Pipe Flow Equation Derivation

[PEDARRIN2]

In the IFGC, they list two equations for finding diameter of a pipe, given pressure drop, volumetric flow, pipe length, and various other factors.

All the equations I find (like in Crane TP 410) either solve for volumetric flow rate or pressure drop, never for pipe diameter.

I took a look in the appendix of the IFGC and also NFPA 54, Annex B and they show a couple formulas where they pull "D" out of under a square root (under the radical, the D has an exponent of 5, but when pulled out, it has a exponent of 2.623 - not sure how that works). I am able to get the equation for distance by simple algebraic manipulation and taking the 1/2.623 root of both sides so D has an exponent of 1.

The problem is I don't think that is appropriate. It was my understanding you could manipulate exponents like that if they were whole numbers, but not if they had decimals. It was my understanding the non whole number exponents were "approximations" that made sense of real experimentation and allowed to predict how things would be - like the compressed gas equations within certain constraints.

Am I correct that the derivation of the pipe diameter by the IFGC equations is not accurate or am I missing something?

 

[zdas04]

I don't know what IFGC is, but the exponent you're referencing is awfully close to the exponent in Weymouth (that is normally given as 2.667). The Weymouth exponent comes from an empirical value for the transmission factor (i.e., reciprocal of the square root of Fanning friction factor) of 11.18*D^(1/6) and then combining it with D^2.5.

Weymouth has a boundary condition of less than 10 barg and no H2S or CO2.

In answer to your arithmetic question, yes the 2.623 root of D^2.623 is D. Since the 2.623 is really probably 2.622856161106579, it isn't exact, but none of your other numbers are either.

David

 

[PEDARRIN2]

IFGC is International Fuel Gas Code.

In the appendix, they list an equation for low pressure systems (<1.5 psi)

Q=187.3*sqrt(D^5*dH/fba*CR*L);

Q is cubic feet per hour at 60 F and 30" Hg
D is pipe diameter in inches
dH is pressure drop in "w.c.,
fba is base friction factor for air at 60 F
CR is factor for viscosity, density and temperature
L is pipe length in feet.

They then modify it to

Q= 2313*D^2.623 * (dH/CR*L)^0.541

If you separate D on one side, you get something close to what they list in Section 402.4 of the IFGC for determining pipe diameter;

D=Q^0.381/(19.17*(dH/(CR*L))^0.206

I guess my question is can you pull out "D" from the original equation so you can solve for it, or do you have to find (or develop) a separate empirical equation from experimental data that will allow you to determine D without manipulating the exponents of an existing empirical equation?

 

[sspeare]

Please see the attached PDF. A few months ago I wondered what a direct solution for the inside diameter would look like. I played around and got what I call "Equation (2)" on the attached PDF.

I don't manipulate equations like this very often, so I'm curious to see how my math holds up to the scrutiny of the members.

My opinion: the IFGC formulae are accurate. They have stood the test of time, trust me. Hope the attached PDF helps.

 

[PEDARRIN2]

sspeare,

Your derivation matches what is in the IFGC, one in 402.4 and the other in Annex A4.

What makes me reluctant to "trust" derivations like this is I tried to derive pressure drop from the high pressure equation and it fell apart (it worked in some cases and didn't in others). that is why I started to doubt the manipulation of empirical equations to obtain unknowns with non whole or non repeating/rounded off exponents.

I could be totally off base and it is allowable to do this type of manipulation - I was just trying to ask members who have dealt more with the math behind the empirical equations.