Question

A refrigeration system based on reverse Carnot cycle operates between \(-30\,^{\circ}\text{C}\) and \(20\,^{\circ}\text{C}\). The refrigeration capacity of the system is 70 kW. Heat rejected at the condenser to atmosphere (hot reservoir), in kW, is nearest to

• A. 84.40
• B. 14.41
• C. 186.67
• D. 116.67

Hint:

Find the Carnot COP from the reservoir temperatures, get the work input, then add it to the cooling load.

Answer 1

The Correct Option is A

Step 1: Convert temperatures to Kelvin.
The cold reservoir (evaporator) is at \( T_C = -30 + 273 = 243\,K \).
The hot reservoir (condenser) is at \( T_H = 20 + 273 = 293\,K \).

Step 2: Find the Carnot COP for refrigeration.
For a reverse Carnot cycle, \( COP = \dfrac{T_C}{T_H - T_C} = \dfrac{243}{293-243} = \dfrac{243}{50} = 4.86 \).

Step 3: Find the work input from the refrigeration capacity.
\( COP = \dfrac{Q_C}{W} \), so \( W = \dfrac{Q_C}{COP} = \dfrac{70}{4.86} = 14.4\,kW \).

Step 4: Apply the energy balance on the cycle.
Heat rejected at the condenser equals cooling load plus work input: \( Q_H = Q_C + W = 70 + 14.4 = 84.4\,kW \).

Final Answer:
The condenser rejects about 84.4 kW to the atmosphere, matching option (A). Options (C) and (D) wrongly treat 70 kW as the work input instead of the cooling load. \[ \boxed{Q_H \approx 84.40\ kW} \]