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Venturi Meter Flow Measurement Example

Medium DifficultyFE Fluid Mechanics

Determine volumetric flow rate through a Venturi meter using differential manometer reading.

[Venturi meter diagram with manometer - to be added]

Concept

A Venturi meter is a flow measurement device that uses the principle of Bernoulli's equation. As fluid flows through a converging section (throat), velocity increases and pressure decreases. The pressure difference between inlet and throat is measured using a differential manometer. The continuity equation relates velocities to cross-sectional areas Bernoulli's equation relates pressure difference to velocity change A discharge coefficient C_{d} accounts for friction losses (typically 0.95-0.99) Mercury manometers amplify the pressure reading due to mercury's high density

Water flows through a horizontal Venturi meter. The inlet diameter is D_{1} = 300 mm and the throat diameter is D_{2} = 150 mm. A differential manometer connected between the inlet and throat shows a mercury deflection of Δ h = 250 mm. Find: Volumetric flow rate Q through the meter (m³/s)

D_{1} = 300 mm = 0.3 m D_{2} = 150 mm = 0.15 m Δ h = 250 mm = 0.25 m C_{d} = 0.98 (discharge coefficient) S G_{H g} = 13.6 ρ = 1000 kg/m^{3} g = 9.81 m/s^{2}

Calculate the inlet and throat areas: A_{1} = \frac{π D _{1}^{2}}{4} A_{1} = \frac{π ( 0.3 ) ^{2}}{4} = 0.0707 m^{2} A_{2} = \frac{π D _{2}^{2}}{4} A_{2} = \frac{π ( 0.15 ) ^{2}}{4} = 0.0177 m^{2}

For a mercury-water manometer, the pressure difference is: Δ P = (S G_{H g} - 1) ρ_{w a t er} g Δ h Δ P = (13.6 - 1) (1000) (9.81) (0.25) Δ P = 30, 870 Pa The factor (S G_{H g} - 1) accounts for the density difference between mercury and water.

Using the Venturi meter flow equation derived from Bernoulli and continuity: Q_{t h eor e t i c a l} = A_{2} \frac{2Δ P}{ρ ( 1 - ( \frac{A _{2}}{A _{1}} ) ^{2} )} Calculate area ratio: \frac{A _{2}}{A _{1}} = \frac{0.0177}{0.0707} = 0.25 Substitute values: Q_{t h eor e t i c a l} = 0.0177 \frac{2 ( 30 , 870 )}{1000 ( 1 - 0.2 5 ^{2} )} Q_{t h eor e t i c a l} = 0.0177 \frac{61 , 740}{937.5} Q_{t h eor e t i c a l} = 0.0177 \times 8.12 = 0.144 m^{3} / s

The actual flow rate accounts for friction losses using the discharge coefficient: Q_{a c t u a l} = C_{d} \times Q_{t h eor e t i c a l} Q_{a c t u a l} = 0.98 \times 0.144 Q_{a c t u a l} = 0.141 m^{3} / s = 141 L/s

Q = 0.141 m^{3} / s = 141 L/s

Q = C_{d} A_{2} \frac{2Δ P}{ρ ( 1 - ( A _{2} / A _{1} ) ^{2} )}, Δ P = (S G_{man o} - 1) ρ g Δ h Notation: Q = volumetric flow rate, C_{d} = discharge coefficient, A_{1}, A_{2} = areas, Δ P = pressure difference, ρ = fluid density, S G_{man o} = specific gravity of manometer fluid.