Homework Help: Carnot engine with finite reservoirs

1. Feb 13, 2010

homology

1. The problem statement, all variables and given/known data

I'm trying to find the final temperature of a carnot engine with two finite thermal reservoirs. We're told that the heat capacities for both reservoirs are constant (and equal) and to regard the change in each reservoir's temperature during any 1 cycle as negligible.

2. Relevant equations

W=PdV for isothermal expansions/compressions and dE=0 for those portions so the work done by (or on) the gas is equal to the energy supplied by (or to) the reservoir.

P=NKT/V where T will be the temperature of either reservoir depending one whether we're compressing or expanding.

3. The attempt at a solution

If each cycle's effect on the reservoir's temperature is neglible, then you might say its dT, so the dq = CdT, but the energy supplied by heating (or cooling is definitely finite, so its more like q=CdT). Also, while I suspect there's an integral somewhere here, I'm not sure of what the other integration variable is, there are going to be a lot of cycles, we might even model it as an infinite number of them, but how would that parameter come into play? I've thought about viewing this as a volume in P-T-V land, so some double integral over the T-V plane, after all, an integral under one isotherm gives the work from that that isotherm. I know that the isotherms are going to approach each other (and the volumes will drift) so that eventually the engine stalls. This sweeps out a surface over the T-V plane with height P(T,V) at a given point.

I suspect that I'm working way too hard at this, since everything changes by a little bit after each cycle. I don't expect, or want much help, just a tiny nudge to get me going.

2. Feb 13, 2010

Count Iblis

Forget the cycles. You can easily solve the problem by consider the entropy of the system. Now we know that the entropy has to go to zero at zero absolute temperature and then implies that heat capacity cannot realy be constant. But if we take some low reference temperature T0 at which the entropy of one reservoir is S0, then we can take the heat capacity constant for T > T0. The entropy can then be written as:

S(T) = S0 + C Log[T/T0]

where C is the heat capacity.

You can now solve the problem by using conservation of energy and by using the fact that the final entropy must be equal or larger than the intitial entropy, the case of equality is only obtained when operating an ideal Carnot engine.

You can also easily generalize this for the case of N reservoirs each at some different initial temperature.

3. Feb 13, 2010

homology

I'm confused about the absolute zero temperature. How does that factor in since we we have two nonzero temperatures and they will approach an intermediate temperature as the cycle "shrinks."

I've also been thinking about the efficiency of the engine. As the temperatures approach one another the efficiency will approach zero. The problem didn't specify anything about this, but perhaps this helps?

4. Feb 14, 2010

Count Iblis

Well, you can also forget about absolute temperature and simply take the entropy of one reservoir to be:

S = C Log(T)

This will give you the correct answers, but this formula cannot be correct for temperatures approaching T = 0.

Start the problem by considering the following question. You have two reservoirs and some general process leading to energy exchange between the reservoirs happens. Some external work W can be performed in this process. We can specify the state of the system by specifying the two temperatures of the two reservoirs. Suppose then that the inititial temperatures are T1 and T2 and that after the process the final temperatures are T1' and
T2'. You know that both heat capacities are C, so what is then the change in the sum of the internal energies of the two reservoirs? What does the First law of Thermodynamics then tell you about the performed external work?

If someone does not know about entropy, he could think that you could choose bot final temperatures very low and therefore let the system do a huge amount of work. But we know that the final entropy can be the same at best (note that all the heat that is exchanged stays in the system, we're not dumping heat from a reservoir to the environment, rather the heat can only flow from one reservoir to another. So the entropy increase of the system itself is equal to the total entropy increase, the Second Law then tells you that the entropy can only increase or stay the same).

So, you write down the entropy increase in terms of the final and initial temperatures and then demand that it be positive or zero. Maximizing W under these restrictions will give you the desired answer.

5. Feb 14, 2010

homology

S=Cln(T)? Isn't it S=kln(constant*T^(3N/2)V^N)? Stuff's still pretty new to me.

The hot reservoir has a temp change of T1'-T1 and the cold has change of T2'-T2. There's no work done on the reservoirs, so the change in energy of the sum is:

[itex]\frac{1}{2}k(T'_1-T_1)+\frac{1}{2}k(T'_2-T_1)[\latex]

So, some of this energy would have been spent on work, right? A reversible engine could only use 1-T2/T1 percent of it for work?

As for entropy, why would someone mistakenly think that two low temperatures would be a good choice? I would think that two very different temperatures would be better (more enclosed area in the P-V diagram) and better efficiency.

Now with infinite reservoirs, the only change in entropy if in the working substance, but since the cycle is...well a cycle, it returns to its initial value over and over again. But now the entropy of our hot reservoir is decreasing (as it cools) and the entropy of the cold reservoir is increasing (because its warming) and the entropy of the working substance is also increasing right? I figure this must be the case because the change in entropy of the hot reservoir will be larger than that of the cold reservoir (area under the curves, work equals q).

I'm still not seeing how this problem resolves itself, but if you don't mind continuing this conversation I would appreciate it.

Cheers,

Kevin

6. Feb 14, 2010

homology

Sorry about the latex, I thought I just needed the [itex] [\latex] tags?

7. Feb 17, 2010

homology

Resolution!

Okay, I finally grasped the problem with some additional help. Here is what I was getting caught up on. I knew that for a complete cycle of a reversible heat engine that entropy was conserved. What was tripping me up is that we weren't completing a cycle, but rather on a tight spiral. So INITIALLY I figured that I couldn't use entropy. But when I was sketching it out the other day, I realized that even though we don't return to our initial state, the state that we do end up at is connected to our initial state via an adibat. There's no change in entropy along an adibat so I can in fact use conservation of entropy.

So then I know that Q_hot/T_hot = Q_cold/T_cold and I also know that Q = CdT for any given cycle, so I can integrate dS=CdT/T for each reservoir and their sum will be zero (since the change in entropy for the hot reservoir is negative). From here it's just algebra.

Sorry for being thickheaded...as I said its been a while.

Thanks for you help,

Kevin