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: List the reservoir temperatures in Kelvin.
Cold side: $T_C = -30 + 273 = 243\ K$. Hot side: $T_H = 20 + 273 = 293\ K$.

Step 2: Use the heat pump COP instead of the refrigeration COP.
For a reversed Carnot engine, the heating COP is $COP_H = \dfrac{T_H}{T_H - T_C} = \dfrac{293}{50} = 5.86$.
This directly gives the fraction of rejected heat to work input, since $COP_H = Q_H/W$.

Step 3: Relate work to the known cooling load.
The two COPs are linked by $COP_H = COP_C + 1$, where $COP_C = T_C/(T_H-T_C) = 4.86$.
So $COP_H = 4.86 + 1 = 5.86$, confirming the value above. Work input: $W = Q_C / COP_C = 70/4.86 = 14.4\ kW$.

Step 4: Get the rejected heat directly.
$Q_H = COP_H \times W = 5.86 \times 14.4 = 84.4\ kW$, the same figure found from the energy balance.

Final Answer:
Using the heating COP cross checks the earlier balance and lands on option (A) again. \[ \boxed{84.40\ kW} \]