Calculate liquid flow through an orifice using an assumed discharge coefficient of 0.62. The upstream pipe has an actual inside diameter of 4.000 inches, the orifice bore is 2.000 inches, and the measured differential is 5 psi.
1. Calculator Inputs
Select the units below before entering the numbers. These are fixed benchmark inputs for checking the calculator.
| Calculator field | Value | Unit / setting |
|---|---|---|
| Solve for | Flow rate | |
| Discharge coefficient, Cd | 0.62 | dimensionless |
| Pipe inside diameter | 4 | in |
| Orifice bore diameter | 2 | in |
| Liquid density | 62.4 | lbm/ft³ |
| Differential pressure | 5 | psi |
2. Step-by-step Math
Convert the inputs
D = 4 × 0.0254 ≈ 0.102 m d = 2 × 0.0254 ≈ 0.0508 m ρ = 62.4 × 0.45359237 / 0.3048³ = 999.5521145 kg/m³ ΔP = 5 × 6894.757293168 ≈ 34,500 Pa
The 4-inch entry is an actual bore, not an NPS 4 Schedule 40 bore. Keep that distinction when reproducing this example.
Calculate the diameter ratio and orifice area
β = d/D = 0.0508 / 0.1016 = 0.5 1 − β⁴ ≈ 0.938 A = πd²/4 ≈ 0.00203 m²
The finite pipe-to-orifice area ratio is included. Omitting 1 − β⁴ changes the predicted flow.
Calculate and convert flow
Q = Cd A √[2ΔP / (ρ(1 − β⁴))]
Q = 0.62 × 0.00202682991639
× √[2 × 34473.78646584 / (999.5521145 × 0.9375)]
Q = 0.0107790410923 m³/s
Q = 38.804547932 m³/h
Q = 0.0107790410923 / (0.003785411784/60)
Q ≈ 171 US gpmUse the differential pressure corresponding to the meter and coefficient. It is not automatically the permanent pressure loss downstream.
3. Compare Your Answers
Keep full precision during the calculation. The calculator may display fewer digits; compare answers rounded to three significant figures; retain unrounded intermediate values.
| Result | Expected answer | Your answer |
|---|---|---|
| Flow rate | 171 gpm | __________ |
| Flow rate in SI | 38.8 m³/h | __________ |
| Beta ratio | 0.5 | __________ |
| Orifice area | 0.00203 m² | __________ |
Second check
Select Differential pressure and enter 170.851284468 gpm with the same geometry, density and Cd. The result should return 5 psi. At 20 psi, this fixed-Cd model predicts twice the original flow because Q varies with √ΔP.
4. Assumptions and Method
Single-phase incompressible liquid, with Cd supplied rather than calculated from Reynolds number or a meter standard. The example does not establish the correct coefficient or tap arrangement for a real meter. It omits gas expansion, cavitation, flashing and uncertainty analysis.
Equation definitions and implementation: Liquid Orifice Flow Calculator. See also calculation methodology and other worked examples.
