What is the Final Temperature of Hot Body C in a Thermodynamic Engine Cycle?

[Telemachus]

Homework Statement

A thermodynamic engine is operated between two cooler bodies A and B, extracting work until the two cooler bodies reach a common temperature T_f. This work is then used as the input to a heat pump, extracting heat from the cooler pair and heating the hot body C. Find the final temperature T_{fC} C if work is maximum.

The initial temperatures for the cooler bodies A and B are: T_{A0}=300ºK,T_{B0}=350ºK, and for the hot body C: T_{C0}=400ºK

The equation of state for the three bodies is: U=aT

So, this is what I did till now.

U=aT\Rightarrow S=S_0+ln\left(\displaystyle\frac{U}{U_0}\right)

\Delta U_{AB}=U_f-U_0=a(T_{fA}+T_{fB})-a(T_{A0}+T_{B0})

T_{fA}=T_{fB}=T_f

\Delta U_{AB}=U_f-U_0=a(2T_f}-T_{A0}-T_{B0})

W=-\Delta U_{AB}=a(-\left(2T_f}+T_{A0}-T_{B0})

\Delta S_{AB}=2a \ln\left(\displaystyle\frac{T_f}{\sqrt{T_{A0}T_{B0}}}\right)

As the work is maximum: T_f=\sqrt{T_{A0}T_{B0}

So W=a(T_{A0}+T_{B0}-2\sqrt{T_{A0}T_{B0}

How do I get T_{Cf} from here?

Bye there, and thanks for posting.

 

[Andrew Mason]

What is the heat capacity of A, B and C? Are they all the same?

AM

 

[Telemachus]

I don't know, is it a? I think it is because it looks like it from the equation of state: du=adT and I know that du=dQ=C(T)dT right? if this is it, then yes, the heat capacity is the same for the three bodies, because all of them have the same equation of state, but different initial temperatures, and then the heat capacity is the constant a.

I think what I did before is wrong, I think that the entropy for the process AB, when I extract work should make the entropy decrease on the system AB and increase in C at the end.

The work calculus is wrong, isn't it?

Thanks for posting Andrew :)

 

[Andrew Mason]

\Delta S_{AB}=2a \ln\left(\displaystyle\frac{T_f}{\sqrt{T_{A0}T_{B0}}}\right)

This is correct but is not clear to me how you got this.

It can be shown in a much easier way, without having to deal expressly with entropy.

The heat flow from A is Qh. For a Carnot engine (most efficient) the heat flow to the cold reservoir is Qc=-Qh(Tc/Th). So:

Qc/Qh = -Tc/Th

But:

Qh = adTh and Qc = adTc, so

dTc/dTh = -Tc/Th, which can be rewritten

dTc/Tc = -dTh/Th

Integrating both sides:

ln(Tf/Tc) = ln(Th/Tf), which means that:

T_f^2 = T_hT_c so:

Tf = \sqrt{T_hT_c}

Now that you know the final temperature, you can determine the total heat flow from A and the total heat flow to B. The difference is the work that is done.

AM

 

[Andrew Mason]

To calculate the change in temperature of C you have to proceed in much the same way. Assume it is a Carnot heat pump and remember: dQc + W = dQh.

dQc/dQh = -Tc/Th

adTc/Tc = -adTh/Th =>

ln(T_{AB_f}/T_{AB_0}) = ln(T_{C_0}/T_{C_f})

So:

T_{AB_f}T_{C_f} = T_{AB_0}T_{C_0}

Recall that the heat flow to C, Qh = W + Qc

Express that in terms of temperature differences and heat capacity, a.

Then you have two equations with two unknowns (the final temperatures of AB and C), so you should be able to work it out.

AM

 

[Telemachus]

Sorry, I've skipped some steps when posting, I think that's why its not completely clear. But essentially what I did was making the entropy change of AB as the superposition of the entropy change on A plus the entropy change on B.

U=aT

then \frac{1}{T}=\frac{a}{u}\rightarrow ds=\frac{a}{u}du (constant mole number and volume I assume).

And
s=s_0+a \ln\left(\displaystyle\frac{U}{U_0}\right), and having in mind that U_0=aT_0 and so on...

\Delta S_{AB}=a \ln\left(\displaystyle\frac{T_f}{T_{A0}}}\right)+a \ln\left(\displaystyle\frac{T_f}{T_{B0 }}\right)=2a \ln\left(\displaystyle\frac{T_f}{\sqrt{T_{A0}T_{B0 }}}\right)

And as the work is maximum there is no change in entropy for AB, so that's equal zero. I've followed an example on Callen till here, but I wasn't completely sure about if it applied on this exercise, and didn't know how to do the last part, which isn't included on the example.

Thanks you Andrew. Now I'm going to take a while for reading and understanding your post, and if any doubt emerges I'll let you know.

Thank you very much.

 

[Telemachus]

The temperature T_{ABf} isn't the one already calculated? T_{ABf}=\sqrt{T_hT_c}?
Sorry but I get lost with so many things around :P

 

[Andrew Mason]

The temperature T_{ABf} isn't the one already calculated? T_{ABf}=\sqrt{T_hT_c}?
Sorry but I get lost with so many things around :P

No. It is the T of AB after the heat pump has expended all the available work (ie. The W you calculated from the first part.)

AM

 

[Telemachus]

Damn, I'm confused. And then what is this final temperature T_{f}=\sqrt{T_hT_c}? I thought that was the final temperature for AB after the work was expelled.

Thanks andrew. I see I can find anyway the other temperature working with the two equations you gave me, but I'm trying to clarify the ideas on my mind.

Oh, by the way, the work as I calculated it at the beginning is ok?

Thanks for the help.

 

[dinhjeffrey]

hey this is off topic but how do u post the image?

 

[Telemachus]

I've attached it, and then used the vbcode to make it appear. [*img]url image[*/img] without the * off course. Making a quote of the original post you can see ;)

 

[Andrew Mason]

Damn, I'm confused. And then what is this final temperature T_{f}=\sqrt{T_hT_c}? I thought that was the final temperature for AB after the work was expelled.

For the heat pump, you begin with A and B at T_f = \sqrt{T_{A0}T_{B0}}.
You end up with a final temperature for AB = T_{fAB} which you have to work out.

Oh, by the way, the work as I calculated it at the beginning is ok?

Yes. The work is simply the difference between Qh and Qc which is:

W = |a\Delta T_B| - |a\Delta T_A| = a(T_{B0}-T_f) - a(T_f-T_{A0}) = a(T_{A0}+T_{B0}-2T_f)

AM

 

[Telemachus]

Thanks Andrew. So its like A and B still interacting with the heat pump after the work is expelled? and the temperature I calculated at first its just the one that maximizes the work, and then the "initial" temperature for the system conformed by the heat pump and AB? something like that? and then, there is a final temperature for AB after the work is expelled as heat by the heat pump?

I'm sorry for making so many questions and being iterative, I just want to make things clear to me. Thanks for your help.

 

[Andrew Mason]

Thanks Andrew. So its like A and B still interacting with the heat pump after the work is expelled?

I am not sure what you are asking. A and B together form the "cold" reservoir for the heat pump between AB and C (which is the "hot" reservoir initially at 400K).

and the temperature I calculated at first its just the one that maximizes the work, and then the "initial" temperature for the system conformed by the heat pump and AB? something like that? and then, there is a final temperature for AB after the work is expelled as heat by the heat pump?

All you need to do the second part (see my post #5) is plug in the value for W that you found in the first part into the second equation: Qh = W + Qc. You then have two equations with two unknowns (the final temperatures of AB and C) which you can readily solve.

AM