Question

Consider an ideal vapour compression refrigeration cycle working on R-134a refrigerant. The COP of the cycle is 10 and the refrigeration capacity is 150 kJ/kg. The heat rejected by the refrigerant in the condenser is _______ kJ/kg.

• A. 165
• B. 148
• C. 126
• D. 137

Hint:

For any vapor compression cycle, the heat rejected in the condenser is always the sum of the heat absorbed in the evaporator and the work input to the compressor.
Remembering this simple energy balance saves time during calculations.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question focuses on the analysis of an ideal Vapour Compression Refrigeration Cycle (VCRS).
The refrigeration capacity (heat absorbed in the evaporator) and the Coefficient of Performance (COP) of the cycle are given.
We need to determine the heat rejected by the refrigerant in the condenser.

Step 2: Key Formula or Approach:
The Coefficient of Performance (COP) of a refrigeration cycle is defined as the ratio of the refrigeration effect to the work input:
\[ \text{COP} = \frac{q_e}{w_c} \]
where $q_e$ is the refrigeration capacity and $w_c$ is the net work input.
By applying the first law of thermodynamics to the cycle, the heat rejected in the condenser ($q_c$) is given by:
\[ q_c = q_e + w_c \]

Step 3: Detailed Explanation:

• The given refrigeration capacity (or effect) $q_e = 150 \text{ kJ/kg}$.

• The given COP of the cycle is $10$.

• Using the COP formula, we compute the work input to the compressor:
\[ w_c = \frac{q_e}{\text{COP}} = \frac{150}{10} = 15 \text{ kJ/kg} \]

• Next, we determine the heat rejected in the condenser using the energy balance equation:
\[ q_c = q_e + w_c = 150 + 15 = 165 \text{ kJ/kg} \]

Step 4: Final Answer:
The heat rejected by the refrigerant in the condenser is $165 \text{ kJ/kg}$.