The thermal efficiency of an engine used to drive a refrigerator is 30%. The heat input into the engine for each MJ removed from the cold body by the refrigerator is 666.67 kJ. What is the COP of the refrigerator? Ans. COP of Ref. = 5 ; w= 200kJ

The thermal efficiency of an engine used to drive a refrigerator is 30%. The heat input into the engine for each MJ removed from the cold body by the refrigerator is 666.67 kJ. What is the COP of the refrigerator? Ans. COP of Ref. = 5 ; w= 200kJ

Name: The thermal efficiency of an engine used to drive a refrigerator is 30
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Description: The thermal efficiency of an engine used to drive a refrigerator is 30%. The heat input into the engine for each MJ removed from the cold body by the refrigerator is 666.67 kJ. What is the COP of the refrigerator? Ans. COP of Ref. = 5 ; w= 200kJ

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter8: Natural Convection

Section: Chapter Questions

Problem 8.46P

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