Measuring Pressure Drop

[engrkatie]

I am wondering about measuring the pressure drop of a piping system. Right now we are using two piezometer rings, one before the object for which I want to measure the pressure drop and one after. Therefore, I am measuring static pressure before and after the object and calculating the pressure drop by subtracting the two values.

It is bothering me that we are not taking the velocity head into account. Are our measurement method and calculation valid? Why or why not? Can you point me to a reference?

Thank you for your time.

 

[zdas04]

If the pipe is the same diameter and the pressures are similiar (i.e., if the downstream pressure is within about 90% of upstream pressure for an approximately isothermal process) then the two instruments should be reasonably well related to each other (assuming both were calibrated to the same zero).

The value you are reading is total pressure so you're not leaving anything out, but if you were, then the error would cancel since you'd be leaving out of both instruments.

David

 

[engrkatie]

zdas04: I think that I am only measuring static pressure with a piezometer ring, not total pressure. Why do you say that I am measuring total pressure?

 

[zdas04]

Because at the point of measuremnt, the source of the force is irrelevant. Pressure is force per unit area. If it comes from a static head or from momentum it reads on the same scale.

David

 

[Latexman]

I'm not familiar with these, but doesn't it depend on the details of your "rings"? Where is the point of measurement? In the boundary layer, where velocity ~ 0 (static pressure), or out in the flow stream where you have a velocity impinging on the element (total pressure)?

Good luck,
Latexman

 

[engrkatie]

The piezometer ring is measuring the static pressure at the wall of the tube only.

 

[ione]

A piezometer ring should measure static pressure (netted from any velocity effect) or better an average static pressure having more taps (usually 4).

 

[ione]

I think you should use Pitot tubes

 

[77JQX]

Correct me if I'm wrong. Without radial flow, from or toward the wall, isn't the static pressure at the wall the same as the flowing pressure plus energy head at the center of the pipe? Isn't this what the Bernoulli equation all about?

 

[BigInch]

zdas, you must add V^2/2/g to get the total head of the stream.

Engrkatie, if velocity before and after your device remains essentially the same, ((V₁-V₂)^2)/2/g is so small you can neglect it, you wouldn't necessarily have to measure the velocity head too. If there is a significant change in velocity, you should add the velocity head to the static head to get the total.



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