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Finding a numerical value for a partial differential

  1. Oct 2, 2013 #1
    I have a function [tex] Z = f(P,T) [/tex]
    and would like to calculate the partial differentials [tex] \left ( \frac{\partial Z}{\partial P} \right )_T [/tex] and [tex] \left ( \frac{\partial Z}{\partial T} \right )_P [/tex] at values of P and T.

    The function Z is compressibility factor (Lee and Kessler equation of state), P and T refer to pressure and temperature. The equation for Z is quite large;
    [tex] Z = \frac{P_rv_r}{T_r} = 1 + \frac{B}{v_r} + \frac{C}{v_r^2} + \frac{D}{v_r^5}+\frac{c_4}{T_r^3v_r^2}\left ( \beta +\frac{\gamma}{v_r^2} \right )\exp\left ( -\frac{\gamma}{v_r^2} \right ) [/tex]
    T_r and P_r refer to reduced pressures and temperatures:
    [tex] T_r = \frac{T}{T_c},~~P_r = \frac{P}{P_c} [/tex] Other variables are constants and other expressions with T_r and P_r in them.

    The solution procedure is a bit weird, but the gist of it is you solve the right hand side of the Z equality for v_r (which is pseudo variable);
    [tex] \frac{P_rv_r}{T_r} = 1 + \frac{B}{v_r} + \frac{C}{v_r^2} + \frac{D}{v_r^5}+\frac{c_4}{T_r^3v_r^2}\left ( \beta +\frac{\gamma}{v_r^2} \right )\exp\left ( -\frac{\gamma}{v_r^2} \right ) [/tex]
    then use some other relations and definitions to find Z. I have this part working correctly.

    Although the expression for Z is long, I can find the derivatives manually.

    Now, my question is this: can I treat v_r as being a constant when I take these derivatives manually? The value of v_r depends on both P and T when I am solving for it, so I'm not sure that I can.

    Finally, if I do figure this out, how would I calculate the numerical value of:
    [tex] \left ( \frac{\partial Z}{\partial P} \right )_T [/tex]
    Since we are holding T constant, would it be equivalent to
    [tex]\frac{\partial Z}{\partial P} [/tex] , treating T as constant when taking the derivatives?

    Hopefully this makes sense, and my apologies if this isn't the appropriate place to ask.
  2. jcsd
  3. Oct 2, 2013 #2


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    Well, if I've understood you correctly, you have introduced a pseudo-variable v_r=v_r(p,T), so that your expression for Z may be rewritten as Z=Z(v_r(P,T), P,T).
    Am I correct?

    Written like this, a fixed P, and differentiation of Z with respect to T will need to account for TWO effects, the change in Z as the result of change in V_r, plus the "direct" contribution in the change of Z due to T.

    To take a trivial example, set Z=v_r+P+T, v_r=P+2T, you'll see you need BOTH contributions to the total change in Z.
  4. Oct 2, 2013 #3
    What I would suggest would be to write:

    So, [itex]dZ = (-\frac{B}{v_r^2}-2\frac{C}{v_r^3}+...)dv_r[/itex]+ dTr term

    Substitute the second equation into this one, and solve for dZ as a function of dPr and dTr.
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