Thermodynamics - Second Law: 2 Heat Engines Connected Between 3 Metal Blocks

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Homework Statement:

Three blocks of metal have the same mass and specific heat. Their initial temperatures are T, 2T, and 3T. Reversible heat engines are connected between these blocks, but no net work is produced or used. No heat is exchanged with the surroundings. Finally two of the blocks reach the same temperature T1, and the third block reaches T2. Find these temperatures.

Relevant Equations:

First law of thermodynamics
Second law of thermodynamics
Hi,

I posted a similar question recently and gained some insight on these types of problems. However, I am slightly stumped on how to approach this variation of the problem.

So I know that:
- there is no net change in enthalpy of the blocks and the engine as the processes are reversible
- [itex] \Delta S_{surroundings} = 0 [/itex] as no heat is exchanged with surroundings
- the internal energy of the system should not change as it is a cycle

I have made a quick sketch of what I envision system to look like - but perhaps I am wrong. I have assumed that the work extracted from a certain turbine connecting block i to j is used to power the pump between those same blocks.
IMG_8748.jpg

(have left out work terms for simplicity in the drawing)

I tried to start writing equations for [itex] dQ_i [/itex] for each block, but am ending up with lots of big equations (am thinking that surely there is a more elegant way of solving this). For example, if we number the blocks 1, 2, and 3 (with the numbering corresponding to the blocks with the initial termperature as [itex] nT [/itex]), then I am getting the following equation for [itex] dQ_1 [/itex]:
[tex] dQ_1 = mcdT_2 - dW_{21} - mcdT_1 - mcdT_1 + mcdT_3 + dW_{31} [/tex]

doing the same for [itex] dQ_2 [/itex] and [itex] dQ_3 [/itex], I get:
[tex] dQ_2 = mcdT_3 - dW_{23} - 2mcdT_2 + mcdT_1 + dW_{21} [/tex]
[tex] dQ_3 = mcdT_2 + dW_{23} - 2mcdT_3 + mcdT_1 + dW_{31} [/tex]

Then, given that we know that [itex] dS = \frac{dQ}{T} = 0 [/itex], then combining those equations, I get:
[tex] \frac{dQ_1}{T_1} + \frac{dQ_2}{T_2} + \frac{dQ_3}{T_3} = 0 [/tex]
rearranging gives:
[tex] mcdT_1 \left( \frac{1}{T_2} + \frac{1}{T_3} + \frac{-2}{T_1} \right) + mcdT_2 \left( \frac{1}{T_3} + \frac{1}{T_1} + \frac{-2}{T_2} \right) + mcdT_3 \left( \frac{1}{T_2} + \frac{1}{T_1} + \frac{-2}{T_3} \right) + dW_{13} \left( \frac{1}{T_3} - \frac{1}{T_1} \right) + dW_{12} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) + dW_{23} \left( \frac{1}{T_3} - \frac{1}{T_2} \right) = 0 [/tex]

I am not really sure if what to do from here (or if this was the right path to follow). I could integrate, but then I am not sure what to do with all the work terms that I have.

Any help would be greatly appreciated.
 
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It seems to me you are getting bogged down in the details. The key to this problem, in my judgment, is to guess which block reaches T2, subject to the constraints that
1. Since there is no work done and no change in internal energy, so Q1+Q2+Q3=0
2. The sum of the three entropy changes is zero.

Apparently, there is only one choice for the block that reaches T2 that satisfies these constraints. So first assume that the correct block is #1. So, it goes from T to T2, while blocks 2 and 3 go from 2T and 3T to T1, respectively. Determine if this can satisfy the two constraints. If there is no real solution to the equations for T1 and T2, discard this choice and move on to block #2.

Actually, it doesn't matter which block you choose for T2. Any choice gives you the same answers for T1 and T2.
 
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Apparently, there is only one choice for the block that reaches T2 that satisfies these constraints. So first assume that the correct block is #1. So, it goes from T to T2, while blocks 2 and 3 go from 2T and 3T to T1, respectively. Determine if this can satisfy the two constraints. If there is no real solution to the equations for T1 and T2, discard this choice and move on to block #2.

Actually, it doesn't matter which block you choose for T2. Any choice gives you the same answers for T1 and T2.
Thank you very much for your reply. Is this a correct way of proceeding?
We can instead write equations for [itex] Q_1 [/itex], [itex] Q_2 [/itex], [itex] Q_3 [/itex]:
[tex] Q_1 = mc \left( T_{final} - T_{initial} \right) = mc \left( T_2 - T \right) [/tex]
[tex] Q_2 = mc \left( T_1 - 2T \right) [/tex]
[tex] Q_3 = mc \left( T_1 - 3T \right) [/tex]

then, evaluating [itex] Q_1 + Q_2 + Q_3 = 0 [/itex], we get [itex] 6T = 2 T_1 + T_2 [/itex]. Does that look correct?

Then I was wondering: how do we know which temperature to divide by when we are dealing with changes in entropy [itex] ds = \frac{dq}{T} [/itex] ?

Then I could set up another equation and solve those simultaneously - I am just stuck as to whether I should divide by the initial or the final temperature.

Thanks for the help.
 
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Thank you very much for your reply. Is this a correct way of proceeding?
We can instead write equations for [itex] Q_1 [/itex], [itex] Q_2 [/itex], [itex] Q_3 [/itex]:
[tex] Q_1 = mc \left( T_{final} - T_{initial} \right) = mc \left( T_2 - T \right) [/tex]
[tex] Q_2 = mc \left( T_1 - 2T \right) [/tex]
[tex] Q_3 = mc \left( T_1 - 3T \right) [/tex]

then, evaluating [itex] Q_1 + Q_2 + Q_3 = 0 [/itex], we get [itex] 6T = 2 T_1 + T_2 [/itex]. Does that look correct?

Then I was wondering: how do we know which temperature to divide by when we are dealing with changes in entropy [itex] ds = \frac{dq}{T} [/itex] ?

Then I could set up another equation and solve those simultaneously - I am just stuck as to whether I should divide by the initial or the final temperature.

Thanks for the help.
What you do is you integrate: $$dS=\frac{dq}{T}=\frac{mCdT}{T}$$for each block.
 
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What you do is you integrate: $$dS=\frac{dq}{T}=\frac{mCdT}{T}$$for each block.
Okay, so does that mean that my entropy equation would look like: ?
[tex] \int_T^{T_2} \frac{1}{T} \, dT + \int_{2T}^{T_1} \frac{1}{T} \, dT + \int_{3T}^{T_1} \frac{1}{T} \, dT = 0 [/tex]
which I would then go on to solve.

Thanks in advance
 
  • #6
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Okay, so does that mean that my entropy equation would look like: ?
[tex] \int_T^{T_2} \frac{1}{T} \, dT + \int_{2T}^{T_1} \frac{1}{T} \, dT + \int_{3T}^{T_1} \frac{1}{T} \, dT = 0 [/tex]
which I would then go on to solve.

Thanks in advance
Yes. That is the next step.
 
  • #7
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Yes. That is the next step.
Thank you for the help. Just to finish the thread for anyone else who may ever read it, solving those two equations simultaneously yields the following (according to Wolfram Alpha): [itex] T_1 = 1.34730 T [/itex] and [itex] T_2 = 3.3054 T[/itex]
 
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Thank you for the help. Just to finish the thread for anyone else who may ever read it, solving those two equations simultaneously yields the following (according to Wolfram Alpha): [itex] T_1 = 1.34730 T [/itex] and [itex] T_2 = 3.3054 T[/itex]
Your 2nd equation was $$T_1^2T_2=6T^3$$, correct?

I also found a 2nd pair of roots at about T1=2.53T and T2=0.94T
 
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Your 2nd equation was $$T_1^2T_2=6T^3$$, correct?

I also found a 2nd pair of roots at about T1=2.53T and T2=0.94T
Sorry for my late reply. Yes, that was my second equation. That is true actually, there were also those other roots - I just chose mine because those were what the answers were.

However, I have not yet developed an intuition as to why the roots I quoted are the the answers and not this other pair. I will continue to think about it and post on here when I have come up with a satisfactory reason.

I wonder if it has to do with the fact that this alternative set of roots has a temperature which is less than T?
 
  • #10
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Sorry for my late reply. Yes, that was my second equation. That is true actually, there were also those other roots - I just chose mine because those were what the answers were.

However, I have not yet developed an intuition as to why the roots I quoted are the the answers and not this other pair. I will continue to think about it and post on here when I have come up with a satisfactory reason.

I wonder if it has to do with the fact that this alternative set of roots has a temperature which is less than T?
In my judgment, the second set of roots is a valid solution also, just as valid as the first. The third set of roots has a negative T1, so it is not physically realistic.
 

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