Maximizing Heat Pump Efficiency with Constant Entropy

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The discussion centers on maximizing heat pump efficiency through the lens of thermodynamics, specifically the second law. Participants explore the implications of an isolated system where heat exchange occurs, emphasizing that heat flows from higher to lower temperatures until equilibrium is reached. The conversation highlights the need for understanding entropy and energy conservation to determine the maximum temperature one object can achieve. Confusion arises regarding the process of heat exchange and the equations necessary to solve the problem, with suggestions to consider the system's behavior over time. Ultimately, the focus remains on the constraints of thermodynamic principles to find the maximum possible temperature in the system.
  • #31
haruspex said:
No, I mean the point at which there is no gradient. When no more heat can be pumped into B, the temperatures of A and C will be equal, right? That gives you your third equation.
From these you can extract a cubic in one unknown. Now, solving a cubic is a lot easier if you happen to know one root. Here's where the problem setter has been kind to you - you do already have one solution to it. (Hint)

I got a cubic in one unknown,
2T_{A1}^3-1000T_{A1}^2+32\times 10^6=0
Honestly, I can not figure out that one root and had to use a calculator. The calculator gave me 2 positive roots, 256.16 and 400. The second root indicates that no heat is extracted from A, I don't see how this could have been obvious to me. :(

Corresponding to these two values of ##T_{A1}##, I get ##T_{B1}## equal to 200 and 487.67. Is 487.67 the correct answer?

haruspex said:
...the max temp achievable will correspond to no change in entropy..
I haven't been able to understand why the total change in entropy is constant? How did you even bring this relation? Have I missed out anything in my notes? If so, please give me a link which explains this.
 
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  • #32
Pranav-Arora said:
I got a cubic in one unknown,
2T_{A1}^3-1000T_{A1}^2+32\times 10^6=0
Honestly, I can not figure out that one root and had to use a calculator. The calculator gave me 2 positive roots, 256.16 and 400. The second root indicates that no heat is extracted from A, I don't see how this could have been obvious to me. :(
You had three equations for the three temperatures: known product, known sum, and two of them the same. The initial conditions met all these criteria, so must be a solution.
Corresponding to these two values of ##T_{A1}##, I get ##T_{B1}## equal to 200 and 487.67. Is 487.67 the correct answer?
That's what I got, as I recall. (Did it in my head while cycling.)
I haven't been able to understand why the total change in entropy is constant?
You were asked for the max temp that could be achieved (in principle). If entropy increases that will reduce your ability to pump heat up the gradient, so to get the max take entropy as constant.

This was a pretty problem. I hope you enjoyed the journey as much as I did.
 
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  • #33
Thank you haruspex for your input and patience. It was fun discussing the problem here, got to learn something new. :smile:

haruspex said:
If entropy increases that will reduce your ability to pump heat up the gradient, so to get the max take entropy as constant.

This cleared all my doubts, thank you again.
 

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