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Carnot engine with a magnetic auxiliary system

  1. Jun 29, 2013 #1

    fluidistic

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    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. jcsd
  3. Jun 30, 2013 #2

    TSny

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    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##.
     
  4. Jun 30, 2013 #3

    fluidistic

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    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.
     
  5. Jun 30, 2013 #4

    TSny

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    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
  6. Jun 30, 2013 #5

    fluidistic

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    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.
     
  7. Jun 30, 2013 #6

    TSny

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    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
  8. Jun 30, 2013 #7

    fluidistic

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