Converting pressure to flow rate

Click For Summary
SUMMARY

This discussion focuses on converting differential pressure measurements from a 6-inch diameter pipe into an air flow rate in cubic feet per minute (cfm). The calculations utilize the Bernoulli equation, specifically V = (2P/d)^0.5 for velocity, where P is the differential pressure and d is the air density at standard temperature and pressure (STP). Despite following the correct formulas, the user encounters unexpectedly high flow rate results, suggesting a misunderstanding of airflow dynamics in low-flow systems. The discussion emphasizes the importance of accurately interpreting flow rates in relation to expected outputs from different systems.

PREREQUISITES
  • Understanding of the Bernoulli equation
  • Knowledge of fluid dynamics principles
  • Familiarity with unit conversions (e.g., inch WC to Pa)
  • Basic concepts of airflow measurement (cfm)
NEXT STEPS
  • Research the effects of laminar vs. turbulent flow on airflow calculations
  • Learn about the impact of friction and thermal losses in low-flow systems
  • Explore methods for measuring and interpreting airflow in HVAC systems
  • Investigate the relationship between differential pressure and flow rate in various pipe sizes
USEFUL FOR

Engineers, HVAC technicians, and anyone involved in fluid dynamics or airflow measurement will benefit from this discussion, particularly those working with low-flow systems and differential pressure measurements.

Phobos
Staff Emeritus
Science Advisor
Gold Member
Messages
1,956
Reaction score
7
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?
 
Physics news on Phys.org
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.
 

Similar threads

Replies
17
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
10
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K