1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook