Converting pressure to flow rate

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Homework Help Overview

The discussion revolves around converting differential pressure measurements from a pipe into an air flow rate, specifically focusing on a low-flow system with assumptions of laminar, incompressible flow. Participants are exploring the relationship between pressure, velocity, and flow rate using the Bernoulli equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the Bernoulli equation to derive flow rate from pressure measurements but questions the validity of their results. Other participants raise points about the relevance of flow rate expectations and suggest considering the physical behavior of air flow.

Discussion Status

Participants are actively engaging with the problem, with some offering insights into the expected flow rates and questioning the assumptions made by the original poster. There is a mix of agreement and differing perspectives on the interpretation of the results, particularly regarding the expected air flow from the pipe.

Contextual Notes

There are constraints related to the resolution of the pressure meter and the assumptions about flow characteristics, such as neglecting friction and thermal losses. The original poster's results seem inconsistent with their expectations based on the physical setup.

Phobos
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Trying to convert differential pressure (inch WC) measurements from a 6-inch diameter pipe into an air flow rate (cfm).

Here's what I got so far...

I'm assuming laminar, imcompressible flow with negligible friction, head, or thermal losses (this is a very low-flow system). From the Bernoulli equation...

V = (2P/d)^0.5

V = velocity
P = differential pressure
d = air density @ STP

area = A = pi(r^2)

flowrate = Q = VA

Seems straightforward enough, but when I plug in my pressure readings, I'm getting too high of a result for Q (I get a result I'd expect for a fan, and not the dribbling of air I'm actually getting from the pipe).

I think I have the units converted correctly, so am I missing something in the velocity equation?
 
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I don't know anything about the topic at hand, but unless you're in 'not even wrong' territory (which it doesn't look like.) You should probably also post your differential pressure, and the flow rate that you got.
 
P = 0.001 inch WC (this is the resolution of my meter...which I am taking as the "detection limit"...I've been getting readings higher than that)
r = 3 inches
d = 0.075 pound/cubic foot

results
V = 126.6 ft/min
A = 0.2 ft^2
Q = 24.8 cfm (cubic feet per minute)

I would expect 24 cfm from a fan, not a pipe from which there is no discernable air flow.


or in metric...
P = 0.2486 Pa
r = 0.0762 m
d = 1.202 kg/m3
...
V = 0.643 m/s
A = 0.018 m2
Q = 0.012 m3/s (which converts to the same cfm as above)
 
Last edited:
"cfm" is offending your intuition --- translate to linear velocity and watch a smoke marker in the air stream.
 
The shorthand I use for this is sqrt(p)*4005=v. So your work checks out. I agree with Bystander - 24.8cfm isn't a whole lot. It is more than "no discernable airflow" though. Its about what a typical 80mm computer case fan gets you on medium power.
 
To find the velocity of the fluid flow, multiply the differential pressure by two and divide this number by the density of the flowing material. For example, if the differential pressure is 20 pounds per square inch and the density of the fluid is 80 pounds per cubic foot, the velocity of the fluid is: 2 x (20 psi) / (80 lb/ft3) = 4 feet per second.
 

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