# Pressure drop across a small section of pipe.

1. Jun 29, 2011

### tectactoe

Hello everyone.

I have natural gas (methane) flowing through some piping, and the last little part of pipe is where I'm concerned.

The gas goes through a drawn aluminum tube, about 5.08 mm ID (it's a very small tube). The tube has three 90 degree bends where it ejects out of an orifice. It is about 0.2286 m length total. The pressure right before the tubing measures 7"wc.... but I really want to know if there's a pressure drop across this bendy little piece of tube. Problem is I can not put a gauge at the end of the tube, before the orifice.

There's reason to believe (from other data) that there may be as big as 1"wc pressure drop across this tube, but I'd like to prove that with technical calcs, and at least get a reasonable close number. Some of the values needed for this were hard to find, but I'll give you what I have

d = 0.00508 m
pipe roughness = 0.0000015 m
L = 0.2286 m
density = 0.667 kg/m^3
dynamic viscosity = 0.0000103 Pa*s

The problem comes with the velocity of the gas... not sure how to find it. We have a flow meter, which is actually what we are testing to see if it's reading correctly, which will currently read between 12.79-12.98 SLPM, which gives about 10.25 - 10.39 m/s

I'm stuck because even if we assume those velocities are correct and proceed to the equations, we still have an unknown friction factor and pressure drop.... PLUS, I'm not sure how exactly to take into account the 90 degree bends... I tried looking up equivilant lengths, but they don't account for pipes that are as small as mine.

Any insight would be great/

Thank you

Last edited: Jun 29, 2011
2. Jun 29, 2011

### W R-P

Well your pipe is quite short so I think you can neglect friction losses.
I think you should apply Bernoulli's equation between the inlet and outlet(since the pressures are known), however the losses in the system will be due to the bends.
The general form of the pipe fitting losses are : K(v^2)/2g , where K is a fitting loss coefficient.
for 90 degree bends, K= 0.9 ["Fluid Mechanics" 4th edition, J.F Douglas et al,Table 10.2 pg 367]

The only variable then, will be the average flow velocity,v, which can be taken as the inlet velocity (or 0.99 V1).
Hope this helps.
William

3. Jun 29, 2011

### tectactoe

so if

hL = K(v2/2g)

where
K = 0.9
v = 10.25 m/s
g = 9.81 m/s2

this would yield hL = 4.82 m

so each 90 bend in this pipe is equal to almost 5 m of head?? that seems kind of extreme, but I guess I have no argument why that wouldn't be the case... I find it strange that this way, the diameter of the pipe is completely left out... are you sure that's okay?

thanks for the help

4. Jun 29, 2011

### W R-P

I thought u were looking for the velocity.
If you take the inlet as point1 and outlet(orifice) as point2
1) The average velocity can be taken as 0.99x the inlet velocity (V= 0.99 x V1).
2) The velocity a the orifice will be a function of the average flow velocity ie (pipe area/orifice area)x(average flow velocity)==> V2= (A1/A2 x V) .[these diameters can be measured to calculate the areas]
3) The pressures at the inlet and outlet are known (i'm assuming you know them)

So with a little arithmetic you should have an equation in terms of the average flow velocity V, and solve for it.

I hope this is much clearer.

5. Jun 30, 2011

### tectactoe

I'm sorry, perhaps I just wasn't being clear enough. My first post is kind of all over the map...

I don't know the pressure at the orifice (exit), thats what I want to know. But I am also not sure of the velocity, which becomes a problem with trying to figure out pressure drop. I have a flow meter that (supposedly) gives some SLPM readings and the best I can do for now is assume those to be correct, in order to get a velocity.

Even then, I'm still having difficultly believing the pressure drop calculations I'm getting.

I supposed I could try to figure out a way to measure pressure drop across that pipe.... but it will not be an easy task, measuring pressure at the exit.

6. Jun 30, 2011

### Phrak

what do you mean by "across"? This means you want the pressure profile as a function of radial displacement from the center line. Don't you mean along the tube?

7. Jun 30, 2011

### chaos_zzy

As i know, the 90 bend loss coefficient is linked with r/d(r=bend radius, d=pipe diameter):
r/d=1, loss coefficient k=0.5
r/d=2, k=0.35
r/d=3, k=0.3

8. Jun 30, 2011

### W R-P

I've written down what I was trying to explain (I assumed the pipe is horizontal,so z1=z2):
http://postimage.org/image/2ydg1exwk/
From the description you've given me,it looks like you're discharging to the atmosphere..if you are, then it will be easy to find the flow velocity.P2 will go to zero, P1 will be 7"wc(1.8 kPa). What i didn't know was the orifice diameter(so i couldn't find A2).So try playing around with the final equation..ANY OTHER IDEAS WOULD BE WELCOME
P.S.
12.98 SLM is a MASS FLOW RATE,not volumetric. I found 1 SLPM for methane to be 1.2195 x 10^-5 kg/s. This gives a velocity around 11.7 m/s with the pipe diameter given.

9. Jul 1, 2011

### tectactoe

WRP, thank you for your help and sorry for all of the confusion... let me show you what I've obtained. You can show me anything I may have messed up, but the numbers are starting to agree....

Taking 1 to be in the pipe and 2 to be at the orifice exit

$\frac{v_{1}^{2}}{2} + gz_{1} + \frac{P_{1}}{\rho} = \frac{v_{2}^{2}}{2} + gz_{2} + \frac{P_{2}}{\rho}$

Using gauge pressure, P2 goes to zero and P1 = 1744 Pa (7"wc).

z1 goes to zero. z2 = 0.0445 m (one of the 90 bends is upward, close to the orifice.

Then I have this equation....

$v_{2} = \frac{A_{1}}{A_{2}}v_{1}$

...is this correct? I am having a little trouble understand when exactly to use average velocity or simply v1.

Anyway, I used that second equation to substitue for v2 in the first equation, and then solved for v1, since that would be the only unknown. I got...

$v_{1} = \sqrt{\frac{2(gz_{2}-\frac{P_{1}}{\rho})}{(1-\frac{A_{1}^{2}}{A_{2}^{2}})}}$

where A1 = 0.0000203 m2
and A2 = 0.00000308 m2 (orifice diameter is 1.98 mm)

This gives me v1 = 11.1 m/s, very close to your calculated value of 11.7 m/s.

Could you please let me know where you found that SLM to kg/s conversion factor? I was looking all over trying to find decent SLM converstions, and I thought I found one for SLM to L/min, given gas density, but it didn't seem to work. So if you could show me where you found that number, I'd be sooo grateful.

From here, I should be able to find the pressure right before the orifice, yes? And I'm still not sure I'm understanding the differences between when to use average velocity or not... but we are making progress.

Again thank you for all the help !

10. Jul 1, 2011