Darcy / Colebrook-White Equations

[wayneorton]

Hi, I want to calculate pipe capacities at certain grades.

Am I correct that I need to calculate f using the Colebrook-White Equation and then substitue it into Darcy's equation?

If so is there a fomulae that solves the Colbrook-White equation in terms of f=..... ? Your comments would be very much appreciated.

[Tom Hill]

You could merely look up the applicable Darcy friction factors on the Moody Diagram without having the calculation them. The moody diagram is readily available is most hydraulics textbooks. I believe it was first published by Lewis F. Moody in ASME Transactions, November 1944 in his article entitled "Friction Factors for Pipe Flow". I use the one copied in Crane TP410.

If you still wish to calculate the Darcy friction factor using the Colebrook-White equation, then I suggest putting it into a spreadsheet. It is an implicit iterative calculation requiring an initial guess (usually around 0.02), and then solving successive iterations until the desired convergence is achieved. I personally find that about 4 or 5 iterations is sufficient. However, I set up my spreadsheets to run about 8 to 10 iterations just to make sure.

Original Colebrook White formula as published:

1/sqrt(f) = -2 * log10((e/3.7D) + ((2.51/Re*sqrt(f))))

Note that "f" occurs on both sides of the equation.
Revise terms to solve for f:

f = 1/ [ 4 * log10((e/3.7D) + ((2.51/Re*sqrt(f))))]^2

Use an initial guess on the right hand side (rhs) of f=0.02,
then solve for new f on the left hand side (lhs).
Take resulting new f from lhs and plug back into rhs.
solve again for new f on lhs. Do this over and over until
consecutive values of f are identical to 5 decimal places.

Good Luck...

[Thomas Gaudich]

regarding your formula:

f = 1/ [ 4 * log10((e/3.7D) + ((2.51/Re*sqrt(f))))]^2

it seems be incorrect. my transformation was:

(1/sqrt(f))^2=(-2log10((e/3.7D) + ((2.51/Re*sqrt(f)))))^2 <=>

1/f=[-2 * log10((e/3.7D) + ((2.51/Re*sqrt(f))))]^2 <=>


f=1/[-2 * log10((e/3.7D) + ((2.51/Re*sqrt(f))))]^2 or

f = 1/ 4 *[ log10((e/3.7D) + ((2.51/Re*sqrt(f))))]^2

if i'm wrong, please correct me, it's been a while since i worked with these formulas. The solving method you've proposed works with these formulas. There's however another way to solve this, using the Goalseek tool in Microsoft Excel. The way i do it is:

first i transform the colebrook formula in:

1=-2*sqrt((f))*log10((e/3.7D) + ((2.51/Re*sqrt(f))))

then, in cell A, i write the formula

=-2*sqrt((f))*log10((e/3.7D) + ((2.51/Re*sqrt(f))))

and in cell B i put a starting number for f

then, with goalseek, make excel calculate the formula in A to value 1 by changing the B cell

[Tom Hill]

I think your formulation is correct. I misplaced the left square bracket. Sorry 'bout that.

I like your suggestion for using goal seek in Excel. However, if one has multiple pipes within the same hydraulic system within the same spreadsheet, then the manual keystrokes involved in performing many goalseek operations can become tedious.

I sometimes have need of performing this same calculation on "gunbarrel" crude oil pipelines, in which the temperature, and hence the fluid viscosity, is changing continuously down the length of the pipeline. In this case, the pipeline is broken into many contiguous segments for pressure drop calculations. I find that the iterative approach, although more messy numerically, is actually quicker than using multiple goalseek operations.

[Thomas Gaudich]

Hello! I agree that calculating everytime you want to find the value is not practical. Since you have to do it several times using diferent values, your sistem works very well. In my worksheet i use this only one time, with a button associated to a macro in VisualBasic, so i only have to clic the button.
Anyway, i´ve made some 'hand' calculations regarding this two methods to find the 'f' value and i've noticed that your method is more accurate (i'm talking about the fifth decimal place) than with 'normal' goalseek. Regarding this, i've multiplied, by a factor of 1000, my calculations, to make the goalseek more accurate.
regards
thomas

[Angelo]

Dear Experts(Thomas Gaudich & Tom Hill )



colebrook-white equation
I need to know what the friction factor will be when:

diameter of pipe is 1m.
roughness of pipe0.4mm
steady flow rate in cubic metres per second is 2
fluid density is 1000kgm^-3
dynamic fluid viscosity is 1x10^-3

i am a student programmer and need to check the answer of my program.



PLEASE HELP!

THANKS

[Thomas Gaudich]

Dear Sir: thank you for calling me an expert, but unfortunatly i'm far from that. About your question, you should check some literature about this problem, because, with your data, your're probably within the limits of a laminar flow. In this conditions, the friction factor is independent of the roughness of the pipe. As far as i know, in laminar flow conditions, the friction factor is given by f=64/Re, being Re the Reynolds Number. With your data, the Reynold's number is about 2546, wich is near to the limits of laminar flow (see, for example, Poiseuille's formula). As conclusion, i don't think you can calculate the factor by Colebrook's equation, in this conditions. But, as i first said, i'm no expert, so you should check this. Anyway, i've used your values in my worksheet and it returned the values of:
Laminar Flow
'f' calculated by the above formula = 0.0251
Turbulent Flow
Rough Pipes
'f' calculated by Colebrook's equation for rough pipes: 0.2720
'f' calculated by Karman-Prandtl's equation for rough pipes: 0.2674
Smooth Pipes
'f' calculated by Karman-Prandtl's equation for smooth pipes: 0.0133
'f' calculated by Nikuradse's equation for smooth pipes: 0.0376
'f' calculated by Blasius's equation for smooth pipes: 0.0445
Hope this helps and please let me know if you find any errors in this values.

[Tom Hill]

Response for Angelo,
I'm flattered, but IMHO still no expert. Although the units of viscosity are omitted from your post, I think that Mr. Gaudich's calcs look reasonable and your hydraulics scenario is in the critical flow regime between laminar and transition to turbulent. From a programming standpoint, you might consider determining the friction factor in the critical flow regime by use of a cubic spline curve fit. It is fairly easy to fit a cubic spline through four data points using Legendre Polynomial of third order. The four data points can be defined as follows: (RE1, ff1), (RE2, ff2), (RE3, ff3), (RE4, ff4) where RE values represent four Reynolds numbers at values of say 2000, 2400, 3800, and 4000. The ff values represent the friction factor values corresponding to those Reynolds Numbers. Use 64/RE1 to determine ff1 and assign the same value to ff2 as well. Use Colebrook White Equation to determine ff4 and assign the same value to ff3. This will adequately define fours points through which a cubic polynomial will fit. I realize that this method does NOT provide for continuity of the slope of the ff curve at RE=2000 and RE=4000, but it's close enough. A mathematical purist might choose the four knowns to solve as the values of ff at RE=2000 and RE=4000, and use the first derivative of FF=64/RE at RE=2000 as the third equation, and first derivative of the Colebrook-White Equation at RE=4000 as the fourth equation to determine the applicable third order polynomial. This is far too messy. Besides that, the derivative of the Colebrook-White equation woud would likely have to be solve numerically rather than analytically all of which is a severe waste of time.
Tom

[Angelo]

good day ,
I took your advice and did a little research.

i wrote the program that works out the friction factor using approximation fd= 0.25/(log10(k/(3.7d) + 5.74/(re^0.9))^2 for re >= 4000.

with this answer as my initial guess i used derivatives to calculate the friction factor. reducing processing time by more 2/3 rd

thanx for your help

[rrbc]

Dear all,

I developed a VBA function for evaluating Colebrook White's equation that might be helpfull to someone. Please, find the code bellow.

Best regards.


'VBA function by Ricardo Carvalho (Lisbon, Portugal)
'February, 2003, E-mail: goldmundo@clix.pt

'Colebrook-White formula
Function J(U, D, k, v As Double) As Double
J = 0.0055 * U ^ 2 / (2 * 9.81 * D) * (1 + (2000 * k / D + 1000000# / (U * D / v)) ^ (1 / 3))
dJ = 100: delta = 0.0000000001
While Not Abs(dJ / J) < delta
f1 = J - U ^ 2 / (8 * 9.81 * D) / Application.Log(k / (3.7 * D) + (2.51 * v) / (D * (2 * 9.81 * D * J) ^ 0.5)) ^ 2
f2 = (J + delta) - U ^ 2 / (8 * 9.81 * D) / Application.Log(k / (3.7 * D) + (2.51 * v) / (D * (2 * 9.81 * D * (J + delta)) ^ 0.5)) ^ 2
dJ = f1 * delta / (f2 - f1)
J = J - dJ
Wend
End Function

[rebis]

hello,

i've been trying to use goalseek in excel to find f , but cannot get the paramaters setup correct to do so.
are there any good tutorials about that can help a newbie to goalseek.

thanks in advance

[usama217]

hello all for colebrook equations i found some books that tells that colebrook equation is
-4 *log(e/(3.7*d)+1.134/(re*f^.5))=1/(f^.5)
so i need some one to tell me the diffrance
thank you all
bye

[carel]

err try this :

http://web.cecs.pdx.edu/~gerry/nmm/c...rtSolution.pdf

they discuss a few methods for solving the eqn. also the version of the colebrook eqn. that Usama217 asked about seems wrong I'd use the one they discussed above if i was you, that is

1/sqrt(f) = -2 * log10((e/3.7D) + ((2.51/Re*sqrt(f))))

seems correct plus i've seen this in two text books adn another lecturers notes so i'd believe this to be more correct it's possible the one Usama217 quotes is a manipulated version of the one i quote above but i can't see how :)

[rroach02]

I use the following as an initial guess in a Newton iteration to the Colebrook-White eqn for hydraulically smooth pipes:

f = 0.0079284 + 0.5966*Re^(-0.35703) + 6.3776*Re^(-0.71406)

The formula gives f correct to two significant digits over the range of [10^3, 10^6]. The Newton solver finds the remaining accuracy to many more digits in fewer than 5 iterations if I need the precision.

[rroach02]

The approximation will indeed be excellent if one uses the correct power on the second term (dang, mis-typed). The correct form is:

f = 0.0079284 + 0.5966*Re^(-0.35703) + 6.3776*Re^(-1.07109)

Profuse apologies. If anyone wishes to have the code for the Newton solver (I chickened out on the analytic derivative and used a finite difference which works just as well) I will be pleased to post it. The original is in QuickBasic, but it is quite short and obvious.