CHE 302
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Problem set #3 Solution
1) Determine the quality of a two-phase liquid–vapor mixture of
(a) H2O at 20°C with a specific volume of 7 m3/kg.
(b) Propane at 15 bar with a specific volume of 0.0077 m3/kg.
(c) Refrigerant 134a at 60°C with a specific volume of 0.009 m3/kg.
(d) Ammonia at 1 MPa with a specific volume of 0.0510 m3/kg.
Solution
(a) H2O at 20°C with a specific volume of 7 m3/kg.
(b) Propane at 15 bar with a specific volume of 0.0077 m3/kg.
(c) Refrigerant 134a at 60°C with a specific volume of 0.009 m3/kg
(d) Ammonia at 1 MPa with a specific volume of 0.0510 m3/kg.
2) A closed system consists of a two-phase liquid–vapor mixture of H2O in equilibrium at 300°F.
The quality of the mixture is 0.80 (80%) and the mass of saturated vapor present is 2 lb.
Determine
(a) the mass of saturated liquid present, in lb,
(b) the total volume of the system, in ft3.
Solution
(a) the mass of saturated liquid present, in lb,
quality x = mg/m =
mg
mg  m f
 mf =
1  0.8
(2) = 0.5 lb
0.8
b) the total volume of the system, in ft3.
V = mv = (2 + 0.5)(5.18) = 12.95 ft3
3) Three lb of saturated water vapor, contained in a closed rigid tank whose volume is 13.30 ft3,
is heated to a final temperature of 400°F. Determine
(a) the pressure at the initial state, in lbf/in.2
(b) the pressure at the final state, in lbf/in.2
Solution
(a) the pressure at the initial state, in lbf/in.2
v1 = v2 = 13.30/3 = 4.4333 ft3/lb, from Table A-3E, P1 = Psat = 100 psia
(b) the pressure at the final state, in lbf/in.2
T = 400°F, v2 = 4.4333 ft3/lb  P2 = 110.8 psia superheated vapor
4) Determine the values of the specified properties at each of the following conditions.
(a) For Refrigerant 134a at T = 60°C and v = 0.072 m3/kg, determine p in kPa, and h in kJ/kg.
(b) For ammonia at p = 2 bar and v = 0.005 m3/kg, determine T in °C and u in kJ/kg.
(c) For saturated liquid and vapor mixture of Refrigerant 22 at T = -10°C and u = 200 kJ/kg,
determine p in bar and v in m3/kg.
Solution
(a) For Refrigerant 134a at T = 60°C and v = 0.072 m3/kg, determine p in kPa, and h in kJ/kg.
p = 358.4 kPa, h = 451.7 kJ/kg.
(b) For ammonia at p = 2 bar and v = 0.005 m3/kg, determine T in °C and u in kJ/kg.
T = − 18.86 °C and u = 101 kJ/kg.
(c) For saturated liquid and vapor mixture of Refrigerant R-22 at T = -10°C and u = 200 kJ/kg,
determine p in bar and v in m3/kg
p = 3.54 bar,
u uf
200  32.74
= 0.8791
ug  u f
223  32.74
v = vf + x(vg − vf) = 0.0007588 + (0.8791)(0.06534 − 0.0007588) = 0.05753 m3/kg
x=
=
5) Refrigerant 134a is compressed with no heat transfer in a piston–cylinder assembly from 30
lbf/in.2, 20°F to 160 lbf/in.2 The mass of refrigerant is 0.04 lb. For the refrigerant as the system,
work done on the system is W = 0.56 Btu. Kinetic and potential energy effects are negligible.
Determine the final temperature, in °F.
U = m(u2 − u1) = Q + W = W  u2 = u1 + Q/m = 161.2 + 0.56/0.04 = 175.2 Btu/lb
Refrigerant 134a at 160 lbf/in.2 and u2 = 175.2 Btu/lb  T = 121.6oF
6) A system consisting of 2 lb of water vapor, initially at 300°F and occupying a volume of 20
ft3, is compressed isothermally to a volume of 9.05 ft3. The system is then heated at constant
volume to a final pressure of 120 lbf/in.2 During the isothermal compression there is energy
transfer by work of magnitude 90.8 Btu into the system. Kinetic and potential energy effects are
negligible. Determine the heat transfer, in Btu, for each process.
State 1: T1 = 300oF, v1 = 20/2 = 10 ft3/lb, u1 = 1104 Btu/lb
State 2: T2 = 300oF, v2 = 9.05/2 = 4.525 ft3/lb, u2 = 849.5 Btu/lb
State 3: P3 = 120 psia, v3 = 9.05/2 = 4.525 ft3/lb, u3 = 1166 Btu/lb
Process 1-2: U12 = Q12 + W12  Q12 = m(u2 − u1) − W12
Q12 = (2)(849.5 − 1104) − 90.8 = − 600 Btu
Process 2-3: U23 = Q23 + W23  Q23 = m(u3 − u2) = (2)(1166 − 849.5) = 633 Btu
7) A 0.2 ft3 aluminum bar with density of 169 lb/ft3, initially at 200°F, is placed in an open tank
together with 4 ft3 of liquid water, initially at 70°F with specific volume = 0.01605 ft3/lb. For the
water and the bar as the system, determine the final equilibrium temperature, in °F, ignoring heat
transfer between the tank and its surroundings. Heat capacity of aluminum and water are Ca =
0.216 Btu/lboF and Cw = 0.998 Btu/lboF
U = 0  Ua + Uw = 0
Let Tf is the final equilibrium temperature
maCa(Tf − Ta) + mwCw(Tf − Tw) = 0  Tf =
ma CaTa  mwCwTw
ma Ca  mwCw
mw = 4 ft3/0.01605 ft3/lb = 249.2 lb
ma = (2 ft3)(169 lb/ft3) = 33.8 lb
Tf =
ma CaTa  mwCwTw (33.8)(0.216)(200  460)  (249.2)(0.998)(70 46 0)
=
ma Ca  mwCw
(33.8)(0.216)  (249.2)(0.998)
Tf = 533.7oR = 73oF
8) Determine the volume, in m3, occupied by 40 kg of nitrogen (N2) at 17 MPa, 180 K.
V = 0.103 m3