ASEN 3113 Thermodynamics and Heat Transfer s1

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 1.00MPa 1.20 MPa Enthalpy, Entropy Enthalpy, Entropy 40°C 268.68, 0.9066 46.32°C (linear inter) 275.95, 0.9294 given(46.32°C) 270.99, 0.9023 50°C 280.19, 0.9428 50°C 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 1.16 MPa 1.20 MPa Enthalpy, Entropy Enthalpy, Entropy Enthalpy, Entropy 46.32°C 275.95, 0.9294 46.32°C 271.98, 0.9077 46.32°C 270.99, 0.9023 50°C 280.19, 0.9428 50°C 276.45, 0.9217 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 ˙ QL  m˙ h1  h4   15tons *3.517kW / ton  m˙ 246 114kJ / kg  m˙  0.40kg / s

(b) The compressor power requirement

Win  m˙ h2  h1   0.40kg / s276  246kJ / kg  12kW

(c) Calculate COP

h1  h4 COPR  h2  h1 246 114 kJ   kg COPR   4.4 276  246 kJ   kg

(d) The heat rejected in the condenser ˙ QH  m˙ h2  h3   0.4kg / s276 114kJ / kg ˙  QH  64.8kW 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 ° ° ° Supercooling  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 0.80MPa 0.90 MPa Enthalpy, Entropy Enthalpy, Entropy Sat Temp (31.33°C) 264.15, 0.9066 35.53°C (linear inter) 268.75, 0.9215 given(35.53°C) 266.18, 0.9054 40°C 273.66, 0.9374 40°C 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 0.877 MPa 0.90 MPa Enthalpy, Entropy Enthalpy, Entropy Enthalpy, Entropy 35.53°C 268.75, 0.9215 35.53°C 266.77, 0.9091 35.53°C 266.18, 0.9054 40°C 273.66, 0.9374 40°C 271.80, 0.9253 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

h1  h4 COPR  h2  h1 250.1 91.5 kJ   kg COPR   8.39 269.0  250.1 kJ   kg 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

V 2 V 2 V 2 V 2 q  w  h  h  2 1  CpT  T  1  T  T  1 12 12 2 1 2 2 1 2 2 1 2Cp 220m / s2 T  250K   T  274K 2 21.003kJ / kg * K  2

 1.401  T   1  274K  0.401  2  P2  P1    48.52kPa   P2  66.8kPa  T1   250K 

Process 2-3 is an isentropic compression of an ideal gas in a compressor

P3  rp P2   1466.8kPa  P3  936kPa

 1 0.401  P   936kPa 1.401  3    T3  T2    274K   T3  583K  P2   66.8kPa 

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  1440K  583K  274K  T5  1131K

 1.401  T   1 1131K 0.401  5    P5  P4    936kPa   P5  403kPa  T4  1440K  Process 5-6 isentropic expansion of an ideal gas in a nozzle (b) answer  1 0.401  P   48.52kPa 1.401  6    T6  T5    1131K   T6  617K  P5   403kPa  V 2 V 2 V 2 q  w  h  h  6 5  CpT  T   6  V  2CpT  T  56 56 6 5 2 6 5 2 6 5 6

V6  21.003kJ / kg * K 1131K  617K 

 V6  1015m / s

Propulsive efficiency (c) answer

wp V6 V1 V1 1015m / s  850m / s850m / s  p    qin CpT4  T3  1.003kJ / kg * K1440K  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