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Callen's problem 4.6-7, thermodynamics

  1. Apr 16, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Hi guys. I'm basically clueless on what's going on, I get 2 different results while I should get the same.
    The problem reads "Each of two bodies has a heat capacity at constant volume given by ##C_v=nc_v=A+2BT## where A and B are known constants (both positive).
    If the 2 bodies are initially at temperatures ##T_1=200K## and ##T_2=400K## and if a reversible work source is available, what are the maximum and minimum final common temperatures to which the 2 bodies can be brought? What is the maximum amount of work that can be transferred to the reversible work source? It is to be assumed throughout that each body is maintained at constant volume.


    2. Relevant equations
    1st law of Thermodynamics: ##dU=\delta Q + \delta W## where U is the internal energy of the system I'm considering, Q is the heat going toward the system and W is the work done on the system.
    Others.

    3. The attempt at a solution
    First I want to calculate the minimum final common temperature. It occurs when the work done BY the system is maximum and it corresponds to a zero change in entropy in the whole system. I.e. ##\Delta S_{\text{Total}}=\Delta S_1+\Delta S_2 + \underbrace {\Delta S _{\text{heat engine}}}_{\text{=0 because the engine works with cycles}}=\Delta S_1+ \Delta S_2##.
    Now I use the equation ##dS=\frac{dQ}{T}## and that ##dQ=nc_vdT## for both bodies. I get that ##\Delta S= \int _{T_1}^{T_f} nc_v \frac{dT}{T}+\int _{T_2}^{T_f} nc_v \frac{dT}{T}=nA\ln \left ( \frac{T_f^2}{T_1T_2} \right ) +2nB[2T_f-(T_1+T_2)]## where I skipped some arithmetical steps. Now the minimum ##T_f## can be found by setting ##\Delta S=0## and isolating ##T_f## (which I don't think is possible, so a numerical method should be used). So far so good I've virtually determined the ##T_f## that they ask me and it cannot be written in terms of elemental functions.

    Now if I use a different way to solve the problem I get a different answer.
    If I consider the whole system 2 bodies+heat engine as my system, I get that ##\Delta U=W## where W is the work done on the system. And delta U is the change in internal energy of the 2 bodies + heat engine but the change in internal energy of the heat engine is 0 because it works with cycles.
    So my goal is to get the work done by the system, in other words, ##-W##.
    The change in internal energy of the first body is worth ##dU_1=dQ_1+\underbrace {dW_1}_{=0 \text{ because V=constant}}=nc_vdT_1 \Rightarrow \Delta U_1=n[A(T_f-T_1)+B(T_f^2-T_1^2)]##. In the same fashion I get ##\Delta U_2##. So ##\Delta U=\Delta U_1+\Delta U_2 =n[2AT_f-A(T_1+T_2)+2BT_f ^2 - B(T_1^2+T_2^2)]## which is worth the work done on the system (because of the 1st Law of Thermodynamics and using the fact that the whole system is isolated and therefore Q=0).
    Therefore the work done by the system is ##-W=W'=n[B(T_1^2+T_2^2)-2BT_f^2+A(T_1+T_2)-2AT_f]##.( The maximum amount of work is done when ##T_f## is minimum so if I could isolate ##T_f## above and plug it here I'd answer the question of the problem.)
    Here I can think of W' as a function of ##T_f## only. In such case I want to see what value of ##T_f## extremizes W'. Derivating W' with respect to ##T_f## and setting that expression equal to 0, I found ##T_f=\frac{A}{2B}## and the second derivatives gives me ##-4B<0## thus W' is maximum and therefore ##T_f## corresponds to a minimum. But it is not equal to the previous one I had found! I really don't understand what's going on. I would like someone to comment on this.

    To continue the problem, if I set ##W'=0## (where the prime denotes the work done BY the system, not the derivative!) I should get ##T_f## maximum because no work is being done by the heat engine and so the heat flows from body 1 to body 2 until both temperatures are equal. In that case I get ##T_{fmax}=-\frac{A}{B}+\sqrt{\left ( \frac{A}{B} \right ) ^2 +2 [T_1^2+T_2^2 + \frac{A(T_1+T_2)}{B}] }##. I wonder if this result is good.
     
    Last edited: Apr 16, 2013
  2. jcsd
  3. Apr 18, 2013 #2

    Andrew Mason

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    Your approach is correct for determining minimum Tf. But the n factor should not be there ie. should be:

    [tex]\Delta S = A\ln \left ( \frac{T_f^2}{T_1T_2} \right ) +2B[2T_f-(T_1+T_2)][/tex]

    which simplifies to:

    [tex]\Delta S = 2A\ln(T_f) - A(\ln 400 + \ln 200) + 4BT_f - 2B(600)[/tex]

    If ΔS = 0,:

    [tex]2A\ln(T_f) + 4BT_f = A(\ln 400 + \ln 200) + 1200B[/tex]

    I agree that you would need to use numerical methods to solve for Tf. That gives you the minimum Tf.

    The maximum Tf though is much easier. The question says that a reversible engine is available - it doesn't say you have to use it. What is Tf if you just let the heat flow irreversibly from one body to the other?

    How is the work done in the first situation (minimum Tf) related to the difference in the internal energies of the bodies at maximum Tf and at minimum Tf?

    AM
     
  4. Apr 18, 2013 #3

    fluidistic

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    I see, thank you very much. On my draft I had not put the n but when copying here I thought I had to put it for some reason that I will check out when I've time.
    Yeah I know, this is basically what I've done. I've found an expression for the work done by the machine: ##-W=W'=n[B(T_1^2+T_2^2)-2BT_f^2+A(T_1+T_2)-2AT_f]## (I have to check out whether the n should be here). When I set W'=0 I found out the maximum T_f: ##T_{fmax}=-\frac{A}{B}+\sqrt{\left ( \frac{A}{B} \right ) ^2 +2 [T_1^2+T_2^2 + \frac{A(T_1+T_2)}{B}] }##, which is as we both said no work is done by the engine and what happens is only the flow of heat from body 1 to body 2 until both temperatures are equal.
    But what I found strange is when I think of W' as a function of T_f, then I derivated W' and equated it to 0 and found out that the minimum T_f does not correspond to the value we had found earlier.
    Ok I'll check this out as soon as I can.
    Thanks a lot for all.
     
  5. Apr 18, 2013 #4

    Andrew Mason

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    For irreversible heat flow resulting in maximum Tf:

    [tex]\Delta U_1 = \int C_vdT = \int (A + 2BT)dT = A(T_f-200) + B(T_f^2-200^2) = -\Delta U_2 = -A(T_f-400) - B(T_f^2-400^2)[/tex]

    So Tf max. is the solution to this quadratic equation:

    [tex]AT_f + BT_f^2 - (300A + 20000B) = 0[/tex]

    The work obtained in the first case should be equal to the heat flow required to raise both bodies from the minimum Tf to the maximum Tf. Your approach (calculating the difference in changes of internal energy in body 1 going from T1 to Tfmin and in body 2 going from T2 to Tfmin) should give the same result. Either way it is a rather ugly looking calculation!

    AM
     
  6. Apr 18, 2013 #5

    fluidistic

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    Yeah I also had to solve a quadratic in T_f. I kept the positive temperature, the other T_f being negative.
    I asked a TA at my university what's wrong between my 2 results for the minimal T_f. He told me that I cannot conclude that if T_f maximizes W' (in my expression ##W'=[B(T_1^2+T_2^2)-2BT_f^2+A(T_1+T_2)-2AT_f] ## ) then T_f is the minimal one. Because this would be ignoring the fact that T_f has a minimum value set by the 2nd Law of Thermodynamics, which I already found using ##\Delta S=0##.
    In other words maximizing my W' would ignore that ##\Delta S \geq 0##. So basically my question is answered and the problem is solved.
    And yeah, I put an extra n in my first post as you pointed out!
    Thanks a lot for all.
     
  7. Apr 19, 2013 #6

    Andrew Mason

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    I see what you were trying to do. I thought you were using the minimum Tf in the expression for ΔU (ie. W = -(ΔU1 + ΔU2) = - Cv(Tf-T1) - Cv(Tf-T2)). You were simply trying to find the value of Tf that maximizes the value of W in that expression. As your instructor pointed out, Wmax (min. Tf) is not the maximum value of W in that expression. The maximum value of W in W = -(ΔU1 + ΔU2) is where ΔU2 = 0 (ie. body 1 ends up at T2 and all the heat flow from 1 is converted to work).

    AM
     
  8. Apr 19, 2013 #7

    fluidistic

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    Ah I see! So this would correspond to an efficiency of 1, right? Which is of course greater than Carnot's engine, thus impossible.
     
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