Volume translated Peng-Robinson equation of state

  • Thread starter maistral
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  • #1
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
 

Answers and Replies

  • #2
dRic2
Gold Member
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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:
  • #3
maistral
237
16
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 :(
 
  • #4
dRic2
Gold Member
875
233
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

$$ log(φ) = \hat Z-1-log( \hat Z-B) - \frac A {2\sqrt{2}B}log \frac {\hat Z + B(1 + \sqrt{2})}{\hat Z + B(1 - \sqrt{2})} $$
$$ ... = Z + T^*-1-log(Z + T^*-B) - \frac A {2\sqrt{2}B}log \frac {Z + T^* + B(1 + \sqrt{2})}{Z + T^* + B(1 - \sqrt{2})} $$
 
  • #5
maistral
237
16
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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