7 replies to this topic

Hey guys,

first time poster. Had a real quick question:

I am working on a model of a pipeline and the problem is to figure out what the roughness of the pipe is. I know that if you have the flow rates, elevation, and pressure at the inlet and outlets you just need to plug it into the Bernoulli Equation with a couple other equqtions. There is an old topic on this board here:
http://www.cheresour...s-in-pipephase/

that this question is answer but I am not quite understanding it. Could someone explain it to me and show me which equations I should be looking at? I know it's a calculation that you need to work backwards into just confused how to come about this.

Thanks!

[Art Montemayor]

Paytong:

What, precisely, is it that you "are not quite understanding"? Latexman spelled it out very specifically in the thread you quote:

"you know (i.e accurately measure) the flow rate, elevations, and pressures at two points (or the pressure drop), and you can characterize the pipe and fittings between these two points into the Bernoulli equation with the Darcy-Weisbach equation representing the frictional loss and the Colebrook equation representing the friction factor, you can back-calculate the "apparant" relative roughness."

Have you thoroughly studied and worked with the Bernoulli, Darcy-Weisbach, and Colebrook equations? Are you familiar with fluid flow? How many fluid flow problems have you resolved?

If you don't understand what Latexman has written, then please so state. No one can help you if you don't admit your lack of knowledge and at what level. To help you, we need your help in describing your handicap.

Sorry I made that a whole lot more confusing than it is lol. Okay, so I am solving for the reynolds number and the friction coefficient, but in the thread i posted he said you need to back into it using the colebrook equation. Plugging in the Reynold's formula into the Colebrook I get

1/((∆p*D/L*2/(ρ Q/A)) )=- 2〖log〗_10 ((∈⁄D_H )/3.7+2.51/((QD_h)/(μ/ρ A))(∆p*D/L*2/(ρ Q/A)) )

Can you not just solve for the thickness (Epsilon)? Why do you have to back into it? That was the question

Also, when solving the Darcy-Weishbach equation isnt the delta p negative so the coefficient of friction would be negative?

[Art Montemayor]

Paytong:

I understand that you want to identify what is an existing friction factor in an existing pipeline. Latexman meant, I believe, that you should understand the basic relationships involved in fluid flow. The Colebrook equation is basically the identity of the friction factor. Unfortunately, you can’t solve this equation directly; it has to be solved by trial-and-error methods. But that is not what you seem to be looking for.

I think (although you haven’t responded to my direct question: “What, precisely, is it that you "are not quite understanding"?”) that you are looking to identify what is the friction factor affecting an existing, operating pipeline on which you know the operating pressure drop, the fluid, the fluid characteristics and the operating temperature. If that is the case – and the fluid is a liquid – then you can apply the Darcy-Weisbach equation, much as Latexman stated:

h_f = f (L/D) (v²/2g)

where,

h_f = the head loss due to friction;
L = the length of the pipe;
D = the hydraulic diameter of the pipe (for a pipe of circular section, this equals the internal diameter of the pipe);
V = the average velocity of the fluid flow, equal to the volumetric flow rate per unit cross-sectional wetted area;
g = the local acceleration due to gravity;
f = a dimensionless coefficient called the Darcy friction factor. It is usually plotted on a Moody diagram.

The pressure drop is certainly not negative – unless you have backwards flow. Simply solve for the head loss due to friction and convert that into a pressure drop.

If I have not guessed correctly at what you want or need, then please resort to correct, logical, and specific communications in English, indicating exactly what you need or want to do, but can’t.

That is exactly what I am looking for. I know the pressure drop, the Length, the Diameter, the average velocity, and the density. I am using the formula:

dP = f * (L/D) * (ρV²)/2

dP = Pressure loss
f = Darcy Friction Factor
L = Length
D = Diameter
ρ = density
V = average velocity

I can now solve for f correct? Simple enough rewriting gives

f = dP * D/L * 2/ρV²

This is how I would solve this correct?

Then the next thing I would want to find is the Roughness of the pipeline. I would then just plug in all of my givens into the Colebrook equation and solve for the roughness correct?

[Art Montemayor]

Paytong:

For someone who is supposed to be a student, you don't pay much attention.

You keep asking the questions, but you don't answer mine: "Have you thoroughly studied and worked with the Bernoulli, Darcy-Weisbach, and Colebrook equations? Are you familiar with fluid flow? How many fluid flow problems have you resolved?"

If you had, I could have told you the Colebrook is an IMPLICIT equation. If you want to "just plug in all of my givens into the Colebrook equation and solve for the roughness", then please be my guest.

Good Luck.

[djack77494]

I can now solve for f correct? Simple enough rewriting gives

f = dP * D/L * 2/ρV²

This is how I would solve this correct?

paytong,
If you are asking, "Have I correctly rearranged the Darcy Weisbach Equation such that I can solve for f?", then the answer is yes, you have. I think Art's point is that you can never solve for epsilon in a similar manner. You can not arrive at an equation of epsilon = xxxx. However, as long as you can assemble an equation with only one unknown value, then you can solve for that unknown, whether explicitly or implicitly. Please review your math references for specifics on solving the various types or equations if that is your problem, or else ask very specific questions as necessary.

Edited by djack77494, 01 July 2010 - 10:24 AM.