Carnot Engine with finite reservoirs

[gfd43tg]

Homework Statement

A Carnot engine operates between two finite heat reservoirs of total heat capacity CtH and CtC.
a) Develop an expression relating TH to TC at any time.
b) Determine an expression for the work obtained as a function of CtH ,CtC , TH and
initial temperatures TH0 and TC0.
c) What is the maximum work obtainable? This corresponds to infinite time, when
the reservoirs obtain the same temperature.
In approaching this problem, use the differential form of Carnot s equation,
dQh/dQc=-TH/TC

and a differential energy balance for the engine,
dW - dQH - dQc = 0
Here QH and QC refer to the reservoirs.

Homework Equations

The Attempt at a Solution

I am having trouble with this problem really even getting started. I can't figure out how to use the hints to find T_H related to T_c through time. I gave it an attempt, but I am pretty sure my solution is wrong.

 

[rude man]

Well, since no one else has responded thus far:

I guess this is to be viewed as a "differential Carnot engine" meaning changes in p and V are differentially small. In other words, on a p-V diagram it would be a teensy-tiny cycle, run an infinite number of times (except the reservoirs aren't constant-temperature reservoirs) ... until the two reservoirs have the same temperature.

So write the differential forms given as hints.
I haven't worked this out myself; will see if there is any response.

 

[gfd43tg]

Yes, I wrote my differential form, but not sure if it's correct.

 

[rude man]

For part (a), use the Carnot engine equation. Cross-multiply & you have an easy-to-solve diff. eq. Solve for T2(T1). I use T1 for Thot and T2 for Tcold. Initial reservoir temperatures are T10 and T20.

for (b), use the 1st law equation combined with the result of part a. Integrate dW to get W.

part c is really just part b integrated until the two reservoir temperatures are the same. Get this final temp. from part a again.

 

[rude man]

OP, still out there? Don't give up, I think I have it figured out.

 

[gfd43tg]

Yes. Here is my new solution. Let's compare. The algebra is ugly

 

[rude man]

OK, it's late my time, I have verified your part (a). Of course, that's the easy part.

I'll finish up tomorrow.

Looks like you're on very solid ground.

 

[rude man]

Just got up, but let me start with:

I have dW = -dQ_h - dQ_c = -C_h dTh - C_c dTc
where remember dQ_h and dT_h < 0 and dQ_c and dT_c > 0.

So just integrate each of the right-hand terms in their own variable:
W = -C_h∫_Th0^T_hdT_h - C_c∫_Tc0^T_cdT_c,

keeping the T's separatet until after doing the trivial integrations w/r/t/ the respective T;
then substituting Tc(Th) from part (a) for the second term on the right.

For part (c) just substitute T_f for T_h and T_c. You get T_f from part (a). T_f is the final temperature of the two "reservoirs".

Main thing here is to keep track of the signs!
How's that sound?
.

 

[gfd43tg]

I was told for part C that you need everything in terms of the initial hot and cold temperatures.

 

[rude man]

I was told for part C that you need everything in terms of the initial hot and cold temperatures.

That's what you get. You previously found T_c = T_c0(T_h/T_h0)^(-Ch/Cc).

So let T_h = T_c = T_f and solve for T_f(T_h0, T_c0).

 

[rude man]

Well, since no one else has responded thus far:

I guess this is to be viewed as a "differential Carnot engine" meaning changes in p and V are differentially small. In other words, on a p-V diagram it would be a teensy-tiny cycle, run an infinite number of times (except the reservoirs aren't constant-temperature reservoirs) ... until the two reservoirs have the same temperature.

So write the differential forms given as hints.
I haven't worked this out myself; will see if there is any response.

Just in case anyone drops in late here, my description of the p-V diagram was incorrect. It would look like a vertical sliver, with differentially small excursions along the isothermals but of course the adiabatics would have to go from T1 to T2 & back. The isothermals would gradually approach each other and finally merge, at which point the diagram is just a point on isotherm Tf.