Compressibility Factor Pressure For Calculating Fan Power

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Discussion Overview

The discussion revolves around the appropriate type of pressure—static or total—used in the compressibility factor, KP, for calculating fan power. It involves theoretical considerations and practical applications in engineering contexts, particularly in relation to fan performance and flow calculations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether total or static pressure should be used in the compressibility factor for fan power calculations, citing conflicting sources, including Howden's Fan Engineering book and a PDF from Howden.
  • Another participant suggests that the choice of pressure type may depend on the velocities produced by the fan, indicating that gases can be treated as incompressible at low velocities (Mach number less than 0.3).
  • A follow-up question seeks clarification on whether treating the gas as incompressible means the KP factor can be ignored, while also noting a requirement to treat the gas as compressible per work standards.
  • A participant confirms that the previous statement about incompressibility is correct but does not provide further resolution on the pressure type issue, instead referencing an external article on compressible flow.

Areas of Agreement / Disagreement

Participants express differing views on whether to use total or static pressure in the compressibility factor, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There are limitations regarding the assumptions made about gas behavior at different velocities and the definitions of total and static pressures, which are not fully explored in the discussion.

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What is the correct type of pressure (static or total) used in the compressibility factor, KP, when calculating fan power? Howden's Fan Engineering book seems to indicate total pressures should be used, but I also have a PDF from Howden that indicates static pressures should be used. Online searching has shown mixed answers. For my calculation, I did it both ways and the difference was small enough to be considered negligible by engineering judgement, however, I would still like to know the correct calculation.

The equations I am using are:

H = (Q*PT*KP) / (6356*NT)

Where,

H = fan shaft power [hp]
NT = fan total efficiency [%]
Q = fan inlet volumetric flow [acfm]
PT = fan total pressure [in. w.c.]
6356 is a conversion factor
KP = compressibility factor [dimensionless]

KP = \gamma/(\gamma-1) * [(p2/p1)^((\gamma-1)/\gamma)-1] / ((p2/p1)-1)

Where,

p2 = total or static pressure at the fan outlet [in. w.c.]
p1 = total or static pressure at the fan inlet [in. w.c.]
 
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It's going to depend on what velocities are produced by the fan. In flow calculations, a gas can be treated like an incompressible fluid when the flow velocity is below 0.3 M (for air at sea lever, 0.3 M is about 100 m/s).
 
1.) Are you saying that if my gas velocities have a Mach # less than 0.3 I may treat the gas as incompressible with negligible difference in results, and the KP factor does not need to be considered?

2.) Per my work standards I'm required to treat it as compressible. So even if #1 above is correct, I'm still stuck trying to figure out if I should use total or static pressures in the compressibility equation.

3.) A small addendum to the compressibility factor equation shown in blue:

KP = γ/(γ-1) * [(p2/p1)^((γ-1)/γ)-1] / ((p2/p1)-1)

Where,

p2 = total or static absolute pressure at the fan outlet [in. w.c.]
p1 = total or static absolute pressure at the fan inlet [in. w.c.]​
 

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