Thermally insulated container with alternating series of walls

[Meden Agan]

Start at the closed end and work backwards.

Mhm, I'm not sure I understand that hint. If you could be more specific, please?

 

[Chestermiller]

Mhm, I'm not sure I understand that hint. If you could be more specific, please?

In my analysis in post #28, I started out with the equation for the final pair of cells, and set the dimensionless displacement of the insulated end $\xi_4$ equal to zero. Then I took the equations for the previous set of cells, and solved for the dimensional displacement $\xi_2$ in terms of the displacement in the one pair of cells further towards the open end. Would it help if I showed the algebra I used for solving the equations in post #28?

 

[Meden Agan]

In my analysis in post #28, I started out with the equation for the final pair of cells, and set the dimensionless displacement of the insulated end $\xi_4$ equal to zero. Then I took the equations for the previous set of cells, and solved for the dimensional displacement $\xi_2$ in terms of the displacement in the one pair of cells further towards the open end. Would it help if I showed the algebra I used for solving the equations in post #28?

Sure it would help. But I did something similar in post #27. I got the same result as you.
Post #27 is a special case of the generalization of post #22, case of $2k$ cells. I cannot get a closure, though. Were you able to get a closed form for the general case of $2k$ cells?

 

[Chestermiller]

Sure it would help. But I did something similar in post #27. I got the same result as you.
Post #27 is a special case of the generalization of post #22, case of $2k$ cells. I cannot get a closure, though. Were you able to get a closed form for the general case of $2k$ cells?

I'll provide my algebraic solution to the equations in post #28 later today.

But, in the meantime, remember that the problem statement calls for the results only for a monatomic gas with Cv=3R/2 and $\gamma=5/3$. I suggest you use this in your analysis to simplify the equations and to simplify the job of documenting the derivations.

 

[Meden Agan]

I'll provide my algebraic solution to the equations in post #28 later today.

But, in the meantime, remember that the problem statement calls for the results only for a monatomic gas with Cv=3R/2 and $\gamma=5/3$. I suggest you use this in your analysis to simplify the equations and to simplify the job of documenting the derivations.

Sure, I'll simplify my algebra. I'm quite aware as to how to obtain $T_0$ for the cases of $2$ cells and $4$ cells. But I'm not about how to get $T_0$ in the case of $2k$ cells.
May I ask you to post your final result for $2k$ cells, if you got it?

 

[Chestermiller]

Sure, I'll simplify my algebra. I'm quite aware as to how to obtain $T_0$ for the cases of $2$ cells and $4$ cells. But I'm not about how to get $T_0$ in the case of $2k$ cells.
May I ask you to post your final result for $2k$ cells, if you got it?

I haven’t done it for the general case yet. I thought they doing it for four shells would be sufficient to give us an idea of how to do that. I’ll take a shot of doing it for the general case in a little while.

 

[Chestermiller]

When I went to 2k cells, or even only 2k=6 cells, the math got very unwieldy and there was not a simple algebraic solution as for the case of 2 or 4 cells. This nonlinear problem would, in my opinion, have to be solved numerically. Of course, some software automatically handles recursive implicit relations even if the math is complicated.

 

[Meden Agan]

When I went to 2k cells, or even only 2k=6 cells, the math got very unwieldy and there was not a simple algebraic solution as for the case of 2 or 4 cells. This nonlinear problem would, in my opinion, have to be solved numerically. Of course, some software automatically handles recursive implicit relations even if the math is complicated.

Mhm, I had this bad feeling. Perhaps if we apply the polytropic process equation to the last section of partitions? How about it?

 

[Chestermiller]

Mhm, I had this bad feeling. Perhaps if we apply the polytropic process equation to the last section of partitions? How about it?

This is going to boil down to a set of non-linear algebraic equations describing a split boundary value problem. It will be solved using a non-linear equation solver. The unknowns are the displacements $x_i$ and the Temperatures of the cells.

 

[Meden Agan]

This is going to boil down to a set of non-linear algebraic equations describing a split boundary value problem. It will be solved using a non-linear equation solver. The unknowns are the displacements $x_i$ and the Temperatures of the cells.

May I ask you to use such a non-linear equation solver and tell me what result do you get?

 

[Chestermiller]

May I ask you to use such a non-linear equation solver and tell me what result do you get?

I no longer have access to software of this type, because I have been retired 22 years now. I'm not willing to subscribe to such software out of my own pocket. Do you have access to the IMSL library or even less extensive software?

 

[Meden Agan]

I no longer have access to software of this type, because I have been retired 22 years now. I'm not willing to subscribe to such software out of my own pocket. Do you have access to the IMSL library or even less extensive software?

Sadly, no. I think I'll give up this problem for a while.

 

[Chestermiller]

Sadly, no. I think I'll give up this problem for a while.

https://www.mathworks.com/help/optim/ug/fsolve.html

 

[Meden Agan]

https://www.mathworks.com/help/optim/ug/fsolve.html

Mhm, I have no idea how to use it.
According to the official solution, the final result is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_0}\right)^{f_n},$$ where $f_n =\dfrac{2}{7-\dfrac{(4+\sqrt{15})^2+(4+\sqrt{15})^{2n}}{(4+\sqrt{15})+(4+\sqrt{15})^{2n \, + \,1}}}$.
It sounds too nice to be true.

 

[Chestermiller]

Mhm, I have no idea how to use it.
According to the official solution, the final result is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_0}\right)^{f_n},$$ where $f_n =\dfrac{2}{7-\dfrac{(4+\sqrt{15})^2+(4+\sqrt{15})^{2n}}{(4+\sqrt{15})+(4+\sqrt{15})^{2n \, + \,1}}}$.
It sounds too nice to be true.

I have no idea where they got that result. It looks like some cancelation is possible between the numerator and denominator. How does this result compare with our results for n=1 and n=2? Have you checked?

 

[Meden Agan]

I have no idea where they got that result. It looks like some cancelation is possible between the numerator and denominator. How does this result compare with our results for n=1 and n=2? Have you checked?

Our result for $n=0$ is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{1/3}.$$
Their result for $n=0$ is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{2/3}.$$

For $n=1$, the expressions are very different.

 

[Chestermiller]

Our result for $n=0$ is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{1/3}.$$
Their result for $n=0$ is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{2/3}.$$

For $n=1$, the expressions are very different.

I meant for two shells and four shells

 

[Meden Agan]

$$4+\sqrt{15}=(\sqrt{5}+\sqrt{3})^2=3(\sqrt{\gamma }+1)^2$$

OK...? How is that useful?

Minor typo: $${\color{red}{2}} \, \left(4+\sqrt{15}\right)= \left(\sqrt{5}+\sqrt{3}\right)^2=3 \, \left(\sqrt{\gamma}+1 \right)^2.$$

 

[Chestermiller]

OK...? How is that useful?

Minor typo: $${\color{red}{2}} \, \left(4+\sqrt{15}\right)= \left(\sqrt{5}+\sqrt{3}\right)^2=3 \, \left(\sqrt{\gamma}+1 \right)^2.$$

I can't use this equation editor. They have now changed it into something that now sucks.

 

[Chestermiller]

$$4+\sqrt{15}=\frac{8+2\sqrt{15}}{2}=\frac{(\sqrt{5}+\sqrt{3})^2}{2}$$