Equivalent length, L/D 2

[PaulLag]

Hello

Hope everybody is fine.

PLease, I'd like to ask a clarification concerning a simple concept, but that is seems to me odd.

Following is the topic.

When considering a return bend, I got that

(L/D)eq = 50.

This means that L = D*50

when considering two diameters,
D1
and
D2
with
D1>D2

This means that

L1 = D1*50
L2 = D2*50

But therefore

L1 > L2

I can't explain why.
Logic would say

if diameter is bigger,
- with same fluid flow, velocity is lower, so that pressure would should be lower.

This formula seems to say the opposite.

I understand this is a simple question, but I am kind of tangled with this up.

Please, can anybody help me ?

Thanks

 

[BigInch]

First of all equivalent length is to calculate pressure drop, not pressure. Once you understand the full implications of that, the rest should be simple. I suggest that you're not getting the theory, so let's do some practical examples and see if they make sense.

Flow rate, and a few other parameters, determines the pressure drop of any diameter pipe.
Say that with a 6in diameter pipe, the flow rate is whatever it is that you have a pressure drop in a straight pipe of 0.32 psi/100ft
With a Return, L = 50 x 6" = 25ft
The pressure drop through the Return is therefore 6ft * 0.32 psi / 100ft = 0.08 psig

Staying with the same 6" pipe, but reducing flow rate, say to some flow rate that gives you 0.16 psi/100ft
then the pressure loss through the return becomes only 0.16 psi/100ft * 25ft = 0.04 psig

OK, so what were those flow rates and velocities? Looking at this chart, for a 6" pipe

http://www.advantageengineering.com/fyi/166/advantageFYI166.php?view=Yes

You can see that they were Q=0.72 cfs, V=3.61 fps, and Q=0.5 cfs, V=2.5 fps respectively.

Now let's change pipe diameter to 3in.
But we'll stay with the same flow rate and velocity of Q=0.72 cfs, V=3.61 fps.
In the chart, stay on the line with the flow rate of 0.72 cfs, look over and under the 3" diameter pipe.
Pressure drop for 100 ft of straight 3" diameter pipe is 9.89 psi/100 ft
L = 50*3 = 150 in =
What's the pressure drop through the return? 9.89psi/100ft * 12.5ft = 1.24 psig
What was that equivalent length? 3in * 50 = 150 inches = 12.5 ft
You said, if diameter is bigger,
- with same fluid flow, velocity is lower, so that pressure would should be lower.
But what we've found above is that your formula does not calculate pressure drop, that's found from Darcy's formula, or from the tables for a given diameter of pipe. For the 6" pipe, Pressure Drop is so very much lower, even using the longer equivalent length of 25 ft (2X the 3" L=12.5 ft) pressure drop is only around 10% that of the 3"
Does it make sense now?
You try it with Q = 0.5 cfs

Learn from the mistakes of others. You don't have time to make them all yourself.

 

[BigInch]

CORRECTION
Flow rate, and a few other parameters, determines the pressure drop of any diameter pipe.
Say that with a 6in diameter pipe, the flow rate is whatever it is that you have a pressure drop in a straight pipe of 0.32 psi/100ft
With a Return, L = 50 x 6" = 25ft
The pressure drop through the Return is therefore 25ft * 0.32 psi / 100ft = 0.08 psig


Learn from the mistakes of others. You don't have time to make them all yourself.

 

[bimr]

Regarding: "if diameter is bigger,
- with same fluid flow, velocity is lower, so that pressure would should be lower.

This should read:

if diameter is bigger,
- with same fluid flow, velocity is lower, so that pressure drop is lower and the residual pressure is higher.

 

[katmar]

What you wrote is correct. Quoting from your original post, and adding in the correction suggested by BigInch :

if diameter is bigger,
- with same fluid flow, velocity is lower, so that pressure drop would be lower.

Although it is correct, it is a bit more complicated than just saying "the pressure drop would be lower". It needs a bit of math to explain it properly.

The pressure drop through a pipe is described by the Darcy-Weisbach Equation which is

[Δ]P_pipe = [ƒ] L/D x (v²/ 2)

and for the bend

[Δ]P_bend = [ƒ] (L/D)_eq x (v²/ 2)

Where
[Δ]P is pressure drop in pressure units (not head)
[ƒ] = Moody friction factor
L = pipe length
D = pipe inside diameter
(L/D)_eq = equivalent length of bend
v = average fluid velocity (volumetric flow / cross sectional area)

I will keep the math simple by assuming D₁ is exactly double D₂. For the same flow rate, the velocity in the pipe of D₁ will be 1/4 of that in D₂. Since (L/D)_eq is the same for both bends, and assuming that the Reynolds number is high enough that the friction factor can be assumed "constant", the pressure drop in the larger bend will be 1/16 of that in the smaller one.

But in the equation for the pipe pressure drop L/D changes with diameter, so to make the pressure drop in the pipe of the larger diameter also exactly 1/16 of that in the smaller diameter pipe we have to double L to make L/D the same for the 2 pipes. And this is exactly what your original post said - i.e. if you double the diameter you double the length of pipe that would give the equivalent pressure drop to the bend.

OK, so BigInch has shown that this is true by example and I have shown that the equations we use predict the same thing but neither of us have exlained why we need a longer length of the larger diameter pipe.

The reason is that the pressure drop through a bend is (almost) independent of the wall friction. The pressure drop is caused by the change in direction and the eddies that are caused by that and this makes the pressure drop proportional to the kinetic energy in the fluid. Hence the v² factor. However, the pressure drop in straight pipe is heavily infuenced by wall friction. As the pipe increases in diameter the circumference increases in proportion with the diameter (i.e. pi x D) but the cross sectional area increases with the square of the diameter. As the diameter increases the ratio of wall area to flow area decreases, thus decreasing the impact of the wall friction.

In summary the pressure drop through the bend is proportional to v² but in the pipe it is proportional to v²/ D so in a larger diameter pipe you need a longer length to give the same pressure drop, ie a longer equivalent length.

Equivalent length questions come up fairly often so I put together a white paper on it, but I did not include this explanation and maybe I need to update that paper. If you are interested in a bit more info on L/D and K values etc you can find the paper at

http://www.katmarsoftware.com/articles/pipe-fitting-pressure-drop.htm

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[katmar]

I see that I managed to leave the density out of the Darcy-Weisbach formula. It's early on a Sunday morning and I am not fully awake yet so I hope you can all forgive me. It does not affect the logic of what I wrote.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"