Refrigeration Cycle

The shown figure describes a refrigeration cycle using R-12a and its associated enthalpies. What is the coefficient of performance (COP)?

Expand Hint
For a refrigeration cycle, the coefficient of performance is:
$$$COP_{ref}=\frac{h_1-h_4}{h_2-h_1}$$$
where $$h$$ is the enthalpy.
Hint 2
For heat pumps and refrigeration cycles, $$h_4=h_3$$ .
For a refrigeration cycle, the coefficient of performance is:
$$$COP_{ref}=\frac{h_1-h_4}{h_2-h_1}$$$
where $$h$$ is the enthalpy.

For heat pumps and refrigeration cycles, $$h_4=h_3$$ . Thus,
$$$COP_{ref}=\frac{(400-250)}{(440-400)}\cdot \frac{kJ/kg}{kJ/kg}=\frac{150}{40}=3.75$$$
3.75
Time Analysis See how quickly you looked at the hint, solution, and answer. This is important for making sure you will finish the FE Exam in time.
  • Hint: Not clicked
  • Solution: Not clicked
  • Answer: Not clicked

Similar Problems from FE Sub Section: Coefficient of Performance
583. Heat Pump Cycle
586. Fridge Cycle
Similar Problems from FE Sub Section: Refrigeration
583. Heat Pump Cycle
586. Fridge Cycle

Similar Problems from FE Section: Basic Cycles
298. Carnot Cycle
454. Heat Pump
457. Heat Pump COP
583. Heat Pump Cycle
586. Fridge Cycle
602. Carnot Fridge
Similar Problems from FE Section: Common Thermodynamic Cycles
583. Heat Pump Cycle
586. Fridge Cycle