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Thin Perforated Plate Head Loss

Pressure drop across a thin perforated plate in a duct (t/d < 0.015, Re > 10⁵).

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Every K = 1 component costs one velocity head — about 60 Pa at 10 m/s in air — and fan power rises with the cube of the flow you push through it.

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Thin Perforated Plate Head Loss

Calculates the pressure-drop loss coefficient and pressure drop across a thin perforated plate (plate thickness-to-hole-diameter ratio < 0.015) in a duct. The high Reynolds number (Re > 10⁵) regime is assumed.

Free Area Ratio

FAR=Total AreaOpen (Free) Area

The FAR must be entered directly. For a plate with circular holes of diameter d on a square pitch p:

FAR=4π(pd)2

Loss Coefficient

For thin plates (thickness / hole diameter < 0.015) at Re > 10⁵:

K=FAR2(0.7071FAR+(1FAR))2

This expression is derived from sharp-edged orifice contraction theory. The factor 0.707 arises from the vena-contracta contraction coefficient for a sharp-edged opening (Cc ≈ 0.61–0.64), and represents viscous contraction of the jet as it enters each hole.

Pressure Drop

Δp=K21ρV2

where ρ is the air density and V is the mean velocity in the duct upstream of the plate.

Validity

  • Plate thickness / hole diameter < 0.015 (thin plate assumption).
  • Reynolds number Re > 10⁵ (high-Re, inertia-dominated regime).

References

  • Idel'chik, I.E. and Fried, E., Flow Resistance, a Design Guide for Engineers, Taylor & Francis, Washington D.C., 1989, p. 260.
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Frequently asked questions

Pressure drop across a thin perforated plate in a duct (t/d < 0.015, Re > 10⁵). Enter your inputs and press Calculate — the worked solution shows every step of the method with your numbers substituted in.

K expresses a component's pressure drop in velocity heads: Δp = K·½ρV². A K of 1 means the component destroys exactly the kinetic energy of the approaching flow. It is dimensionless, so the same K applies at any flow rate — the pressure drop then scales with velocity squared.

Compute each component's Δp at its local velocity and add them, together with straight-duct friction. Where the duct area is constant you can equivalently sum the K values first. The result is the static pressure the fan must develop at that flow rate.