vzeos,
Yes - Crane's use of the expression
K=f
T x Constant
is exactly my objection.
I accept that they had a lot of legacy data that needed to be converted, but calling the conversion factor f
T is what has introduced the confusion. Crane noticed that the friction factor in pipe, and the K value for fittings such as bends, varied with the diameter of the pipe at the same rate (assuming turbulent flow). It was convenient to link the K values to the friction factors because they (i.e the friction factors) were well documented. Strictly the changes in friction factor and K values are
consequenses of the change in pipe size but Crane have made it appear that the change in the friction factor is the
cause and the change in the K value is the consequense of this.
It may seem that I am splitting hairs with this definition, but the confusion amongst experienced engineers in this thread is proof of my claims.
I am not sure what you mean by "My problem with this methodology is that (L/D) is not a constant but K is." Do you mean that K is a constant in Crane's methodology, or that K is a constant in fact? The fact is that for turbulent flow in a given pipe size K is more constant than (L/D).
Your diligent re-processing of pleckner's example proves the case. Given that we have turbulent flow and a fixed pipe size, we can take K values as virtually constant. However, when we use the (L/D) method we are saying that a fitting is
equivalent to a certain length of pipe, and if the characteristics of the pipe (i.e. its friction factor) changes then of course the length of the pipe that would give rise to an equivalent pressure drop must change too.
For example, water flowing at 2 m/s through a standard radius 90 degree 4" bend (K=0.24) will result in a pressure drop of 480 Pascal. This pressure drop will be virtually the same irrespective of the material or roughness of the bend, provided that the bends are geometrically identical. Pleckner also stated this (22 Dec 06, 18:37) - "..we all agree that we do not adjust the K value for the fitting according to the actual pipe friction factor..." vzeos, I noticed that you do not agree with this and I think this is something you need to re-examine. You stated (23 Dec 06, 10:12) "If you are using cast iron pipe or plastic pipe or any other pipe material, you need to use an f
T value appropriate to your pipe diameter and your pipe roughness." This is probably our main point of disagreement. Please re-read my post of 22 Dec 06, 15:51 where I described the elements that make up the pressure drop through a bend, or read the original in Crane 410 page 2-12.
If the pipe attached to this bend was commercial steel pipe with a roughness of 0.05 mm then that pipe would have a pressure drop of 366 Pascal/meter. The bend is therefore equivalent in pressure drop to 480/366 = 1.311 meter of this pipe. Since the ID of Sch40 pipe is 0.102 m the L/D ratio for this pipe is 1.311/0.102 = 12.86
However, if the attached pipe was highly polished with a roughness of 0.0015 mm then the pressure drop in the pipe would drop to 307 Pascal/meter. The pressure drop through the bend is now equivalent to 480/307 = 1.564 meter of the polished pipe and the L/D ratio becomes 1.564/0.102 = 15.33
This is exactly what you have done in your re-work of Pleckner's example. You have taken the K values as a given (i.e. constant) and workeded out the length of the actual pipe that is truly equivalent to that K value by applying the friction factor in that pipe, so the calculation has to come back to the same answer.
I'm afraid I have to disagree with Pleckner's analysis of your calculation where he says "This is an interesting phenomenon but it is totally 100% technically wrong ." What you have proven is no phenomenon - it is a bog standard hydraulic calculation and it is 100% totally technically defensible.
Well, it's Christams Eve so we should take time out to celebrate what we do agree on. I wish you all a peaceful time over the holidays.
regards
Harvey
Katmar Software
Engineering & Risk Analysis Software