Question

A cyclic heat engine operates between a source temperature of 800°C and a sink temperature of 30°C. The least rate of heat rejection per kW net output of engine will be nearly
1. 0.2 kW 
2. 0.4 kW 
3. 0.6 kW 
4. 0.8 kW 

Answer 1

Correct Answer - Option 2 : 0.4 kW 

Concept:

Reversible heat engine cycle will reject least rate of heat (QL) per kW net output (W) of engine.

For reversible heat engine,

\(\eta = \frac{W}{{{Q_H}}} = \frac{{{T_H} - {T_L}}}{{{T_H}}} = 1 - \frac{{{T_L}}}{{{T_H}}}\)

Calculation:

Given T_H = 800°C = 1073 K, T_L = 30°C = 303 K, W = 1 kW,

\(\frac{W}{{{Q_H}}} = 1 - \frac{{{T_L}}}{{{T_H}}} \Rightarrow \frac{1}{{{Q_H}}} = 1 - \frac{{303}}{{1073}}\)

⇒ Q_H = 1.39 kW

⇒ Q_L = Q_H – W = 1.39 – 1 = 0.39 kW ≈ 0.4 kW