--- title: "Venturi meter discharge and numerical" subject: "Fluid Mechanics" chapter: 7 key: "semester_4/fluid_mechanics_mechanical:venturi-meter-derivation-numerical" source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/venturi-meter-derivation-numerical site: BCT Engineering --- # Venturi meter discharge and numerical Fluid Mechanics · Chapter 7: Flow Measurement - Asked in: year not given, 5+3 marks - Video: [Venturimeter](https://www.youtube.com/watch?v=yCsgtPIBUHg) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/venturi-meter-derivation-numerical - Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-7 ## Question (a) Derive the expression for discharge through a horizontal Venturi meter. Differentiate between a Venturi meter and an orifice meter. (b) A Venturi meter has an inlet diameter of 300 mm and a throat diameter of 150 mm. It is connected to a differential U-tube mercury manometer (specific gravity 13.6) which reads 20 cm. The pipe carries water and the coefficient of discharge is 0.98. Find the discharge. ## Short answer Discharge Q = Cd a1 a2 / sqrt(a1² - a2²) sqrt(2gh), with h = x(Sm/Sf - 1). Using a1=0.0707 m², a2=0.0177 m², h=2.52 m, Cd=0.98 gives Q = 0.126 m³/s. ## Answer #### (a) Venturi meter A Venturi meter consists of a converging cone, a short throat and a diverging cone. The pressure drop between inlet (section 1) and throat (section 2) is used to find the discharge. ```text 1 2 | converging | throat diverging ==\__ _____ __/== \__________/ \____________/ p1 ---| |--- p2 ``` Apply Bernoulli's equation between 1 and 2 (horizontal, ideal fluid): $$ \frac{p_1}{\rho g} + \frac{V_1^2}{2g} = \frac{p_2}{\rho g} + \frac{V_2^2}{2g} $$ Let $h = \dfrac{p_1 - p_2}{\rho g}$ (differential head). Continuity: $a_1V_1 = a_2V_2$, so $V_1 = V_2\,a_2/a_1$. $$ h = \frac{V_2^2}{2g}\left[1 - \left(\frac{a_2}{a_1}\right)^2\right] \quad \Rightarrow \quad V_2 = \frac{a_1}{\sqrt{a_1^2 - a_2^2}}\sqrt{2gh} $$ Theoretical discharge $Q_{th} = a_2V_2$. Including the coefficient of discharge $C_d$ (about 0.96 to 0.99) for losses: $$ Q = C_d\,\frac{a_1a_2}{\sqrt{a_1^2 - a_2^2}}\sqrt{2gh} $$ For a differential mercury manometer with reading $x$: $h = x\left(\dfrac{S_m}{S_f} - 1\right)$. | Basis | Venturi meter | Orifice meter | |---|---|---| | Construction | Cone, throat, diffuser | Thin plate with sharp hole | | Head loss | Small (pressure mostly recovered) | Large (no recovery) | | $C_d$ | High (0.96 to 0.99) | Low (about 0.6 to 0.65) | | Cost and size | Costly, long | Cheap, compact | | Use | Large flows, where energy loss matters | Small pipes, low cost | #### (b) Numerical $$ \begin{aligned} a_1 &= \frac{\pi}{4}(0.3)^2 = 0.07069\ \text{m}^2, \quad a_2 = \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2 \\ h &= x\left(\frac{13.6}{1} - 1\right) = 0.2 \times 12.6 = 2.52\ \text{m of water} \end{aligned} $$ $$ \begin{aligned} Q &= 0.98 \times \frac{0.07069 \times 0.01767}{\sqrt{0.07069^2 - 0.01767^2}} \times \sqrt{2 \times 9.81 \times 2.52} \\ &= 0.98 \times \frac{1.2492\times10^{-3}}{0.06844} \times 7.032 \\ &= 0.1258\ \text{m}^3/\text{s} \end{aligned} $$ Velocity at the inlet = $0.1258/0.07069 = 1.78\ \text{m/s}$, and at the throat = $7.12\ \text{m/s}$. **Answer: $Q = 0.126\ \text{m}^3/\text{s} = 126\ \text{L/s}$** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is the discharge equation of a horizontal Venturi meter? A. Q = C_d a1 a2 sqrt(2gh) / sqrt(a1^2 + a2^2) B. Q = C_d a2 sqrt(2gh) C. Q = C_d a1 a2 sqrt(2gh) / sqrt(a1^2 - a2^2) D. Q = C_d (a1 - a2) sqrt(2gh) Answer: C. Bernoulli and continuity between inlet and throat give Q = C_d a1 a2 sqrt(2gh)/sqrt(a1^2 - a2^2). (Key point: Venturi discharge formula) 2. How does a Venturi meter compare with an orifice meter? A. Lower C_d and greater head loss B. Higher C_d (0.96-0.99) and much smaller head loss, but costlier C. Same head loss but cheaper D. Cheaper and compact with better pressure recovery Answer: B. A Venturi recovers most pressure and has high C_d, but is costly and long; an orifice has large loss and C_d about 0.6. (Key point: Venturi vs orifice meter) 3. For a mercury (SG 13.6) differential manometer reading x = 0.2 m with water in the pipe, what is h? A. 0.2 x 13.6 = 2.72 m B. 0.2 x 12.6 = 2.52 m of water C. 0.2 x 14.6 = 2.92 m D. 0.2 m Answer: B. h = x(S_m/S_f - 1) = 0.2 x 12.6 = 2.52 m of water. (Key point: Manometer head) 4. With a1 = 0.07069 m^2, a2 = 0.01767 m^2, C_d = 0.98 and h = 2.52 m, what is Q? A. About 0.126 m^3/s B. About 1.26 m^3/s C. About 0.0126 m^3/s D. About 0.5 m^3/s Answer: A. Q = 0.98 x (1.2492e-3/0.06844) x 7.032 = 0.1258 m^3/s. 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