Volume translated Peng-Robinson equation of state

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SUMMARY

The discussion centers on the volume-translated Peng-Robinson equation of state (VTPR EOS) and its application in calculating fugacity coefficients. Participants clarify that the volume correction term, denoted as ##t##, is a constant added to the volume, leading to the substitution ##V' = V + t##. The integration of the fugacity expression is confirmed to be with respect to ##d(V+c)##, and the standard fugacity coefficient equation can be adapted for VTPR EOS. The confusion arises from the interchange of VEXP and VEOS in the referenced paper, which complicates the understanding of the equations.

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maistral
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Hi. Please excuse my ignorance but this entire volume translation formulas for EOS confuses me to no end.

Could someone tell me how the volume-translated Peng-Robinson exactly works? How do I calculate the fugacity expression of VTPR? Do I integrate the V + c terms against dV or do i integrate it entirely with respect to d(V + c)?

According to this paper (Tsai, J-C., Chen, Y-P.: Application of a volume-translated Peng-Robinson equation of state on vapor-liquid equilibrium calculations, 1997):
https://ibb.co/fGeMhe

If by comparison the fugacity equation is the same with the original PR EOS then I assume that they integrated the volume-translated equation with respect to d(V+c)? What I did to verify is to numerically integrate the volume-translated PREOS and the result did not equal the result of the integrated equation (did it in MATLAB, as seen here)

Or am I misunderstanding something? Please help. Thank you!
https://ibb.co/k1Gv8K
 
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What is ##t## ? Is it a sort of constant ?

Ok, I Wikipedia(d) it and I understand it is a correction on the volume and it is a constant. So just substitute ##V' = V + t## so ##Z'## will be:
##Z' = \frac{PV'}{RT} = Z + T^*##
and now you have a standard PR EoS. You can use the standard expression for ##log(φ)## with ##Z'## instead of ##Z##
 
Last edited:
dRic2 said:
What is ##t## ? Is it a sort of constant ?

Ok, I Wikipedia(d) it and I understand it is a correction on the volume and it is a constant. So just substitute ##V' = V + t## so ##Z'## will be:
##Z' = \frac{PV'}{RT} = Z + T^*##
and now you have a standard PR EoS. You can use the standard expression for ##log(φ)## with ##Z'## instead of ##Z##
Hi! Thanks for replying.

Sadly, I got even more confused. I hope you could guide me; maybe I could begin by understanding the meaning of V in the equation of state itself. Is it Vexp or VEOS? This is so confusing, sorry :(
 
I'm not very familiar with this topic, I'll tell you what I figured it out by looking at my book on thermodynamics and some wikipedia.

First, How do you calculate the fugacity coefficient ##φ##? As you did, you use the following formula (1):

$$log(φ) = \int_0^P (Z-1) \frac {dP} P$$

with (2)

$$ Z = \frac {PV}{RT}$$

Standard PR EOS looks like this (3):

$$ P = \frac {RT}{V-b} - \frac {a α(T)} {V^2 + 2bV - b^2} $$

and we have an analytical solution for the integral of equation 1. My book gives the following solution(4):

$$ log(φ) = Z-1-log(Z-B) - \frac A {2\sqrt{2}B}log \frac {Z + B(1 + \sqrt{2})}{Z + B(1 - \sqrt{2})} $$

But, sadly, we are not working with a standard PR (equation 3) - instead we are working with VTPR EoS (https://en.wikipedia.org/wiki/VTPR) (5):

$$ P = \frac {RT}{V + c - b} - \frac {a α(T)} {(V+c)^2 + 2b(V+c) - b^2} $$

where ##c## is a constant. So, here is the trick: use the substitution ##\hat V = V + c## so you can re-write equation 5 like this (6):

$$ P = \frac {RT}{ \hat V - b} - \frac {a α(T)} {( \hat V)^2 + 2b(\hat V) - b^2} $$

Which is exactly like the standard PR EoS (equation 3)! This means you can use the result found above (4)

Then equation 2 becomes (7):

$$ \hat Z = \frac {P \hat V} {RT} = \frac {P(V+c)}{RT} = \frac {PV} {RT} + \frac {Pc} {RT} = Z + T^*$$

Where I used the definition of ##T^* = \frac {Pc} {RT}## found in the paper you attached.

This means that

$$ log(φ) = \hat Z-1-log( \hat Z-B) - \frac A {2\sqrt{2}B}log \frac {\hat Z + B(1 + \sqrt{2})}{\hat Z + B(1 - \sqrt{2})} $$
$$ ... = Z + T^*-1-log(Z + T^*-B) - \frac A {2\sqrt{2}B}log \frac {Z + T^* + B(1 + \sqrt{2})}{Z + T^* + B(1 - \sqrt{2})} $$
 
dRic2 said:
I'm not very familiar with this topic, I'll tell you what I figured it out by looking at my book on thermodynamics and some wikipedia.

First, How do you calculate the fugacity coefficient ##φ##? As you did, you use the following formula (1):

$$log(φ) = \int_0^P (Z-1) \frac {dP} P$$

with (2)

$$ Z = \frac {PV}{RT}$$

Standard PR EOS looks like this (3):

$$ P = \frac {RT}{V-b} - \frac {a α(T)} {V^2 + 2bV - b^2} $$

and we have an analytical solution for the integral of equation 1. My book gives the following solution(4):

$$ log(φ) = Z-1-log(Z-B) - \frac A {2\sqrt{2}B}log \frac {Z + B(1 + \sqrt{2})}{Z + B(1 - \sqrt{2})} $$

But, sadly, we are not working with a standard PR (equation 3) - instead we are working with VTPR EoS (https://en.wikipedia.org/wiki/VTPR) (5):

$$ P = \frac {RT}{V + c - b} - \frac {a α(T)} {(V+c)^2 + 2b(V+c) - b^2} $$

where ##c## is a constant. So, here is the trick: use the substitution ##\hat V = V + c## so you can re-write equation 5 like this (6):

$$ P = \frac {RT}{ \hat V - b} - \frac {a α(T)} {( \hat V)^2 + 2b(\hat V) - b^2} $$

Which is exactly like the standard PR EoS (equation 3)! This means you can use the result found above (4)

Then equation 2 becomes (7):

$$ \hat Z = \frac {P \hat V} {RT} = \frac {P(V+c)}{RT} = \frac {PV} {RT} + \frac {Pc} {RT} = Z + T^*$$

Where I used the definition of ##T^* = \frac {Pc} {RT}## found in the paper you attached.

This means that

$$ log(φ) = \hat Z-1-log( \hat Z-B) - \frac A {2\sqrt{2}B}log \frac {\hat Z + B(1 + \sqrt{2})}{\hat Z + B(1 - \sqrt{2})} $$
$$ ... = Z + T^*-1-log(Z + T^*-B) - \frac A {2\sqrt{2}B}log \frac {Z + T^* + B(1 + \sqrt{2})}{Z + T^* + B(1 - \sqrt{2})} $$
Wow, thanks. Actually I tried going back to the (Z-1)/P dP integral and worked from there. You killed off a lot of doubts.

Apparently the paper derives the equations very poorly, and the paper interchanged VEXP and VEOS which confused me even more. Thank you!
 

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