Project ForgeSign up free
Thermodynamics
First Law of Thermodynamics
GATE 2009
Medium

In an ideal vapour compression refrigeration cycle, the specific enthalpy of refrigerant (in kJ/kg) at the following states is given as: Inlet of condenser: 283, Exit of condenser: 116, Exit of evaporator: 232. The COP of this cycle is

  • 2.27
  • B2.75
  • C3.27
  • D3.75

Explanation

COP=Refrigerating effect/Work input=(h_evap,exit - h_cond,exit)/(h_comp,exit - h_evap,exit)=(232-116)/(283-232)=116/51≈2.27.

Source: GATE ME 2009 (2009)

This question is part of a 15-day revision cycle — sign in and Project Forge will bring it back the day before you'd forget it.

Start revising free