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Refrigerator (Thermodynamics)

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img21.imageshack.us/img21/1197/questionof.jpg [Broken]


    3. The attempt at a solution

    For the first part, if we assume that the cycle is reversible from the 2nd law we have

    [itex]\Delta S = \frac{\Delta Q}{T}+S_{gen} = 0[/itex]

    And here is the Clasius Inequality

    [itex]S_{gen} \geq 0 \implies \frac{\Delta Q}{T} \leq 0[/itex]

    [itex]\frac{Q_x}{T_1}-\int^{T_2}_{T_1} \frac{\Delta Q}{T} \leq 0[/itex]

    Since ΔQ = dT

    [itex]\frac{Q_x}{T_1}-\int^{T_2}_{T_1} \frac{dT}{T} = \frac{Q_x}{T_1}- \ln \frac{T_2}{T_1} \leq 0[/itex]

    [itex]T_2 \geq T_1e^{\frac{Q_x}{T_x}}[/itex]

    Since we do not know the temprature values we can't make numerical calculation of the lower limit for the final temperature of y. But did I derive the correct equation?

    For the second part of the equation, to find the minimum value of work that needs to be supplied to the heat pump I tried to use the 1st Law;

    Q − W = U = 0 → W = mcΔT - U

    How can we use this to show the above expression for Wmin?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 24, 2012 #2
    We do not know if the cycle is reversible or irreversible. But I assumed that it is reversible because if it is not then more and more work is required due to the entropy generated.
     
    Last edited: Mar 24, 2012
  4. Mar 24, 2012 #3
    Okay, I figured out part (1) :smile:, but I'm still having some trouble working out the minimum work.

    The entropy change of x is

    [itex]\Delta S_x = \int^{T_2}_{T_1} C_p \frac{dT}{T} = C_p \ln \frac{T_2}{T_1}[/itex]

    The entropy change of y is

    [itex]\Delta S_y = \int^{T_2’}_{T_1} C_p \frac{dT}{T} = C_p \ln \frac{T_2’}{T_1}[/itex]

    The total entropy change is

    [itex]\Delta S_{tot} =\Delta S_x + \Delta S_y = C_p \ln \frac{T_2}{T_1} + C_p \ln \frac{T_2’}{T_1}[/itex]

    And we know from the entropy principal that ΔS ≥ 0, so

    [itex]\left( C_p \ln \frac{T_2}{T_1} + C_p \ln \frac{T_2’}{T_1} \right) \geq 0[/itex]

    [itex]C_p \ln \frac{T_2T_2’}{T_1^2} \geq 0[/itex]

    The minimum value of T2’ would be

    [itex]C_p \ln \frac{T_2T_2’}{T_1^2} = 0 = \ln 1[/itex]

    [itex]\therefore T_2’ = \frac{T_1^2}{T_2}[/itex]

    Okay now the minimum work:

    Heat removed from x to cool it is Q=Cp (T1-T2) and the heat added to y is Q+W=Cp (T2'-T1) so

    W= Cp(T2'-T1)-Cp(T1-T2)=Cp(T'2+T2-2T1)

    Substituting the minimum temprature we found we end up with

    [itex]W_{min} = C_p \left( \frac{T_1^2}{T_2} +T_2-2T_1 \right) = C_p \frac{(T_1-T_2)^2}{T_2}[/itex]

    But the expression I need is:

    [itex]W_{min} = \frac{mc (T_1-T_2)^2}{T_2}[/itex]

    What could we do to arrive at that expression? :confused:
     
    Last edited: Mar 25, 2012
  5. Mar 25, 2012 #4
    The only way I can have:

    [itex]W_{min} = C_p \frac{(T_1-T_2)^2}{T_2} = \frac{mc (T_1-T_2)^2}{T_2}[/itex]

    is if Cp is equal to mc. But how could this be?
     
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