (13/2)O2(g) Æ 4CO2(g) + 5H2O(g) 
 
For all subsequent calculations, the basis is one mole of butane. The heat of reaction is 
calculated as shown in Part (a) 
 
 [ ]J 10657.2 6298, ×−=∆ rxnH 
 
The moles of nitrogen and oxygen in the feed stream are calculated according to the method in 
Part (a). The compositions are 
 
Streams 
104 HCn 2On 2Nn 2COn OHn 2 
1 (Inlet) 1 6.5 24.5 0 0 
2 (Outlet) 0 0 24.5 4 5 
 
The Pc data listed in Part (a) can also be used for this reaction since there is no remaining 
butane. 
 
 K 23762 =T 
(c) 
The combustion reaction for pentane is 
 
 C5H12(g) + 8O2(g) Æ 5CO2(g) + 6H2O(g) 
 
The basis is one mole of pentane. The heat of reaction is calculated as shown in Part (a). 
 
 [ ]J 10272.3 6298, ×−=∆ rxnH 
 
The moles of nitrogen and oxygen in the feed stream are calculated according to the method in 
Part (a). The compositions are listed below 
 
Streams 
125 HCn 2On 2Nn 2COn OHn 2 
1 1 8 30.1 0 0 
2 0 0 30.1 5 6 
 
Substitution of the values into the expression used to find 2T and subsequent evaluation results 
in 
 
 K 23822 =T 
 
The adiabatic flame temperatures are nearly identical in all three cases. 
 101
2.56 
The equation for the combustion of methane is 
 
 CH4(g) + 2O2(g) Æ CO2(g) + 2H2O(g) 
 
Using Equation 2.72 
 
 ⎥⎦
⎤⎢⎣
⎡×−=∆ °
mol
J 1002.8 5rxnh 
 
The basis for this problem is 
 
 ( ) mol 1
14
=CHn 
 
Also, let ξ represent the fractional conversion of methane. Therefore, the composition of the 
product gas leaving the reactor is 
 
 
( ) ( )
( ) ( )
( )
( )
( ) mol 2
mol 
mol 52.7
mol 12
mol 11
2
2
2
2
2
2
2
2
2
4
ξ
ξ
ξ
ξ
=
=
=
−=
−=OH
CO
N
O
CH
n
n
n
n
n
 
 
Furthermore, the heat of reaction is calculated as follows 
 
 ( ) [ ]J 1002.8 5298, ×−=∆ ξrxnH 
 
From Table A.2.2 
 
Species A 310×B 610×C 510−×D 910×E 
CH4 1.702 9.081 -2.164 0 0 
O2 3.639 0.506 0 -0.227 0 
CO2 5.457 1.045 0 -1.157 0 
H2O 3.470 1.45 0 0.121 0 
N2 3.280 0.593 0 0.04 0 
 
After substitution of the outlet composition values, heat capacity data, and the heat of reaction 
into the following equation 
 
 102
 ( ) ( ) 01273
298
2298, =+∆ ∫ ∑
i
iPirxn dTcnH 
 
integration provides an equation with one unknown: ε . Solving the equation gives 
 
 ξ =0.42 
 
Since the fractional conversion is 0.42, 58% of the methane passed through the reactor unburned. 
 
 103
2.57 
For the entire cycle, 
 
 [ ]kJ 50
0
12
31231211
−=∆∴
=∆+∆+∆=∆
U
UUUU
 
 
From state 1 to state 2 
 
 [ ]kJ 40012
121212
−=∴
+=∆
Q
WQU
 
 
From state 2 to state 3 
 
 [ ]kJ 023
232323
=∴
+=∆
Q
WQU
 
 
From state 3 to state 1 
 
 [ ]kJ 25031
313131
−=∴
+=∆
W
WQU
 
 
Hence, the completed table is 
 
Process [ ]kJ U∆ [ ]kJ W [ ]kJ Q 
State 1 to 2 -50 -400 350 
State 2 to 3 800 800 0 
State 3 to 1 -750 -250 -500 
 
To determine if this is a power cycle or refrigeration cycle, look at the overall heat and work, 
11W and 11Q . 
 
 
[ ]
[ ]kJ 150
kJ 150
31231211
31231211
−=++=
=++=
QQQQ
WWWW
 
 
Since work is done on the system to obtain a negative value of heat, which means that heat is 
leaving the system, this is a refrigeration cycle. 
 
 
 104
2.58 
Refer to the graph of the Carnot cycle in Figure E2.20. From this graph and the description of 
Carnot cycles in Section 2.9, it should be clear that state 3 has the lowest pressure of all 4 states, 
and state 1 has the highest pressure. States 1 and 2 are at the higher temperature. States 3 and 4 
have the lower temperature. Since both the temperature and pressure are known for states 1 and 
3, the molar volume can be calculated using 
 
 
P
RTv = 
 
The table below summarizes the known thermodynamic properties. 
 
State [ ]K T [ ]bar P [ ]/molm 3v 
1 1073 60 0.00149 
2 1073 
3 298 0.2 0.124 
4 298 
 
For each step of the process, potential and kinetic energy effects can be neglected. The step from 
state 1 to state 2 is a reversible, isothermal expansion. Since it is isothermal, the change in 
internal energy is 0, and the energy balance becomes 
 
 1212 WQ −= 
 
From Equation 2.77, 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
2
112 ln P
PnRTW 
 
where 12W is the work done from state 1 to state 2. The value of 2P is not known, but 
recognizing that the process from state 2 to 3 is an adiabatic expansion provides an additional 
equation. The polytropic relationship can be employed to find 2P . A slight modification of 
Equation 2.89 provides 
 
 constPvk = 
 
From the ideal gas law 
 
 
v
RTP = 
 
Combining this result with the polytropic expression and noting that R is constant, allows the 
expression to be written as 
 105
 
 constTvk =−1 
 
Therefore, 
 
 1
1
1
3
2
3
2
−− ⎟⎟⎠
⎞
⎜⎜⎝
⎛= kkv
T
Tv 
 
Substituting the appropriate values (k=1.4) gives 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
mol
m 00504.0
3
2v 
 
Applying the ideal gas law 
 
 [ ]bar 7.17
2
2
2 == v
RTP 
 
Now, 12W can be calculated. 
 
 [ ]kJ 89.1012 −=W 
 
Calculation of 34W follows a completely analogous routine as calculation for 12W . The 
following equations were used to find the necessary properties 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=⎟⎟⎠
⎞
⎜⎜⎝
⎛= −−
mol
m 0367.0
31
1
1
1
4
1
4
kkv
T
Tv 
 [ ]bar 675.0
4
4
4 == v
RTP 
 
Now the following equation can be used 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
3
4
334 ln P
PnRTW 
 
which gives 
 
 [ ]kJ 01.334 =W 
 
 106
For an adiabatic, reversible process, Equation 2.90 states 
 
 [ ]121 TTk
nRW −−= 
 
This equation will be used to calculate the work for the remaining processes. 
 
 [ ] [ ]kJ 11.16
1 2323
−=−−= TTk
nRW 
 [ ] [ ]kJ 11.16
1 4141
=−−= TTk
nRW 
 
To find the work produced for the overall process, the following equation is used 
 
 41342312 WWWWWnet +++= 
 
Evaluating this expression with the values found above reveals 
 
 [ ]kJ 88.7−=netW 
 
Therefore, 7.88 kilojoules of work is obtained from the cycle. 
The efficiency of the process can be calculated using Equation 2.98: 
 
 72.0
89.10
88.7 ===
H
net
Q
Wη 
 
since [ ]kJ 89.1012 =−= WQH . Alternatively, if we use Equation E2.20D. 
 
 
H
C
T
T−= 1η 
 
where 43 TTTC == and 21 TTTH == . Upon substitution of the appropriate values 
 
 72.0=η 
 
 
 107
2.59 
Since this is a refrigeration cycle, the direction of the cycle described in Figure 2.17 reverses. 
Such a process is illustrated below: 
 
State 2 State 1
T1 
P1
T2=T1 
P2
Constant TH
QH
Constant TC
QC
Isothermal 
expansion
State 3 State 4
T4=T3 
 P4
T3 
P3
Isothermal 
compression
Well - insulated
State 1 State 4
T1 
P1
T4 
P4
Adiabatic 
compression
State 3
T3 
P3
State 2
T2 
P2
Adiabatic 
expansion
 
 
States 1 and 2 are at the higher temperature. States 3 and 4 have the lower temperature. Since 
the both the temperature and pressure are known for states 2 and 4, the molar volume can be 
calculated using 
 
 
P
RTv = 
 
The following table can be made 
 
State [ ]K T [ ]bar P [ ]/molm 3v 
1 1073 
2 1073 60 0.00149 
3 298 
4 298 0.2 0.124 
 
For each step of the process, potential and kinetic energy effects can be neglected. The process 
from state 1 to state 2 is a reversible, isothermal expansion. Since it is isothermal, the change in 
internal energy is 0, and the energy balance becomes 
 
 1212 WQ −= 
 
 108
From Equation 2.77, 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
2
12 ln P
PnRTW H 
 
where 12W is the work done from state 1 to state 2. The value of 1P is not known, but 
recognizing that the process from state 4 to 1 is an adiabatic compression provides an additional 
relation. The polytropic relationship can be employed to find 1P . A slight modification of 
Equation 2.89 provides 
 
 constPvk = 
 
From the ideal gas law 
 
 
v
RTP = 
 
Combining this result with the polytropic expression and noting that R is constant allows the 
expression to be written as 
 
 constTvk =−1 
 
Therefore, 
 
 
1
1
1
4
1
4
1
−− ⎟⎟⎠
⎞
⎜⎜⎝
⎛= kkv
T
Tv 
 
Substituting the appropriate values (k=1.4) gives 
 
 ⎥⎦
⎤⎢⎣
⎡=⎟⎟⎠
⎞
⎜⎜⎝
⎛= −−
mol
m 00504.0
31
1
1
4
1
4
1
k
kv
T
Tv 
 [ ]bar 7.17
1
1
1 == v
RTP 
 
 
Now, 12W can be calculated. 
 
 [ ]kJ 9.1012 =W 
 
Calculation of 34W follows a completely analogous routine as the calculation for 12W . The 
following equations were used to find the necessary properties: 
 109
 
 ⎥⎦
⎤⎢⎣
⎡=
mol
m 0367.0
3
3v 
 
Applying the ideal gas law 
 
 [ ]bar 675.0
3
3
3 == v
RTP 
 
Now the following equation can be used 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
3
4
34 ln P
PnRTW C 
 
which gives 
 
 [ ]kJ 0.341 −=W 
 
Equation 2.90 can be used to determine the work for adiabatic, reversible processes. This 
equation will be used to calculate the work for the remaining processes. 
 
 [ ] [ ]kJ 11.16
123
−=−−= HC TTk
nRW 
 [ ] [ ]kJ 11.16
141
=−−= CH TTk
nRW 
 
To find the work produced for the overall process, the following equation is used 
 
 41342312 WWWWWnet +++= 
 
Evaluating this expression with the values found above reveals[ ]kJ 88.7=netW 
 
Therefore, 7.88 kJ of work is obtained from the cycle. The coefficient of performance is defined 
in Equation 2.99 as follows 
 
 
net
C
W
QCOP = 
 
where CQ is the equal to 34Q . From the energy balance developed for the process from state 3 
to state 4 
 110
 
 [ ]kJ 01.33434 =−= WQ 
 
Therefore, 
 
 [ ][ ] 382.0kJ 88.7
kJ 01.3 ==COP 
 
 111
2.60 
 
(a) 
The Pv path is plotted on log scale so that the wide range of values fits (see Problem 1.13) 
logP
v
1
100
0.075
2
3
4
log v
 
(b) 
The work required to compress the liquid is the area under the Pv curve from state 3 to state 4. 
Its sign is positive. The power obtained from the turbine is the area under the curve from state 1 
to 2. Its sign is negative. The area under the latter curve is much larger (remember the log 
scale); thus the net power is negative. 
 
(c) 
First, perform a mass balance for the entire system: 
 
 mmmmm &&&&& ==== 4321 
 
Since no work is done by or on the boiler, the energy balance for the boiler is 
 
 HQhmhm &&& =− 41 ˆˆ 
 
Similarly, the energy balance for the condenser is 
 
 CQhmhm &&& =− 23 ˆˆ 
 
To find the necessary enthalpies for the above energy balances, we can use the steam tables: 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 5.34241ˆh (520 ºC, 100 bar) 
 
 112
 ⎥⎦
⎤⎢⎣
⎡=⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡+⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡=
kg
kJ 2.2334
kg
kJ 8.257490.0
kg
kJ 77.16810.0ˆ2h 
 (sat. liq at 7.5 kPa) (sat. vap. at 7.5 kPa) 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 77.168ˆ3h (sat. liquid at 0.075 bar) 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 81.342ˆ4h (subcooled liquid at 80 ºC, 100 bar) 
 
Now, we can calculate the heat loads: 
 
 ( ) [ ]kW 308169
kg
kJ 81.342
kg
kJ 5.3424kg/s 100 =⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡=HQ& 
 
 ( ) [ ]kW 216543
kg
kJ 2.2334
kg
kJ 68.771kg/s 100 −=⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡=CQ& 
 
(d) 
Use Equation 2.96: 
 
 0=+ netnet QW && 
 
From Part (c), we know 
 
 [ ] [ ] [ ]kW 91626kW 216543kW 308169 =−=netQ& 
 
Therefore, 
 
 [ ]kW 91626−=netW& 
 
(e) 
Using the results from Parts (c) and (d): 
 
 [ ][ ] 297.0kW 081693
kW 91626 ==η