Name:
HW #05
Due: 03/06/2012
ES 346
Solution Set
4-7
1 of 12
Problem:
A piston–cylinder device initially contains 0.07 m3 of nitrogen gas at 130 kPa and
120°C. The nitrogen is now expanded polytropically to a state of 100 kPa and
100°C. Determine the boundary work done during this process.
Solution:
From Tbl A-2:
R = 0.2968 kJ
kg ⋅ K
Assuming the nitrogen is an Ideal Gas, determine the mass of the
gas in the system and the volume of the system after the expansion
(state 2):
(130 kN m )(0.07 m ) kNkJ⋅ m 
N2
130 kPa
120°C
3
2
m=
P1V1
=
= 0.07802 kg
RT1  0.2968 kJ

(120
+
273
K)

kg ⋅ K 

mRT2
=
V2 =
P2
3
(100 + 273 K)
(0.07802 kg) 0.2968 kPa ⋅ m
kg ⋅ K 

= 0.08637 m 3
100 kPa
For a polytropic process:
P1V1n = P2V 2n
⇒ (130 kPa)(0.07 m 3 ) n = (100 kPa)(0.08637 m 3 ) n
⇒ n = 1.249
The boundary work for a polytropic process is determined as:
P2V 2 − P1V1
1− n
(100 kPa)(0.08637 m 3 ) − (130 kPa)(0.07 m 3 )  kJ 
=
= 1.86kJ

3 
1 − 1.249
 kPa ⋅ m 
Wb =
Ans
hw05soln.docx
HW #05
Due: 03/06/2012
Name:
Solution Set
ES 346
4-10
2 of 12
Problem:
A mass of 5 kg of saturated water vapor at 300 kPa is heated at constant
pressure until the temperature reaches 200°C. Calculate the work done by the
steam during this process.
Solution:
Assume that the process is quasi-equilibrium.
Note that the pressure remains constant during this process. We can use Tbl A-4
thru A-6 to obtain the specific volumes at the initial and the final states. We can
also plot this information on a P-v Process Diagram as shown:
P1 = 300 kPa 
3
 ⇒ v 1 = v g = 0.60582 m /kg
Sat. vapor 
P
(kPa)
300
1
2
P2 = 300 kPa 
3
⇒v 2 = 0.71643 m /kg
T2 = 200°C 
V
From its definition, the boundary work is determined to be:
2
Wb ,out = ∫ P dV = P (V 2 − V1 ) = mP (v 2 − v 1 )
1
 1 kJ 

= (5 kg)(300 kPa)(0.71643 − 0.60582) m 3 /kg
3 
 1 kPa ⋅ m 
= 165.78kJ
Ans
The positive sign indicates that work is done by the system (work output).
hw05soln.docx
Name:
HW #05
Due: 03/06/2012
Solution Set
ES 346
4-25
3 of 12
Problem:
1-kg of water that is initially at 90°C with a quality of 10 percent occupies a
spring-loaded piston–cylinder device, such as that in Fig. P4–25. This device is
now heated until the pressure rises to 800 kPa and the temperature is 250°C.
Determine the total work done during the process in kJ.
Solution:
P
Assume that the process is in quasi-equilibrium.
2
800 kPa
The initial state is given to be a
saturated mixture at 90°C.
1
From Tbl A-4:
P1 = 70.183 kPa
v
v 1 = v f + x(v g − v f )
= 0.001036 m
= 0.23686 m
3
kg
+ (0.10)(2.3593 − 0.001036) m
3
kg
3
kg
At the final state conditions of 800 kPa and 250°C, from Tbl A-6:
3
v 2 = 0.29321 m kg
The work done is equal to the area under the process line 1-2:
P1 + P2
m(v 2 − v 1 )
2
(70.183 + 800)kPa
 1 kJ 
=
(1 kg)(0.29321 − 0.23686)m 3 
3 
2
 1 kPa ⋅ m 
= 24.52kJ
Wb ,out = Area =
Ans
hw05soln.docx
Name:
HW #05
Due: 03/06/2012
4-34
4 of 12
Solution Set
ES 346
Problem:
Saturated water vapor at 200°C is isothermally condensed to a saturated liquid in
a piston–cylinder device. Calculate the heat transfer and the work done during
this process, in kJ/kg.
Solution:
Assume that the cylinder is stationary and thus the kinetic and potential energy
changes are zero. Also assume that there are no work interactions involved
other than the boundary work. Finally, assume that the thermal energy stored in
the cylinder itself is negligible and that the compression or expansion process is
quasi-equilibrium.
Consider the contents of the cylinder as the system. This is a closed system
since no mass enters or leaves. The energy balance for this stationary, closed
system can be expressed as:
E in − E out
14
24
3
Net energy transfer
by heat, work, and mass
=
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
Wb ,in − Qout = ∆U = m(u 2 − u1 )
(since KE = PE = 0)
Qout = Wb ,in − m(u 2 − u1 )
From Tbl A-4:
T
m3
T1 = 200°C  v 1 = v g = 0.12721
kg

x1 = 1
 u1 = u g = 2594.2 kJ kg
P1 = P2 = 1554.9 kPa
2
1
3
v
m
T2 = 200°C  v 2 = v f = 0.001157
kg

kJ
x2 = 0
 u 2 = u f = 850.46 kg
hw05soln.docx
Name:
HW #05
Due: 03/06/2012
Solution Set
ES 346
4-34
5 of 12
The work done during this process is:
2
wb,out = ∫ P dV = P (v 2 − v 1 )
1
= (1554.9 kPa)(0.001157 − 0.12721) m
= − 196.0 kJ
3
 1 kJ 


kg  1 kPa ⋅ m 3 


kg
That is:
Ans
wb,in = 196.0 kJ
kg
Substituting into the energy balance equation:
q out = wb ,in − (u 2 − u1 ) = wb ,in + u fg
= 196.0 kJ
kg
+ 1743.7 kJ
kg
= 1940 kJ
Ans
kg
hw05soln.docx
Name:
HW #05
Due: 03/06/2012
ES 346
Solution Set
4-44
6 of 12
Problem:
Saturated R-134a vapor at 100°F is condensed at constant pressure to a
saturated liquid in a closed piston–cylinder system. Calculate the heat transfer
and work done during this process, in Btu/lbm.
Solution:
Use the following assumptions for this solution:
1. The cylinder is stationary and thus the kinetic and potential energy changes
are zero.
2. There are no work interactions involved other than the boundary work.
3. The thermal energy stored in the cylinder itself is negligible.
4. The compression or expansion process is quasi-equilibrium.
The system will consist of the contents of the cylinder. This is a closed system
since no mass enters or leaves. The energy balance for this stationary, closed
system can be expressed as:
E in − E out
14
24
3
Net energy transfer
by heat, work, and mass
=
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
Wb ,in − Qout = ∆U = m(u 2 − u1 )
R-134
100°F
(since KE = PE = 0)
Q
Qout = Wb ,in − m(u 2 − u1 )
From Tbl A-11E, the properties of the refrigerant at the initial and final states are:
3
T1 = 100°F  v 1 = v g = 0.34045 ft lbm
⇒
x1 = 1
u1 = u g = 107.45 Btu

lbm
3
v 2 = v f = 0.01386 ft lbm
u 2 = u f = 44.768 Btu
lbm
T2 = 100°F 
 ⇒ P1 = P2 = 138.93 psia
x2 = 0

u fg = 62.683 Btu
lbm
h fg = 71.080 Btu
lbm
T
2
1
v
hw05soln.docx
Name:
HW #05
Due: 03/06/2012
ES 346
Solution Set
4-44
7 of 12
The boundary work done during this process is:
2
wb,out = ∫ Pdv = P(v 2 − v 1 )
1
3
= (138.93psia)(0.01386 − 0.34045)ft


1 Btu

 = −8.396 Btu
lbm 5.404 psia ⋅ ft 3 
lbm


The negative sign indicates that this work is actually work done on the system.
Therefore, for our energy balance considerations, we change it to read:
wb,in = 8.396 Btu
Ans
lbm
Substituting into energy balance equation gives:
q out = wb ,in − (u 2 − u1 ) = wb ,in + u fg
= 8.396 + 62.683 = 71.08 Btu
Ans
lbm
Note that since the process is assumed to be quasi-equilibrium, the heat transfer
may also be determined from:
− q out = h2 − h1
q out = h fg = 71.080 Btu
lbm
since ∆U + Wb = ∆H.
hw05soln.docx
Name:
HW #05
Due: 03/06/2012
Solution Set
ES 346
4-66
8 of 12
Problem:
Nitrogen gas to 20 psia and 100°F initially occupies a volume of 1 ft3 in a rigid
container equipped with a stirring paddle wheel. After 5000 lbf·ft of paddle wheel
work is done on the nitrogen, what is its final temperature?
Solution:
Assume that Nitrogen is an ideal gas since it is at a high temperature and low
pressure relative to its critical point values of 126.2 K (227.1 R) and 3.39 MPa
(492 psia). Assume that the kinetic and potential energy changes are negligible.
Finally, assume that constant specific heats at room temperature can be used for
nitrogen.
From Tbl A-1E & A-2Ea:
cv = 0.177 Btu
lbm ⋅ ° R
3
R = 0.3830 psia ⋅ ft
Nitrogen
20 psia
100°F
Wpw
lbm ⋅ ° R
Consider the nitrogen as the system. This is a closed system since no mass
crosses the boundaries of the system. The energy balance for this system can
be expressed as:
E in − E out
14
24
3
=
Net energy transfer
by heat, work, and mass
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
Wpw,in = ∆U = mcv (T2 − T1 )
(since KE = PE = 0)
Using the Ideal Gas Law, the mass of nitrogen is:
P1V
(20 psia)(1 ft 3 )
=
= 0.09325 lbm
RT1 (0.3830 psia ⋅ ft 3
)(560 °R)
lbm ⋅ ° R
Substituting into the energy balance:
,
Wpw,in = mcv (T2 − T1 )
m=
(5000 lbf ⋅ ft)
(
)
1 Btu
= (0.09325 lbm) 0.177 Btu
(T − 560)°R
lbm ⋅ °R 2
778.17 lbf ⋅ ft
⇒ T2 = 949° R = 489° F
hw05soln.docx
Ans
HW #05
Due: 03/06/2012
Name:
ES 346
Solution Set
4-126
9 of 12
Problem:
A mass of 5 kg of saturated liquid–vapor mixture of water is contained in a
piston–cylinder device at 125 kPa. Initially, 2 kg of the water is in the liquid phase
and the rest is in the vapor phase. Heat is now transferred to the water, and the
piston, which is resting on a set of stops, starts moving when the pressure inside
reaches 300 kPa. Heat transfer continues until the total volume increases by 20
percent. Determine (a) the initial and final temperatures, (b) the mass of liquid
water when the piston first starts moving, and (c) the work done during this
process. Also, show the process on a P-v diagram.
Solution:
Assume that the process is quasi-equilibrium.
(a) Initially the system is a saturated mixture at 125 kPa pressure, and thus the
initial temperature is found in Tbl A-5 to be:
T1 = Tsat@125 kPa = 105.97°C
Ans
The total initial volume is:
V1 = m f v f + m g v g
3
 + (3kg )1.3750 m 3 
= (2kg ) 0.001048 m
kg
kg 



= 4.127 m 3
P
2
3
1
v
hw05soln.docx
HW #05
Due: 03/06/2012
Name:
ES 346
4-126
Solution Set 10 of 12
The total and specific volumes at the final state are:
V 3 = 1.2V1 = (1.2 )(4.127m 3 ) = 4.953 m 3
V3
3
4.953 m 3
v3 =
=
= 0.9905 m
kg
5 kg
m
Interpolating from Tbl A-6:
,


3
 T3 = 373.6°C
m
v3 = 0.9905
kg 
P3 = 300 kPa
Ans
(b) When the piston first starts moving, P2 = 300 kPa and V2 = V1 = 4.127 m3.
The specific volume at this state is:
v2 =
V2
m
=
3
4.127 m 3
= 0.8254 m
kg
5 kg
which is greater than vg = 0.60582 m3/kg at 300 kPa. Thus no liquid is left in the
cylinder when the piston starts moving.
Ans
(c) No work is done during process 1-2 since V1 = V2. The pressure remains
constant during process 2-3 and the work done during this process is:
3
Wb = ∫ P dV = P2 (V 3 − V 2 )
2
 1 kJ 
 = 247.6kJ
= (300 kPa )(4.953 − 4.127 )m 3 
3 
 1 kPa ⋅ m 
Ans
hw05soln.docx
Name:
HW #05
Due: 03/06/2012
ES 346
4-155
Solution Set 11 of 12
Problem:
Two 10-ft3 adiabatic tanks are connected by a valve. Initially, one tank contains
water at 450 psia with 10 percent quality, while the second contains water at 15
psia with 75 percent quality. The valve is now opened, allowing the water vapor
from the high-pressure tank to move to the low-pressure tank until the pressure
in the two becomes equal. Determine the final pressure and the final mass in
each tank.
Solution:
Assume that the system is stationary and thus the kinetic and potential energy
changes are zero and that there are no heat or work interactions involved.
Consider both tanks to be the system. This is a closed system since no mass
enters or leaves. The energy balance for this stationary, closed system can be
expressed as:
E in − E out
14
24
3
=
Net energy transfer
by heat, work, and mass
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ∆U = U 2 − U 1 = 0
U1 = U 2
m1, A u1, A + m1, B u1, B = m2, A u 2, A + m2, B u 2, B
where the high-pressure and low-pressure tanks are denoted by A and B,
respectively.
From Tbl A-5E, the specific volume in each tank can be found:
3
3
3
P1, A = 450 psia  v 1, A = v f + xv fg = 0.01955 ft
+ (0.10)(1.0324 − 0.01955) ft
= 0.12084 ft
lbm
lbm
lbm

Btu
Btu
Btu
x1, A = 0.10
+ (0.10)(683.52)
= 504.02
 u1, A = u f + xu fg = 435.67
lbm
lbm
lbm
hw05soln.docx
HW #05
Due: 03/06/2012
Name:
4-155
Solution Set 12 of 12
ES 346
3
3
3
P1, B = 15 psia  v 1, B = v f + xv fg = 0.01672 ft
+ (0.75)(26.297 − 0.01672) ft
= 19.727 ft
lbm
lbm
lbm

Btu
Btu
x1, B = 0.75  u1, B = u f + xu fg = 181.16 Btu
+ (0.75)(896.52)
= 853.55
lbm
lbm
lbm
The total mass contained in both tanks is:
VA VB
10 ft 3
10 ft 3
m=
+
=
+
= 83.26 lbm
3
v 1, A v 1, B 0.12084 ft 3
19.727 ft
lbm
lbm
Similarly, the initial total internal energy contained in both tanks is:
U 1 = m1, A u1, A + m1, B u1, B
= (82.75lbm)(504.02 Btu
lbm
) + (0.5069lbm)(853.55 Btu
lbm
) = 42,155 Btu
The specific internal energy and the specific volume are:
u1 = u 2 =
U 42,155 Btu
=
= 506.3 Btu
lbm
m 83.26 lbm
3
20 ft 3
= 0.2402 ft
v1 = v 2 = =
lbm
m 83.26 lbm
V
Now, the final state is fixed. The pressure in this state may be obtained by
iteration in Tbl A-5E after estimating the starting point from the Mollier Diagram:
P2 = 313 psia
Ans
Since the specific volume is now the same in both tanks, and both tanks have the
same volume, the mass is equally divided between the tanks at the end of this
process,
m 83.26 lbm
m 2, A = m 2, B = =
= 41.63lbm
2
2
hw05soln.docx
Ans