PROBLEM 3.5 Determine the phase or phases in a

PROBLEM 3.5
Determine the phase or phases in a system consisting of H2O at the following conditions and
sketch the p-v and T-v diagrams showing the location of each state.
(a)
(b)
(c)
(d)
(e)
p = 100 lbf/in.2, T = 327.86oF
p = 100 lbf/in.2, T = 240oF
T = 212oF, p = 10 lbf/in.2
T = 70oF, p = 20 lbf/in.2
p = 14.7 lbf/in.2, T = 20oF
(a) p = 100 lbf/in.2, T = 327.86oF
100 lbf/in.2
p
Two-phase
liquid-vapor
mixture
T
327.86oF
100 lbf/in.2
327.86oF (Table A-3E)
v
v
(b) p = 100 lbf/in.2, T = 240oF
p
T<Tsat@p
sub-cooled liquid
100 lbf/in.2
T
327.86oF
.
100 lbf/in.2
327.86oF (Table A-3E)
240oF
v
.
240oF
v
Problem 3.5 (Continued)
(c) T = 212oF, p = 10 lbf/in.2
p<psat@T
superheated
vapor
p
T
14.7 lbf/in.2
10 lbf/in.2
.
14.7 lbf/in.2
.
o
212oF (Table A-3E)
10 lbf/in.2
212 F
v
v
(d) T = 70oF, p = 20 lbf/in.2
p>psat@T
sub-cooled liquid
p
.
T
20 lbf/in.2
0.3632 lbf/in.2 (Table A-2E)
20 lbf/in.2
0.3632 lbf/in.
70oF
.
2
v
2
70oF
v
o
(e) p = 14.7 lbf/in. , T = 20 F
p>psat@T
solid
(T is below the triple
point temperature)
p
.
14.7 lbf/in.
T
14.7 lbf/in.2
0.0505 lbf/in.2 (Table A-5E)
2
0.0505 lbf/in.2
20oF
v
.
20oF
v
PROBLEM 3.9
COMMENT: As the pressure increases, the difference in specific volume between saturated
vapor and saturated liquid decreases. At the critical pressure, the two states
coincide and the difference is zero.
PROBLEM 3.20
PROBLEM 3.94
PROBLEM 3.102
PROBLEM 3.102
PROBLEM 3.111
As shown in Fig. 3.111, a piston-cylinder assembly fitted with a paddle wheel contains air,
initially at p1 = 30 lbf/in.2, T1 = 540oF, and V1 = 4 ft3. The air undergoes a process to a final state
where p2 = 20 lbf/in.2, V2 = 4.5 ft3. During the process, the paddle wheel transfers energy to the
air by work in the amount 1 Btu, and there is energy transfer to the air by heat in the amount of
12 Btu. Assuming ideal gas behavior, and neglecting kinetic and potential energy effects,
determine for the air (a) the temperature at state 2, in oR, and (b) the energy transfer by work
from the air to the piston, in Btu.
KNOWN: Data are provided for air contained in a piston-cylinder assembly fitted with a paddle
wheel.
FIND: For the process of the air, find the temperature at the final state and the energy transfer
by work to the piston.
SCHEMATIC AND GIVEN DATA:
Q = - 12 Btu
Wpw = - 1 Btu
Wpist = ?
ENGINEERING MODEL: 1. The air is the
closed system. 2. Kinetic and potential energy
effects are negligible. 3. The ideal gas model
applies for the air.
ANALYSIS: (a) Using the ideal gas equation of state
m=
=
= 0.324 lb
T2 =
=
= 750oR
and
(b) Noting that W = Wpw + Wpist and ΔU = m(u2 – u1) the energy balance reduces as follows.
ΔKE + ΔPE + ΔU = Q – (Wpw + Wpist)
Wpist = Q – Wpw – m(u2 – u1)
From Table A-22E: u1 = 172.43 Btu/lb and u2 = 128.25 Btu/lb. Thus
Wpist = ( 12 Btu) – ( 1 Btu) – (0.324 lb)(128.25 – 172.43)Btu/lb = 3.31 Btu (out)
PROBLEM 3.117
As shown in Fig. P3.117, 20 ft3 of air at T1 = 600 oR, 100 lbf/in.2 undergoes a polytropic
expansion to a final pressure of 51.4 lbf/in.2 The process follows pV1.2 = constant. The work is
W = 194.34 Btu. Assuming ideal gas behavior for the air, and neglecting kinetic and potential
energy effects, determine
(a) the mass of air, in lb, and the final temperature, in oR.
(b) the heat transfer, in Btu.
KNOWN: Air undergoes a polytropic process in a piston-cylinder assembly. The work is
known.
FIND: Determine the mass of air, the final temperature, and the heat transfer.
SCHEMATIC AND GIVEN DATA:
T1 = 600oR
p1 = 100 lbf/in.2
V1 = 20 ft3
P2 = 51.4 lbf/in.2
Q
Air
W = 194.34 Btu
pV1.2 = constant
p
100
ENGINEERING MODEL: 1. The air is a closed system.
2. Volume change is the only work mode. 3. The process is
polytropic, with pV1.2 = constant and W = 194.34 Btu.
4. Kinetic and potential energy effects can be neglected.
pv1.2 = constant
.1
T1
51.4
ANALYSIS: (a) The mass is determined using the ideal gas equation of state.
m=
=
= 9.00 lb
To get the final temperature, we use the polytropic process, pV1.2 = constant, to evaluate V2 as
follows.
V2 =
Now
=
(20 ft3) = 34.83 ft3
.2
T2
v
PROBLEM 3.117 (CONTINUED)
T2 =
= 537oR
=
Alternative solution for T2
The work for the polytropic process can be evaluated using W =
constant, and incorporating the ideal gas equation of state, we get
W=
. For the process pV1.2 =
=
Solving for T2 and inserting values
T2 =
+ T1 =
+ (600oR) = 537oR
(b) Applying the energy balance; ΔKE + ΔPE + ΔU = Q – W. With ΔU = m(u2 – u1), we get
Q = m(u2 – u1) + W
From Table A-22E: u(600oR) = 102.34 Btu/lb and u(537oR) = 91.53 Btu/lb. Thus,
Q = (9.00 lb)(91.53 – 102.34)Btu/lb + (194.34 Btu) = 97.05 Btu (in)
PROBLEM 3.132