Venturi Flow Rate Calculator
Enter upstream area, throat area, pressure difference and density to compute Venturi-tube flow rate Q=A₂√(2ΔP/(ρ(1−(A₂/A₁)²))). A₁=0.01, A₂=0.001, ΔP=5000 Pa, ρ=1000 → Q≈0.00318 m³/s, v₂≈3.18 m/s.
Input Data
Results
At a glance:A Venturi tube measures pipe flow via Bernoulli's principle: at the constricted throat the fluid speeds up and static pressure drops, so measuring the pressure drop yields the flow rate. From continuity A₁v₁=A₂v₂ and Bernoulli P₁+½ρv₁²=P₂+½ρv₂² (assuming no loss) the ideal flow is Q=A₂√(2ΔP/(ρ(1−(A₂/A₁)²))), where A₁, A₂ are the upstream and throat areas, ΔP=P₁−P₂ the pressure difference, ρ the density and v₂=Q/A₂ the throat velocity. In practice a discharge coefficient Cd (0.95-0.99) multiplies Q because of friction and flow separation. Example: A₁=0.01, A₂=0.001 (ratio 0.1), ΔP=5000 Pa, ρ=1000: denominator 1−0.01=0.99, Q=0.001×√(10000/990)≈0.001×3.178≈0.00318 m³/s and v₂=Q/A₂≈3.18 m/s.
Formula
Ideal flow: Q = A₂√(2ΔP/(ρ(1−(A₂/A₁)²)))
Throat velocity: v₂ = Q/A₂
Real flow: multiply by discharge coefficient Cd (≈0.95-0.99).
$$Q = A_2\sqrt{\frac{2\Delta P}{\rho\left(1-\left(\frac{A_2}{A_1}\right)^2\right)}}, \quad v_2 = \frac{Q}{A_2}$$How to Use
- Enter the upstream area A₁ and throat area A₂ (m²).
- Enter the pressure difference ΔP (Pa) and fluid density ρ (kg/m³).
- The tool returns the volumetric flow Q (m³/s and L/s) and throat velocity v₂.
Case Studies
Water main metering
A₁=0.01, A₂=0.001, ΔP=5000 Pa, ρ=1000 kg/m³.
Q≈0.00318 m³/s (≈3.18 L/s), throat velocity v₂≈3.18 m/s.
Multiply by Cd≈0.98 for the real discharge.
Venturi fertilizer injector
A pressure drop at the throat sucks fertilizer concentrate into irrigation water.
The same ΔP-flow relation sets the injection rate.
Widely used in drip and sprinkler fertigation.
FAQ
Why does a Venturi measure flow?
Continuity forces the velocity up at the throat, Bernoulli drops the static pressure there, and the measured ΔP is uniquely related to Q by Q=A₂√(2ΔP/(ρ(1−(A₂/A₁)²))).
What is the discharge coefficient Cd?
A correction (≈0.95-0.99) for friction and flow separation so that real Q=Cd×ideal Q. It depends on Reynolds number and geometry but is near constant in the turbulent range.
Does it work for gases?
Yes, but if ΔP is a large fraction of the absolute pressure, use the compressible form (with an expansibility factor). For small pressure ratios the incompressible formula is fine.
Why is A₂ smaller than A₁?
The area ratio A₂/A₁ creates the throttling that produces a measurable pressure drop; typical ratios are 0.1-0.3. As A₂→A₁, ΔP→0 and the meter loses sensitivity.
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.