      3 votes

# A New Simplified, Accurate, Method For Colebrook Equations

A New Simplified, Accurate, method for Colebrook equations
by Harrell Geron, NRCS Civil Engineer

This is the most simple and accurate solution for the Darcy Friction Factor with Excel. It works with the popular Colebrook-White equations. After learning this method, you can design your own procedures.

Here are four Colebrook-White equations written to work in Excel.

1/sqrt(f)=-2*Log(Rr/3.7+2.51/Re*1/sqrt(f))
1/sqrt(f)=1.14-2*Log(Rr+9.35/Re*1/sqrt(f))
1/sqrt(f)=1.14+2*Log(1/Rr)-2*Log(1+(9.3/(Re*Rr)*1/sqrt(f)))
1/sqrt(f)=1.74-2*Log(2*Rr+18.7/Re*1/sqrt(f))

Steps----------------

If you don't know values for Re and Rr, use these ...
Re=200000 and Rr=0.04
If you don't do Excel, read last paragraph

1. Select one of these equations and paste it into an Excel workbook,
2. Edit the equation, remove the left side up to "="
3. Replace the Re with the Reynolds Number, Replace Rr with the relative roughness.
4. Replace the "1/sqrt(f)" near right end with the number 3.
5. The result of this will provide an initial value for 1/sqrt(f).
6. Copy that Excel cell to a cell just beneath that initial value.
7. Edit that 2nd cell. Replace that "3" with a point and click to 1st cell.
8. Copy that second cell down to about 20 rows. The result will begin repeating the same value.
9. That repeating number is the value of 1/sqrt(f), it usually is solved by step 7 on average.
10. Write one more equation that is 1 divided by the square of that repeating value.

The last step will give the Darcy f factor with at least 15 decimals of accuracy. To check it, use it in the right side of the equation with your computed f inside the "1/sqrt(f).

That result will be the same as your repeating value that was 1/sqrt(f) if it does not equal it, then you made an error some where, An average of 7 steps will achieve an accuracy of 15 digits (the maximum digit accuracy for Excel).

After learning this you can write a VBA code or make an Excel template that will work well.

How does this work? The complex equation has one small group of variables on both the left and right side.
We guessed 3 for it's value, and compute the right side. The result is our next guess to get a new result. Do it again, Each result will be the next guess for "1/.sqrt(f)" In an average of seven loops, the solution is found to 15 decimal places. Those complex approximation usually get only 3 decimal places right. It is easy to check the results and find my looping guess is always correct. You will find that your initial guess makes little difference. If you guess 100, not 3, the number of loops will be about same, but the answer IS EXACTLY the same.

Download Excel Spreadsheets related to this method here: http://www.cheresour...white-equation/

• 1 Harrell
Are these two equations the same?

1/sqrt(f)=1.14-2*Log(Rr+9.35/Re*1/sqrt(f))
1/sqrt(f)=1.14+2*Log(1/Rr)-2*Log(1+(9.3/(Re*Rr)*1/sqrt(f)))

Actually the second one is "neater" like this...
1/sqrt(f)=1.14-2*Log(Rr+9.3/Re*1/sqrt(f))

And then we see the difference is one has 9.35 and the other 9,3

Each of the 4 different equations is very small.
Actually I found one more, it is...

1/sqrt(f)=-0.869*Ln(Rr/3.7+2.523/Re*1/sqrt(f))

Each one of these I rearranged slightly to put the 1/sqrt(f) at the beginning and end to help users find the 1/sqrt(f) with no difficulty. Also you should be sure to get the parenthesis right, or Excel will say there's a problem.
• 0

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