SUBJECT: THERMODYNAMIC COURSE: II ASSI.LACTURE: NATIQ ABBAS Example 2:- Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and -10°C at a rate of 0.05 kg/s and leaves at 0.8 MPa and 50°C. The refrigerant is cooled in the condenser to 26°C and 0.72 MPa and is throttled to 0.15 MPa. Disregarding any heat trans fer and pressure drops in the connecting lines between the components; determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, and (b) the coefficient of performance of the refrigerator. Solution: - P 0.14 MPa T--10 C h 246.36 kJ/kg P 0.8 MPa T 50 C h2 = 286.69 %3D AVERSIT P= 0.72 MPa T-26 C- - hg = 87.83 kJ/kg !! h3 =h, =87.83 KJ/kg h4 = h3 (throttling) h4 = 87.83 kJ/kg !! 0.S MPa 50°C 0.72 MPa 26 C 0.15 MPa 0.14 MPa -10°C SAMARRA RING

SUBJECT: THERMODYNAMIC COURSE: II ASSI.LACTURE: NATIQ ABBAS Example 2:- Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and -10°C at a rate of 0.05 kg/s and leaves at 0.8 MPa and 50°C. The refrigerant is cooled in the condenser to 26°C and 0.72 MPa and is throttled to 0.15 MPa. Disregarding any heat trans fer and pressure drops in the connecting lines between the components; determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, and (b) the coefficient of performance of the refrigerator. Solution: - P 0.14 MPa T--10 C h 246.36 kJ/kg P 0.8 MPa T 50 C h2 = 286.69 %3D AVERSIT P= 0.72 MPa T-26 C- - hg = 87.83 kJ/kg !! h3 =h, =87.83 KJ/kg h4 = h3 (throttling) h4 = 87.83 kJ/kg !! 0.S MPa 50°C 0.72 MPa 26 C 0.15 MPa 0.14 MPa -10°C SAMARRA RING

Refrigeration and Air Conditioning Technology (MindTap Course List)

8th Edition

ISBN:9781305578296

Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson

Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson

Chapter45: Domestic Refrigerators And Freezers

Section: Chapter Questions

Problem 2RQ: The operating condition for the single compressor in a household refrigerator is the lowest box...

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An ammonia refrigeration compressor takes its suction from the evaporator at a temperature of -20oF (-28.9oC) and a quality of 95%. The compressor discharges at a pressure of 100 psi (689.5 kPa), liquid ammonia leaves condenser at 50oF (10oC). Properties of Ammonia: At 689.6 kPa (Tsat = 13.3oC)      h = 1,568 kJ/kg; s = 5.7676 kJ/kg.K At 10oC:      hf = 246.531 kJ/kg At -28.9oC by interpolation      hf = 68.5 kJ/kg; hg = 1424.5 kJ/kg sf = 0.49366; sg = 6.0452Find the heat absorbed by the evaporator in kJ/kg.
An ammonia refrigeration compressor takes its suction from the evaporator at a temperature of -20oF (-28.9oC) and a quality of 95%. The compressor discharges at a pressure of 100 psi (689.5 kPa), liquid ammonia leaves condenser at 50oF (10oC). Properties of Ammonia: At 689.6 kPa (Tsat = 13.3oC)      h = 1,568 kJ/kg; s = 5.7676 kJ/kg.K At 10oC:      hf = 246.531 kJ/kg At -28.9oC by interpolation      hf = 68.5 kJ/kg; hg = 1424.5 kJ/kg             sf = 0.49366; sg = 6.0452Find the COP. a. 4 b. 4.25 c. 5 d. 5.25
An ammonia refrigeration compressor takes its suction from the evaporator at a temperature of -20oF (-28.9oC) and a quality of 95%. The compressor discharges at a pressure of 100 psi (689.5 kPa), liquid ammonia leaves condenser at 50oF (10oC). Properties of Ammonia: At 689.6 kPa (Tsat = 13.3oC)      h = 1,568 kJ/kg; s = 5.7676 kJ/kg.K At 10oC:      hf = 246.531 kJ/kg At -28.9oC by interpolation      hf = 68.5 kJ/kg; hg = 1424.5 kJ/kg      sf = 0.49366; sg = 6.0452 Determine the quality of refrigerant upon entering to the evaporator.