Question
Download Solution PDFIn a refrigeration system, the work input is 20 kJ/kg while 80 kJ/kg of heat is rejected out of the system. The COP of the system will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The coefficient of performance is the ratio of heat extracted in the refrigerator to the work done on the refrigerant.
COP (Coefficient of Performance) of the ideal refrigerator
\(COP = \frac{{{Q_1}}}{W_R} = \frac{{{T_1}}}{{{T_2} - {T_1}}}\)
where T2 = Temperature of the hot reservoir, T1 = Temperature of the cold reservoir, Q1 = Heat rejected from the cold reservoir, WR = Work required to pump the heat out.
Calculation:
Given:
WR = 20 kJ/kg, Q2 = 80 kJ/kg,
Q2 = Q1 + WR
Q1 = 80 - 20 = 60 kJ/kg
\(COP = \frac{{{Q_1}}}{W_R} =\frac{60}{20}=3\)
Last updated on May 28, 2025
-> SSC JE ME Notification 2025 will be released on June 30.
-> The SSC JE Mechanical engineering application form will be available from June 30 to July 21.
-> SSC JE 2025 CBT 1 exam for Mechanical Engineering will be conducted from October 2 to 31.
-> SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.