Question

Heat in the amount of 150 KJ is tranferred directly from a hot reservoir at Th = 550k to two cooler reservoirs at T1 = 350k and T2 = 250k. The surrounding temperatures is TS = 300K. If the heat transferred to the reservoir at T1 is half that tranferred to the reservoir at T2, calculate:

a. the entropy generation in KJ / K

b. the lost work

**Please show me the formula and steps how to do this problem please.

Answer 1

Given that

heat transferred to reservoir at T1 = half of heat transferred to reservoir at T2

Q1 = Q2/2

Total heat transfer

Q = Q1 + Q2

Q = Q2/2 + Q2

-150 = 1.5 Q2

Q2 = - 100 kJ

Q1 = - 50 kJ

Entropy change of hot reservoir

Sh = Q/Th = (-150 kJ)/(550 K) = - 0.273 kJ/K

Entropy change of cool reservoirs

Sc = S1 + S2 = Q1/T1 + Q2/T2

= 50/350 + 100/250

= 0.543 kJ/K

Part a

Entropy generation Sg = Sh + Sc = - 0.273 + 0.543

= 0.270 kJ/K

Part b

Lost work

W = Sg Ts = 0.270 x 300 = 81 kJ