A real refrigerator takes in 700 KJ of heat at a temperature of 270 C and gives

Question

A real refrigerator takes in 700 KJ of heat at a temperature of 270 C and gives off heat at 300 C. The heat given off by the refrigerator is fed into a Carnot heat engine. THe refrigerator performs at 55% of the performance of a Carnot refrigerator.

a/ Determine the work required by the refrigerator.

b/ If 800 KJ of heat (Q3) is removed from the Carnot heat engine, at what temperature (T3) does the Carnot heat engine reject heat? What is the entropy of generation for the Carnot heat engine ( in KJ/K)

Explanation / Answer

TL = 270 C = 270 + 273 = 543 K

TH = 300 C = 300 + 273 = 573 K
carnot refrigerator performance = QL / W = QL / (QH - QL)
= TL / (TH-TL)
= 543 / (573-543)
= 543 / 30 = 18.1
real refrigerator performance = 0.55 x 18.1 = 9.955
a) => QL_real/ W_real = 9.955
=> 700 kJ / W_real = 9.955
   => W_real = 700/9.955 = 70.316 kJ
b) Q given to carnot = QH of refrigerator
   QL / QH = TL/TH => QH = 700 x 573/543 = 738.674 kJ
and it occurs at 573 K
Q removed from carnot = 800 kJ and it occurs at T3
so Qremoved/ Q recieved = T3/ TH => T3 = 800/738.674 x 573
   = 620.571 K = 347.571 C
entropy generation = Qrecieved / TH - Qremoved / T3 = 738.674 / 573 - 800/ 620.571
= 0

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