ASEN 3113 Thermodynamics and Heat Transfer
Homework #5
Assigned: October 12, 2006
Due: October 18, 2006 (noon)
(Total points noted in each section; must clearly show equations with values and units, drawings,
assumptions, etc.)
1. (20 points, 15 points for equations, 5 points for correct answer)
A cold storage room is being used to store strawberries at 1°C. The evaporator and condenser
temperatures are -2 and 45°C, respectively. The refrigeration load is 15 tons (1 ton of
refrigeration load = 3.517 kW). Assume the refrigerant is R-134a. Determine the following:
(a) The refrigerant flow rate
(b) The compressor power requirement
(c) COP
(d) The heat rejected in the condenser
Solution:
First determine the enthalpies in the refrigerator. Assume the fluid entering the compressor is
-2°C and a saturated vapor.
h1= 246.06 kJ/kg ; s1 = 0.9201 kJ/kg*K
Assume the temperature exiting the condenser is 45°C. Look up the values for point 3 at
saturated liquid properties.
h3= 113.75 kJ/kg ; P3 = 1.160 MPa
Using the saturation pressure at point 3 and the entropy at point 2 (s1 = s2), use the superheated
tables to look up the enthalpy at point 2.
The superheated tables give enthalpy at 1.00 MPa and 1.20 MPa; must interpolate between these
tables to create a table of enthalpy and entropy at P3=P2 at 1.16 MPa. Because we are searching
for an entropy of 0.9201 kJ/kg*K, the tables of 1.00 MPa and 1.20 MPa show that the entropy
we are searching for is between 40°C and 50°C. Because the saturation temperatures of 1.00
MPa and 1.20 MPa are different, we must first interpolate the 1.00 MPa table to match the
saturation temperature of 46.32°C as the first temperature in our new 1.16 MPa table. You must
match the temperatures when linear interpolating between two pressures in the superheated table.
Table A-10
40°C
46.32°C (linear inter)
50°C
1.00MPa
Enthalpy, Entropy
268.68, 0.9066
275.95, 0.9294
280.19, 0.9428
1.20 MPa
Enthalpy, Entropy
given(46.32°C)
50°C
270.99, 0.9023
275.52, 0.9164
Now linear interpolate between 1.00 MPa and 1.20 MPa to create a new table at 1.16 MPa for
the two temperatures 46.32°C and 50°C. Linear interpolating between the two pressures gives
the following results.
Find enthalpy at point 2 using entropy at point 1 and pressure at point 3. Use the superheated
tables A10. Linear interpolate between 1.00 MPa and 1.20 MPa to create a new table at 1.16
MPa. Looking at the tables, the entropy we desire is between 46.32°C and 50°C.
1.00 MPa
Enthalpy, Entropy
46.32°C 275.95, 0.9294
50°C
280.19, 0.9428
1.16 MPa
Enthalpy, Entropy
46.32°C 271.98, 0.9077
50°C
276.45, 0.9217
1.20 MPa
Enthalpy, Entropy
46.32°C 270.99, 0.9023
50°C
275.52, 0.9164
Now linear interpolate between 46.32°C and 50°C in the 1.16 MPa using the value of entropy to
find enthalpy. The value for entropy is known, 0.9201 kJ/kg*K. After linear interpolating, the
following results are obtained.
P2 = 1.16 MPa
Enthalpy, Entropy, Temperature
275.94 kJ/kg, 0.9201 kJ/kg*K, 49.6 °C
To find the enthalpy at point 4, Assume that the enthalpy is the same as point 3.
After the previous calculations, the following enthalpies will be used for the rest of the analysis.
h1 = 246.06 kJ/kg
h2 = 275.94 kJ/kg
h3 = 113.75 kJ/kg
h4 = 113.75 kJ/kg
(a) Use the refrigeration load to determine the refrigerant flow rate
Q L  m h1  h4   15tons * 3.517kW / ton  m 246  114kJ / kg
 m  0.40kg / s
(b) The compressor power requirement
 h2  h1   0.40kg / s276  246kJ / kg  12kW
Win  m
(c) Calculate COP
h  h4
COPR  1
h2  h1
COPR 
246  114 kJ kg
276  246 kJ kg
 4.4
(d) The heat rejected in the condenser
Q H  m h2  h3   0.4kg / s276  114kJ / kg
 Q  64.8kW
H
2. (20 points, 15 points for equations, 5 points for correct answer)
For a refrigeration system using R -134a, the evaporator temperature is -5°C, however the
refrigerant entering the compressor is at 5°C. The condenser temperature is 35°C, but the
refrigerant entering the expansion valve is at 30°C.
(a) What is the amount of superheating and subcooling?
(b) If the refrigeration load is 3 tons, calculate the COP for this system.
Solution:
(a) Superheating is the difference in temperature of fluid entering compressor and evaporator.
Supercooling is the difference in temperature of fluid entering condenser and entering expansion
valve.
Superheating  Tentering _ compressor  Tevaporator  5 C   5 C  10  C


Supercooli ng  Tcondenser  Tentering _ exp ansion_ valve  35 C  30  C   5 C
(b) First assume that the temperature of the fluid entering the compressor is a saturated vapor at
5°C.
Look up enthalpy and entropy at point 1 (entering compressor)
h1=250.1 kJ/kg ; s1 = 0.9164 kJ/kg*K
Assume isentropic expansion in compressor, s1 = s2 and assume pressure is constant through the
condenser. Assume that at the exit of the condenser, the fluid is a saturated liquid at 35°C. Look
up enthalpy and saturation pressure at point 3.
h3 = 98.78 kJ/kg; Saturation Pressure @ 3 = 0.887MPa
Using the saturation pressure at point 3 and the entropy at point 2 (s1 = s2), use the superheated
tables to look up the enthalpy at point 2.
The superheated tables give pressures at 0.80 MPa and 0.90 MPa; must interpolate between these
tables to create a table of enthalpy and entropy at P3=P2 at 0.877MPa. Because we are searching
for an entropy of 0.9164 kJ/kg*K, the tables of 0.80 MPa and 0.90MPa show that the entropy we
are searching for is between the saturation temperature and 40°C in both tables. Because the
saturation temperatures of 0.80 MPa and 0.90 MPa are different, we must first interpolate the
0.80 MPa table to match the saturation temperature of 35.53°C as the first temperature in our
new 0.877 MPa table. You must match the temperatures when linear interpolating between two
pressures in the superheated table.
Table A-10
Sat Temp (31.33°C)
35.53°C (linear inter)
40°C
0.80MPa
Enthalpy, Entropy
264.15, 0.9066
268.75, 0.9215
273.66, 0.9374
0.90 MPa
Enthalpy, Entropy
given(35.53°C)
40°C
266.18, 0.9054
271.25, 0.9217
Now linear interpolate between 0.80 MPa and 0.90 MPa to create a new table at 0.877MPa for
the two temperatures 35.53°C and 40°C. Linear interpolating between the two pressures gives
the following results.
0.80MPa
Enthalpy, Entropy
35.53°C 268.75, 0.9215
40°C
273.66, 0.9374
0.877 MPa
Enthalpy, Entropy
35.53°C 266.77, 0.9091
40°C
271.80, 0.9253
0.90 MPa
Enthalpy, Entropy
35.53°C 266.18, 0.9054
40°C
271.25, 0.9217
Now linear interpolate between 35.53°C and 40°C in the 0.877 MPa using the value of entropy
to find enthalpy. The value for entropy is known, 0.9164 kJ/kg*K. After linear interpolating,
the following results are obtained.
P2 = 0.877 MPa
Enthalpy, Entropy, Temperature
269.04 kJ/kg, 0.9164 kJ/kg*K, 37.5 °C
To find the enthalpy at point 4, use the saturated liquid properties at 30°C. Searching in table A8
gives the following enthalpy
h4 = 91.49 kJ/kg
Using the previously calculated values for enthalpy, calculate COP
h1 = 250.1 kJ/kg
h2 = 269.04 kJ/kg
h3 = 98.78 kJ/kg
h4 = 91.49 kJ/kg
COPR 
COPR 
h1  h4
h2  h1
250.1  91.5 kJ kg
269.0  250.1 kJ kg
 8.39
3 (20 points, 15 points for equations, 5 points for correct answer)
A turbojet aircraft is flying with a velocity of 220 m/s at an altitude of 5800 m. The pressure
ratio across the compressor is 14, and the temperature at the turbine inlet is 1440 K. Assume
ideal operations for all components and ideal gas constant specific heats at the altitude
temperature. Determine the following:
(a) The pressure at the turbine exit
(b) The velocity of the exhaust gases
(c) The propulsive efficiency
Solution:
First determine the air properties at the specified altitude.
Temperature = -22.7°C (250 K), Pressure = 48.52 kPa, Cv = 0.716 kJ/kg*K,
Cp = 1.003 kJ/kg*K, γ = 1.401
Process 1-2 is an isentropic compression of an ideal gas in a diffuser. For convenience, we can
assume that the aircraft is stationary and the air is moving toward the aircraft with a velocity of
220 m/s. Ideally, the air will leave the diffuser with a negligible velocity V2=0. Using first law,
we obtain
V22  V12
V12
V12
q12  w12  h2  h1 
 Cp T2  T1  
 T2  T1 
2
2
2Cp
T2  250 K 
T
P2  P1  2
 T1




220m / s 2
21.003kJ / kg * K 
 1
 T2  274 K
 274 K 
 48.52kPa

 250 K 
1.401
0.401
 P2  66.8kPa
Process 2-3 is an isentropic compression of an ideal gas in a compressor
P3  rp P2   1466.8kPa  P3  936kPa
P
T3  T2  3
 P2



 1

 936kPa 
 274 K 

 66.8kPa 
0.401
1.401
 T3  583K
Process 4-5 isentropic expansion of an ideal gas in a turbine (a) answer
wcomp _ in  wturb _ out
h3  h2  h4  h5
CpT3  T2   CpT4  T5 
T5  T4  T3  T2  1440 K  583K  274 K  T5  1131K
T
P5  P4  5
 T4




 1
 1131K 
 936kPa

 1440 K 
1.401
0.401
 P5  403kPa
Process 5-6 isentropic expansion of an ideal gas in a nozzle (b) answer
P
T6  T5  6
 P5



 1

 48.52kPa 
 1131K 

 403kPa 
0.401
1.401
 T6  617 K
V62  V52
V2
 CpT6  T5   6  V6  2CpT5  T6 
2
2
V6  21.003kJ / kg * K 1131K  617 K 
q 56  w56  h6  h5 
 V6  1015m / s
Propulsive efficiency (c) answer
wp
V  V1 V1
1015m / s  850m / s 850m / s
p 
 6

qin CpT4  T3  1.003kJ / kg * K 1440 K  583K 
  p  16.4%
4 (20 points, 15 points for equations, 5 points for correct answers)
Consider a steam power plant operating as an ideal rankine cycle. The condenser pressure is
8kPa. The turbine can operate at two pressures, 18 MPa or 4 MPa. Determine for each of the
turbine pressures:
(a) Mass flow rate of steam
(b) Heat transfer rate of the working fluid passing through the condenser and boiler.
(c) Thermal efficiency
5 (20 points, 15 points for equations, 5 points for correct answers)
(a) A steam power plant is operating as a regenerative vapor power cycle with one closed
feedwater heater that tap off between the first and second turbine stages. The turbine inlet
conditions are pressure 12 Mpa, temperature 510 C. Condenser pressure is 6 kPa. Boiler inlet
temperature is 170 C. Work of the cycle is 320 MW. Determine:
(a) Thermal efficiency
(b) Mass flow rate into the turbine first stage
Assumptions:
(1) Each component is analyzed as a control volume at steady state.
(2) All processes are internally reversible except fro the expansion through the trap
(throttling process) and in the closed feedwater heater.
(3) Turbines, pump, and feedwater heater operate adiabatically.
(4) Kinetic and potential energy functions are negligible.
(5) Condensate exists in the condenser and closed heater as saturated liquid at the respective
pressures.
1 bar = 100kPa