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ODE, step change, Heat transfer

  1. Apr 23, 2005 #1
    Hi all.

    I am doing some work with temperature equations. I have a book that gives an equation and then manipulates it. However, I can not follow what the author does, so can anyone help:

    He starts with:

    dT/dt = Q/MC - (T-O)/RMC

    (in the following I use the text "delta" to represent lower case delta)

    He then says he solves this for delta T (change in temperature), due to a step change in source temperature (delta O).

    The equation then becomes:

    delta T(t) = delta O (1- e^(-t/RMC))

    I use this 2nd equation in my work, but just cant follow how the author jumps from the 1st eqn to this one.

    I 'solve' the original ODE - but it comes no where near what he gets.

    Any help would be great.
  2. jcsd
  3. Apr 24, 2005 #2


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    This is what I get:



    I get:


    with K the integration factor.

    Then the differential of T with respect to O is:

    [tex]dT=\frac{dT}{dO}\Delta O[/tex]

    [tex]dT=\Delta O[/tex]

    See, not happening for me. Perhaps someone can help us.
  4. Apr 26, 2005 #3
    Thanks for having a look.

    Thats pretty much what I get..

    From: [tex] \frac{dT}{dt} = \frac{Q}{MC} - \frac{1}{RMC} (T- \theta)[/tex]

    I then diff with respect to theta, and get what you get,

    I cant understand how he, and I quote...

    "This equation can be solved for the change in temperature [tex] \delta T[/tex] due to a change in the temperature of the medium [tex]\delta\theta[/tex].

    The result for a unit step change in [tex] \theta [/tex] is:

    [tex] \delta T(t) = \delta \theta (1-e^{\frac{-t}{RMC}}) [/tex]"

    I just can't get this (although I know that the first equation is correct, and so is the final equation - the 2nd equaion I have used a lot in my work, and it is correct).

    Anyone else got any ideas?

    (PS just realised the form can use latex).
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