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Heat Conduction Problem

  1. Nov 6, 2004 #1
    Hello friends

    I have a problem and also know its answer but I have some trouble figuring it out. Here goes

    Two vessels filled with different liquids are at temperatures [tex]T_{1}[/tex] and [tex]T_{2}[/tex] respectively. They are joined by a metal rod of length l; area of cross-section A and thermal conductivity k. The masses and specific heats are [tex]m_{1}[/tex], [tex]m_{2}[/tex] and [tex]s_{1}[/tex], [tex]s_{2}[/tex] respectively. If [tex]T_{1} > T_{2}[/tex], calculate the time when the temperature difference between two liquids is halved, assuming there is no radiation loss by the liquids and the rod to the surroundings.

    Answer: [tex]t = \frac{lm_{1}m_{2}s_{1}s_{2}}{kA(m_{1}s_{1} + m_{2}s_{2})}ln2[/tex]

    My problem is that I keep getting a similar answer but without the ln2 because I am setting up a wrong equation. If T1 > T2, then heat should flow from m1 to m2 through the rod and so in time dt, the temperature of m1 should drop to T_{A}-dT whereas that of the mass m2 should rise to T_{B}+dT. There was another variation of the problem where the temperature of m1 was constant and for that very reason the problem was simple. Here it seems that my differential equations/reasoning are wrong. Please offer some suggestions as to how this problem can be approached.

    Thanks and cheers
    Vivek
     
  2. jcsd
  3. Nov 6, 2004 #2

    Doc Al

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    While the heat lost by one liquid equals the heat gained by the other, the two sides do not have the same change in temperature since their masses and specific heats are different.
     
  4. Nov 6, 2004 #3
    Oh yes Doc, that was actually a typo and I just discovered it. It wasn't quite a flaw in my reasoning as on paper I took dT_A and dT_B. But I can't get the thing to work still :cry:
     
  5. Nov 6, 2004 #4

    Clausius2

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    Let's see...

    i) Conservation of Energy for vessel 1:

    [tex]m_1 s_1 \frac{dT_1}{dt}=-q''A[/tex]

    ii) Conservation of Energy for vessel 2:

    [tex]m_2 s_2 \frac{dT_2}{dt}=q''A[/tex]

    iii) Heat transfer:

    [tex] q''=-k\frac{dT}{dx}=-k\frac{T_1-T_2}{l}[/tex]

    The problem here is to find a relation between T1 and T2.

    Let me think of it some time. Meanwhile you could post your starting equations, I mean those you have said are wrong.
     
  6. Nov 6, 2004 #5

    Doc Al

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    I'll point you in the right direction. Start with Fourier's law of thermal conductivity:
    [itex]dQ/dt = \frac{kA}{l} \Delta T[/itex], where [itex]\Delta T[/itex] is the temperature difference. Now, to keep from going nuts, I will use a different symbol (X) to represent the temperature difference [itex]\Delta T[/itex], so [itex]dQ/dt = \frac{kA}{l} X[/itex].

    The next step is to write dQ in terms of dX. I'll leave that to you. (Think: for every bit of heat that flows, one temperature gets cooler, the other hotter, thus reducing the difference in temperature.)

    (Edit: I had mistakenly called that equation for thermal conductivity Newton's law of cooling. My mistake! :blushing: )
     
    Last edited: Nov 6, 2004
  7. Nov 6, 2004 #6

    Clausius2

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    Please do not post that words here for the sake of the forum. That you have posted is not the Newton Law of Cooling. It is the same thing as to say:

    start with 2nd Newton's law:

    [tex] \nabla x E=0[/tex]

    ??
     
  8. Nov 6, 2004 #7

    Doc Al

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    You are correct. That equation for thermal conductivity is Fourier's law, not Newton's law of cooling. Sorry! (I'll go back and fix it.)

    No matter what you call it, that's where you should begin in solving this problem. :smile:
     
  9. Nov 6, 2004 #8
    Well, these are the equations I wrote too (though slightly differently). The problem is how to solve this system in general for the temperatures as a function of time. I think you missed the area out of your last equation which should be

    [tex]q''=-k\frac{dT}{dx}=-kA\frac{T_1-T_2}{l}[/tex]

    Also, the first two equations are of the form dQ = mcdT (or dQ/dt = mcdT/dt = q''. The area doesn't come into them. A = area of cross-section). Okay you've used q'' in a different way. But here's what I have done: (in my computations q'' = dQ/dt and not q''A = dQ/dt)

    Differentiating the heat transfer equation with respect to time, we get
    [tex]\frac{dq''}{dt}=\frac{-kA}{l}\frac{d(T_{1}-T_{2})}{dt}[/tex]

    [tex]\frac{dq''}{dt} = \frac{-kA}{l}(-\frac{q''}{m_{1}s_{1}} - \frac{q''}{m_{2}s_{2}})[/tex]

    which gives
    [tex]\frac{dq''}{q''} = \frac{kA}{l}(\frac{1}{m_{1}s_{1}} + \frac{1}{m_{2}s_{2}})dt[/tex]

    Now I think I can integrate both sides to get the ln on the left side. But what about the boundary conditions on the left hand side? Is the following expression correct?

    [tex]\ln(\frac{q''}{q''_{0}}) = \frac{kA}{l}(\frac{1}{m_{1}s_{1}} + \frac{1}{m_{2}s_{2}})\Delta t[/tex]

    For the given problem

    [tex]q'' = -kA\frac{T_{1,final}-T_{2,final}}{l}[/tex]
    [tex]q''_{0} = -kA\frac{T_{1,initial}-T_{2,initial}}{l}[/tex]


    Thanks (both DocAl and clausius) and cheers

    Vivek
     
    Last edited: Nov 6, 2004
  10. Nov 6, 2004 #9
    Hi

    One small clarification:

    the equation for heat transfer is actually

    [tex]q'' = \frac{kA}{l}(T_{1}-T_{2})[/tex]

    the minus sign removed as T_{1}>T_{2}. And yes I get the right answer.

    Thanks for your help guys.

    Cheers
    Vivek
     
  11. Nov 7, 2004 #10

    Clausius2

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    q'' is the heat power per unit surface (W/m^2), it is for that reason that it has the sign '' above. It is the usual notation in heat engineering books (see Incropera) So I haven't missed anything because units match each other.

    Anyway, I haven't understood you very well, do you have solved the problem yet?.
     
  12. Nov 7, 2004 #11

    Clausius2

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    Please, Don't take it as an offence, it was well intended. :wink:
     
  13. Nov 8, 2004 #12
    Hello Clausius

    Yes I understand your notation now though I must admit I haven't used it before so I wasn't sure of it initially :-). Yes I have solved the problem. Thanks

    Cheers
    Vivek
     
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