I have a function [tex] Z = f(P,T) [/tex](adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Finding a numerical value for a partial differential

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