Help a College Student With Pipe Flow Calculations

[dan-g]

a newbie here so not sure if this is in the right place!
i'm in the middle of a paper that I'm doing for college, investigating the effects of roughness on air flow.
i set up 2 pipes, 1 rough and 1 smooth (800mm long, 36mm diameter), in a wind tunnel, measured the velocity of the air entering and exiting the pipe. i'v got a bit of a problem with the calculations though, i can't seem to find a formula that will give me the air speed at the end of the pipes? any input'd be great! thanks

 

[SteamKing]

I'm confused. You have measured the velocity entering and exiting the pipe. Don't you want to compare the exit velocities of both pipes against the roughness?

 

[dan-g]

well I'm trying to compare the velocity difference of say the smooth pipe in the wind tunnel to a theoretical velocity difference of the smooth pipe if you get me?!

 

[dan-g]

bit more info might help!
pipe is 800mm long, 36mm diameter and absolute roughness is 0.000045m
using wind speed entering the pipe of 10m/s, what is the wind speed exiting the pipe??
so far i'v got reynolds no to be 23454.5455, friction factor as 0.0207 and frictional head loss to be 2.34745m

would that be right?? i need to calculate the velocity of the air exiting the pipe now??

 

[rcgldr]

Note that mass flow is constant throughout a pipe, so unless the air is compressing or expanding, in a constant diameter pipe velocity will be nearly constant. The friction in a constant diameter pipe usually reduces pressure, and velocity could increase a tiny amount depending on how much the air's density is reduced.

 

[dan-g]

Note that mass flow is constant throughout a pipe, so unless the air is compressing or expanding, in a constant diameter pipe velocity will be nearly constant. The friction in a constant diameter pipe usually reduces pressure, and velocity could increase a tiny amount depending on how much the air's density is reduced.

would that be true for an open ended pipe though? as the inlet is closer to the source of the air

 

[rcgldr]

Note that mass flow is constant throughout a pipe, so unless the air is compressing or expanding, in a constant diameter pipe velocity will be nearly constant. The friction in a constant diameter pipe usually reduces pressure, and velocity could increase a tiny amount depending on how much the air's density is reduced.

Would that be true for an open ended pipe though? As the inlet is closer to the source of the air.

During startup, there could be a tiny amount of accumulation of mass within the pipe, but this can't be sustained (the amount of mass within the pipe can't continue to increase indefinitely). Once a steady flow is achieved, then mass flow at all points within the pipe is constant. Any restriction within the pipe reduces that overall constant rate of mass flow at all points within the pipe

 

[dan-g]

from the experiment these are the results that i got, which shows that there a clear velocity drop though??
smooth pipe
inlet velocity(m/s) outlet velocity(m/s) velocity drop(m/s)
4.03 ...... 2.07 ...... 1.97
5.63 ...... 2.4 ...... 3.23
7.3 ......3.63 .....3.67
10.3 ...... 4.37......5.93
11.67...... 4.8.....6.87

rough pipe
inlet velocity(m/s) outlet velocity(m/s) velocity drop(m/s)
4.17 .......1.92 ......2.25
6 ........2.4 .....3.6
7.73........3.83 ...... 3.9
10.6........ 4.6 ...... 6
12.1 ...... 5.17 ...... 6.93

 

[rcgldr]

from the experiment these are the results that i got, which shows that there a clear velocity drop though?

So assuming that outlet velocity is about 1/2 inlet velocity, and that the density of the air in the pipe is not doubled, then this means more air is going into the pipe than coming out of the pipe. Either the pipe is storing air, or it's making the air dissappear (such as convertion of mass into energy). How could this be possible? How are you measuring the speed of the air in the pipe?

If your wind tunnel can generate thin vertical pulses of smoke, and if you can use a clear piece of pipe for this test, then you could video the smoke pulses and play back in slow motion to observe what's going on inside the pipe.

 

[dan-g]

the air is measured using a anemometer similar to ...http://www.ebay.co.uk/itm/Advanced-Digital-Handheld-Wind-Speed-Meter-Anemometer-Beaufort-Scale-vwlocity-/170976656131?pt=UK_Sound_Vision_Other&hash=item27cf00bb03

i know there's some discrepancies in the results due some air flowing around the pipe, no way of getting smoke through the tunnel I'm afraid

 

[rcgldr]

the air is measured using a anemometer ...

How big is the anemometer compared to the diameter of the pipe (it could be restricting the flow). Are you able to actually place the anemometer well inside the pipe? Do you have access to two anemometers small enough that both could be placed well inside the pipe so that anemometer placement wouldn't be a factor in measuring relative flow? If the pipes are too small, you could drill large holes near the ends of the pipe and place the anemometer through the hole, and plug up the remainder of the hole with a rag to prevent flow through the hole.

 

[dan-g]

the fan blades of the anemometer are about 40mm so you there is a problem there! havnt really got the time to get any smaller 1's as the paper has to be handed up in 4 days but i think i'l retest it with a few alterations made to measure just the flow through the pipe, thanks for the help, i'l be back fr more with the new results!

 

[dan-g]

right so until i can retest this i thought i'd look over the calculations, just looking for verification really

smooth
L=0.8m ∅=0.036m ε=0.000045m v=10m/s ρ=1.29kg/m^3 μ=0.0000198kg/ms
giving reynolds number= 23454.5455
friction factor (0.316/Re^.25)=0.0255
then using ΔP=f(L/D)((ρv^2)/2) = 36.5997Pa

rough
L=0.8m ∅=0.036m ε=0.000198m v=10m/s ρ=1.29kg/m^3 μ=0.0000198kg/ms
giving reynolds number= 23454.5455
friction factor f= 〖[1.14+2 log_10(D/ε) ]〗^(-2)=0.034
then using ΔP=f(L/D)((ρv^2)/2) = 48.744Pa

giving a 33% increase in efficiency when smooth??

Are these correct??

thanks for the help!

 

[rcgldr]

I'm not sure what you're trying to calculate, but I wouldn't focus much on the air speed. Assuming that the density of the air within the pipe doesn't change much, then the velocity through the pipe is nearly constant. The velocity through the smooth pipe could be different than the velocity through the rough pipe.

I'm not sure what you're trying to accomplish, but the main difference between the two pipes should be the pressure near the exit point of the pipes, but I'm not sure how you would measure this very accurately without some type of pressure sensor. You'd need to create static ports (flush mounted tube so that it doesn't protude) in the sides of the pipes and make sure that the wind tunnel flow was not leaking into or out from the port connection.

 

[haruspex]

I suggest that the more flow-resisting pipe (whether that be the rough or the smooth) will create a greater backpressure at the inlet. To the extent that air can bypass the inlet, more should do so for the rough pipe, leading to a lower mass velocity. Pairing up the five measurements suggests the smooth pipe offers the greater resistance. So here's a question: what is the difference in the set-up between the 4.03m/s into the smooth pipe and the 4.17m/s into the rough pipe other than the pipe? Is the fan working just as hard in both?
As for air pressure, shouldn't that drop along the length of the pipe, leading to a faster linear velocity at the exit than at the inlet? I struggle to understand how the linear velocity can decrease if the pipe is constant diameter.

 

[dan-g]

really confused now!
the purpose of the whole thing is to compare the airflow through a rough and a smooth pipe. I'm only going to compare the calculations with the calculations and the test results with test results.
from my calculations above i'v got a greater pressure drop from the rougher pipe which is what i expected, are the numbers working out though or is it even possible!?? have i used the correct procedure to obtain the pressure drop over the length of the pipe or am i completely way off??
i'v a new set up to eliminate the flow around the pipe and just focus on the flow directly through the pipe and i'l test that tomorrow so ignore the test numbers for now!

 

[haruspex]

the purpose of the whole thing is to compare the airflow through a rough and a smooth pipe. I'm only going to compare the calculations with the calculations and the test results with test results.
from my calculations above i'v got a greater pressure drop from the rougher pipe which is what i expected, are the numbers working out though or is it even possible!?? have i used the correct procedure to obtain the pressure drop over the length of the pipe or am i completely way off??
i'v a new set up to eliminate the flow around the pipe and just focus on the flow directly through the pipe and i'l test that tomorrow so ignore the test numbers for now!

Previously it was velocity drop, now you're saying pressure drop, which makes much more sense. You should expect to see a drop in pressure but a rise in linear speed.

 

[dan-g]

Previously it was velocity drop, now you're saying pressure drop, which makes much more sense. You should expect to see a drop in pressure but a rise in linear speed.

it was only possible to measure the velocity with what i had in college but the calculations are all based on pressure drop, sorry about that!

 

[haruspex]

it was only possible to measure the velocity with what i had in college but the calculations are all based on pressure drop, sorry about that!

OK. If you manage to rerun it with no bypass, I believe you should be able to infer the pressure drop from the rise in linear speed. Assuming it's isothermal, linear speed * pressure should be a constant (since they both correspond to the rate of mass flow). But maybe adiabatic is nearer the truth, which makes it a bit more complicated.
But... I've just had an awkward thought. Maybe the dynamics are more complex than I've supposed. It's well known that the water level downstream of a sluice gate rises for a distance. This is because the momentum of the water is great enough to oppose increasing backpressure. So the pressure in the pipe might not fall steadily from the inlet, but rise initially. Would it be possible to check the flow rate at, say, 1/4 of the way along the pipe?

 

[dan-g]

OK. If you manage to rerun it with no bypass, I believe you should be able to infer the pressure drop from the rise in linear speed. Assuming it's isothermal, linear speed * pressure should be a constant (since they both correspond to the rate of mass flow). But maybe adiabatic is nearer the truth, which makes it a bit more complicated.
But... I've just had an awkward thought. Maybe the dynamics are more complex than I've supposed. It's well known that the water level downstream of a sluice gate rises for a distance. This is because the momentum of the water is great enough to oppose increasing backpressure. So the pressure in the pipe might not fall steadily from the inlet, but rise initially. Would it be possible to check the flow rate at, say, 1/4 of the way along the pipe?

so a pressure drop is what I'm looking for then(which will show as a velocity increase?), to prove that the rough walls restrict flow or am i picking that up wrong?
the system should be fairly close to completely isothermal, well close enough for the test that i could neglect it for now.
it would be fairly difficult to measure 1/4 of the way down the pipe unless i cut it and make some sort of collar or something for it, not really possible with the time scale i'v left!
thanks for the help so far

 

[dan-g]

just doing a bit more on this and...
dynamic pressure=1/2*ρv^2
@10m/s...=1/2*1.29*10^2
...=64.5

64.5-36.5997=27.9003...(36.5997 is the pressure drop due to friction)

27.9003=1/2*1.29*v^2
0.645v^2=27.9003
v^2=43.2567
v=√43.2567
v=6.57695m/s

does that not suggest a drop in velocity or am i just substituting in my own version of reality for all that!??

 

[haruspex]

just doing a bit more on this and...
dynamic pressure=1/2*ρv^2

Try p = c - ρv²/2

 

[dan-g]

Try p = c - ρv²/2

where c=?? constant or what??

 

[haruspex]

where c=?? constant or what??

Yes. Given that there's no change in gravitational PE of the air flow, it's constant for the flow. This is just Bernoulli's equation, right?

 

[dan-g]

well it turns out that for some strange reason the wind tunnel sucks instead of blows! iv new results that arent affected by flow around the pipe
inlet is the measurement taken furthest from the fan/source

smooth pipe

inlet velocity outlet velocity velocity increase
1.7......4.5.....2.8
2.6......6.8.....4.2
3.3......8.9.....5.6
4.7......11.4......6.7
5.1......12.6......7.5

rough pipe

inlet velocity outlet velocity velocity increase
1.6......4.6......3
2.5......6.8.....4.3
3.4......9.4......6
4.3......11.8......7.5
5.......13.4......8.4

 

[rcgldr]

inlet velocity outlet velocity velocity increase
1.7......4.5.....2.8

It's unlikely that air density is reduced by a factor of 3 from flowing through a pipe suspended in a wind tunnel, and if the density isn't being reduced by a factor of 3, then you have magical pipes that output more air than they input. Assuming you don't have magical pipes, there's an issue with how you are measuring the wind velocities. If you can't put the velocity sensor well within the pipe, then I suspect some sort of turbulence issue near the ends of the pipe and/or the velocity sensor.

 

[dan-g]

Try p = c - ρv²/2

i'm slightly confused with that though, doesn't that give an equation with two variables then??

 

[rcgldr]

Try p = c - ρv²/2

i'm slightly confused with that though, doesn't that give an equation with two variables then??

Yes, it's a relationship between pressure and velocity of air within a streamline when there no losses. In your case this doesn't apply because the purpose of the rough pipe is to induce losses since friction of the pipe will convert the energy related to pressure into heat. What's common with Bernoulli, is that mass flow within the pipe is constant, and assuming that pipe diameter is constant, then

ρ v = c

where ρ is density, v is velocity, and c is a constant.

 

[dan-g]

It's unlikely that air density is reduced by a factor of 3 from flowing through a pipe suspended in a wind tunnel, and if the density isn't being reduced by a factor of 3, then you have magical pipes that output more air than they input. Assuming you don't have magical pipes, there's an issue with how you are measuring the wind velocities. If you can't put the velocity sensor well within the pipe, then I suspect some sort of turbulence issue near the ends of the pipe and/or the velocity sensor.

so i can't just put the results down to magic!?
all magic aside then can anything be concluded from the test?? i would imagine that the error is with the air just as it exits the pipe but seeing as i only have to compare the rough with the smooth would the error not be common to both cases?
the only air flow that is being measured is the air going in and out of the pipe as everything else has been blocked off

 

[haruspex]

so i can't just put the results down to magic!?
all magic aside then can anything be concluded from the test?? i would imagine that the error is with the air just as it exits the pipe but seeing as i only have to compare the rough with the smooth would the error not be common to both cases?
the only air flow that is being measured is the air going in and out of the pipe as everything else has been blocked off

Without seeing the exact set up and procedure, it's hard to comment further. rcgldr and I agree that the velocity cannot really be dropping to the extent shown, if at all. If the anemometer is poor (needs lubricating?) it might be drawing too much power from the airflow, so will report a lower speed when the air pressure is low. But this still seems a quite inadequate explanation.

 

[rcgldr]

You've got a core issue trying to measure the wind speed, which should not be chaging much within the pipes. If you could get a clear pipe, perhaps you could try using soap bubbles instead of pulsed smoke, but since your wind tunnel is drawing air (as opposed to blowing it), you'll need to make sure that the soap bubbles won't be a problem. You'll need to place some tape on the far side of the pipe at regular intervals so you can get an ideal of the speed of the flow within the pipe. Note that the bubbles will move at various speeds depending on location within the pipe, so you'll have to average out the results.

This is what you'd really need, a pulse smoke generator as shown in this video: