Solutions – Properties of Pure Substances
1. An ammonia compressor has an inlet volume flow rate of 10 ft3/h of saturated vapor at a
pressure of 10 psia. The outlet flow rate, at a pressure of 100 psia and a temperature of
300oF, has the same mass flow rate as the inlet. What is the outlet volume flow rate?
Applying the equation that the volume flow rate equals the mass flow rate times the specific
volume and the statement that the inlet and outlet mass flow rates are the same gives
in
in
out
out
in
out
inin
outout
in
out
outoutoutininin V
v
v
V
v
v
vm
vm
V
V
vmVvmV
From the ammonia tables we find vin = vg(10 psia) = 25.80599 ft3/lbm and vout = v(100 psia, 300oF)
= 4.6929 ft3/lbm. Substituting these values and the inlet volume flow rate of 10 ft3/h into the
equation above gives the desired result.
h
ft
lb
ft
lb
ft
V
v
v
V
m
m
in
in
out
out
3
3
3
10
80599.25
6929.4
h
ft
Vout
3
819.1
2. If ammonia has a pressure of 10 psia and an enthalpy of 160 Btu/lbm, what are its
temperature and specific volume?
At a pressure of 10 psia the values of hf and hg are, respectively, 61.26 Btu/lbm and 659.12
Btu/lbm. Since hf < h < hg, we are in the mixed region so that the temperature is the saturation
temperature at the given pressure of 10 psia. Thus T = –41.33oF.
In the mixed region the specific volume is found from the quality: v = vf + x(vg – vf). We find the
quality from the enthalpy then use the quality to find the specific volume as follows:
1652.0
86.597
26.61160
)10(
)10(
m
mm
fg
f
lb
Btu
lb
Btu
lb
Btu
psiah
psiahh
x
mmm
fgf
lb
ft
lb
ft
lb
ft
psiavpsiavxpsiavv
333 02319.080599.25
1652.0
02319.0
)10()10()10(
m
lb
ft
v3
281.4
3. What would your answer to question 1 be if you used the ideal gas law with R = 0.6301
psia·ft3/lbm·R for ammonia? (Use the ammonia tables to determine the temperature of the
inlet ammonia.)
From the analysis of problem 1 we have the following result:
in
in
out
out V
v
v
V
. For ideal gas
behavior, v = RT/P. We are given the outlet temperature of Tout = 300oF = (300 + 459.67)R =
759.67 R. Since the initial state is a saturated vapor, the initial temperature is the saturation
temperature at the initial pressure of 10 psia. From the tables this temperature is found as Tin =
Tsat(Pin = 10 psia) = –41.33oF = 418.34 R.
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