pitot tube velocity measurement

[blitz97]

Hello all.

I have an application where I need to measure the velocity of water that is stirring in a tank at the wall a certain depth below the surface. I have mounted a handmade pitot tube and pressure tap side by side against the wall underwater and have connected those to simple air line coming out of the water over the wall of the tank and than back down the outside of wall of the tank that connect to a manometer on the side of the tank. See attached sketch.

When stagnat, water fills the two pressure taps and up the air lines up to the water surface. When stirring, the water level in the tube elevates, pushing the air between the water in the air line and the oil in the manometer to give us a differential pressure reading. Using Bernoullis equation and the pressure being equal at both points I solve for V to get V = sqrt(((2(densityoil-densitywater)*g*h)/densitywater). h being the height of the manometer column. Using this to calculate the velocity of the flow produces a value I feel is innacurate. Do I need to take into account any of the other elevations somehow? Am I thinking about this correctly? Any thoughts would be appreciated in regards to this topic.

 

[BigInch]

"up the air lines" What air lines? I don't think that you should have air in the line.

Independent events are seldomly independent.

 

[bimr]

Have no idea as to the manner that you will calibrate such a device.

Why don't you just rent one of the propeller meters?

http://www.erdco.com/hp302.htm

 

[berkshire]

You can do it with air but the U tube has to be inverted not as shown. you would also need a sloping ramp because you are not going to get much differential pressure. You may be better off using a colored mineral oil instead.
B.E.

 

[blitz97]

I am using a mineral oil in the manometer...just have air trapped between the water and the oil that I have no way or purging. Thus causing calibration issues as mentione....

@BIMR: tried to find a propeller meter in the past but they were all too big. This one from ERDCO looks great but I called them and they no longer make such a thing....

 

[berkshire]

If it is just a tiny bubble it should not make a difference. If you have a large slug of air, then you will definitely need to purge the line. One way is to disconnect the line on the tank pitot side, attach a small funnel to the line, fill it with water ,then raise the line until the air bubble goes up into the funnel, then lower the funnel and the line back into the water, and remake your connection. Then do the same thing to the static side.
Hope this helps.
B.E.

 

[zdas04]

Oh the things we ask Mr. Bernoulli to do. The equation assumes constant density so you can't compare two different fluids.

Bernoulli's equation does work here, but not in the way you trying to use it. The density of the oil is important in determining the pressure, but not the velocity directly (i.e., use the height change to get a pressure and then carry that number into Bernoulli).

If condition 1 is at rest (v=0) then:

P1/[ρ](water)=V2^2/2+P2/[ρ](water) and

V2=((P1-P2)/[ρ](water))^0.5 (be really careful of the units here, that messes up this equation almost as often as misapplying it)

I would run my tube from the pitot tube to the closed side of the monometer and leave the other leg open (your static leg doesn't help here). Then
P1-P2= [ρ](oil)*[Δ]h*g

in one step:
V2=(([ρ](oil)*[Δ]h*g)/[ρ](water))^0.5 (the densities are ratioed, not subtracted)


David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.

 

[anning]

One possible solution might be to run both tubes up the side of the tank and end somewhere above the level of water inside the tank and have them both open to atmosphere. That way the reference pressure for both tubes is atmospheric and no calibration is required. The water column connected to the static port should be the at the same height as the water level in the tank and if the difference in height between the two columns is h then the velocity should then be v = sqrt(2*g*h).

This might not be practical because if the velocity isn't within a certain range h might be so small that you can't measure it accurately enough or so large that you need an extension ladder to reach the top of the water column. If the latter is the case, you can get around this by making the tube connected to the dynamic port first bend down then rise up with a denser liquid in the bottom of the U-bend. The calculation becomes a little more complex in this case. In this scenario, like the simpler one I described above, leave both ends of your manometer tubes open to atmosphere.

 

[bimr]

The standard propellers are 2-Inch diameter. Other sizes are available. Plus they are claiming !% accuracy.

http://www.swoffer.com/pdf/3000bulet12.pdf

 

[zdas04]

Eliminating the monometer would give you a height at very close to the level in the tank and would be more a function of gas entrainment than velocity. I think that using a monometer and a trapped air column is the right way to go. Calibration is not necessary since you are only concerned about dP not an absolute value.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.

 

[blitz97]

Thanks everyone for the feedback. I am just not getting any results from this pitot tube setup that make sense. I feel the vertical aspect of this setup is adversely affecting our manometer results as different water levels inthe tank prodcue different manometer readings. The submersible propellers for velocity measurement mentioned got me thinking and I have made some paddles coming off a free rotating shaft and submerged this in our water tank. Am getting some free spinning rpm counts from this that are making more sense.

 

[Latexman]

That's not surprising to me. A stirred tank is supposed to create chaos! And it can change mixing patterns as the level changes. You have not mentioned any details about the mixing. Is the impeller axial flow, radial flow, mixed flow? Is there more than one impeller? Are they different? Is there baffles? With no baffles, mixing will not be efficient, so I assume you have baffles. Baffles change the flow direction of the contents from circular to bottom-to-top, or vice versa.

Good luck,
Latexman

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[IRstuff]

Just how far apart are the two ports? From your drawing it looks like you expect to have a pressure drop between the two ports, which seems a bit speculative to me, unless you've got cyclonic speeds in this tank. A typical pitot tube setup has the reference port at what should be zero velocity; by the same token, one of your two ports ought to always be in stagnant fluid.



TTFN
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