Enter all but one of the discharge coefficient, density of the fluid, cross-sectional area of the orifice, and velocity of the fluid into the calculator to determine the pressure drop across the orifice; this calculator can also evaluate any of the variables given the others are known.
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Orifice Pressure Drop Formula
Orifice pressure drop is the reduction in static pressure that occurs when fluid accelerates through a restriction. This value is commonly used to estimate flow resistance, compare restriction sizes, and understand how much differential pressure is required to move fluid through an opening.
The most direct definition of pressure drop is the difference between upstream and downstream pressure:
\Delta P = P_1 - P_2
A standard incompressible-flow relation for an orifice is:
Q = C_d A \sqrt{\frac{2 \Delta P}{\rho}}Rearranging the equation to solve for pressure drop gives:
\Delta P = \frac{\rho}{2}\left(\frac{Q}{C_d A}\right)^2If you know fluid velocity instead of volumetric flow rate, use the flow conversion below:
Q = A V
Substituting velocity into the pressure-drop form gives:
\Delta P = \frac{\rho V^2}{2 C_d^2}In practical use, this means pressure drop increases with fluid density and rises rapidly as velocity increases. If flow rate stays the same, a smaller orifice area forces velocity upward, which increases the differential pressure.
Variable Definitions
| Variable | Meaning | Common Units |
|---|---|---|
| ΔP | Pressure drop across the orifice | Pa, kPa, bar, psi |
| P1 | Upstream pressure | Pa, kPa, bar, psi |
| P2 | Downstream pressure | Pa, kPa, bar, psi |
| C or Cd | Discharge coefficient used to account for contraction and real-flow losses | Dimensionless |
| ρ | Fluid density | kg/m3, lb/ft3 |
| A | Cross-sectional area of the orifice opening | m2, ft2, in2 |
| V | Fluid velocity through the opening | m/s, ft/s |
| Q | Volumetric flow rate | m3/s, ft3/s |
How to Use the Calculator
- Enter the fluid density in a unit system consistent with your area, velocity, and pressure units.
- Enter the discharge coefficient for the orifice geometry being analyzed.
- Enter the orifice area. For a circular opening, compute area from diameter first.
- Enter the fluid velocity through the orifice. If you only know flow rate, convert it to velocity using area.
- Calculate the missing value and confirm the result is in the expected pressure unit.
For a circular orifice, area is found from diameter with:
A = \frac{\pi d^2}{4}If your known value is flow rate rather than velocity, convert first with:
V = \frac{Q}{A}What Changes the Pressure Drop?
- Velocity: Pressure drop scales with the square of velocity, so even a modest increase in speed can produce a much larger differential pressure.
- Density: Denser fluids create a larger pressure drop at the same velocity.
- Orifice size: For the same flow rate, a smaller opening increases velocity and therefore increases pressure drop.
- Discharge coefficient: A lower coefficient indicates greater real-world losses and a higher required differential pressure for the same flow.
- Flow regime and geometry: Edge shape, plate thickness, Reynolds number, and installation details can all affect the result.
Example
Suppose a fluid with density 1000 kg/m3 flows through an orifice area of 2.0 × 10-4 m2 at a volumetric flow rate of 0.0012 m3/s, and the discharge coefficient is 0.62.
V = \frac{0.0012}{2.0 \times 10^{-4}} = 6 \text{ m/s}\Delta P = \frac{1000 \cdot 6^2}{2 \cdot 0.62^2} \approx 4.68 \times 10^4 \text{ Pa}The estimated pressure drop is therefore about 46.8 kPa.
Important Assumptions
- The calculation assumes steady flow through the restriction.
- It is most useful when density does not change significantly across the orifice.
- The discharge coefficient should match the actual orifice geometry as closely as possible.
- Measured upstream and downstream pressures should be taken at consistent locations.
- Detailed design for gases or high-accuracy metering may require additional corrections for compressibility, expansibility, and installation effects.
Common Questions
Why does pressure drop increase so quickly?
Because velocity is squared in the governing relationship. Doubling velocity can increase pressure drop by roughly a factor of four.
Do I need area if I already know velocity?
Area is still useful for checking geometry and converting between velocity and flow rate. If flow rate is the known input, area is required to determine velocity first.
Can this calculator be used for gases?
It can provide a first estimate, but gases with larger pressure changes often need more advanced treatment because density may vary across the orifice.
