314 ENGINEERING THERMODYNAMICS
DHARM
M-therm\Th6-1.PM5(i)The entropy produced during heat transfer ;
(ii)The decrease in available energy after heat transfer.
Solution. Refer Fig. 6.7.Fig. 6.7
Temperature of source, T 1 = 1000 K
Temperature of system, T 2 = 500 K
Temperature of atmosphere, T 0 = 300 K
Heat received by the system, Q = 7200 kJ/min.
(i)Net change of entropy :
Change in entropy of the source during heat transfer= −TQ 1 = − 10007200 = – 7.2 kJ/min-K
Change in entropy of the system during heat transfer= TQ 2 =^7200500 = 14.4 kJ/min-K
The net change of entropy, ∆S = – 7.2 + 14.4 = 7.2 kJ/min-K. (Ans.)
(ii)Decrease in available energy :
Available energy with source
= (1000 – 300) × 7.2 = 5040 kJ