Flare Line Back Pressure Calculation 1

[10815L]

Hi alls,
Flare line Back pressure calculating for different segments with equation 21, my answer is coming wrong, can any one please corrrect me.
Eq:1/M^2((p1/p2)^2-1)-Ln(P1/p2)^2
Mac=.4003
P1=.5904
P2=1.5
fl/d=.95
my answer is coming very high(33.5) while answer in example is 1.08
Thanks

 

[Latexman]

1.729

Good luck,
Latexman

 

[Latexman]

10815L,

Please take a look at your equation in the original post above, it is not the same as the equation you sent me in a personal conversation:

fl/D=1/(M2)^*((p1/p2)^2-1)-Ln(p1/p2)^2

Therefore, my answer above is not correct. I'm also not sure if the equation you sent me in the personal conversation is correct, i.e. "1/(M2)^?*((p1/p2)^2-1)"

Please take great care to give the right equation; a readily available reference, if possible; and a detailed explanation of what you are doing/hoping to accomplish so we can all be on the same page. I get the feeling you were in a hurry, or something else, when you wrote the original post.

Good luck,
Latexman

 

[10815L]

Hi Latexman,
1st why i want this calculation: I'm going to tie-in spearator with existing facility and need to calculate the pressure drop between existing and new line.
2nd: i attached arcticla from chemical engineering.com and in this article equation is used for isothermal flow which is same in api521.
But in this article equation may be not correct, further please read this article and expalin in details.
Thanks for your quick response.

 

[Latexman]

f = 0.0152
Leq = 17.8
D = 0.3
fL/D = 0.9019
M2 = 0.4003
P2 = 1.5

I got P1 = 1.5914. The article said P1 = 1.6234 (P1/P2 = 1.08226). I'm within 2% of their answer. Check your work.

Good luck,
Latexman

 

[10815L]

Latexman,
Then its mean, problem will be solved with first part of equation only and 2nd part of euqation no meaning becuase in final step 2nd part of equation is not coming in picture.
My confusion is in 2nd part of equation.
Thanks
10815L

 

[Latexman]

Do you mean:

fL/D = 1/(M2)^2*((P1/P2)^2-1) - LN(P1/P2)^2

First part

Second part is the natural logarithm of the pressure ratio squared.

Good luck,
Latexman

 

[Latexman]

I see what you mean on that "second part". When using Excel SOLVER to calculate P1, my paraentheses were not correct yesterday. They are now.

Corrected P1 = 1.6222 Essentially the same answer as example.

See attached spreadsheet.

Good luck,
Latexman

 

[10815L]

Sir,
my answer is same your in excel -6.4 but not able to get 1.6 positive, my problem is this part of problem but claculating the pressure drop with AGPA book equation for deltap for 100m is ok.
If you explain in more detail is very helpful to understand it.
Thanks

 

[Latexman]

Show your work that has the problem.

Good luck,
Latexman