Chapter 3
1. A 1.8-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the
rest is in the vapor form. Determine (a) the pressure of the steam, and (b) the quality of the saturated
mixture
Properties At 220°C vf = 0.001190 m3/kg and vg = 0.08609 m3/kg (Table A-4).
Analysis (a) Two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. The
pressure of the steam is the saturation pressure at the given temperature. Then the pressure in the
tank must be the saturation pressure at the specified temperature,
P = Tsat @220°C = 2320 kPa
(b) The total mass and the quality are determined as
mf =
Vf
1/3 × (1.8 m3 )
=
= 504.2 kg
v f 0.001190 m3/kg
mg =
Vg
2/3 × (1.8 m3 )
=
= 13.94 kg
v g 0.08609 m3/kg
Steam
1.8 m3
220°C
mt = m f + mg = 504.2 + 13.94 = 518.1 kg
x=
mg
mt
=
13.94
= 0.0269
518.1
(c) The density is determined from
v = v f + x(v g − v f ) = 0.001190 + (0.0269)(0.08609) = 0.003474 m 3 /kg
ρ=
1
v
=
1
= 287.8 kg/m 3
0.003474
2. A piston–cylinder device contains 0.85 kg of refrigerant- 134a at _10°C. The piston that is free to
move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now,
heat is transferred to refrigerant-134a.
Analysis (a) The final pressure is equal to the initial pressure, which is determined from
P2 = P1 = Patm +
mp g
πD 2 /4
= 88 kPa +
⎞
(12 kg)(9.81 m/s 2 ) ⎛
1 kN
⎜
⎟ = 90.4 kPa
2
2 ⎟
⎜
π (0.25 m) /4 ⎝ 1000 kg.m/s ⎠
(b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10°C and at the
final state of 90.4 kPa and 15°C are (from EES)
v1 = 0.2302 m3/kg
h1 = 247.76 kJ/kg
3
v 2 = 0.2544 m /kg h2 = 268.16 kJ/kg
The initial and the final volumes and the volume change are
V1 = mv 1 = (0.85 kg)(0.2302 m 3 /kg) = 0.1957 m 3
3
V 2 = mv 2 = (0.85 kg)(0.2544 m /kg) = 0.2162 m
3
ΔV = 0.2162 − 0.1957 = 0.0205 m 3
(c) The total enthalpy change is determined from
ΔH = m(h2 − h1 ) = (0.85 kg)(268.16 − 247.76) kJ/kg = 17.4 kJ/kg
R-134a
0.85 kg
10°C
Q
3. Consider a sealed can that is filled with refrigerant- 134a. The contents of the can are at the room
temperature of 25°C. Now a leak develops, and the pressure in the can drops to the local atmospheric
pressure of 90 kPa. The temperature of the refrigerant in the can is expected to drop to (rounded to the
nearest integer)
(a) 0°C
(b) -29°C (c) -16°C
(d) 5°C
(e) 25°C
Answer (b) -29°C
T1=25 "C"
P2=90 "kPa"
T2=TEMPERATURE(R134a,x=0,P=P2)
"Some Wrong Solutions with Common Mistakes:"
W1_T2=T1 "Assuming temperature remains constant"
4. Complete this table for H2O:
T, °C
50
120.21
250
110
P, kPa
12.352
200
400
600
v, m3 / kg
4.16
0.8858
0.5952
0.001051
Phase description
Saturated mixture
Saturated vapor
Superheated vapor
Compressed liquid
5. Complete this table for refrigerant-134a:
T, °C
-8
30
-12.73
80
P, kPa
320
770.64
180
600
v, m3 / kg
0.0007569
0.015
0.11041
0.044710
Phase description
Compressed liquid
Saturated mixture
Saturated vapor
Superheated vapor
6. A 0.5-m3 vessel contains 10 kg of refrigerant-134a at _20°C. Determine (a) the pressure, (b) the
total internal energy, and (c) the volume occupied by the liquid phase.
Analysis (a) The specific volume of the refrigerant is
v=
V
m
=
0.5 m 3
= 0.05 m 3 /kg
10 kg
At -20°C, vf = 0.0007362 m3/kg and vg = 0.14729 m3/kg (Table A-11). Thus the tank contains
saturated liquid-vapor mixture since vf < v < vg , and the pressure must be the saturation pressure at
the specified temperature,
P = Psat @ − 20o C = 132.82 kPa
(b) The quality of the refrigerant-134a and its total internal energy are determined from
x=
v −v f
v fg
=
0.05 − 0.0007362
= 0.3361
0.14729 − 0.0007362
R-134a
10 kg
-20°C
u = u f + xu fg = 25.39 + 0.3361× 193.45 = 90.42 kJ/kg
U = mu = (10 kg)(90.42 kJ/kg) = 904.2 kJ
(c) The mass of the liquid phase and its volume are determined from
m f = (1 − x)mt = (1 − 0.3361) × 10 = 6.639 kg
V f = m f v f = (6.639 kg)(0.0007362 m3/kg) = 0.00489 m 3
7. Determine the specific volume, internal energy, and enthalpy of compressed liquid water at 100°C
and 15 MPa using the saturated liquid approximation. Compare these values to the ones obtained
from the compressed liquid tables.
Analysis Compressed liquid can be approximated as saturated liquid at the given temperature. Then
from Table A-4,
T = 100°C ⇒ v ≅ v f @ 100°C = 0.001043 m 3 /kg (0.72% error)
u ≅ u f @ 100°C = 419.06 kJ/kg
h ≅ h f @ 100°C = 419.17 kJ/kg
(1.02% error)
(2.61% error)
From compressed liquid table (Table A-7),
v = 0.001036 m 3 /kg
P = 15 MPa ⎫
u = 414.85 kJ/kg
T = 100°C ⎬⎭
h = 430.39 kJ/kg
The percent errors involved in the saturated liquid approximation are listed above in parentheses.
8. A piston–cylinder device contains 0.8 kg of steam at 300°C and 1 MPa. Steam is cooled at constant
pressure until one-half of the mass condenses.
(a) Show the process on a T-v diagram.
(b) Find the final temperature.
(c) Determine the volume change.
Analysis (b) At the final state the cylinder contains saturated liquid-vapor mixture, and thus the final
temperature must be the saturation temperature at the final pressure,
(Table A-5)
T = Tsat@1 MPa = 179.88°C
(c) The quality at the final state is specified to be x2 = 0.5. The
specific volumes at the initial and the final states are
P1 = 1.0 MPa ⎫
3
⎬ v = 0.25799 m /kg
T1 = 300 o C ⎭ 1
P2 = 1.0 MPa
x2 = 0.5
H2O
300°C
1 MPa
(Table A-6)
⎫
⎬ v 2 = v f + x2v fg
⎭ = 0.001127 + 0.5 × (0.19436 − 0.001127)
= 0.09775 m3/kg
T
1
2
Thus,
ΔV = m(v 2 − v 1 ) = (0.8 kg)(0.09775 − 0.25799)m 3 /kg = −0.1282 m 3
v
9. A rigid tank contains water vapor at 250°C and an unknown pressure. When the tank is cooled to
150°C, the vapor starts condensing. Estimate the initial pressure in the tank.
Analysis This is a constant volume process (v = V /m = constant), and the initial specific volume is
equal to the final specific volume that is
v 1 = v 2 = v g @150°C = 0.39248 m 3 /kg (Table A-4)
T °C
since the vapor starts condensing at 150°C.
Then from Table A-6,
1
T1 = 250°C
⎫
⎬ P = 0.60 MPa
v1 = 0.39248 m3/kg ⎭ 1
H2O
250
T1= 250°C
P1 = ?
150
2
v
10. A piston–cylinder device initially contains steam at 3.5 MPa, superheated by 5°C. Now, steam
loses heat to the surroundings and the piston moves down hitting a set of stops at which point the
cylinder contains saturated liquid water. The cooling continues until the cylinder contains water at
200°C. Determine
(a) the initial temperature,
(b) the enthalpy change per unit mass of the steam by the time the piston first hits the stops, and
(c) the final pressure and the quality (if mixture).
Analysis (a) The saturation temperature of steam at 3.5 MPa is
Tsat@3.5 MPa = 242.6°C (Table A-5)
Then, the initial temperature becomes
T1 = 242.6+5 = 247.6°C
Also,
P1 = 3.5 MPa ⎫
⎬h1 = 2821.1 kJ/kg
T1 = 247.6°C ⎭
(Table A-6)
Steam
3.5 MPa
Q
(b) The properties of steam when the piston first hits the stops are
P2 = P1 = 3.5 MPa ⎫ h2 = 1049.7 kJ/kg
⎬
3
x2 = 0
⎭ v 2 = 0.001235 m /kg
(Table A-5)
Then, the enthalpy change of steam becomes
Δh = h2 − h1 = 1049.7 − 2821.1 = -1771 kJ/kg
(c) At the final state
v 3 = v 2 = 0.001235 m3/kg ⎫⎪ P3 = 1555 kPa
T3 = 200°C
⎬
⎪⎭ x3 = 0.0006
(Table A-4 or EES)
The cylinder contains saturated liquid-vapor mixture with a small mass of vapor at the final state.
11. A 400-L rigid tank contains 5 kg of air at 25°C. Determine the reading on the pressure gage if the
atmospheric pressure is 97 kPa.
Assumptions At specified conditions, air behaves as an ideal gas.
Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1).
Analysis Treating air as an ideal gas, the absolute pressure in the tank is determined from
P=
mRT
V
=
(5 kg)(0.287 kPa ⋅ m 3 /kg ⋅ K)(298 K)
0.4 m 3
Pg
= 1069.1 kPa
Thus the gage pressure is
Pg = P − Patm = 1069.1 − 97 = 972.1 kPa
Air
400 L
25°C
12. Determine the specific volume of superheated water vapor at 10 MPa and 400°C, using (a) the
ideal-gas equation, (b) the generalized compressibility chart, and (c) the steam tables. Also determine
the error involved in the first two cases.
Properties The gas constant, the critical pressure, and the critical temperature of water are, from
Table A-1,
R = 0.4615 kPa·m3/kg·K,
Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis (a) From the ideal gas equation of state,
v=
RT (0.4615 kPa ⋅ m 3 /kg ⋅ K)(673 K)
=
= 0.03106 m 3 /kg (17.6% error)
P
(10,000 kPa)
(b) From the compressibility chart (Fig. A-15),
10 MPa
P
⎫
=
= 0.453 ⎪
Pcr 22.06 MPa
⎪
⎬ Z = 0.84
673 K
T
⎪
TR =
=
= 1.04
⎪⎭
Tcr 647.1 K
PR =
Thus,
v = Zv ideal = (0.84)(0.03106 m 3 /kg) = 0.02609 m 3 /kg (1.2% error)
(c) From the superheated steam table (Table A-6),
P = 10 MPa
T = 400°C
} v = 0.02644 m /kg
3
H2O
10 MPa
400°C