EGR 334 Thermodynamics: Homework 08
Problem 3: 49
A closed rigid tank is filled with water. Initially the tank hold 9.9 ft 3 saturated vapor and 0.1 cu. ft saturated liquid,
each at 212 deg F. The water is heated until the tank contains only saturated vapor. For the water, determine a) the
quality at the initial state, b) the temperature at the final state in deg F, and c) the heat transfer in Btu, Kinetic and
potential energy effects can be ignored.
----------------------------------------------------------------------------------------------------------------------------- ------------State 1: V1 = 9.9 ft3
State 2: Saturated vapor.
V1liq = 0.1 ft3
T 1 = 212 oF
Constant volume process: V1  V2  9.9  0.1  10.0 ft 3
Look up Saturation values at T= 212 deg. F.
vf = 0.01672
uf = 180.1 Btu/lbm
Mass of saturated liquid:
Vliq  mliq v f
mliq 

Vliq
vf

p 1 = 14.7 psi
vg = 26.80
ug = 1077.6 Btu/lbm
0.1 ft 3
 5.981 lbm
0.01672 ft 3 / lbm
Q
Mass of saturated vapor:
Vvapor  mvapor v g

mvapor 
Vvapor
vg

9.9 ft 3
 0.3694 lbm
26.80 ft 3 / lbm
Total mass of system:
m  mliq  mvapor  5.981  0.3694  6.350 lbm
Quality of initial state:
x1 
mvapor
mliquid  mvapor

mvapor
mtotal

0.3694
 0.05817  5.8%
6.350
Then
u1  u f  x(ug  u f )  180.1  0.05817(1077.6  180.1)  232.3 Btu / lbm
At state 2, all mass is saturated vapor or
v2  vsat _ 2 
V2
10 ft 3

 1.5748 ft 3 / lbm
mtotal 6.350 lbm
Look up saturated vapor conditions for this specific volume off the saturated vapor table A-2E.
Note that at T = 410 vg = 1.6743 ug = 1117.6
T= 420 vg = 1.502
ug = 1118.3
using interpolation:
T2  410 1.5748  1.6743

420  410 1.502  1.6743
and
T2  415.8 o F

u2  1117.6
1.5748  1.6743

1118.3  1117.6 1.502  1.6743

u2 = 1118.0 Btu/lbm
For a system which doesn't undergo a change of volume the work done by the system is 0.
W 0
1st Law of thermodynamics:
U  Q  W
Q  U  U 2  U1  m(u2  u1 )  6.350lbm (1118.0  232.5)Btu / lbm  5623.0Btu
EGR 334 Thermodynamics: Homework 08
Problem 3:55
A piston cylinder assembly containing water, initially a liquid at 50 deg F. undergoes a process at a constant
pressure of 20 psi to a final state where the water is a vapor at 300 deg. F. Kinetic and potential energy effects are
negligible. Determine the work and heat transfer, in BTU per lb, for each of three parts of the overall process;
a) from the initial liquid state to saturated liquid state, b) from saturated liquid to saturated vapor, and c) from
saturated vapor to the final vapor state all at 20 psi.
----------------------------------------------------------------------------------------------------------------------------- -----------State 1: T1 = 50 deg F.
State 2: Sat. liq.
State 3: sat. vapor
State 4: T 2=300 deg F.
p1 = 20 psi
p2 = 20 psi
Find spec. volume and internal energy at each of the four states.
State 1: v1 ≈ vf (T=50 deg)
State 2: v2 = vf (p=20psi)
State 3: v3 = vg(p=20psi)
= 0.01602 ft3/lb
= 0.01683 ft3/lb
= 20.09 ft3/lb
u1 ≈ uf (T=50 deg)
= 18.06 Btu/lb
u2 = uf (p=20psi)
=196.19 Btu/lb
State 4: v2 = v(T300,p20)
= 22.36 ft3/lb
u3 = ug(p=20psi)
= 1082.0 Btu/lb
u2 = u(T300,p20)
= 1108.7 Btu/lb
State 1 to State 2:
U 2  U1  Q12  W1 2
since the liquid is treated as incompressible:
V  0
W12  0
Q12 / m  (u2  u1 )  196.19  18.06  178.13 Btu / lbm
State 2 to State 3: occurs as a constant pressure process so
W23   pdV  p (V3  V2 )
or
2
12in
1Btu
= 74.3 Btu/lbm
W23 / m  p(v3  v2 )  (20lb / in )(20.09  0.01683) ft / lbm
1 ft 778.17 ft  lb
2
3
Q23 / m  (u3  u2 )  W23 / m  (1082.0  196.19)  74.3Btu / lbm  960.11Btu / lbm
State 3 to State 4: occurs as a constant pressure process so
W3 4   pdV  p (V4  V3 )
or
2
12in
1Btu
= 8.40 Btu/lbm
W34 / m  p(v4  v3 )  (20lb / in )(22.36  20.09) ft / lbm
1 ft 778.17 ft  lb
2
3
Q34 / m  (u4  u3 )  W34 / m  (1108.7  1082.0)  8.4Btu / lbm  35.1Btu / lbm
Summary:
W12 / m  0
Q12 / m  178.13 Btu / lbm
W23 / m  74.3Btu / lbm
W34 / m  8.4 Btu / lbm
Q23 / m  960.11Btu / lbm
Q34 / m  35.1Btu / lbm
EGR 334 Thermodynamics: Homework 08
Problem 3:68
Water contained in a piston cylinder assembly initially at 300 deg F, a quality of 90% and a volume of 6 ft 3 is heated
at constant temperature to saturate vapor. If the rate of heat transfer is 0.3 Btu/s, determine the time in minutes for
this process of the water to occur. Kinetic and potential energy effects are negligible.
----------------------------------------------------------------------------------------------------------------------------- -----------State 1: T1 = 300 deg F.
State 2: sat. vapor
x1 = 90%
T 2 = 300 deg F.
V1 = 6 ft3
Heat Rate: Q dot = 0.3 Btu/s
For T = 300 deg.
p = 66.98 psi
vf = 0.01745 ft3/lbm
uf = 269.5 Btu/lbm
vg = 6.472 ft3/lbm
ug = 1100.0 Btu/lbm
Find the internal energy at each state:
State 1:
v1  v f  x(vg  v f )  0.01745  0.90(6.472  0.01745)  5.827 ft 3 / lb
u1  u f  x(u g  u f )  269.5  0.9(1100  269.5)  1016.95 Btu / lbm
State 2:
v2  6.472 ft 3 / lb
u2  1100 Btu / lbm
Find amount of mass:
m
V1
6 ft 3

 1.0297lbm
v1 5.827 ft 3 / lb
Volume of State 2:
1st Law of Thermo:
U  Q  W
where:
U  1.0297lbm (1100  1016.95)Btu / lbm  85.5Btu
Q  Q t  (0.3Btu / s) t
W   pdV  p(V2  V1 )  p m(v2  v1 )
2
12in
1Btu
 (66.98lb / in )(1.0297lbm)(6.472  5.827) ft / lbm
 8.23Btu
1 ft 778.17 ft  lb
2
therefore:
U  Q  W
85.5Btu  (0.3Btu / s )t  8.23Btu
(85.5  8.23) Btu
t
 312.4s  5.21min
0.3Btu / s
3
EGR 334 Thermodynamics: Homework 08
Problem 3:78
One lbm of water contained in a piston cylinder assembly undergoes the power cycle shown in Fig. 3.78. For each of
the four processes, evaluate the work and heat transfer each in Btu. For the overall cycle, evaluate the thermal
efficiency.
----------------------------------------------------------------------------------------------------------------------------- --------------mass = 1 lbm
State 1: p1 = 700 psi
saturated liquid.
State 2: p 2=700 psi
saturated vapor
State 3: p3= 70 psi mixture
v3 = v2
State 4: p4 = 70 psi
v4 = v1
p
mixture
1
700 psi
Start by looking up intensive properties at the four states:
State 1: T1= Tsat at p = 700 psi.
from table A-3E: T1= 503.23 oF
v1 = vf(p=700) = 0.02051 ft3/lbm
u1 = uf(p=700) = 488.9 Btu/lbm
70 psi
4
State 2: T2= Tsat at p = 700 psi.
from table A-3E: T1= 503.23 oF
v2 = vg(p=700) = 0.656 ft3/lbm
u2 = ug(p=700) = 1117.0 Btu/lbm
v1=v4
State 3: p3= 70 psi
( from table A-3E: p=70 psi then vf=0.01748 ft3/lbm vg= 6.209 ft3/lbm
uf = 272.6 Btu/lbm ug = 1100.6 Btu/lbm
v2 = v3 = 0.656 ft3/lbm
therefore quality is
x3 
v3  v f
vg  v f

0.656  0.01748
 0.1031  10.31%
6.209  0.01748
and then
u3  u f  x(ug  u f )  272.6  0.1031(1100.6  272.6)  358.0Btu / lbm
State 4:
v4 = v1 = 0.02051 ft3/lbm
therefore quality is
x4 
v4  v f
vg  v f

0.02051  0.01748
 0.000489  0.0489%
6.209  0.01748
and then
u4  u f  x4 (u g  u f )  272.6  0.000489(1100.6  272.6)  273.0 Btu / lbm
Process 1 to 2:
U  Q  W
where:
U12  m(u2  u1 )  1lbm(1117  488.9) Btu / lbm  628.1Btu
W1 2   pdV  p1 (V2  V1 )  p1m(v2  v1 )
2
 (700lb f / in2 )(1lbm)(0.656  0.02051) ft 3 / lbm
12in
1Btu
 82.3Btu
1 ft 778.17lb f  ft
Q12  U  W  628.1  82.3  710.4Btu
Process 2 to 3:
U  Q  W where: U 23  m(u3  u2 )  1lbm(358.0  1117) Btu / lbm  759Btu
W23  0
Q23  U  W  759  0  759Btu
2
1
3
v2=v3
v
Process 3 to 4:
U  Q  W
where:
U 34  m(u4  u3 )  1lbm(273  358) Btu / lbm  85Btu
W3 4   pdV  p3 (V4  V3 )  p3 m(v4  v3 )
2
12in
1Btu
 (70lb f / in )(1lbm)(0.02051  0.656) ft / lbm
 8.23Btu
1 ft 778.17lb f  ft
2
3
Q34  U  W  85  8.23  93.23Btu
Process 4 to 1:
U  Q  W where: U 41  m(u1  u4 )  1lbm(488.9  273) Btu / lbm  215.9 Btu
W41  0
Q41  U  W  215.9  0  215.9Btu
Heat input into the system:
Qin  Q12  Q41  710.4  215.9  926.3Btu
Heat dumped to the environment:
Qout  Q23  Q34  759  93.23  852.2Btu
Work done by system:
Wcycle  82.3  8.23  74.07 Btu
thermal efficiency:

Wcycle
Qin

74.07
 0.07996  8.0%
926.3