359539134 Solution Manual Engineering and Chemical Thermodynamics Milo D Koretsky pdf

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Chapter 1 Solutions 
Engineering and Chemical Thermodynamics 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Wyatt Tenhaeff 
Milo Koretsky 
 
Department of Chemical Engineering 
Oregon State University 
 
koretsm@engr.orst.edu 
 
 
 
 
 2 
1.2 
An approximate solution can be found if we combine Equations 1.4 and 1.5: 
 
molecular
keVm =
2
2
1  
molecular
kekT =2
3 
 
m
kTV 3≈∴

 
 
Assume the temperature is 22 ºC. The mass of a single oxygen molecule is kg 1014.5 26−×=m . 
Substitute and solve: 
 
 [ ]m/s 6.487=V

 
 
The molecules are traveling really, fast (around the length of five football fields every second). 
 
Comment: 
We can get a better solution by using the Maxwell-Boltzmann distribution of speeds that is 
sketched in Figure 1.4. Looking up the quantitative expression for this expression, we have: 
 
 dvvv
kT
m
kT
mdvvf 22
2/3
2
exp
2
4)(





−




=
π
π 
 
where f(v) is the fraction of molecules within dv of the speed v. We can find the average speed 
by integrating the expression above 
 
[ ]m/s 4498
)(
)(
0
0 ===
∫
∫
∞
∞
m
kT
dvvf
vdvvf
V
π

 
 
 
 
 
 
 
 3 
1.3 
Derive the following expressions by combining Equations 1.4 and 1.5: 
 
 
a
a m
kTV 32 =

 
b
b m
kTV 32 =

 
 
Therefore, 
 
 
a
b
b
a
m
m
V
V
=
2
2


 
 
Since mb is larger than ma, the molecules of species A move faster on average. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 4 
1.4 
We have the following two points that relate the Reamur temperature scale to the Celsius scale: 
 
 ( )Reamurº 0 C,º 0 and ( )Reamurº 80 C,º 100 
 
Create an equation using the two points: 
 
 ( ) ( ) Celsius º 8.0Reamurº TT = 
 
At 22 ºC, 
 
 Reamurº 6.17 =T 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 5 
1.5 
 
(a) 
After a short time, the temperature gradient in the copper block is changing (unsteady state), so 
the system is not in equilibrium. 
 
(b) 
After a long time, the temperature gradient in the copper block will become constant (steady 
state), but because the temperature is not uniform everywhere, the system is not in equilibrium. 
 
(c) 
After a very long time, the temperature of the reservoirs will equilibrate; The system is then 
homogenous in temperature. The system is in thermal equilibrium. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 6 
1.6 
We assume the temperature is constant at 0 ºC. The molecular weight of air is 
 
 kg/mol 0.029g/mol 29 ==MW 
 
Find the pressure at the top of Mount Everest: 
 
 ( ) ( ) [ ]( )( )
( ) 





















⋅
−
=
K 73.152
Kmol
J 314.8
m 8848m/s .819kg/mol029.0expatm 1P 
 kPa 4.33atm 330.0 ==P 
 
Interpolate steam table data: 
 
 Cº 4.71=satT for kPa 4.33=satP 
 
Therefore, the liquid boils at 71.4 ºC. Note: the barometric relationship given assumes that the 
temperature remains constant. In reality the temperature decreases with height as we go up the 
mountain. However, a solution in which T and P vary with height is not as straight-forward. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 7 
1.7 
To solve these problems, the steam tables were used. The values given for each part constrain 
the water to a certain state. In most cases we can look at the saturated table, to determine the 
state. 
 
(a) Subcooled liquid 
Explanation: the saturation pressure at T = 170 [o
(b) Saturated vapor-liquid mixture 
C] is 0.79 [MPa] (see page 508); 
Since the pressure of this state, 10 [bar], is greater than the saturation 
pressure, water is a liquid. 
Explanation: the specific volume of the saturated vapor at T = 70 [oC] is 5.04 
[m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 508); Since the volume 
of this state, 3 [m3/kg], is in between these values we have a saturated vapor-
liquid mixture. 
(c) Superheated vapor 
Explanation: the specific volume of the saturated vapor at P = 60 [bar] = 6 [MPa], 
is 0.03244 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 511); Since 
the volume of this state, 0.05 [m3
(d) Superheated vapor 
/kg], is greater than this value, it is a vapor. 
Explanation: the specific entropy of the saturated vapor at P = 5 [bar] = 0.5 
[MPa], is 6.8212 [kJ/(kg K)] (see page 510); Since the entropy of this state, 
7.0592 [kJ/(kg K)], is greater than this value, it is a vapor. In fact, if we go to the 
superheated water vapor tables for P = 500 [kPa], we see the state is constrained 
to T = 200 [oC]. 
 
 
 
 8 
1.8 
From the steam tables in Appendix B.1: 
 
 








=
kg
m 003155.0ˆ
3
criticalv ( )MPa 089.22 C,º 15.374 == PT 
 
At 10 bar, we find in the steam tables 
 
 








=
kg
m 001127.0ˆ
3
sat
lv 
 








=
kg
m 19444.0ˆ
3
sat
vv 
 
Because the total mass and volume of the closed, rigid system remain constant as the water 
condenses, we can develop the following expression: 
 
 ( ) satvsatl
critical vxvxv ˆˆ1ˆ +−= 
 
where x is the quality of the water. Substituting values and solving for the quality, we obtain 
 
 0105.0=x or 1.05 % 
 
A very small percentage of mass in the final state is vapor. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 9 
1.9 
The calculation methods will be shown for part (a), but not parts (b) and (c) 
 
(a) 
Use the following equation to estimate the specific volume: 
 
( ) ( ) ( ) ( )[ ]Cº 250 MPa, 8.1ˆCº 250 MPa, 0.2ˆ5.0Cº 250 MPa, 8.1ˆCº 250 MPa, 9.1ˆ vvvv −+= 
Substituting data from the steam tables, 
 
 ( )








=
kg
m 11821.0Cº 250 MPa, 9.1ˆ
3
v 
 
From the NIST website: 
 ( )








=
kg
m 11791.0Cº 250 MPa, 9.1ˆ
3
NISTv 
 
Therefore, assuming the result from NIST is more accurate 
 
 % 254.0%100
ˆ
ˆˆ
Difference % =





×
−
=
NIST
NIST
v
vv
 
 
(b) 
Linear interpolation: 
 ( )








=
kg
m 13284.0Cº 300 MPa, 9.1ˆ
3
v 
 
NIST website: 
 ( )








=
kg
m 13249.0Cº 300 MPa, 9.1ˆ
3
NISTv 
 
Therefore, 
 
 % 264.0Difference % = 
 
(c) 
Linear interpolation: 
 ( )








=
kg
m 12406.0Cº 270 MPa, 9.1ˆ
3
v 
 
 10 
Note: Double interpolation is required to determine this value. First, find the molar volumes at 
270 ºC and 1.8 MPa and 2.0 MPa using interpolation. Then, interpolate between the results from 
the previous step to find the molar volume at 270 ºC and 1.9 MPa. 
 
NIST website: 
 ( )








=
kg
m 12389.0Cº 270 MPa, 9.1ˆ
3
NISTv 
 
Therefore, 
 
 % 137.0Difference % = 
 
With regards to parts (a), (b), and (c), the values found using interpolation and the NIST website 
agree very well. The discrepancies will not significantly affect the accuracy of any subsequent 
calculations. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 11 
1.10 
For saturated temperature data at 25 ºC in the steam tables, 
 
 





=





=
kg
L 003.1
kg
m 001003.0ˆ
3
sat
lv 
 
Determine the mass: 
 
 [ ] kg 997.0
kg
L 003.1
L 1
ˆ
=






== sat
lv
Vm 
 
Because molar volumesof liquids do not depend strongly on pressure, the mass of water at 25 ºC 
and atmospheric pressure in a one liter should approximately be equal to the mass calculated 
above unless the pressure is very, very large. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 12 
1.11 
First, find the overall specific volume of the water in the container: 
 
 








==
kg
m 2.0
kg 5
m 1ˆ
33
v 
 
Examining the data in the saturated steam tables, we find 
 
 satv
sat
l vvv ˆˆˆ << at bar 2=
satP 
 
Therefore, the system contains saturated water and the temperature is 
 
 Cº 23.120== satTT 
 
Let lm represent the mass of the water in the container that is liquid, and vm represent the mass 
of the water in the container that is gas. These two masses are constrained as follows: 
 
 kg 5=+ vl mm 
 
We also have the extensive volume of the system equal the extensive volume of each phase 
 
 Vvmvm satvv
sat
ll =+ ˆˆ 
 ( ) ( ) 333 m1/kgm8857.0kg/m 001061.0 =+ vl mm 
 
Solving these two equations simultaneously, we obtain 
 
 kg 88.3=lm 
 kg12.1=vm 
 
Thus, the quality is 
 
 224.0
kg 5
kg 12.1
===
m
mx v 
 
The internal energy relative to the reference state in the saturated steam table is 
 
 satvv
sat
ll umumU ˆˆ += 
 
From the steam tables: 
 
 13 
 [ ]kJ/kg 47.504ˆ =satlu 
 [ ]kJ/kg 5.2529ˆ =satvu 
 
Therefore, 
 
 ( ) [ ]( ) ( ) [ ]( )kJ/kg 529.52kg .121kJ/kg 47.504kg 88.3 +=U 
 kJ 4.4790=U 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 14 
1.12 
First, determine the total mass of water in the container. Since we know that 10 % of the mass is 
vapor, we can write the following expression 
 
 ( ) ( )[ ]satvsatl vvmV ˆ1.0ˆ9.0 += 
 
From the saturated steam tables for MPa 1=satP 
 
 








=
kg
m 001127.0ˆ
3
sat
lv 
 








=
kg
m 1944.0ˆ
3
sat
vv 
 
Therefore, 
 
( ) ( )
















+
















=
+
=
kg
m 1944.01.0
kg
m 001127.09.0
m 1
ˆ1.0ˆ9.0 33
3
sat
v
sat
l vv
Vm 
kg 9.48=m 
 
and 
 
 ( ) ( )( )kg/m 1944.0kg 9.481.0ˆ1.0 3== satvv vmV 
 3m 950.0=vV 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 15 
1.13 
In a spreadsheet, make two columns: one for the specific volume of water and another for the 
pressure. First, copy the specific volumes of liquid water from the steam tables and the 
corresponding saturation pressures. After you have finished tabulating pressure/volume data for 
liquid water, list the specific volumes and saturation pressures of water vapor. Every data point 
is not required, but be sure to include extra points near the critical values. The data when plotted 
on a logarithmic scale should look like the following plot. 
 
The Vapor-Liquid Dome for H2O
0.0010
0.0100
0.1000
1.0000
10.0000
100.0000
0.0001 0.001 0.01 0.1 1 10 100 1000
Specific Volume, v (m3/kg)
Pr
es
su
re
, P
 (M
Pa
) Vapor-Liquid VaporLiquid
Critical Point
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 16 
1.14 
The ideal gas law can be rewritten as 
 
 
P
RTv = 
 
For each part in the problem, the appropriate values are substituted into this equation where 
 
 



⋅
⋅
=
Kmol
barL 08314.0R 
 
is used. The values are then converted to the units used in the steam table. 
 
(a) 






=













==



=
−
kg
m 70.1
L
m 10
mol
kg 018.0
mol
L 7.30
ˆ
mol
L 7.30
33
3
MW
vv
v
 
 
For part (a), the calculation of the percent error will be demonstrated. The percent error will be 
based on percent error from steam table data, which should be more accurate than using the ideal 
gas law. The following equation is used: 
 
 %100
ˆ
ˆˆ
Error% 





×
−
=
ST
STIG
v
vv
 
 
where IGv is the value calculated using the ideal gas law and STv is the value from the steam 
table. 
 
 %62.1%
kg
m 6729.1
kg
m 6729.1
kg
m 70.1
Error%
3
33
=
































−








= 
 
This result suggests that you would introduce an error of around 1.6% if you characterize boiling 
water at atmospheric pressure as an ideal gas. 
 
 
 17 
 
(b) 
 








=



=
kg
m 57.3ˆ
mol
L 3.64
3
v
v
 
 =Error% 0.126% 
 
For a given pressure, when the temperature is raised the gas behaves more like an ideal gas. 
 
(c) 
 








=



=
kg
m 0357.0ˆ
mol
L 643.0
3
v
v
 
 %Error = 8.87% 
 
(d) 
 








=



=
kg
m 0588.0ˆ
mol
L 06.1
3
v
v
 
 %Error = 0.823% 
 
 
The largest error occurs at high P and low T. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 18 
1.15 
The room I am sitting in right now is approximately 16 ft long by 12 feet wide by 10 feet tall – 
your answer should vary. The volume of the room is 
 
 ( )( )( ) 33 m 4.54ft 1920ft 10ft 16ft 12 ===V 
 
The room is at atmospheric pressure and a temperature of 22 ºC. Calculate the number of moles 
using the ideal gas law: 
 
 ( )( )
( )K 15.295
Kmol
J 314.8
m 4.54Pa 1001325.1 35










⋅
×
==
RT
PVn 
 mol 2246=n 
 
Use the molecular weight of air to obtain the mass: 
 
 ( ) ( )( ) kg 2.65kg/mol 029.0mol 27.2246 === MWnm 
 
 
That’s pretty heavy. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 19 
1.16 
First, find the total volume that one mole of gas occupies. Use the ideal gas law: 
 
 
( )
( ) 





=
×










⋅
==
mol
m 0244.0
Pa 101
K 15.293
Kmol
J 314.8 3
5P
RTv 
 
Now calculate the volume occupied by the molecules: 
 
 




=





×





= 3
3
4
molcule one
of Volume
 moleper 
molecules ofNumber 
rNv A
occupied π 
 ( ) 


 ×






 ×
= −
310
23
m 105.1
3
4
mol 1
molecules 10022.6
πoccupiedv 
 








×= −
mol
m 1051.8
3
6occupiedv 
 
Determine the percentage of the total volume occupied by the molecules: 
 
 % 0349.0%100Percentage =








×=
v
voccupied 
 
A very small amount of space, indeed. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 20 
1.17 
Water condenses on your walls when the water is in liquid-vapor equilibrium. To find the 
maximum allowable density at 70 ºF, we need to find the smallest density of water vapor at 40 ºF 
which satisfies equilibrium conditions. In other words, we must find the saturation density of 
water vapor at 40 ºF. From the steam tables, 
 
 








=
kg
m 74.153ˆ
3
sat
vv (sat. water vapor at 40 ºF = 4.44 ºC) 
 
We must correct the specific volume for the day-time temperature of 70 ºF (21.1 ºC ) with the 
ideal gas law. 
 
 
3.294
K 294.3
6.277
K 6.277, ˆˆ
T
v
T
vsatv = 
 ( )








=
















==
kg
m 163
kg
m 74.153
K 6.277
K 3.294ˆˆ33
K 6.277,
6.277
3.294
K 294.3
sat
vvT
Tv 
 
Therefore, 
 
 





×== − 3
3
m
kg 1013.6
ˆ
1ˆ
sat
vv
ρ 
 
If the density at 70 ºF is greater than or equal to [ ]33 kg/m 1013.6 −× , the density at night when the 
temperature is 40 ºF will be greater than the saturation density, so water will condense onto the 
wall. However, if the density is less than [ ]33 kg/m 1013.6 −× , then saturation conditions will not 
be obtained, and water will not condense onto the walls. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 21 
1.18 
We consider the air inside the soccer ball as the system. We can answer this question by looking 
at the ideal gas law: 
 
 RTPv = 
 
If we assume it is a closed system, it will have the same number of moles of air in the winter as 
the summer. However, it is colder in the winter (T is lower), so the ideal gas law tells us that Pv 
will be lower and the balls will be under inflated. 
 
Alternatively, we can argue that the higher pressure inside the ball causes air to leak out over 
time. Thus we have an open system and the number of moles decrease with time – leading to the 
under inflation. 
 
 
 22 
1.19 
First, write an equation for the volume of the system in its initial state: 
 
Vvmvm satvv
sat
ll =+ ˆˆ 
 
Substitute ( )mxml −= 1 and xmmv = : 
 
 ( )[ ] Vvxvxm satvsatl =+− ˆˆ1 
 
At the critical point 
 
 Vvm critical =ˆ 
 
Since the mass and volume don’t change 
 
 ( ) satvsatl
critical vxvxv ˆˆ1ˆ +−= 
 
From the steam tables in Appendix B.1: 
 
 








=
kg
m 003155.0ˆ
3
criticalv ( )MPa 089.22 C,º 15.374 == PT 
 
At 0.1 MPa, we find in the steam tables 
 
 








=
kg
m 001043.0ˆ
3
sat
lv 
 








=
kg
m 6940.1ˆ
3
sat
vv 
 
Therefore, solving for the quality, we get 
 
 00125.0=x or 0.125 % 
 
99.875% of the water is liquid. 
 
 23 
1.20 
As defined in the problem statement, the relative humidity can be calculated as follows 
 
 
saturationat water of mass
 waterof massHumidity Relative = 
 
We can obtain the saturation pressure at each temperature from the steam tables. At 10 [o
[kPa] 10.122.19.0 =×=Waterp
C], the 
saturation pressure of water is 1.22 [kPa]. This value is proportional to the mass of water in the 
vapor at saturation. For 90% relative humidity, the partial pressure of water in the vapor is: 
 
 
 
This value represents the vapor pressure of water in the air. At 30 [o
[kPa] 12.263.55.0 =×=Waterp
C], the saturation pressure of 
water is 4.25 [kPa]. For 590% relative humidity, the partial pressure of water in the vapor is: 
 
 
 
Since the total pressure in each case is the same (atmospheric), the partial pressure is 
proportional to the mass of water in the vapor. We conclude there is about twice the amount of 
water in the air in the latter case. 
 24 
1.21 
 
(a) 
When you have extensive variables, you do not need to know how many moles of each substance 
are present. The volumes can simply be summed. 
 
 ba VVV +=1 
 
(b) 
The molar volume, v1
ba
ba
tot nn
VV
n
Vv
+
+
== 11
, is equal to the total volume divided by the total number of moles. 
 
 
 
We can rewrite the above equation to include molar volumes for species a and b. 
 
 b
ba
b
a
ba
a
ba
bbaa v
nn
nv
nn
n
nn
vnvnv
+
+
+
=
+
+
=1 
 bbaa vxvxv +=1 
 
(c) 
The relationship developed in Part (a) holds true for all extensive variables. 
 
 ba KKK +=1 
 
(d) 
 
 bbaa kxkxk +=1 
 
 
 
 
 
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
 
 
∆ ?h 1
P
 
 
∆ ?h 2
 
 
∆ ?h 3
 
∆ ?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 25ˆCº 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. H2






=
kg
kJ 87.104ˆ ,
2
satl
OHh
O vapor at 25 ºC) 
 (sat. H2






=
kg
kJ 3.2442qˆ
O liquid at 25 ºC) 
 
which upon substitution gives 
 
 
 
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 shownbelow. 
 
 
 
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. 
 
 
 
 
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: 
[ ]( ) [ ]( ) 





=





−





=−=∆
kg
kJ 15.314
kg
kJ 87.104
kg
kJ 02.419C25at ˆC100at ˆˆ o1,
o
2, 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
kg 5
=




=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 07923.0kg
m 02641.0kg) 3(ˆ =
















== vmV 
 
Now 2V and 2v are found as follows 
 
 [ ]36312 m 4536.0Pa 100.2
J 748740m 07923.0 =
×
−
−=−=
EP
WVV 
 [ ][ ] 





===
kg
m 1512.0
kg 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 020.0kg
m 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 specific internal 
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 == 5.111
5.1 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: 
 
 
 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: 
 
 
 29 
 
Now Q can be solved. 
 
 [ ] [ ] [ ]kJ 1662kJ 6580kJ 4918 −=−=−∆= WUQ 
 
 
(c) 
This part asks us to design
WU =∆
 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 
 
 
 
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
=








−−−= ∫
iv
v
i vd
v
vPvvPmW 
 
Substitutingin 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: 
 
 
 
 
 
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






=
kg
m 043.0ˆ
3
iv
 to the above equation. 
 
Solution A: 
 
 
 
 31 
which gives 
 
 [ ]bar 8.70=iP 
 
This solution is graphically shown below: 
 
 
 
Solution B 
 
 





=
kg
m 0762.0ˆ
3
iv 
 
which gives 
 
 [ ]bar 0.30=iP 
 
This solution is graphically shown below: 
 
 
 32 
2.19 
 
(a) 
Force balance to find k: 
 
Piston
Fspring=kx Fatm=PatmA
Fgas=PgasA
Fmass=mg
 
 
 Pgas = Patm +
mg
A −
kx
A 
since ∆V=Ax 
 Pgas = Patm +
mg
A +
k∆V
A2
 
 
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 +
mg
A +
k∆V
A2
 
 
 
 ∫ 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 [m 3]
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 m
3
0.05 m3
 
 
 
 
n
 ⇒ n = 1.35 
 
Now using the force balance 
 
Pgas,int = Patm +
mg
A +
k∆V
A2 
 
with the above equation yields: 
 
 Pint = Pi
Vi
V int( )
n
= Patm +
mg
A +
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 ×10
5 N
m2
 
 
 
 
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 [m 3]
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
∫∫ ==
f
i
f
i
V
V
V
V
gasrevC dV
V
constdVPW 35.1,
 = const, we have the following 
expression: 
 
 
 
Calculate the constant from the initial state 
 35 
 
 ( )( ) 335.13 5 1075.1m05.0Pa101. ×=×=const 
 
Therefore, 
 
WC,rev = 1.75 ×10
3
.05
.03
∫ dV
V1.35
 = − 1.75×10
3
0.35 V
−0.35
.05
.03
 [ ]J2800−= 
 
 
 
 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 wasused 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: 
 
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 [o 250 [C] oC] 
v 0.44779 [m3 0.11144 [m/kg] 3/kg] 
u 3475.35 [kJ/kg] 2679.58 [kJ/kg] 
V 0.01 m 0.05 m3 3 
 
 
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: 
 
 [ ][ ] 





==
+
+
=
kg
m 30.0
kg 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: 
 
 
 
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
wu =∆
 in 
the bubble) remains constant. Assuming the process is adiabatic and potential and kinetic energy 
effects are negligible, the energy balance is 
 
 
 
Expressions for work and internal energy can be substituted to provide 
 
 ( ) ( )1212 vvPdvPTTc EEv −−=−=− ∫ 
 
where cv = cP – R. Since CO2
( ) 





−−=





−−=−
1
12
2
1
1
2
2
12 P
TPTR
P
T
P
TRPTTc Ev
 is assumed an ideal gas, the expression can be rewritten as 
 
 
 
where the equation was simplified since the final pressure, P2, is equal to the external pressure, 
PE






+=





+
1
2
12 11 Pc
RPT
c
RT
vv
. Simplifying, we get: 
 
 
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. H2
00125.0=x
O at P = 1 bar) 
 
Now the quality can be found 
 
 
 
Thus, the quality of the water is 0.125%. 
 
(c) 
To determine the required heat input, 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 (H2








=








=
kg
m 1.2506ˆ
kg
m 33.417ˆ
3
1
3
1
v
l
u
u
O at its critical point) 
 (sat. H2
[ ] [ ]J 1010.5kJ 6.5102 6×==∆U
O at P=1 bar) 
 
Evaluation of the expression reveals 
 
 
 
 
 
 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, 
 
 






−
+=
t
c
h
surr
veCTT 1 
 
where C1
time
Te
m
pe
ra
tu
re
T0
Tsurr
 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. 
 
 
 
 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, 
 
 






−
+=
t
c
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
s
in
ininoutout
sys
WQhmhm
dt
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. H2






=
kg
kJ 2.3177iˆnh
O vapor at 1 MPa) 
 
 (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
l
 
 
Mass 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
outoutoutout
um
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
s
in
ininoutout
sys
WQhmhm
dt
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 thesystem. 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

 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

 
 
From the problem statement and the assumptions made, 
 
 



=



=
s
m 0
s
kg 200
2V
m


 
 
Therefore, 
 
 


=
s
m 101V

 
 
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, )(2
1 VVMWee airkk

−=− 
 
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

 π 
 
The energy balance is now 
 
 62 
( ) ( )










−+








++++







= ∫ − 21223222
2
2
2
2 )(
2
1
4
2
1
VVMWdTETDTCTBTARVD
RT
PW air
T
T
s


 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: 
 
 
 
To solve for nWs  / we need a first law balance. With negligible eK and eP
( ) sWQhhn  ++−= 210
, the 1st law for a 
steady state process becomes: 
 
 
 
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 2
1ˆˆˆˆ VVeehh KK

−=−=− 
 
Since the steam outlet velocity is much greater than the velocity of the inlet, the above 
expression is approximately equal to 
 
 ( )2212 2
1ˆˆ Vhh

−=− 
 
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