College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
Solutions to First Quiz – 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
Vin  m in vin Vout  m out vout
Vout m out vout vout


m in vin
vin
Vin
v
Vout  out Vin
vin
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.
Vout
4.6929 ft 3
v
lbm
10 ft 3
 out Vin 
vin
25.80599 ft 3 h
lbm
1.819 ft 3

Vout 
h
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:
160 Btu 61.26 Btu

h  h f (10 psia )
lbm
lbm
x

 0.1652
597.86 Btu
h fg (10 psia )
lbm


v  v f (10 psia )  x v g (10 psia )  v f (10 psia ) 
 25.80599 ft 3 0.02319 ft 3 
0.02319 ft 3

 0.1652

lbm
lbm
lbm


Jacaranda (Engineering) 3519
E-mail: lcaretto@csun.edu
Mail Code
8348
v
4.281 ft 3
lbm
Phone: 818.677.6448
Fax: 818.677.7062
Solutions to first quiz
ME 370, L. S. Caretto, Fall 2010
Page 2
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: Vout 
vout 
Vin . For ideal gas
vin
behavior, v = RT/P. We are given the outlet temperature of T out = 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 T in =
Tsat(Pin = 10 psia) = –41.33oF = 418.34 R.
Using the basic equation from problem 1, the ideal gas equation of state, and the problem data
we find the outlet volume flow rate as follows:
Vout
RTout
10 ft 3



10
psia
759
.
67
R
v
P
P T V
h
 out Vin  out Vin  in out in 
RTin
100 psia 418.34 R 
vin
PoutTin
Pin
1.816 ft 3
Vout 
h