An air conditioner operates on the vapor-compression refrigeration cycle with refrigerant-134a as the refrigerant. The air conditioner is used to keep a space at 21°C while rejecting the waste heat to the ambient air at 37°C. The refrigerant enters the compressor at 180 kPa superheated by 2.7°C at a rate of 0.06 kg/s and leaves the compressor at 1200 kPa and 60°C. R-134a is subcooled by 6.3°C at the exit of the condenser. Determine (a) the rate of cooling provided to the space, in Btu/h, and the COP, (b) the isentropic efficiency and the exergy efficiency of the compressor, (c) the exergy destruction in each component of the cycle and the total exergy destruction in the cycle, and (d) the minimum power input and the second-law efficiency of the cycle.
(a)
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The rate of cooling provided to the space.
The coefficient of performance of the air conditioner.
Answer to Problem 118RP
The rate of cooling provided to the space is
The coefficient of performance of the air conditioner is
Explanation of Solution
Show the T-s diagram as in Figure (1).
Express the refrigeration load.
Here, specific enthalpy at state 1 and 4 is
Express the rate of work input.
Express the coefficient of performance of the air conditioner.
Conclusion:
Perform the unit conversion of pressure at state 1 from
Refer to Table A-13, obtain the saturation temperature at a pressure of 0.18 MPa as
Calculate the temperature at state 1.
Substitute
Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to pressure and inlet temperature of
Here, specific enthalpy and entropy is
Write the formula of interpolation method of two variables.
Here, the variables denote by x and y is specific entropy at state 2 and specific enthalpy at state 2s respectively.
Show the specific enthalpy at state 2s corresponding to specific entropy by referring the table A-13 at pressure of 1.2 MPa and specific entropy of
Specific entropy at state 2 |
Specific enthalpy at state 2s |
0.9268 | 278.28 |
0.9484 | |
0.9615 | 289.66 |
Substitute the Table (1) values in Equation (IV) to get 285.34 kJ/kg.
Thus, the specific enthalpy at state 2s is,
Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to exit state of pressure and temperature of
Here, specific enthalpy and entropy at exit state are
Refer to Table A-13, obtain the saturation temperature at a pressure of 1.2 MPa as
Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to state 3 of pressure and temperature of
Here, specific enthalpy and entropy at state 3 are
Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to state 4 of pressure and specific enthalpy of
Substitute 0.06 kg/s for
Thus, the rate of cooling provided to the space is
Substitute 0.06 kg/s for
Substitute 2.670 kW for
Thus, the coefficient of performance of the air conditioner is
(b)
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The isentropic efficiency of the compressor.
The exergy efficiency of the compressor.
Answer to Problem 118RP
The isentropic efficiency of the compressor is
The exergy efficiency of the compressor is
Explanation of Solution
Express the isentropic efficiency of the compressor.
Here, specific enthalpy at state 2s is
Calculate the reversible power for the compressor.
Here, ambient air temperature is
Calculate the exergy efficiency for the compressor.
Conclusion:
Substitute 285.34 kJ/kg for
Thus, the isentropic efficiency of the compressor is
Substitute 0.06 kg/s for
Substitute 2.428 kW for
Thus, the exergy efficiency of the compressor is
(c)
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The exergy destruction in each component of the cycle.
The total exergy destruction in the cycle.
Answer to Problem 118RP
The exergy destruction in each component of the cycle are
The total exergy destruction in the cycle is
Explanation of Solution
Compressor
Express the entropy generation for cycle 1-2.
Express the exergy destruction for the cycle 1-2.
Condenser
Express the entropy generation for cycle 2-3.
Here, the rate of heat rejected is
Calculate the value of
Substitute
Express the exergy destruction for the cycle 2-3.
Expansion valve
Express the entropy generation for cycle 3-4.
Express the exergy destruction for the cycle 3-4.
Evaporator
Express the entropy generation for cycle 4-1.
Here, temperature at refrigeration load is
Express the exergy destruction for the cycle 4-1.
Calculate the total exergy destruction.
Conclusion:
Substitute 0.06 kg/s for
Substitute
Substitute 0.06 kg/s for
Substitute
Substitute 0.06 kg/s for
Substitute
Substitute 0.06 kg/s for
Substitute
Thus, the exergy destruction in each component of the cycle are
Substitute 0.2426 kW for
Thus, the total exergy destruction in the cycle is
(d)
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The minimum power input of the cycle.
The second law efficiency of the cycle.
Answer to Problem 118RP
The minimum power input of the cycle is
The second law efficiency of the cycle is
Explanation of Solution
Express the exergy of the heat transferred from the low-temperature medium.
Here, minimum power input of the cycle is the exergy of the heat transferred from the low-temperature medium.
Calculate the second law efficiency of the cycle.
Calculate the total exergy destruction in the cycle.
Conclusion:
Substitute 8.213 kW for
Thus, the minimum power input to the cycle is
Substitute 0.4470 kW for
Thus, the second law efficiency of the cycle is
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Chapter 11 Solutions
THERMODYNAMICS-SI ED. EBOOK >I<
- A gas refrigeration system utilizing air as the working fluid possesses a pressure ratio of 3. The air is directed into the compressor at 5°C. The high pressure air is then cooled down to 40°C by rejecting heat to the ambien environment. The refrigerant exits the turbine at 85°C and subsequently absorbs heat from the cooling area prior to its entry into the regenerator. The mass flow rate of air is 0.45 kg/s. Given that the iClosetropic efficiencies are 82 percent for the compressor and 87 percent for the turbine, and using constan specific heats at ambient temperature, find: (a) The effectiveness of the regenerator. (b) The rate at which heat is extracted from the cooling area. (c) The Coefficient of Performance (COP) of the cycle.arrow_forwardA commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at –30°C by rejecting its waste heat to cooling water that enters the condenser at 18°C at a rate of 0.25 kg/s and leaves at 26°C. The refrigerant enters the condenser at 1.2 MPa and 65°C and leaves at 42°C. The inlet state of the compressor is 60 kPa and –34°C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine the theoretical maximum refrigeration load for the same power input to the compressor. Water 26°C ↑ 18°C 1.2 MPа 42°C 65°C Condenser Expansion Win valve Evaporator Compressor 60 kPa -34°Carrow_forwardA 5 ton refrigeration system operates on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid. The refrigerant enters the compressor as a saturated vapor at 360 kPa and is compressed to 1.6 MPa with a compressor outlet temperature of 70°C. Also, it enters the throttling valve as a saturated liquid. Determine the quality of the refrigerant at the end of the throttling process isentropic efficiency of the compressor determine the coefficient of performance Draw the cycle on a T-s diagram determine the 2 and determine the compressor input power with respect to the saturation lines [Note: A "ton" of refrigeration is a term used in the air conditioning industry. Historically, it is the rate of cooling necessary to freeze one ton (2000 lbm) of water in 24 hours. Using the "latent heat" of fusion, the mass and the time, this cooling rate works out to be 12,000 Btu/h = 3.517 kW.]arrow_forward
- A 5 ton refrigeration system operates on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid. The refrigerant enters the compressor as a saturated vapor at 360 kPa and is compressed to 1.6 MPa with a compressor outlet temperature of 70°C. Also, it enters the throttling valve as a saturated liquid. Determine the quality of the refrigerant at the end of the throttling process ( determine the isentropic efficiency of the compressor determine the coefficient of performance ), and determine the compressor input power. Draw the cycle on a T-s diagram with respect to the saturation lines [Note: A "ton" of refrigeration is a term used in the air conditioning industry. Historically, it is the rate of cooling necessary to freeze one ton (2000 lbm) of water in 24 hours. Using the "latent heat" of fusion, the mass and the time, this cooling rate works out to be 12,000 Btu/h = 3.517 kW.]arrow_forwardConsider a cooling device operating in the ideal vapor compression refrigeration cycle using R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 180 kPa and exits at 800 kPa and 50°C. It leaves the condenser as a saturated liquid at 800 kPa. What is the efficiency coefficient of the cooling device?arrow_forwardConsider a 332 kJ/min refrigeration system that operates on an ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 140 kPa and is compressed to 800 kPa. Determine the power input to the compressor in kW.arrow_forward
- A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at -30°C by rejecting its waste heat to cooling water that enters the condenser at 18°C at a rate of 0.25 kg/s and leaves at 26°C. The refrigerant enters the condenser at 1.2 MPa and 65°C and leaves at 42°C. The inlet state of the compressor is 60 kPa and -34 °C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet (b) the refrigeration load in kW (c) the COP of the refrigerator (d) the theoretical maximum refrigeration load in kW for the same power input to the compressor. To obtain enthalpy values on the cooling water use steam tables.arrow_forwardA refrigerating system has a load of 50 kJ/s uses ammonia as working fluid. The refrigerated space is kept at a temperature of –30°C. Condensing pressure is at 1.0 MPa. The ammonia refrigerant exits the evaporator with 5-degree superheat and at 160 kPa. Superheated refrigerant is discharged from the compressor at a temperature of 140°C. Wire drawing at the compressor suction is 5 kPa and the compressor discharge is 20 kPa. If the overall mechanical efficiency is 82%, determine the power input to the compressor.arrow_forwardA refrigerated room is kept at -27°C by a vapor-compression cycle with R-134a as the refrigerant. Heat is rejected to cooling water that enters the condenser at 16°C at a rate of 0.22 kg/s and leaves at 23 °C. The refrigerant enters the condenser at 1.2 MPa and 65 °C and leaves at 42°C. The inlet state of the compressor is 60 kPa and -34 °C and the compressor is estimated to gain a net heat of 150 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor.arrow_forward
- Consider a gas refrigeration system with air as the working fluid. The pressure ratio is 5.5. Air enters the compressor at 0°C. The high-pressure air is cooled to 35 °C by rejecting heat to the surroundings. The refrigerant leaves the turbine at -95°C and then it absorbs heat from the refrigerated space before entering the regenerator. The mass flow rate of air is 0.55 kg/s. Assuming isentropic efficiencies of 90% for both the compressor and the turbine, determine (a) the effectiveness of the regenerator, (b) the rate of heat removal from the refrigerated space, and (c) the COP of the cycle. Also, determine (d) the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle. In this cycle, take the compressor and turbine inlet temperatures to be 0 and 35 °C, respectively, and use the same compressor and turbine efficiencies. Use constant specific heat for air at room temperature with Cp = 1.005 kJ/kg K and k = 1.4.arrow_forwardConsider a refrigrator that operates on the vapor compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 70 kPa, and exits at 1200 kPa and 90°C, and leaves the condenser as saturated liquid at 1200 kPa. The coefficient of performance of this refrigrator isarrow_forwardRefrigerant-134a enters the compressor of a cooling system as superheated vapor at 0.18 MPa and 0°C with a flow rate of 0.15 kg/s. It exits the compressor at 0.8 MPa and 60°C. Post compression, the refrigerant is cooled in the condenser to 28°C and 1.4 MPa. Subsequently, it's throttled to 0.16 MPa. Neglecting any heat transfer and pressure drops in the pipelines between the components, represent the cycle on a T-s diagram concerning saturation lines. Calculate: (a) The rate of heat extraction from the cooling area and the energy input to the compressor. (b) The isentropic efficiency of the compressor. (c) The Coefficient of Performance (COP) of the cooling system.arrow_forward
- Refrigeration and Air Conditioning Technology (Mi...Mechanical EngineeringISBN:9781305578296Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill JohnsonPublisher:Cengage Learning
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