This is chemical engineering question. 1. Heat in the amount of 150 kJ is transf

Question

This is chemical engineering question.

1. Heat in the amount of 150 kJ is transferred directly from a hot reservoir at Tr= 550 K to two cooler reservoirs at T-350 K and T2 = 250 K. The surroundings temperature is 300 K. If the heat transferred to the reservoir at T: is 50 kJ and that transferred to the reservoir at T2is 100 k a) Calculate the entropy generation in k]/K. b) Calculate the lost work. c) How could the process be reversible? I /2) I 12] 2. An inventor has devised a complicated nonflow process in which 1 mol of air is the working fluid. The net effects of the process are claimed to be: A change in state of the air from 523.15 K and 3 bar to 353.15 K and 1 bar. The production of 1800 J of work. The transfer of an undisclosed amount of heat to a heat reservoir at 30°C. Determine whether the claimed performance of the process is consistent with thermodynamic laws. I /15] Assume that air is an ideal gas for which Cp = (7/2R

Explanation / Answer

Ans 1

Part a

Change in the entropy of the hot reservoir

SH = Q / TH

= -150 kJ/550K

= - 0.273 kJ/K

Change in Entropy of the cool reservoirs

= S1 + S2

= (Q1/ T1) + (Q2/ T2)

= (50kJ/350 K) + (100kJ/250 K)

= 0.543 kJ/K

Entropy generation

Sg =SH + (S1+S2)

= - 0.273 + 0.543

= 0.270 kJ/K

Sg > 0, the process is irreversible

Part b

Lost work

Wlost = Sg x T

= 0.270 kJ/K x 300 K

= 81 kJ

Part c

For the process to be reversible, all three reservoirs temperatures will remain same or temperature difference will be differential

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