Pressure drop across a small section of pipe.

[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

EDIT: I had the meter readings wrong, I've updated it.

 

[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

 

[tectactoe]

so if

h_L = K(v²/2g)

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

this would yield h_L = 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

 

[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.

 

[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), that's 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.

 

[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?

 

[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

 

[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.

 

[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 !

 

[W R-P]

Your equations look correct but YOU FORGOT TO ADD THE LOSSES.lol
Have a look http://postimage.org/image/2ydg1exwk/
V1 is the velocity JUST before it enters the pipe.In the pipe the flow takes a parabolic profile when it is fully developed,due to boundary layer conditions,so velocity is maximum at the center of the pipe-this is the average flow velocity.This is normally taken as 0.99V1(or you can assume they are equal)...V2 is the velocity at the centre of the orifice,the area has reduced so V2 will definitely be > V1. The head losses are usually in terms of the average flow velocity,that's why i distinguished b/n the V and V1.
I hope this clears your first question.

The SLPM table i got was from http://www.solidworks-apac.com/2010/06/22/standard-lpm-or-slpm-is-actually-a-mass-flow-rate/
if you look on the second sheet you will find the data for different gases so i took methane's data and pasted it in the previous worksheet.Then edit the pressure value(change to 1744 Pa), the '1 SLPM' value for methane will have a corresponding kg/s value. (I realized 1 SLPM differs for different gases)
I hope all this is making sense now...
You're on track so let me know if you need help with anything else.