# Carnot engine with a magnetic auxiliary system

1. Jun 29, 2013

### fluidistic

1. The problem statement, all variables and given/known data
I would like some assistance to solve the following problem.
A magnetic system satisfies Curie's law $M=nDB/T$, has a specific heat capacity at constant magnetization $C_M=\text{constant}$. It is used in a Carnot engine that operates between temperatures $T_h$ and $T_c$ ($T_c<T_h$).
M is the magnetization and B is the external magnetic field.
1)Sketch a qualitative (B,M) diagram of a complete cycle.
2)Calculate the work done by the engine after 1 cycle.
3)Calculate the efficiency of the engine.

2. Relevant equations
$C_M=\frac{T}{n} \left ( \frac{\partial S}{\partial T} \right ) _{M,n}=\left ( \frac{\partial U}{\partial T}\right ) _{M,n}$ (1)

3. The attempt at a solution
1)I've done the sketch, I don't think there's anything particular about it.
2)This is where I'm stuck.
From equation (1) I've determined that $S(T,M,n)=C_Mn \ln T+f(M,n)$ and that $U(T,M,n)=C_MT+g(M,n)$. Not sure this can help.
I know that the work done is the area enclosed by the sketch in the B-M diagram, namely $W=\oint B dM$.
I know that in a Carnot cycle there are 2 adiabatic and 2 isothermal processes. Also the 1st law of Thermodynamics states that $\Delta U =Q+W$ so after a cycle $\Delta U=0$ and so $Q=-W$. In other words the heat toward the auxiliary system is equal to the work done BY the system. And since there are 2 adiabatic processes (no heat is being absorbed by the system), I get that the work done after 1 cycle is equal to the heat absorbed by the system during the 2 isothermal processes.
Now if I think of M as a function of B, T and n and assuming that n is constant for the auxiliary system then $dM=\frac{nDdB}{T}-\frac{nDBdT}{T^2}$. Note that for the 2 isotherms, $dM=\frac{nDdB}{T}$.
I also know that $dW=BdM$ and so for the isotherms, $dW=\frac{nDBdB}{T}$. I know that there's 1 isotherm at a temperature of $T_h$ and the other at a temperature of $T_c$. My problem is that I don't know what are the limits of the integral if I integrate this expression. So I can't really get W via this expression.
I don't really know how to proceed further.
Any tip is appreciated!

2. Jun 30, 2013

### TSny

Let $\small (M_1, B_1)$, $\small (M_2, B_2)$, $\small (M_3, B_3)$, and $\small (M_4, B_4)$ be the states on the $\small B$-$\small M$ diagram corresponding to the "corners" of the Carnot cycle. 1→2 is isothermal at $T_h$, 2→3 is adiabatic (reducing $\small T$ to $\small T_c$), etc.

So, for the leg 1→2, your limits of integration for $\small W=-\int B dM$ will be from $\small M_1$ to $\small M_2$, etc.

You want to get expressions for the total work done in the cycle, and for the heat added during the cycle. These expressions will be in terms of the $\small M_i$'s, the hot and cold temperatures, and the constants $\small n, D, C_M$.

3. Jun 30, 2013

### fluidistic

Oh I see, thanks.
I didn't know I could assume $M_i$'s to be known.
Following what I had done I reach that $W=\frac{1}{2nD}[T_h(M_1^2-M_2^2)-T_c(M_3^2-M_4^2)]$ (work done ON the system so to answer the question I'd have to set -W to get the work done BY the system). If that's the correct answer then I guess I'm done for part 2). If it's wrong I'm going to post what I did to reach this.

4. Jun 30, 2013

### TSny

That's close to what I get, except for signs. The work done by the system is $\small -\int{BdM}$ [Edited to replace the symbol H by B ]. So, I think your first term $\small \frac{1}{2nD}T_h(M_1^2-M_2^2)$ represents the positive work done by the system as $\small M_1$ is reduced isothermally to $\small M_2$ (with $\small M_2 < M_1$). The work done by the system in going from 3 to 4 (with $\small M_4 > M_3$) should be negative.

The signs are confusing because you need to reduce M to get positive work done by the system (which is opposite to P-V work where V should be increased to get positive work done by the system).

Last edited: Jun 30, 2013
5. Jun 30, 2013

### fluidistic

So basically I've got the signs wrong?
I know that the total work done BY the system should be positive. My answer to the question would be $W'=-W=\frac{1}{2nD}[T_h(M_2^2-M_1^2)-T_c(M_4^2-M_3^2)]$. I believe that it's positive but I'm not 100% sure.

6. Jun 30, 2013

### TSny

I agree with your second term (which yields negative work done by the system in going from 3 to 4), but I don't agree with the sign of your first term (which would also give negative work in going from 1 to 2). As I see it, $\small M_2 < M_1$ and $\small M_4 > M_3$.

Work done by the system in going isothermally from a to b is $\small W = -\int_a^b{BdM} = -\frac{T}{2nD}(M_b^2-M_a^2)$.

Last edited: Jun 30, 2013
7. Jun 30, 2013

### fluidistic

You are right TSny, I understand my error now. Thanks a lot.