Question

Evaluate the COP of a refrigeration system having work input of 120 kJ/kg and refrigeration effect generated is 300 kJ/kg

• A. 2.1
• B. 1.5
• C. 2.5
• D. 2.8

Hint:

To solve performance problems, always remember: $\text{COP} = \frac{\text{Refrigeration Effect}}{\text{Work Input}}$. Ensure both parameters have identical units before performing the division.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The performance of a refrigeration system is evaluated using the Coefficient of Performance (COP).
The COP is a dimensionless parameter defined as the ratio of the desired refrigeration effect to the work input required to drive the cycle.
A higher COP indicates a more efficient refrigeration system, as it achieves greater cooling with less electrical energy consumption.
Key Formula or Approach:
The formula used to calculate the Coefficient of Performance is:
\[ \text{COP} = \frac{\text{Refrigeration Effect } (Q_c)}{\text{Work Input } (W_{\text{in}})} \]

Step 2: Detailed Explanation:
In this problem, the key thermodynamic parameters are provided:
The refrigeration effect, which is the heat removed from the refrigerated space, is given as:
\[ Q_c = 300 \text{ kJ/kg} \]
The work input, which is the energy required by the compressor to complete the cycle, is given as:
\[ W_{\text{in}} = 120 \text{ kJ/kg} \]
Substituting these values into the COP formula:
\[ \text{COP} = \frac{300}{120} \]
\[ \text{COP} = 2.5 \]
This means that for every kilowatt of electrical power supplied to the compressor, the system extracts 2.5 kilowatts of heat from the cold storage area.
This is a standard operating efficiency value for typical vapor compression refrigeration cycles used in the dairy industry.

Step 3: Final Answer
The Coefficient of Performance (COP) of the refrigeration system is 2.5.