CFM converted to m/s wind speed

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SUMMARY

The discussion focuses on converting airflow from CFM to m/s for a setup involving two 42" drum fans producing 19,000 CFM each in a 10'D x 5'L tube. The calculated velocity at the tube's end is approximately 2.5 m/s, derived from the formula v = Q/a, where Q is the airflow and a is the cross-sectional area. The conversation emphasizes the importance of fan performance curves and the effects of pressure changes and back-pressure on airflow efficiency. Recommendations include considering a single fan for better efficiency and contacting a fan manufacturer for precise specifications.

PREREQUISITES
  • Understanding of airflow calculations, specifically CFM to m/s conversions.
  • Familiarity with fan performance curves and their impact on airflow.
  • Knowledge of fluid dynamics principles, including pressure changes and drag.
  • Basic geometry for calculating cross-sectional areas of tubes.
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  • Research fan performance curves for 42" drum fans to understand their efficiency.
  • Learn about airflow dynamics in duct systems to optimize fan placement.
  • Investigate the effects of tube diameter on airflow velocity and pressure.
  • Explore the design considerations for using multiple fans versus a single larger fan.
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Engineers, HVAC professionals, and anyone involved in airflow optimization and fan system design will benefit from this discussion.

BigApple2625
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Can you help with this question? I have a solid tube 10'D x 5'L with 2 42" drum fans in one end producing about 19,000CFM each. I need to know what the velocity will be at the end of the tube in M/S. Also, will the fans be more efficient if there is a seal around the opening where the fan housings are at?
 
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As a 'ballpark' figure if the fans are putting out 19,000 ft^3/min and that volume is going through a 0.5 ft^2 hole (ie pi * 5/12^2) then you will get 19,000 ft^3/min / 0.5ft^2 = 40,000 ft/min flow

In reality pressure changes, drag, back-pressure and non-linear flow will complicate matters
 
40,000 ft/min? Wholly mackerel! That's about M=.6

How about this:

v=\frac{Q}{a}

v=\frac{2*19,000}{.25*\pi*10^2}

v = 484 \frac{ft}{min}

v = 2.5 \frac{m}{s}

Like was mentioned though, the doubling up of fans will have an affect on the flow characteristics so this would be the absolute best you would ever see. In reality you would see less than this.
 
Oops - a bit of a Spinal-Tap stonehenge there

I mean, who decided that ' was feet and " was inches ?
 
If your doubling up your fans, pressure will nearly double but the volume will remain the same (with respect the performance curve), so you then need to work out a new fan performance curve. Also the one fan will need to be rotating clockwise the other anti-clockwise to avoid swirl.

Unless you have the performance curves for the fans, the you will just be guessing or your need to test them.
 
Last edited:
Turv said:
If your doubling up your fans, pressure will nearly double but the volume will remain the same, so you then need to work out a new fan performance curve. Also the one fan will need to be rotating clockwise the other anti-clockwise to avoid swirl.

Unless you have the performance curves for the fans, the you will just be guessing or your need to test them.
Would you recommend using a single fan instead of the 2 42" units and if so, what size would I have to use in order to match the output of the 2 units? My goal is to increase the velocity to approximately 7-9 m/s using no more than 2 HP. Maybe using a smaller tube would accomplish this?
 
Without the fan curves it is really pointless going through calculations, because they may not do that size, contact a fan company and give them the details you require.
 
mgb_phys said:
Oops - a bit of a Spinal-Tap stonehenge there

I mean, who decided that ' was feet and " was inches ?

Excellent reference.
 

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