Refrigerant-134a enters the condenser of a residential heat pump at 200 kPa and 35°C at a rate of 0.161 kg/s and leaves at 200 kPa as saturated liquid. If the compressorconsumes 2.5 kW of power, determine (a) the COP of the heat pump and (b) the rate of heat absorption from outside airCOP = 10 657 & QL = 10 998 kWCOP = 6 346 & QL = 15 758 kWCOP = 12 658 & QL = 12 667 kWCOP= 15 773 & QL = 36.931 kW Intensive properties per unit mass are called specific properties.TrueFalse

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Refrigerant-134a enters the condenser of a residential heat pump at 200 kPa and 35°C at a rate of 0.161 kg/s and leaves at 200 kPa as saturated liquid. If the compressor consumes 2 5 kW of power determine (a) the COP of the heat pump and bị the rate of heat absorption from outside air COP = 10 657 & QL = 10 998 kW COP = 6 346 & QL = 15 758 KW COP = 12 658 & QL = 12 667 kW COP= 15 773 & QL= 36.931 kW

Refrigerant-134a enters the condenser of a residential heat pump at 200 kPa and 35°C at a rate of 0.161 kg/s and leaves at 200 kPa as saturated liquid. If the compressor consumes 2 5 kW of power determine (a) the COP of the heat pump and bị the rate of heat absorption from outside air COP = 10 657 & QL = 10 998 kW COP = 6 346 & QL = 15 758 KW COP = 12 658 & QL = 12 667 kW COP= 15 773 & QL= 36.931 kW

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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