Question

An irreversible heat engine operates between ${47}^{\circ }C\text{and}{637}^{\circ }C.$ The rate of heat rejection per 2 kW of net output can be

• A. 0.50 kW
• B. 0.75 kW
• C. 1.08 kW
• D. 1.25 kW

Answer 1

The correct option is D 1.25 kW

${\eta }_{max}={\eta }_{rev}=1-\frac{T_2}{T_1}=1-\frac{47+273}{637+273}$

$=1-\frac{320}{910}$

$\frac{W_{net}}{Q_{\text{rejected}}}=\frac{59}{91}$

$Q_s=\frac{91}{59}\times 2=\frac{182}{59}kW$

$Q_{\text{rejected}}=Q_s-W_{\text{net}}=\frac{182}{59}-2$

$=\frac{64}{59}=1.08kW$

$\text{For irreversible heat engine}{Q}_{\text{rej}}>1.08kW.$

So only option (d) can be a possible solution out of the given options.