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?
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)
"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.