A heat pump working on a reversed Carnot cycle takes in energy from a reservoir maintained at 3ºC and delivers it to another reservoir where temperature is 77ºC. The heat pump drives power for it's operation from a reversible engine operating within the higher and lower temperature limits of 1077ºC

Question

A heat pump working on a reversed Carnot cycle takes in energy from a reservoir maintained at 3ºC and delivers it to another reservoir where temperature is 77ºC. The heat pump drives power for it's operation from a reversible engine operating within the higher and lower temperature limits of 1077ºC and 77ºC. For 100 kJ/s of energy supplied to the reservoir at 77ºC, estimate the energy taken from the reservoir at 1077ºC. [U.P.S.C. 1994]

Accepted Answer

Arrangement for heat pump and heat engine operating together is shown here. Engine and pump both reject heat to the reservoir at 77ºC (350 K).
For heat engine.

[latex]\begin{aligned}η_{E}&= 1 - \frac{350}{1350} = \frac{W}{Q_{1}}\0.7407&= \frac{Q_{1} - Q_{2}}{Q_{1}}\0.7407&= 1 - \frac{Q_{2}}{Q_{1}}\Q_{2}&= 0.2593 Q_{1}\ ext{For heat pump}\qquad\qquad\qquad\qquad&\COP_{HP}&= \frac{Q_{4}}{Q_{4} - Q_{3}}\COP_{HP}&= \frac{350}{350 - 276} = \frac{Q_{4}}{Q_{4} - Q_{3}}\\Longrightarrow \quad\qquad \qquad\qquad \qquad \qquad Q_{4}&= 1.27 Q_{3}\ ext{Work output from engine} &= ext{ Work input to pump} \Q_{1} – Q_{2}&= Q_{4} – Q_{3} \Rightarrow Q_{1} - 0.2593 Q_{1} = Q_{4} - \frac{Q_{4}}{1.27}\ ext{Also it is given that }Q_{2} + Q_{4}& = 100\end{aligned}\ ext{Substituting }Q_{2} ext{ and } Q_{4} ext{ as function of }Q_{1} ext{ in following expression,}\\begin{aligned}\quad \qquad \qquad \qquad\qquad \qquad Q_{2} + Q_{4}&= 100\0.2593 Q_{1} + \frac{Q_{1}}{0.287}&= 100\Q_{1}&= 26.71 kJ\end{aligned}\ ext{Energy taken by engine from reservoir at 1077ºC}\ = extbf{26.71 kJ}[/latex]

Question 4.13: A heat pump working on a reversed Carnot cycle takes in ener...

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