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Work done by a expanding and contracting wire over a Carnot cycle

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data

    Basically I have the following problem. Given the equation of state [itex]\sigma = \frac{b}{T}\frac{L - L_0}{L_0}[/itex] where b and [itex]L_0[/itex] are positive constants, calculate the work done over a Carnot cycle by a wire with this equation of state.

    2. Relevant equations

    I already have the work done over the isoterm curves. But when I try to calculate the work on the adiabatic processes I just get stuck over and over again.

    3. The attempt at a solution

    I use the first law of thermodynamics

    [itex] du = \delta Q + \delta W[/itex]

    also we know that the work for this system is the following

    [itex]\delta W = - \tau dL[/itex]

    and in a book of basic thermodynamics I have found that

    [itex]dT = -\frac{T}{c_\tau}\left( \frac{\partial L}{\partial T} \right)_\tau d\tau[/itex]

    Any ideas?
    Last edited: Feb 23, 2014
  2. jcsd
  3. Feb 24, 2014 #2


    User Avatar

    Staff: Mentor

    Hi Uriel. Welcome to PF!

    Can you give more details about what you've tried?

    What is ##\tau##?
  4. Feb 24, 2014 #3
    [itex] \tau [/itex] is the [itex] \sigma [/itex] in the equation of state, it's linear tension, I change the name of the variable because [itex] \sigma [/itex] might be confused by strains and that's not the case, I'm sorry for that.

    Using the first law on the isotherm process we got that the internal energy doesn't change, therefore

    [itex] d u = 0[/itex]

    and using this [itex] \delta W = -\delta Q [/itex]

    now we know that the work is computed with the integral [itex] \delta W = - \int \tau dL [/itex]
    so it's really simple to calculate, for the case on the two isotherm processes.

    My problem arises when I try to calculate the work performed over the adiabatic cycles. I have two options, integrate over the path of the adiabatic curve from one state to the other, or I need to know the internal energy of the system and using that plug in the temperature of each state and calculate the difference.
    That is the awful part; given the equation [itex] dT = - \frac{T}{c_\tau} \left( \frac{\partial L}{\partial T} \right)_\tau d \tau [/itex]

    I need to know L, but this is easy, [itex] L = \frac{L_0 T}{b}\tau + L_0 [/itex],

    next I calculate the partial derivative of L with respect to T leaving [itex] \tau [/itex] constant.

    So [itex]\left( \frac{\partial L}{\partial T} \right)_\tau = \frac{L_0 \tau}{b} [/itex]

    then [itex]dT = - \frac{T}{c_\tau} \left( \frac{\partial L}{\partial T} \right)_\tau d \tau = - \frac{T}{c_\tau} \frac{L_0 \tau}{b} d \tau[/itex].

    Reordering the terms in the differential equation we get

    [itex] \frac{dT}{T} = - \frac{L_0}{c_\tau b} \tau d \tau [/itex]

    solving the differential equation we get this

    [itex] ln \left( \frac{T_f}{T_i} \right) = - \frac{L_0}{c_\tau b} \left( \frac{\tau_f^2 - \tau_i^2}{2} \right) [/itex]

    And that's all I got, I don't have the slightest idea what to do next.
  5. Feb 22, 2015 #4
    In which book did you see that problem?
  6. Feb 22, 2015 #5
    I'm not sure it was a book problem. I think the professor just made it up, but she was very fond of the book of Callen.
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