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Venturi Meter Principle

A venturi meter measures discharge by converting pressure head into velocity head at a throat. The pressure drop is read using pressure taps or a differential manometer.

$$Q_{th}=A_2\sqrt{\frac{2g h}{1-(A_2/A_1)^2}}$$$$Q=C_d Q_{th}$$

Differential Manometer Head

Convert the manometer reading to equivalent head of the flowing liquid before using the venturi equation.

$$h=x\left(\frac{S_m}{S_f}-1\right)$$

Here $x$ is the manometer deflection, $S_m$ is the manometer-fluid specific gravity, and $S_f$ is the flowing-fluid specific gravity.

Venturi Discharge from Mercury Reading

A 100 mm by 50 mm venturi carries water. A mercury differential manometer reads 180 mm. Use $C_d=0.98$. Find the discharge.

$$h=0.18(13.6-1)=2.268\text{ m water}$$$$A_1=0.007854,\quad A_2=0.001963\text{ m}^2$$$$Q_{th}=0.001963\sqrt{\frac{2(9.81)(2.268)}{1-(0.001963/0.007854)^2}}=0.01351\text{ m}^3/\text{s}$$$$Q=0.98(0.01351)=0.01324\text{ m}^3/\text{s}$$

Answer: $Q=0.01324\text{ m}^3/\text{s}$.

Pitot Tube Velocity

A Pitot tube converts velocity head into pressure head. With coefficient $C$, the measured velocity is corrected from the theoretical value.

$$V=C\sqrt{2gh}$$

For differential manometers, convert the reading first into equivalent head of the flowing liquid.

Problem: Pitot Tube with Mercury Differential Manometer

A Pitot-static tube is inserted in a water main. The attached mercury-water differential manometer reads a deflection of 60 mm of mercury. If the Pitot tube coefficient is C = 0.98, find the velocity of flow in m/s.

$$h = x\left(\frac{S_{Hg}}{S_w} - 1\right) = 0.060(13.6 - 1) = 0.060(12.6) = 0.756 \text{ m of water}$$ $$V = C\sqrt{2gh} = 0.98\sqrt{2(9.81)(0.756)} = 0.98\sqrt{14.83} = 0.98(3.851) = 3.77 \text{ m/s}$$

Answer: The water velocity at the Pitot tube tip is 3.77 m/s. The mercury manometer amplifies the small velocity head by a factor of 12.6 (the SG difference), making readings practical.

Problem: Venturi Meter in a Vertical Pipe

A venturi meter is installed in a vertical pipe carrying water upward. The pipe inlet (lower tap) has diameter 200 mm and the throat (upper tap) has diameter 100 mm. The throat is 0.50 m above the inlet. A mercury U-tube differential manometer connected between the two taps shows a reading of 120 mm. Use Cd = 0.97. Find the discharge.

For a venturi in a vertical pipe, the differential head formula already accounts for the elevation difference between taps:

$$h = x\left(\frac{S_{Hg}}{S_w} - 1\right) = 0.120(12.6) = 1.512 \text{ m of water}$$ $$A_1 = \frac{\pi(0.20)^2}{4} = 0.03142 \text{ m}^2, \quad A_2 = \frac{\pi(0.10)^2}{4} = 0.007854 \text{ m}^2$$ $$Q_{th} = A_2 \sqrt{\frac{2gh}{1-(A_2/A_1)^2}} = 0.007854\sqrt{\frac{2(9.81)(1.512)}{1-(0.25)^2}}$$ $$Q_{th} = 0.007854\sqrt{\frac{29.67}{0.9375}} = 0.007854(5.628) = 0.04420 \text{ m}^3/\text{s}$$ $$Q = C_d Q_{th} = 0.97(0.04420) = 0.04287 \text{ m}^3/\text{s}$$

Answer: Discharge is 0.04287 m³/s (42.87 L/s). The elevation difference between taps is already embedded in the differential manometer reading, so the formula remains the same as for a horizontal meter.

Problem: Back-Calculating the Discharge Coefficient

A 150 mm × 75 mm venturi meter is calibrated by collecting the actual discharge. When the mercury differential manometer reads 200 mm, the volume collected in a tank in 60 seconds is 1.80 m³. Determine the actual discharge coefficient Cd of this meter.

$$Q_{actual} = \frac{V}{t} = \frac{1.80}{60} = 0.030 \text{ m}^3/\text{s}$$ $$h = 0.200(12.6) = 2.52 \text{ m of water}$$ $$A_1 = \frac{\pi(0.15)^2}{4} = 0.01767 \text{ m}^2, \quad A_2 = \frac{\pi(0.075)^2}{4} = 0.004418 \text{ m}^2$$ $$Q_{th} = A_2\sqrt{\frac{2gh}{1-(A_2/A_1)^2}} = 0.004418\sqrt{\frac{2(9.81)(2.52)}{1-(0.25)^2}}$$ $$Q_{th} = 0.004418\sqrt{\frac{49.44}{0.9375}} = 0.004418(7.263) = 0.03209 \text{ m}^3/\text{s}$$ $$C_d = \frac{Q_{actual}}{Q_{th}} = \frac{0.030}{0.03209} = 0.935$$

Answer: The calibrated discharge coefficient is Cd = 0.935. This is lower than the theoretical 1.0 due to real fluid viscous losses and vena-contracta effects in the meter.

Problem: Orifice Plate Meter — Differential Head and Discharge

An orifice plate with a 60 mm throat is installed in a 120 mm water pipe. A mercury differential manometer shows a deflection of 150 mm. The orifice discharge coefficient is 0.63. Determine the flow rate in L/s and compare the permanent pressure loss concept.

$$h = 0.150(12.6) = 1.890 \text{ m of water}$$ $$A_{or} = \frac{\pi(0.060)^2}{4} = 2.827 \times 10^{-3} \text{ m}^2$$ $$A_{pipe} = \frac{\pi(0.120)^2}{4} = 1.131 \times 10^{-2} \text{ m}^2$$ $$Q_{th} = A_{or}\sqrt{\frac{2gh}{1-(A_{or}/A_{pipe})^2}}$$ $$= 2.827 \times 10^{-3}\sqrt{\frac{2(9.81)(1.890)}{1-(0.25)^2}} = 2.827 \times 10^{-3}(6.296) = 0.01780 \text{ m}^3/\text{s}$$ $$Q = 0.63(0.01780) = 0.01121 \text{ m}^3/\text{s} = 11.21 \text{ L/s}$$

Answer: Discharge is 11.21 L/s. Unlike a venturi, an orifice plate has a large permanent head loss (not recovered downstream) due to flow separation — typically 40–80% of the differential head, making it less efficient but cheaper than a venturi meter.

Problem: Pitot-Static Probe — Stagnation vs Static Pressure

A Pitot-static tube is mounted in a water channel. The static pressure at the tap is 45 kPa gage and the stagnation pressure at the nose is 48.2 kPa gage. Taking the Pitot coefficient C = 1.0 (ideal), find the velocity of the water stream.

The dynamic pressure is the difference between stagnation and static pressures:

$$\Delta p = p_{stag} - p_{static} = 48.2 - 45.0 = 3.2 \text{ kPa}$$ $$\frac{1}{2}\rho V^2 = \Delta p$$ $$V = \sqrt{\frac{2\Delta p}{\rho}} = \sqrt{\frac{2(3200)}{1000}} = \sqrt{6.4} = 2.53 \text{ m/s}$$

Equivalently using head form:

$$h = \frac{\Delta p}{\gamma} = \frac{3200}{9810} = 0.326 \text{ m}$$ $$V = C\sqrt{2gh} = 1.0\sqrt{2(9.81)(0.326)} = 2.53 \text{ m/s}$$

Answer: Water velocity is 2.53 m/s. The Pitot-static tube directly converts the difference between stagnation and static pressures into velocity — no flow restrictions or losses are introduced.