# Replacing Qh with Ql in Etropy equation in Heat Pump

In the heat pump below, the $$T_H$$ is not constant. It is assumes that the process is steady and reversible. So when constructing an entropy equation over the CV (the purple-colored area): $$\dot ms_1+\dot Q_H/T_H+\dot S_{gen}=\dot ms_2$$ Where $$\dot S_{gen}=0$$, why is it allowed to replace the term $$\dot Q_H/T_H$$ with $$\dot Q_L/T_L$$ (The $$T_L$$ is constant by the way)? P.s. This is from the "Fundamentals of Thermodynamics" 8e Claus Borgnakke, Richard E. Sonntag. With the "previous problem" diagram Fig. P7.28 • In the first sentence you say TH is not constant. Then, in parenthesis, you say TH is constant. Which is it? – Chet Miller Oct 6 '18 at 23:44
• Is it possible for you to provide more context exactly as it appears in your book? – Chet Miller Oct 7 '18 at 0:08
• Yes. Done. Post updated. – Jek Denys Oct 7 '18 at 0:15

Your equation describes the entropy balance on the fluid passing through the heat exchanger between points 1 and 2. In this equation, $$Q_H$$ represents the heat received by the heat exchange fluid from the heat pump. The interface between the heat pump and the heat exchanger fluid is assumed to be at a constant temperature $$T_H$$ (say because the working fluid in the heat pump is experiencing a phase change at a constant pressure). So, the rate of entropy transfer between the heat pump and the heat exchanger fluid is $$\dot{Q}_H/T_H$$. And $$\dot{S}_{gen}$$ in the equation is the rate of entropy generation within the heat exchange fluid (as a result of temperature gradients within the heat exchanger fluid). If the heat pump is operating close to ideally, I can see how you could replace $$\dot{Q}_H/T_H$$ by $$\dot{Q}_C/T_C$$ in the equation. However, I can't see how it can be assumed that $$\dot{S}_{gen}$$ in the heat exchanger can be assumed to be zero.
• Ah! Thank you for pointing out the $S_{gen}$. I forgot to mention that the problem states that this is Steady State, Reversible process. Does this help? – Jek Denys Oct 7 '18 at 15:16
• What appears to be indicated is that the heat pump (not the heat exchanger) is operating close to ideal reversibility. It is implied that all the irreversibility (entropy generation) resides within the fluid flowing between points 1 and 2 on the non-heat-pump side of the heat exchanger. If the heat pump is operating close to reversibly, there is no entropy generation within the heat pump, and the entropy transfer from the cold reservoir to the heat pump ($Q_C/T_C$) is equal to the entropy transfer from the heat pump to the heated fluid in the heat exchanger ($Q_H/T_H$). – Chet Miller Oct 7 '18 at 15:40