<![CDATA[Harrell's Blog]]> http://www.cheresources.com/invision/blog/20-harrells-blog/ Wed, 14 Oct 2015 16:05:52 +0000 support@cheresources.com (Cheresources.com Community) IP.Blog 60 The Easy And True Solution Of Colebrook-White Equations http://www.cheresources.com/invision/blog/20/entry-488-the-easy-and-true-solution-of-colebrook-white-equations/ Since the equation starts with "1/sqrt(f)=" and have a another thing of the right like "sqrt(f)". But the equations have not been solved since 1930. May folks have made "approximations", and some are close and some are not right. But I have learned a easy and true solution for different equations.
My solutions are easy and true in Excel.
For the Colebrook-White equations change to f=1/(______)^2 where the ______ is the made right equations.


Rr= 0.02722 (Rr number is at cell B1)
Re= 66,391 (Re number is at cell B2)
Equation 1/sqrt(f)=-2*Log(Rr/3.7+2.51/(Re*sqrt(f))) (Cells A6 to A25 is 1 to 20)
Solution =1/(-2*LOG($B$1/3.7+2.51/($B$2*SQRT(B4))))^2 (This will be at cell B6)
Guess f= 1 (Guess number at cell B5)
1 0.0550474813204902 FALSE (Enter at C6 is "=B5=B6)
2 0.0554204850782122 FALSE (Copy cell B6 and C6 down to places)
3 0.0554188475042326 FALSE
4 0.0554188546575135 FALSE
5 0.0554188546262658 FALSE
6 0.0554188546264023 FALSE
7 0.0554188546264016 FALSE
8 0.0554188546264016 TRUE
9 0.0554188546264016 TRUE
10 0.0554188546264016 TRUE
11 0.0554188546264016 TRUE
12 0.0554188546264016 TRUE
13 0.0554188546264016 TRUE
14 0.0554188546264016 TRUE
15 0.0554188546264016 TRUE
16 0.0554188546264016 TRUE
17 0.0554188546264016 TRUE
18 0.0554188546264016 TRUE
19 0.0554188546264016 TRUE
20 0.0554188546264016 TRUE]]> Wed, 14 Oct 2015 12:53:10 +0000 http://www.cheresources.com/invision/blog/20/entry-488-the-easy-and-true-solution-of-colebrook-white-equations/ <![CDATA[A Different View Of Harrell's Colebrook-White Solution]]> http://www.cheresources.com/invision/blog/20/entry-149-a-different-view-of-harrells-colebrook-white-solution/
It has been tested, and re-tested over and over, and always gets the right answers if you don't make
a weird error.

This may help those who are questioning "Why does it work?"

In the Wikipedia you can read about the Lambert W function. Here's a quote near the beginning...

"The Lambert W relation cannot be expressed in terms of elementary functions. It is useful in combinatorics, for instance in the enumeration of trees. It can be used to solve various equations involving exponentials and also occurs in the solution of delay differential equations, such as y'(t) = a y(t − 1)."

The "a" represents the numbers that don't change as we move toward the answer, like the numbers for Re, Rr, 2,51, 3.7. We are computing "1/sqrt(f)" which is the y in that quote. y(t-1) should by y to the sub (t-1), the y value of y just before the y'(t). To appear simpler to show in Excel I write is like this y2=a*y1. In other words we assume an initial value for y1 and multiply by a, the result is y2, Then do another step where we use the y2 result as y1 and get another y2. The "funnel" that directs the solution is the value of "a".

Colebook-White Equation written for Excel is 1/sqrt(f)=-2*Log(Rr/3.7+2.51/Re*1/sqrt(f)). Above I said about the Lambert W function y2=a*y1. y1 is our initial guess, "a" is the complex part "-2*Log(Rr/3.7+2.51/Re*". .... y1 is part of that result like Lambert showed as "a*y1".

Some argue that this is not a solution, just like pi() is not 3.14159. They said pi() has an infinity of digits following the "3." I would not disagree about the math, but for engineering designs a specific number of decimals places works for specific designs. In Excel we can easily get 15 decimal places which exceeds the needs of all the designs engineers normally do.

The attached XLS version of Excel shows how simple my method is.to find the Darcy Friction Factor.
Please comment and ask questions. Your questions will help me know how to explain it,]]> Sat, 01 Oct 2011 14:06:00 +0000 http://www.cheresources.com/invision/blog/20/entry-149-a-different-view-of-harrells-colebrook-white-solution/ <![CDATA[Harrell's "funnel" To The Colebook-White Solution.]]> http://www.cheresources.com/invision/blog/20/entry-147-harrells-funnel-to-the-colebook-white-solution/

Attached Files

]]> Tue, 27 Sep 2011 22:45:00 +0000 http://www.cheresources.com/invision/blog/20/entry-147-harrells-funnel-to-the-colebook-white-solution/ <![CDATA[How Fast Is Harrell's Colebrook-White Solution?]]> http://www.cheresources.com/invision/blog/20/entry-146-how-fast-is-harrells-colebrook-white-solution/
In those spreadsheets you enter Right side of the selected equation with the Reynolds number and relative
roughness and a "guess" for the "1/sqrt(f)" variable. I suggested 3 for a guess so that the initial results would be the same as mine, just so you could see it you had the equation right. The next step was to copy that initial cell to a cell below and revise the "3" with a "point and click" to the result of the first cell. The next was to copy that revised second cell down about 20 more cells. The results of those cells converge to the value of 1/sqrt(F). Then last step was to divide 1 by the square of that repeating value of 1/sqrt(F), and then you will have the utmost accurate result for Excel, just because Excel can go beyond the 15 or 16 decimal place.

If you had to do this for many different pipes, that would take quite some time. But this manual method was just to demonstrate how my solution works. I did not explain the math, except to those that asked why this works. To explain it simply, understand that an accurate result will be 1/sqrt(F) = 1/sqft(F), But the right side of the equation has other variables and constants and even a Log function. Those different parts including the Log function are the path to the solution.

The guess we entered is just a small part To see what I mean by "small part", try this. Use 300 instead of the 3... one hundred times bigger. How much difference did that make in the result? If you were using Re=200,000 and Rr=0.4 the 3 gives 3.9 and the 300 gives 3.6. That means those other parts of the right side of the equation are pointing us toward the solution. Think of those parts as a funnel pointing toward 1/sqrt(f). Our guess might be off a lot, but the funnel is pointing us toward the correct solution of 1/sqrt(F). If you keep using the result as you next guess you will see the steps are getting smaller and smaller, until the change becomes zero.

Some people say this is not a solution. I don't argue, but I know it is just like the fraction 100/3. What is answer of that division? You might say that it is 33.33, but it is just like long division, One more step would be 33.333. That's closer, but how close do you need to be? Using Excel, you would get 33.333333...a maximum of fifteen 3's. Each step takes you closer and closer... Just like my "solution".

My solution is so simple any programmer could write the code to do the looping until the result becomes nothing, or so small that Excel can't go any further. But how time is needed? Test my code's speed.

Download this "Test 10000.xls" file and you can see how fast it can do 10,000 different solutions... I also added about a dozen other public approximations so you can compare their answers to mine. Also you can enter your own variables of Re and Rr and it will solve those. It should also work in Excel '97, but I don't have Excel 97 to make sure.

Attached Files

]]> Tue, 27 Sep 2011 05:04:00 +0000 http://www.cheresources.com/invision/blog/20/entry-146-how-fast-is-harrells-colebrook-white-solution/ A New Simplified, Accurate, Method For Colebrook Equations http://www.cheresources.com/invision/blog/20/entry-144-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/

]]> Fri, 16 Sep 2011 07:45:00 +0000 http://www.cheresources.com/invision/blog/20/entry-144-a-new-simplified-accurate-method-for-colebrook-equations/