← Fluid Mechanics & Hydraulics · BOI SO Civil Engineer

Chapter 5 of 12

Bernoulli's Equation & Applications

In the BOI SO Civil Engineer syllabus under Fluid Mechanics & Hydraulics · 2 parts

Fluid Dynamics — Euler's & Bernoulli's Equations · Flow Measurement — Venturimeter, Orifices, Mouthpieces, Notches & Weirs

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📑 Contents (15 sections)

Part 1 of 2

Fluid Dynamics — Euler's & Bernoulli's Equations

Last reviewed 16 Sept 2026 · 6 min read

Forces acting on a flowing fluid

Gravity — weight of the fluid.
Pressure — from the surrounding fluid.
Viscous — internal friction.
Turbulent — Reynolds stresses in turbulent flow.
Compressibility and surface tension — usually negligible in hydraulic problems.

Newton's second law with gravity and pressure only gives Euler's equation; adding viscous forces gives the Navier–Stokes equations; adding turbulence terms gives the Reynolds equations.

Euler's equation of motion

For steady flow of an ideal fluid along a streamline:

\frac{d p}{ρ} + g d z + V d V = 0

Bernoulli's equation

Integrating Euler's equation for an incompressible fluid:

FormulaBernoulli's equation

\frac{p}{ρ g} + \frac{V ^{2}}{2 g} + z = constant (along a streamline)

$\frac{p}{ρ g}$ = pressure head (pressure energy per unit weight)
$\frac{V ^{2}}{2 g}$ = velocity (kinetic) head
$z$ = datum (potential) head

Total energy per unit weight (total head) is constant along a streamline.

Assumptions

Ideal fluid — non-viscous (no friction losses).
Steady flow.
Incompressible fluid.
Flow along a streamline (between two points on the same streamline; for irrotational flow, between any two points).
No energy added or removed (no pumps or turbines), no heat transfer.
Velocity uniform across the section (one-dimensional).

Real fluids

For real flow between sections 1 and 2 with head loss $h_{L}$ , pump head $H_{p}$ and turbine head $H_{t}$ :

\frac{p _{1}}{ρ g} + α_{1} \frac{V _{1}^{2}}{2 g} + z_{1} + H_{p} = \frac{p _{2}}{ρ g} + α_{2} \frac{V _{2}^{2}}{2 g} + z_{2} + H_{t} + h_{L}

Kinetic energy correction factor $α$ accounts for non-uniform velocity across the section:

α = \frac{1}{A V ^{3}} \int u^{3} d A

$α$ = 2.0 for laminar flow in pipes; about 1.01–1.1 for turbulent flow (often taken as 1). The momentum correction factor $β = \frac{1}{A V ^{2}} \int u^{2} d A$ is 4/3 for laminar pipe flow and about 1.01–1.04 for turbulent flow.

Energy line and hydraulic gradient line

Total energy line (TEL) — plots total head $\frac{p}{ρ g} + \frac{V ^{2}}{2 g} + z$ along the flow. It always slopes downward in the direction of flow (losses), except where a pump adds energy.
Hydraulic gradient line (HGL) — plots piezometric head $\frac{p}{ρ g} + z$ ; lies below the TEL by the velocity head.
Where the HGL falls below the pipe, pressure is negative (sub-atmospheric) — risk of air entry and cavitation.
At a sudden enlargement, the HGL may rise (velocity head converts to pressure) while the TEL drops by the loss.

Part 2 of 2

Flow Measurement — Venturimeter, Orifices, Mouthpieces, Notches & Weirs

Last reviewed 16 Sept 2026 · 8 min read

Pipe flow meters

All three meters create a pressure difference by contracting the flow; Bernoulli's equation and continuity give the discharge.

FormulaVenturimeter / orifice meter / flow nozzle

Q = C_{d} \frac{a _{1} a _{2}}{a _{1}^{2} - a _{2}^{2}} 2 g h

$a_{1}$ = pipe area; $a_{2}$ = throat (or orifice) area; $h$ = difference in piezometric head between inlet and throat.

With a differential manometer reading $x$ (manometric liquid $S_{m}$ , pipe liquid $S$ ):

h = x (\frac{S _{m}}{S} - 1)

(For heavier manometric liquid; for lighter liquid, $h = x (1 - S_{l} / S)$ .)

Meter	Construction	$C_{d}$ (typical)	Head loss
Venturimeter	Short converging cone (≈ 20–22°), throat, long diverging cone (≈ 5–7°) to recover pressure	0.95–0.99	Low
Flow nozzle	Nozzle inserted in pipe	≈ 0.95–0.99	Intermediate
Orifice meter	Thin plate with a sharp-edged hole (usually half the pipe diameter)	0.60–0.65	High

The throat diameter of a venturimeter is usually between one-third and three-quarters of the pipe diameter (commonly about half).
The divergent cone is long and gradual to avoid flow separation.
For an inclined venturimeter, the differential manometer reading gives $h$ directly — the equation is unchanged (the elevation difference is included in piezometric head).
The venturimeter is costlier and longer; the orifice meter is cheap and compact but wastes more energy.

Other flow-measuring devices

Pitot tube — point velocity: $V = C_{v} 2 g Δ h$ (see Bernoulli's Equation).
Rotameter — float in a tapered tube (variable-area meter).
Current meter — rotating cups or propeller in rivers and canals.
Electromagnetic and ultrasonic meters — no obstruction to flow.

Orifices

An orifice is an opening in the wall or base of a tank with a closed perimeter, whose thickness is small. The jet contracts to a minimum section — the vena contracta — at about half the orifice diameter downstream.

FormulaHydraulic coefficients

Coefficient of contraction $C_{c} = \frac{a _{c}}{a}$ — typically 0.61–0.69 (≈ 0.64)
Coefficient of velocity $C_{v} = \frac{V _{a c t u a l}}{2 g H}$ — typically 0.95–0.99 (≈ 0.97)
Coefficient of discharge $C_{d} = C_{c} \times C_{v}$ — typically 0.61–0.65 (≈ 0.62)
Coefficient of resistance $C_{r} = \frac{loss of head}{actual head}$ ; loss $= H (1 - C_{v}^{2})$

Small orifice: $Q = C_{d} a 2 g H$

$C_{v}$ from jet trajectory: $C_{v} = \frac{x}{4 y H}$ ( $x$ horizontal, $y$ vertical distance from vena contracta)

Large orifices

When the head is not large compared with the orifice depth, velocity varies over the opening:

Q = \frac{2}{3} C_{d} b 2 g (H_{2}^{3/2} - H_{1}^{3/2})

( $b$ = width; $H_{1}$ , $H_{2}$ = heads at top and bottom edges.)

Submerged orifices

Fully submerged: $Q = C_{d} a 2 g H$ , $H$ = difference between upstream and downstream water levels.
Partially submerged: sum of the free portion (large-orifice formula) and the submerged portion.

Time of emptying

FormulaTime to lower the level from

H_{1}

H_{2}

Tank of constant area $A$ through an orifice:

T = \frac{2 A ( H _{1} - H _{2} )}{C _{d} a 2 g}

Completely empty: set $H_{2}$ = 0.

Hemispherical tank (radius $R$ ) and circular horizontal cylinder have their own formulas because the area varies with depth.

Mouthpieces

A mouthpiece is a short tube (length 2–3 times diameter) fitted to an orifice.

Mouthpiece	$C_{d}$
External cylindrical	0.855
Convergent-divergent	≈ 1.0 (loss nearly eliminated)
Borda's (internal / re-entrant), running free	0.50
Borda's, running full	0.707

In an external cylindrical mouthpiece the jet contracts and then expands to fill the tube; the head loss due to sudden enlargement is about $0.5 V^{2} /2 g$ , giving $C_{v} = C_{d}$ = 0.855 (jet leaves full bore, $C_{c}$ = 1).
The pressure at the vena contracta inside an external mouthpiece is below atmospheric (about $0.89 H$ below atmosphere), so the head over it is limited to avoid the liquid vaporising and the flow breaking away (for water with $H_{a t m}$ = 10.3 m and separation at about 2.5 m absolute: $10.3 - 0.89 H \geq 2.5$ → $H \leq$ about 8.8 m).

Notches and weirs

A notch is an opening in the side of a tank or small channel with the liquid surface below its top edge (usually a metal plate). A weir is a larger structure (masonry or concrete) across a river or canal; the principle is the same. The sheet of water flowing over is the nappe; the top edge is the crest (sill).

FormulaDischarge over notches

Rectangular notch/weir (crest length $L$ , head $H$ ):

Q = \frac{2}{3} C_{d} L 2 g H^{3/2}

With $C_{d}$ = 0.62: $Q \approx 1.84 L H^{3/2}$ (SI).

Triangular (V) notch (vertex angle $θ$ ):

Q = \frac{8}{15} C_{d} tan \frac{θ}{2} 2 g H^{5/2}

For $θ$ = 90°, $C_{d}$ = 0.6: $Q \approx 1.417 H^{5/2}$

Trapezoidal notch: rectangular part + triangular part.

Stepped notch: sum of discharges through each rectangular step.

Advantages of the triangular notch

Same expression for all heads (only $H$ is measured).
More accurate for small discharges (greater head for small flow).
$C_{d}$ is fairly constant for all heads.
Ventilation of the nappe is not needed.

Cipolletti weir

A trapezoidal weir with side slopes of 1 horizontal : 4 vertical, which compensate for the reduction of discharge due to end contractions of a rectangular weir. Its discharge is given by the rectangular formula without end-contraction correction: $Q = 1.86 L H^{3/2}$ (SI).

Francis formula (end contractions)

Each end contraction reduces the effective crest length by $0.1 H$ :

Q = 1.84 (L - 0.1 n H) H^{3/2}

$n$ = number of end contractions (2 for a weir with both ends contracted).

Velocity of approach

When the channel upstream is not very large, the approach velocity $V_{a}$ adds a head $h_{a} = V_{a}^{2} /2 g$ :

Q = \frac{2}{3} C_{d} L 2 g [(H + h_{a})^{3/2} - h_{a}^{3/2}]

(solved by trial: first ignore $h_{a}$ , find $V_{a} = Q / channel area$ , then correct.)

Broad-crested, ogee and submerged weirs

Broad-crested weir (crest width $B > H /2$ ): flow is critical on the crest ( $h = 2 H /3$ ); maximum discharge $Q_{ma x} = 1.705 C_{d} L H^{3/2}$ .
Narrow-crested weir ( $B < H /2$ ): treated like a rectangular notch.
Ogee weir — crest profile follows the lower nappe of a sharp-crested weir; discharge by the rectangular formula (used as spillways).
Submerged (drowned) weir — downstream level above the crest: $Q = \frac{2}{3} C_{d 1} L 2 g (H_{1} - H_{2})^{3/2} + C_{d 2} L H_{2} 2 g (H_{1} - H_{2})$ (free portion + drowned portion).

Ventilation of weirs

The space under the nappe of a suppressed (full-width) weir must be ventilated. If air is removed, pressure below the nappe drops, the nappe is drawn towards the weir (depressed or clinging nappe) and discharge increases — so readings become inaccurate.

Error in discharge due to error in head

Device	$\frac{d Q}{Q}$
Rectangular notch ( $Q \propto H^{3/2}$ )	$\frac{3}{2} \frac{d H}{H}$
Triangular notch ( $Q \propto H^{5/2}$ )	$\frac{5}{2} \frac{d H}{H}$
Orifice ( $Q \propto H^{1/2}$ )	$\frac{1}{2} \frac{d H}{H}$

Time of emptying a tank over a weir

Rectangular weir: $T = \frac{3 A}{C _{d} L 2 g} (\frac{1}{H _{2}} - \frac{1}{H _{1}})$

Triangular notch: $T = \frac{5 A}{4 C _{d} tan ( θ /2 ) 2 g} (\frac{1}{H _{2}^{3/2}} - \frac{1}{H _{1}^{3/2}})$

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