Heat engine, maximum work

[briteliner]

Homework Statement

An object maintained at constant volume has heat capacity Cv, which is independent of T. The object is initially at a temperature Th, and a heat reservior at a lower temperature Tc is available. Show that the maximum work that can be extracted when a heat engine operating between the object and the reservior brings the object to the temperature of the reservior is

W = Cv [ (Th - Tc) - Tcln (Th/Tc) ].

Hint, the maximum amount of work is obtained when the process is rversible, i.e., when the entropy change of the universe is zero.

Solve the previous problem if the cold reservior is replaced by an object of heat capacity Cv. Hint: First find the final temperature of the two objects.

Homework Equations

The Attempt at a Solution

W=Qh-Qc
dQ=mCvdT--> Q=mCv(Tf-Ti)
I can see how the first part of the expression came about by integraing dQ and considering Tf as Th and Ti as Tc, but I don't get where the natural log came in since i don't know where a 1/T term would come in...

 

[Andrew Mason]

The Attempt at a Solution

W=Qh-Qc
dQ=mCvdT--> Q=mCv(Tf-Ti)
I can see how the first part of the expression came about by integraing dQ and considering Tf as Th and Ti as Tc, but I don't get where the natural log came in since i don't know where a 1/T term would come in...

Since the object has heat capacity Cv, what happens to its temperature as heat flows from it to the cold reservoir? The expression for the temperature of the object as a function of heat flow (which is measured by the change in temperature multiplied by heat capacity) gives you the logarithmic result.

The 1/T term comes in the expression for dW:

$$dW = (1-T_c/T_h)dQ_h$$

$$dQ_h = CvdT$$

AM

 

[briteliner]

thanks!
what about if the cold reservoir were replaced by an object of heat capacity Cv? would the 2 just come to equilibrium and the work be 0?

 

[Andrew Mason]

thanks!
what about if the cold reservoir were replaced by an object of heat capacity Cv? would the 2 just come to equilibrium and the work be 0?

What would the relationship be between Tc and Th if the heat capacities are the same? (think of the relationship between a change in Tc and a change in Th).

AM

 

[Count Iblis]

You can also derive this by first writing down the expression for the entropy. You then say that in the limit of maximum work the entropy doesn't change. Using this method, it is straightforward to find the general expression of the maximum work you can extract from N objects each with heat capacity Cv that are initially at temperatures T1, T2, T3,..., Tn.

 

[Count Iblis]

Let's derive that general formula.

Instead of two objects, we have N objects each with some arbitrary temperature. We have to deal with two processes: Heat flow and performed work. The heat flow is entirely between the objects and leaves the total internal energy of the N objects together unchanged. The performed work by the system is energy that leaves the system of the N objects. Denote by U_i the total internal energy at the start and by U_f the final total internal energy, we have:

U_f = U_i - W (1)

where W is the work performed by the system of the N objects.

We can compute the internal energy in terms of the heat capacity. Although the heat capacity is assumed to be a constant, this can only be true in some limited temperature range. We'll assume that the heat capacity is constant above some temperature T0 which is smaller than the temperatures of the objects. We can then write:

U_i = N U(T0) + C(T1 - T0) +C(T2- T0) + C(T3 -T0)+...+
C(TN-T0)

Here U(T0) is the internal energy function at temperature T0, C is the heat capacity of the objects and Ti is the temperature of object i. We can rewrite (2) as:

U_i = N[U(T0) - C T0] + C (T1 + T2 + T3+...TN ) (3)


If the final temperature is T_f, then:

U_f = NU(T0) + N C (T_f - T0) =

N[U(T0) - C T0] + N C T_f (4)

We see from (1) that:

W = U_i - U_f = NC [(T1 + T2 + T3+...TN)/N - T_f] (5)

Clearly, the lower the final temperature, the more work will be performed by the system. The fact that the entropy cannot decrease imposes a limit on how low the final temprerature can be. So, if we compute the entropy of an body at temperature T, we can easily find what the maximum work is that can be extracted from the system.

We have:

dS = dQ/T

The heat capacity is a constant C for T > T0, therefore dQ = C dT for
T > T0. This gives:

dS = C dT/T

If we integrate both sides from T0 to T we get:

S(T) - S(T0) = C Log(T/T0) ------->

S(T) = S(T0) + C Log(T/T0)

The initial entropy of the system of three bodies, S_i, is thus given by:

S_i = S(T1) + S(T2) + S(T3) +...S(TN)=

N S(T0) + C[Log(T1/T0) + Log(T2/T0) + Log(T3/T0)+...Log(TN/T0)]

The final entropy, S_f, is:

S_f = N S(T_f) = N S(T0) + NC Log(T_f/T0)

If we demand that S_f >= S_i, we get:

NC Log(T_f/T0) >= C[Log(T1/T0) + Log(T2/T0) +...+ Log(TN/T0)] -->


Log(T_f/T0) >= 1/N [Log(T1/T0) + Log(T2/T0) + ...Log(TN/T0)]---->

T_f >= (T1 T2 T3 ...TN)^(1/N)

Eq. (5) then implies that:


W <= NC [Arithmetic average of initial temperatures - Geometric average of initial temperatures ]

The maximum amount of work that can be extracted is obtained in the limit of a reversible process in which the total entropy does not increase. Then, you have an equals sign in the above equation:

W = NC [Arithmetic average of initial temperatures - Geometric average of initial temperatures ]

 

[Mapes]

Nice!