Venturi Meter Discharge

Q = C_{d} A_{2} \frac{2 ( p _{1} - p _{2} ) / ρ}{1 - ( A _{2} / A _{1} ) ^{2}} = C_{d} A_{2} \frac{2 g Δ h}{1 - β ^{4}}

Incompressible Bernoulli between inlet and throat, with Cd < 1 for losses. β=D2/D1. An orifice meter uses the same form with a smaller Cd (~0.6 vs ~0.98).

$A 2$: Throat area
$C d$: Discharge coefficient
$β$: D2/D1

Step-by-step solved example

D1=100 mm, D2=50 mm, Δh=0.8 m water, Cd=0.98. Q?

Venturi: inlet D1 to throat D2; Δp gives Q.

1. β⁴
β=0.5, β⁴=0.0625, 1−β⁴=0.9375.
2. Q
A2=π×0.025²=0.00196 m², √(2gΔh/0.9375)=√(16.72)=4.09, Q=0.98×0.00196×4.09=0.0079 m³/s.

Answer: Q ≈ 0.0079 m³/s

10 practice questions

0/10 correct

1.Throat velocity is

2.Pressure at the throat is

3.If Δh is mercury manometer, convert to

4.Cd for a well-made venturi is nearest

5.Orifice Cd is nearest

6.Cavitation risk is at the

7.Pitot measures

8.If A2→A1, predicted Q

9.Mass flow is

10.β=0.5, β⁴=