Guest Question.

300 kJ/s of heat is supplied at a constant fixed temperature of 290°C to a heat engine. The heat rejection takes place at 8.5°C. The following results were obtained :

(i) 215 kJ/s are rejected.

(ii) 150 kJ/s are rejected.

(iii) 75 kJ/s are rejected.

Classify which of the result report a reversible cycle or irreversible cycle or impossible results.

Admin

Solution

Heat supplied at 290°C = 300 kJ/s

Heat rejected at 8.5°C : (i) 215 kJ/s, (ii) 150 kJ/s, (iii) 75 kJ/s.

Applying Clausius inequality to the cycle or process. Thus,

(i) \(\sum_{cycle}^{}\frac{\delta Q}{T}=\frac{300}{290+273}-\frac{215}{8.5+273}\)

= 0.5328 – 0.7637 = – 0.2309 < 0.

\(\therefore\) Cycle is irreversible.

(ii) \(\sum_{cycle}^{}=\frac{300}{290+273}-\frac{150}{8.5+273}\)

= 0.5328 – 0.5328 = 0

\(\therefore\) Cycle is reversible.

(iii) \(\sum_{cycle}^{}\frac{\delta Q}{T}=\frac{300}{290+273}-\frac{75}{8.5+273}\)

= 0.5328 – 0.2664 = 0.2664 > 0.

This cycle is impossible by second law of thermodynamics, i.e. Clausius inequality.