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