4.29 Refrigerant 134a flows at steady state through a horizontal tube having an inside diameter
of 0.05 m. The refrigerant enters the tube with a quality of 0.1, temperature of 36oC, and
velocity of 10 m/s. The refrigerant exits the tube at 9 bar as a saturated liquid. Determine
(a) the mass flow rate of the refrigerant, in kg/s.
(b) the velocity of the refrigerant at the exit, in m/s.
(b) the rate of heat transfer, in kW, and its associated direction with respect to the refrigerant.
KNOWN: Refrigerant 134a flows through a tube.
FIND: (a) the mass flow rate of the refrigerant, in kg/s, (b) the velocity of the refrigerant at the
exit, in m/s, and (c) the rate of heat transfer, in kW, and its associated direction with respect to
the refrigerant.
SCHEMATIC AND GIVEN DATA:
Refrigerant 134a
x1 = 0.1
1
T1 = 36oC
V1 = 10 m/s
D = 0.05 m
2
Saturated liquid
p2 = 9 bar
ENGINEERING MODEL:
1. The control volume shown with the schematic is at steady state.
2. For the control volume, Wcv  0 , and pe = 0.
ANALYSIS:
 2
 D V1
A1V1  4
 .
(a) Mass flow rate at the inlet is m1 

v1
v1
Specific volume of the refrigerant at the inlet is determined using quality
v1 = vf1 + x1(vg1 – vf1)
From Table A-10 at T1 = 36oC, vf1 = 0.0008590 m3/kg and vg1 = 0.0223 m3/kg. Solving
v1 = 0.0008590 m3/kg + (0.1)(0.0223 m3/kg – 0.0008590 m3/kg) = 0.003003 m3/kg
Substituting values and solving for mass flow rate give
 
2 m 
 0.05 m  10 
4
 s
m 1   
m3
0.003003
kg
m 1 = 6.54 kg/s
1
(b) From the mass rate balance
2  m
1 
m
A 2V2 A1V1

v2
v1
Since the tube diameter is constant, the areas cancel. Solving for exit velocity yields
v 
V2  V1 2 
 v1 
Specific volume for the saturated liquid refrigerant at the exit is obtained from Table A-11:
v2 = vf2 = 0.0008576 m3/kg
Substituting and solving for the exit velocity give
3

 0.0008576 m 
kg 
 m 
V2  10 

 s 
m3 
0
.
003003

kg 

V2 = 2.86 m/s
(c) The energy rate balance
 [(h1 – h2) + ½ (V12 – V22) + g(z1 – z2)]
0 = Q cv  Wcv  m
simplifies to
 [(h2 – h1) + ½ (V22 – V12)]
Q cv  m
Specific enthalpy of the refrigerant at the inlet is determined using quality
h1 = hf1 + x1hfg1
From Table A-10 at T1 = 36oC, hf1 = 100.25 kJ/kg and hfg1 = 166.15 kJ /kg. Solving
h1 = 100.25 kJ /kg + (0.1)(166.15 kJ /kg) = 116.87 kJ /kg
Specific enthalpy for the saturated liquid refrigerant at the exit is obtained from Table A-11:
h2 = hf2 = 99.56 kJ/kg
2
Substituting values and solving for rate of heat transfer give
2
2


m  m



 2.86   10 
kg  
kJ 
1 kJ
1 N  kW
s  s


Qcv   6.54  99.56 - 116.87 

s 
kg
2
1000 N  m 1 kg  m   kJ 


 
s 2   s 

Q cv = −113.5 kW
Since the heat transfer rate is negative, heat transfer is from the refrigerant.
3