Power Lost on Restriction in Ductwork

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SUMMARY

The discussion focuses on calculating power loss due to pressure drop in air filters within residential central air ductwork. The formula provided is CFM*dP/6356/eff=hp, where CFM represents airflow in cubic feet per minute, dP is the pressure drop in inches of water gauge, and eff is the efficiency of the fan. A typical HVAC fan operates at approximately 65% efficiency. For instance, using 2000 CFM and a pressure drop of 0.2 inches results in a power loss of 0.097 horsepower.

PREREQUISITES
  • Understanding of HVAC systems and airflow dynamics
  • Familiarity with pressure drop measurements in inches of water gauge (wg)
  • Knowledge of fan efficiency ratings and their impact on performance
  • Basic mathematical skills for applying formulas
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  • Research the impact of different air filter types on pressure drop
  • Learn about HVAC fan efficiency ratings and how to improve them
  • Explore advanced airflow calculations in duct design
  • Investigate the relationship between CFM and horsepower in HVAC systems
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HVAC engineers, residential contractors, and anyone involved in optimizing air filtration systems in ductwork.

gm10
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Hello,

I need a formula to calculate how much power is wasted on different types of air filters. A typical application is a residential central air ductwork. A typical pressure drop is 0.1-0.2 inch wg. Typical volume passing through filters is 700-2000 cfm. A typical filter size is 2-3 sq. ft.

Your help is much appreciated in advance.

gm

This is not a homework. :smile:
 
Last edited:
Engineering news on Phys.org
Welcome to PF. The equation is:

CFM*dP/6356/eff=hp

That's flow in CFM, dP in inches w.g. and efficiency of the fan. A typical good HVAC fan is about 65% efficient, so for example, 2000 CFM at 0.2" is 2000*.2/6356/.65=0.097hp
 
russ,

Awesome. Thanks a lot.
 

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