- #1
NEGATIVE_40
- 22
- 0
I have a function Z = f(P,T)
and would like to calculate the partial differentials \left ( \frac{\partial Z}{\partial P} \right )_T and \left ( \frac{\partial Z}{\partial T} \right )_P 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;
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 )
T_r and P_r refer to reduced pressures and temperatures:
T_r = \frac{T}{T_c},~~P_r = \frac{P}{P_c} 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);
\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 )
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:
\left ( \frac{\partial Z}{\partial P} \right )_T
Since we are holding T constant, would it be equivalent to
\frac{\partial Z}{\partial P} , treating T as constant when taking the derivatives?
Hopefully this makes sense, and my apologies if this isn't the appropriate place to ask.
and would like to calculate the partial differentials \left ( \frac{\partial Z}{\partial P} \right )_T and \left ( \frac{\partial Z}{\partial T} \right )_P 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;
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 )
T_r and P_r refer to reduced pressures and temperatures:
T_r = \frac{T}{T_c},~~P_r = \frac{P}{P_c} 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);
\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 )
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:
\left ( \frac{\partial Z}{\partial P} \right )_T
Since we are holding T constant, would it be equivalent to
\frac{\partial Z}{\partial P} , treating T as constant when taking the derivatives?
Hopefully this makes sense, and my apologies if this isn't the appropriate place to ask.