8.2 Flow Measurement: Orifices, Weirs, and Venturi Meters

8.2 Flow Measurement: Orifices, Weirs, and Venturi Meters

Flow measurement is an indispensable core competency for civil engineers engaged in water resources, irrigation, municipal water supply, and stormwater management. In the CELE Hydraulics syllabus, measurement problems are divided into two operational regimes: closed conduit devices (orifices, Venturi meters, Pitot tubes, flow nozzles) and free-surface open channel structures (sharp-crested weirs, broad-crested weirs, flumes). Both categories originate from Bernoulli's principle and continuity, modified by empirical discharge coefficients that account for fluid viscosity and streamline contraction.

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1. Flow Through Standard Orifices and Torricelli's Theorem

An orifice is an opening with a closed perimeter in the wall or bottom of a tank, vessel, or conduit through which fluid discharges under the influence of pressure head.

Torricelli's Theorem and the Vena Contracta

Applying Bernoulli's equation between the free liquid surface of a deep open tank (where $v_1 \approx 0, P_1 = 0$) and a discharging jet open to the atmosphere ($P_2 = 0$) located a vertical depth $h$ below the surface yields Torricelli's Theorem for ideal efflux velocity: $0 + 0 + h = 0 + \frac{v_t^2}{2g} + 0 \implies v_t = \sqrt{2gh}$ In a real sharp-edged circular orifice, fluid converging toward the opening cannot turn sharp $90^\circ$ corners instantaneously due to momentum. Consequently, the streamlines continue to converge downstream of the physical orifice plate, forming a minimum jet cross-sectional area called the vena contracta, located at a distance of approximately $0.5 D$ downstream from the orifice opening.

The Three Fundamental Hydraulic Coefficients

1. Coefficient of Velocity ($C_v$): The ratio of the actual average jet velocity at the vena contracta to the ideal theoretical velocity: $C_v = \frac{v_{\text{act}}}{v_t} = \frac{v_{\text{act}}}{\sqrt{2gh}} \quad (\text{Typical range: } 0.97 \le C_v \le 0.99)$
2. Coefficient of Contraction ($C_c$): The ratio of the cross-sectional area of the jet at the vena contracta ($A_{\text{jet}}$) to the gross geometric area of the orifice opening ($A_o$): $C_c = \frac{A_{\text{jet}}}{A_o} \quad (\text{Typical range for sharp edge: } 0.62 \le C_c \le 0.64)$
3. Coefficient of Discharge ($C_d$): The ratio of the actual volumetric discharge to the theoretical discharge: $C_d = \frac{Q_{\text{act}}}{Q_{\text{ideal}}} = \frac{A_{\text{jet}} v_{\text{act}}}{A_o v_t} = C_c \cdot C_v$ $Q_{\text{act}} = C_d A_o \sqrt{2gh} \quad (\text{Standard sharp-edged orifice: } C_d \approx 0.60\text{ to }0.62)$

Head Loss Across an Orifice

The kinetic energy lost due to fluid shear and turbulence at an orifice is expressed as: $h_L = \frac{v_t^2}{2g} - \frac{v_{\text{act}}^2}{2g} = \left(\frac{1}{C_v^2} - 1\right) \frac{v_{\text{act}}^2}{2g}$

Trajectory of a Horizontal Jet

If an orifice discharges horizontally at height $y$ above the ground, the trajectory coordinates $(x, y)$ of the vena contracta are governed by projectile motion: $x = v_{\text{act}} t, \quad y = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2y}{g}}$ $v_{\text{act}} = \frac{x}{\sqrt{2y/g}} = \frac{x\sqrt{g}}{\sqrt{2y}} \implies C_v = \frac{v_{\text{act}}}{\sqrt{2gh}} = \frac{x}{2\sqrt{y h}}$

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2. Submerged and Large Vertical Orifices

Submerged Orifices

When the discharging jet issues entirely into a downstream liquid pool rather than the open atmosphere, the orifice is fully submerged. Applying Bernoulli's equation between the upstream and downstream liquid surfaces shows that the driving head is the differential surface elevation $\Delta H = H_1 - H_2$: $Q = C_d A_o \sqrt{2g \Delta H}$

Large Vertical Rectangular Orifices

When the vertical height of an orifice ($d$) is large relative to the head ($h < 2d$), the velocity varies significantly from the top edge (head $h_1$) to the bottom edge (head $h_2$). Integrating elemental strips of width $b$ and height $dh$: $dQ = C_d (b \, dh) \sqrt{2gh} \implies Q = \frac{2}{3} C_d b \sqrt{2g} \left(h_2^{3/2} - h_1^{3/2}\right)$

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3. Unsteady Orifice Flow: Time to Empty Tanks

When liquid discharges through an orifice without inflow, the liquid depth decreases over time. Equating the volume of fluid leaving the tank in time $dt$ to the reduction in liquid storage: $- A_s \, dh = Q_{\text{out}} \, dt = C_d A_o \sqrt{2gh} \, dt \implies dt = - \frac{A_s}{C_d A_o \sqrt{2g}} h^{-1/2} \, dh$ where $A_s = A_s(h)$ is the horizontal cross-sectional surface area of the tank at depth $h$.

Prismatic or Cylindrical Tank ($A_s = \text{constant}$)

Integrating between initial head $h_1$ and final head $h_2$: $t = \int_{h_2}^{h_1} \frac{A_s}{C_d A_o \sqrt{2g}} h^{-1/2} \, dh = \frac{2 A_s}{C_d A_o \sqrt{2g}} \left(\sqrt{h_1} - \sqrt{h_2}\right)$ For total drainage ($h_2 = 0$): $t_{\text{drain}} = \frac{2 A_s \sqrt{h_1}}{C_d A_o \sqrt{2g}}$

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4. Closed Conduit Devices: Venturi Meter and Pitot Tube

The Pitot Tube

A Pitot tube consists of an open tube with a $90^\circ$ bend facing directly upstream. At the stagnation point at the tube opening, fluid velocity decelerates to zero ($v = 0$), converting dynamic pressure into stagnation pressure head: $h_{\text{stag}} = h_{\text{static}} + \frac{v^2}{2g}$. $v = \sqrt{2g (h_{\text{stag}} - h_{\text{static}})} = \sqrt{2g \Delta h}$ In a Pitot-static tube, static pressure holes are integrated circumferentially, and an empirical instrument coefficient $C_p \approx 0.98\text{–}1.00$ accounts for geometry: $v = C_p \sqrt{2g \Delta h}$.

The Venturi Meter

A Venturi meter consists of a smooth converging section ($21^\circ$ cone angle), a constricted cylindrical throat, and a gradual diverging expansion cone ($5^\circ\text{ to }7^\circ$) designed to prevent flow separation.

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Writing Bernoulli's equation and continuity between pipe inlet (1) and throat (2): $\frac{P_1}{\gamma} + z_1 + \frac{v_1^2}{2g} = \frac{P_2}{\gamma} + z_2 + \frac{v_2^2}{2g}$ Letting differential piezometric head be $\Delta H = \left(\frac{P_1}{\gamma} + z_1\right) - \left(\frac{P_2}{\gamma} + z_2\right)$, and substituting $v_1 = v_2 (A_2 / A_1)$: $v_2 = \sqrt{\frac{2g \Delta H}{1 - (A_2/A_1)^2}} = \frac{A_1}{\sqrt{A_1^2 - A_2^2}} \sqrt{2g \Delta H}$ $Q_{\text{act}} = C_d \frac{A_1 A_2}{\sqrt{A_1^2 - A_2^2}} \sqrt{2g \Delta H}$ where $C_d$ is the discharge coefficient ($0.96 \le C_d \le 0.99$). When a differential manometer containing gage fluid of specific gravity $SG_m$ connects sections 1 and 2, reading deflection $y$: $\Delta H = y \left(\frac{SG_m}{SG_f} - 1\right)$

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5. Open-Channel Flow Measurement: Weirs

A weir is an overflow structure placed across an open channel to measure discharge as a function of the measured head $H$ above the weir crest.

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6. Worked Example: Venturi Meter with Differential Manometer

Problem Statement: A horizontal Venturi meter is installed in a $200\text{ mm}$ internal diameter municipal water pipeline ($D_1 = 0.200\text{ m}$). The throat diameter is $100\text{ mm}$ ($D_2 = 0.100\text{ m}$). A differential mercury manometer ($SG_m = 13.6$) connected across the piezometer taps indicates a mercury column deflection of $y = 220\text{ mm}$ ($0.220\text{ m}$). The water temperature is $20^\circ\text{C}$ ($SG_w = 1.00$, $\gamma_w = 9.81\text{ kN/m}^3$). If the meter coefficient of discharge is $C_d = 0.980$, calculate (a) the differential head in meters of water, (b) the actual discharge in liters per second, and (c) the throat velocity.

Step-by-Step Solution:

1. Compute differential piezometric head ($\Delta H$): $\Delta H = y \left(\frac{SG_m}{SG_w} - 1\right) = (0.220\text{ m})(13.6 - 1.0) = 0.220 \times 12.6 = 2.772\text{ m of water}$

2. Calculate conduit and throat cross-sectional areas: $A_1 = \frac{\pi D_1^2}{4} = \frac{\pi (0.200)^2}{4} = 0.031416\text{ m}^2$ $A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (0.100)^2}{4} = 0.007854\text{ m}^2$ $\text{Area Ratio: } \frac{A_1}{A_2} = \left(\frac{0.200}{0.100}\right)^2 = 4.0 \implies \left(\frac{A_1}{A_2}\right)^2 = 16.0$

3. Compute actual discharge ($Q_{\text{act}}$): $Q_{\text{ideal}} = \frac{A_2}{\sqrt{1 - (A_2/A_1)^2}} \sqrt{2g \Delta H} = \frac{0.007854\text{ m}^2}{\sqrt{1 - (1/16)}} \sqrt{2(9.81\text{ m/s}^2)(2.772\text{ m})}$ $\sqrt{2g \Delta H} = \sqrt{54.3866} = 7.3747\text{ m/s}$ $\frac{1}{\sqrt{1 - 1/16}} = \frac{1}{\sqrt{15/16}} = \frac{4}{\sqrt{15}} = 1.0328$ $Q_{\text{ideal}} = 0.007854 \times 1.0328 \times 7.3747 = 0.05982\text{ m}^3/\text{s}$ $Q_{\text{act}} = C_d Q_{\text{ideal}} = 0.980 \times 0.05982\text{ m}^3/\text{s} = 0.05862\text{ m}^3/\text{s} = 58.62\text{ L/s}$

4. Calculate throat velocity ($v_2$): $v_2 = \frac{Q_{\text{act}}}{A_2} = \frac{0.05862\text{ m}^3/\text{s}}{0.007854\text{ m}^2} = 7.464\text{ m/s}$

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7. CELE Exam Traps & Common Computational Errors