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) p = 100 lbf/in.2, T = 327.86
oF
(b) p = 100 lbf/in.2, T = 240
oF
(c) T = 212oF, p = 10 lbf/in.
2
(d) T = 70oF, p = 20 lbf/in.
2
(e) p = 14.7 lbf/in.2, T = 20
oF
(a) p = 100 lbf/in.2, T = 327.86
oF
(b) p = 100 lbf/in.2, T = 240
oF
p T
v v
100 lbf/in.2
327.86oF (Table A-3E)
100 lbf/in.2
327.86oF
Two-phase
liquid-vapor
mixture
100 lbf/in.2
327.86oF (Table A-3E)
240oF
p
v v
T
. 327.86
oF
100 lbf/in.2
240oF .
T
PROBLEM 3.19
A tank contains a two-phase liquid-vapor mixture of Refrigerant
22 at 10 bar. The mass of
saturated liquid in the tank is 25 kg and the quality is 60%.
Determine the volume of the tank, in
m3, and the fraction of the total volume occupied by saturated
vapor.
First, determine the specific volume using Eq. 3.2 and data from
Table A-8 at 10 bar.
vx = vx + x (vg vf) = 0.8352 x 10-3
+ (0.6) (0.0236 - 0.8352 x 10-3
) = 0.01449 m3/kg
The total mass is determined from the mass of saturated liquid
and the definition of quality, as
follows.
m = mf / (1 x) = (25 kg) / (1 0.6) = 62.5 kg
Now, the volume is
V = vx m = (0.01449 m3/kg) (62.5 kg) = 0.9056 m
3
The volume occupied by saturated vapor is
Vg = vg (m mf) = (0.0236 m3/kg) (62.5 25)kg = 0.885 m3
fraction occupied by vapor = (0.885)/(0.9056) = 0.977
(97.7%)
Note: Even though the vapor is only 60% of the mixture by mass,
it occupies nearly the entire
volume because the specific volume of saturated liquid is much
smaller than the specific volume
of saturated vapor at this pressure.
R 22
p = 10 bar
x = 0.6 saturated liquid
mf = 25 kg
saturated vapor p = 10 bar
x= 0.6 . T = 23.40oC
T
v
