Hydraulics and Fluid Mechanics is the study of fluids at rest and in motion — from basic fluid properties and pressure measurement, through hydrostatic forces, buoyancy, kinematics and Bernoulli's equation, to viscous and turbulent pipe flow, drag/lift, boundary layer theory, dimensional analysis, weirs, turbines, pumps and open channel flow. Every formula, coefficient value, diagram, solved example and exam-pattern table is included.
After studying this chapter you will be able to:
Prerequisite: Foundation Engineering (seepage, dam spillway and drainage design concepts introduced there rely on the flow, pressure and open-channel principles developed here). Leads to: Engineering Hydrology, which applies these flow-measurement and open-channel principles to rainfall-runoff and river hydraulics.
A fluid is a substance that deforms continuously under the application of any shear stress, however small. Unlike a solid (which deforms by a fixed amount), a fluid flows — velocity continues to increase as long as shear stress is applied.
| Property | Symbol | Definition | SI Unit | Water (20°C) | Air (20°C) |
|---|---|---|---|---|---|
| Density | ρ | Mass per unit volume = m/V | kg/m³ | 998 | 1.204 |
| Specific weight | γ | Weight per unit volume = ρg | N/m³ | 9789 | 11.8 |
| Specific gravity | S or G | ρ_fluid / ρ_water | dimensionless | 1.0 | 0.0012 |
| Dynamic viscosity | μ | \(\tau = \mu\,\dfrac{du}{dy}\); resistance to shear | Pa·s (N·s/m²) | 1.002×10⁻³ | 1.81×10⁻⁵ |
| Kinematic viscosity | ν | μ/ρ; momentum diffusivity | m²/s (Stoke = 10⁻⁴ m²/s) | 1.004×10⁻⁶ | 1.51×10⁻⁵ |
| Bulk modulus | K | \(K = -V\,\dfrac{dP}{dV} = \rho\,\dfrac{dP}{d}\rho\) | N/m² (Pa) | 2.18×10⁹ | — |
| Surface tension | σ | Force per unit length at free surface | N/m | 0.0728 | — |
| Vapour pressure | p_v | Pressure at which liquid boils | Pa (kPa) | 2.34 kPa | — |
Newtonian fluids (water, air, most oils) follow a linear relationship between shear stress and velocity gradient — viscosity μ is constant regardless of the rate of shear. Non-Newtonian fluids do not: the power-law model captures pseudoplastic (shear-thinning) fluids like paint and blood (\(n<1\)) and dilatant (shear-thickening) fluids like cornstarch suspensions and wet sand (\(n>1\)); a Bingham plastic (toothpaste, sewage sludge) additionally requires a minimum yield stress before it will flow at all.
In liquids, viscosity decreases with rising temperature, since intermolecular cohesive forces (the dominant resistance mechanism) weaken as molecules gain thermal energy (Andrade's equation). In gases, viscosity increases with rising temperature, since the dominant resistance mechanism is momentum transfer between faster-moving molecules, which intensifies with temperature (Sutherland's law) — the opposite trend to liquids, and a frequently tested contrast.
The bulk modulus \(K\) measures a fluid's resistance to uniform compression; its reciprocal is the compressibility \(\beta\). The speed of sound in a fluid depends directly on \(K\) and inversely on density, giving roughly 1480 m/s in water and 343 m/s in air. The Mach number (velocity relative to the speed of sound) below about 0.3 justifies treating a compressible fluid (like air) as effectively incompressible for most engineering flow calculations.
Surface tension creates a pressure jump across any curved liquid interface — a droplet has one curved surface pair (giving \(4\dfrac{\sigma}{d}\)), a soap bubble has two films (giving \(8\dfrac{\sigma}{d}\)), and a liquid jet has just one surface (giving \(2\dfrac{\sigma}{d}\)). Capillary rise or depression in a narrow tube depends on the contact angle: water wets glass (\(\theta\approx0°\), giving a rise), while mercury does not wet glass (\(\theta\approx140°\), giving a depression).
Cavitation occurs when local pressure in a flowing liquid falls below the vapour pressure, causing vapour bubbles to form; when these bubbles subsequently collapse in a higher-pressure zone, they generate shock waves that cause erosion, noise, and vibration in hydraulic machinery (pump impellers, turbine runners, control valves). Prevention relies on maintaining sufficient Net Positive Suction Head (NPSH) at critical locations such as a pump inlet.
\( \tau = \mu\,\dfrac{du}{dy} \)
Power law (non-Newtonian): \( \tau = K\left(\dfrac{du}{dy}\right)^n \)
\(n<1\): pseudoplastic; \(n>1\): dilatant; \(n=1\): Newtonian
Bingham plastic: \( \tau = \tau_0 + \mu_B\dfrac{du}{dy} \)
Liquids (Andrade): \( \mu(T) \approx A\,e^{B/T} \) (decreases with T)
Gases (Sutherland): \( \mu(T) \propto T^{0.5\text{–}0.75} \) (increases with T)
Bulk modulus: \( K = -\dfrac{dP}{dV/V} = \rho\dfrac{dP}{d\rho} \), \(\beta = \dfrac{1}{K}\)
Speed of sound: \( c = \sqrt{\dfrac{K}{\rho}} \) (water ≈ 1480 m/s; air ≈ 343 m/s)
Mach number: \( Ma = \dfrac{V}{c} \); \(Ma<0.3\) → incompressible assumption valid
Droplet: \( \Delta P = 4\dfrac{\sigma}{d} \) Bubble: \( \Delta P = 8\dfrac{\sigma}{d} \) Jet: \( \Delta P = 2\dfrac{\sigma}{d} \)
Capillary rise: \( h = \dfrac{4\sigma\cos\theta}{\rho g d} \)
Water-glass: \(\theta\approx0°\) (rise); Mercury-glass: \(\theta\approx140°\) (depression)
Cavitation number: \( \sigma_c = \dfrac{p_{atm}-p_v}{\tfrac12\rho V^2} \)
Cavitation occurs when \(\sigma_c\) falls below a critical value for the geometry.
Given: A fluid layer between two plates 2 mm apart has the top plate moving at 1.5 m/s. If \(\mu = 0.9\ \text{Pa·s}\), find the shear stress (assuming a linear velocity profile).
Solution:
\( \dfrac{du}{dy} = \dfrac{1.5}{0.002} = 750\ \text{s}^{-1} \)
\( \tau = 0.9\times750 = 675\ \text{Pa} \)
Answer: \(\tau = 675\ \text{Pa}\).
Given: A glass tube of diameter 1 mm is dipped in water (\(\sigma=0.0728\ \text{N/m}\), \(\theta\approx0°\)). Find the capillary rise.
Solution:
\( h = \dfrac{4\times0.0728\times\cos0°}{998\times9.81\times0.001} = \dfrac{0.2912}{9.79} \approx 0.0297\ \text{m} \)
Answer: \(h\approx29.7\ \text{mm}\).
Given: A soap bubble of diameter 4 mm has surface tension \(\sigma=0.025\ \text{N/m}\). Find the excess internal pressure.
Solution:
\( \Delta P = 8\dfrac{\sigma}{d} = 8\dfrac{\times0.025}{0.004} = \dfrac{0.2}{0.004} = 50\ \text{Pa} \)
Answer: \(\Delta P = 50\ \text{Pa}\).
Fig. 1.1 — Shear stress vs. rate of shear (rheological classification): Newtonian fluids give a straight line through the origin; pseudoplastic and dilatant fluids curve; Bingham plastics require a yield stress τ₀ before flowing.
Pressure at any point in a static fluid increases linearly with depth, governed by the hydrostatic pressure equation. Pascal's Law states that pressure applied to an enclosed fluid is transmitted equally and undiminished in all directions — the principle behind the hydraulic jack, where a small force on a small piston produces a large force on a large piston in proportion to the area ratio.
| Type | Principle | Range / Use |
|---|---|---|
| Simple U-tube | Heavy gauge fluid (mercury) balances pressure; read height difference | Moderate pressures; most common |
| Inverted U-tube | Light gauge fluid (air or oil) at top; for measuring small differential pressures in liquids | Small pressure differences in water/oil pipes |
| Inclined manometer | Limb inclined at angle θ; amplifies reading by 1/sinθ | Very small pressures; wind tunnel tests |
| Micro-manometer | Large reservoir + narrow tube; small pressure differences | Extremely small ΔP; laboratory |
| Differential manometer | Connects two pressure points; measures difference directly | Across orifices, valves, Venturi meters |
| Bourdon gauge | Coiled metal tube deforms under pressure; mechanical indicator | Industrial; wide pressure range |
The standard procedure: start from one point, add \(\rho gh\) for every step going down, and subtract \(\rho gh\) for every step going up; the algebraic sum between two points gives the pressure difference between them. For gas pressure measurement, the weight of the gas column itself is negligible compared to the gauge liquid, so only the mercury (or other gauge fluid) column need be considered.
Pressure at any point depends only on the depth below the free surface and the fluid density — not on the shape or total volume of the container. This is the hydrostatic paradox: a tall narrow tube and a wide, flat, open container filled to the same depth exert identical pressure at the base, despite holding vastly different volumes of fluid.
Pressure: \( p = F/A \) (Pa); \( 1\ \text{bar}=10^5\ \text{Pa} \); \( 1\ \text{atm}=101.325\ \text{kPa} \)
Hydrostatic equation: \( \dfrac{dp}{dz} = -\rho g \)
Pressure at depth h: \( p = p_0 + \rho gh \)
Hydraulic jack: \( F_1/A_1 = F_2/A_2 \)
Simple U-tube (points A, B; gauge fluid \(\rho_m\), pipe fluid \(\rho_f\)):
\( p_A - p_B = \rho_m g h_m - \rho_f g(z_A-z_B) \)
Gas pressure (gauge = mercury, gas weight negligible): \( p_A = \rho_{Hg}\,g\,h \)
Atmospheric pressure: \( 101.325\ \text{kPa} = 10.33\ \text{m water} = 760\ \text{mm Hg} \)
\( \gamma_{water}\approx9.81\ \text{kN/m}^3 \) (9.79 at 20°C)
\( \gamma_{mercury}=133.4\ \text{kN/m}^3 \) (\(S_{Hg}=13.6\))
Given: A U-tube manometer measures the pressure of water in a pipe using mercury as the gauge fluid. The mercury level difference is 250 mm, and both limbs are at the same elevation as the pipe centreline. Find the gauge pressure in the pipe.
Solution:
\( p_A = \rho_{Hg}\,g\,h_m = 13600\times9.81\times0.25 = 33{,}354\ \text{Pa} \)
Answer: \(p_A\approx33.35\ \text{kPa}\).
Given: A hydraulic jack has a small piston of diameter 20 mm and a large piston of diameter 200 mm. Find the force on the large piston for a 100 N force on the small piston.
Solution:
\( F_2 = F_1\times(A_2/A_1) = 100\times(\dfrac{200}{20})^2 = 100\times100 = 10{,}000\ \text{N} \)
Answer: \(F_2 = 10\ \text{kN}\) — a 100:1 mechanical advantage from the area ratio.
Given: An inclined manometer tube is set at 15° to the horizontal. Find the amplification factor for the reading compared to a vertical tube for the same pressure.
Solution:
Amplification \(=1/\sin\theta = 1/\sin15° = \dfrac{1}{0.2588} \approx 3.86\)
Answer: The inclined reading is about 3.86 times longer than an equivalent vertical reading — improving readability for small pressures.
Fig. 2.1 — Simple U-tube manometer: mercury column height difference h_m balances the pipe fluid pressure; add ρgh going down, subtract going up.
The total hydrostatic force on a submerged plane surface depends only on the depth of the centroid below the free surface and the total area — not on the shape's orientation directly. However, the force does not act at the centroid; it acts at the centre of pressure (CP), which always lies below the centroid because pressure increases with depth (the lower portion of the surface experiences higher pressure and therefore a proportionally greater share of the total force). As submergence depth increases, the CP moves progressively closer to the centroid, since the relative variation in pressure across the surface becomes smaller.
A curved surface is analysed by resolving the resultant hydrostatic force into horizontal and vertical components. The horizontal component equals the force on the vertical projection of the curved surface (computed exactly as for a plane surface). The vertical component equals the weight of the (real or imaginary) volume of fluid directly above the curved surface, and acts through the centroid of that volume. The two components combine vectorially to give the resultant force and its direction.
For a rectangular dam retaining water of depth H, the hydrostatic pressure distribution is triangular, so the resultant force acts at \(\dfrac{H}{3}\) above the base (the centroid of the triangle). This total force, multiplied by its lever arm, gives the overturning moment about the toe, which must be resisted by the moment of the dam's self-weight about the same point, with an appropriate factor of safety.
\( F = \rho g\,\bar y\,A \)
\(\bar y\) = depth of centroid below free surface; \(A\) = area
Centre of pressure: \( y_{cp} = \bar y + \dfrac{I_G}{\bar y A} \) (always \(y_{cp}>\bar y\))
Inclined surface: same \(F\) formula (\(\bar y\)=vertical depth of centroid)
Rectangle (\(b\times d\)): \( I_G = \dfrac{bd^3}{12} \)
Triangle (\(b,h\)): \( I_G = \dfrac{bh^3}{36} \)
Circle (dia. \(D\)): \( I_G = \pi \dfrac{D^4}{64} \)
Horizontal: \( F_H = \rho g\,\bar y_{proj}\,A_{proj} \) (acts at CP of projected area)
Vertical: \( F_V = \rho g\times(\text{volume of fluid above}) \) (acts at centroid of that volume)
Resultant: \( F_R = \sqrt{F_H^2+F_V^2} \), \( \tan\theta = F_V/F_H \)
Total force (rectangular dam, width \(b\)): \( F = \tfrac12\rho g H^2 b \), acting at \(\dfrac{H}{3}\) from base
Overturning moment: \( M_O = F\times \dfrac{H}{3} \)
FOS against overturning: \( FOS = \dfrac{M_{resisting}}{M_O} \ge 1.5 \)
Given: A vertical rectangular gate 2 m wide, 3 m deep, has its top edge at the water surface. Find the total force and the depth of the centre of pressure.
Solution:
\( \bar y = \dfrac{3}{2} = 1.5\ \text{m} \), \( A = 2\times3=6\ \text{m}^2 \)
\( F = 1000\times9.81\times1.5\times6 = 88{,}290\ \text{N} \approx 88.3\ \text{kN} \)
\( I_G = 2\dfrac{\times3^3}{12} = 4.5\ \text{m}^4 \)
\( y_{cp} = 1.5 + \dfrac{4.5}{1.5\times6} = 1.5+0.5=2.0\ \text{m} \)
Answer: \(F\approx88.3\ \text{kN}\), acting at \(y_{cp}=2.0\ \text{m}\) below the surface.
Given: A quadrant gate of radius 2 m and width 1 m holds back water level with the top of the gate. Find \(F_H\) and \(F_V\).
Solution:
Vertical projection is a rectangle \(2\ \text{m}\times1\ \text{m}\), \(\bar y_{proj}=1\ \text{m}\)
\( F_H = 1000\times9.81\times1\times(2\times1) = 19{,}620\ \text{N} \approx 19.6\ \text{kN} \)
Volume above the curved surface (quarter circle prism): \( V = \tfrac{\pi}{4}(2)^2\times1 = 3.14\ \text{m}^3 \)
\( F_V = 1000\times9.81\times3.14 = 30{,}803\ \text{N} \approx 30.8\ \text{kN} \)
Answer: \(F_H\approx19.6\ \text{kN}\), \(F_V\approx30.8\ \text{kN}\).
Given: A rectangular dam retains water to a height of 8 m. Find the total force per metre length and the overturning moment about the toe.
Solution:
\( F = \tfrac12\times1000\times9.81\times8^2 = 313{,}920\ \text{N/m} \approx 313.9\ \text{kN/m} \)
\( M_O = 313.9\times(\dfrac{8}{3}) = 313.9\times2.67 \approx 837.9\ \text{kN·m/m} \)
Answer: \(F\approx313.9\ \text{kN/m}\), \(M_O\approx838\ \text{kN·m/m}\) about the toe.
Fig. 3.1 — Hydrostatic force \(F=\rho g\bar y A\) acts at the centre of pressure (CP), below the centroid G.
Fig. 3.2 — Second moment of area \(I_G\) about the centroidal axis for common shapes; used directly in the \(y_{cp}\) formula.
The buoyant force on a submerged or floating body equals the weight of the fluid it displaces. A floating body reaches equilibrium when the buoyant force exactly balances its own weight — the fraction of the body's volume that is submerged equals the ratio of the body's density to the fluid's density. A body whose density exceeds the fluid's density cannot achieve this balance while fully submerged, so it sinks; a body of lower density floats with only a partial fraction submerged.
When a floating body tilts slightly, its centre of buoyancy (the centroid of the displaced volume) shifts because the submerged shape changes. The metacentre M is the point where a vertical line through the new (shifted) centre of buoyancy intersects the original vertical line through the centre of gravity. If M lies above the centre of gravity G, the resulting couple rights the body back to equilibrium (stable); if M lies below G, the couple instead increases the tilt (unstable, capsizing).
The metacentric height \(GM\) is the distance between the metacentre and the centre of gravity — positive (M above G) means stable equilibrium, with a larger GM giving a "stiffer" (faster-righting, but less comfortable) vessel. \(GM\) can be computed geometrically from the vessel's waterplane geometry (\(BM = \dfrac{I_{WL}}{V_{sub}}\)), or measured experimentally on an actual vessel by shifting a known weight a known distance and measuring the resulting small angle of tilt.
\( F_B = \rho_f\,g\,V_{sub} \)
For a floating body: \( F_B = W \) → \( \rho_f g V_{sub} = \rho_b g V_{total} \)
Fraction submerged: \( V_{sub}/V_{total} = \dfrac{\rho_{body}}{\rho_{fluid}} = \dfrac{S_{body}}{S_{fluid}} \)
\( GM = BM - BG \) (M above G: \(GM>0\) → stable)
\( BM = \dfrac{I_{WL}}{V_{sub}} \)
\(I_{WL}\) = 2nd moment of waterplane area about longitudinal centroidal axis
\( I_{WL} = \dfrac{LB^3}{12} \), \( V_{sub} = LBd \)
\( BM = \dfrac{B^2}{12d} \)
\( GM = OB + BM - OG = \dfrac{d}{2} + \dfrac{B^2}{12d} - OG \)
\( GM = \dfrac{W_s\,x}{W\tan\theta} \)
\(W_s\)=shifted weight; \(x\)=shift distance; \(\theta\)=measured tilt angle
Given: A block of wood has specific gravity 0.65. Find the fraction of its volume submerged when floating in fresh water.
Solution:
\( V_{sub}/V_{total} = \dfrac{S_{body}}{S_{fluid}} = \dfrac{0.65}{1.0} = 0.65 \)
Answer: 65% of the block's volume is submerged.
Given: A rectangular pontoon, \(L=12\ \text{m}\), \(B=4\ \text{m}\), floats at a draught \(d=1.5\ \text{m}\). The centre of gravity is 1.2 m above the keel (\(OG=1.2\ \text{m}\)). Find GM and comment on stability.
Solution:
\( BM = \dfrac{B^2}{12d} = \dfrac{16}{12\times1.5} = \dfrac{16}{18} = 0.889\ \text{m} \)
\( OB = \dfrac{d}{2} = 0.75\ \text{m} \)
\( GM = OB+BM-OG = 0.75+0.889-1.2 = 0.439\ \text{m} \)
Answer: \(GM\approx0.44\ \text{m}>0\) — the pontoon is stable.
Given: On a ship weighing 5000 kN, a weight of 30 kN is shifted 6 m across the deck, causing a 2° list. Find the metacentric height.
Solution:
\( GM = \dfrac{W_s x}{W\tan\theta} = \dfrac{30\times6}{5000\times\tan2°} = \dfrac{180}{5000\times0.0349} = \dfrac{180}{174.5} \)
Answer: \(GM\approx1.03\ \text{m}\).
Fig. 4.1 — Metacentre M: when a body tilts, the buoyancy line through the new centre of buoyancy B' intersects the original vertical at M; stable if M is above G, unstable if below.
| Classification | Criterion |
|---|---|
| Steady / Unsteady | \(\partial \dfrac{V}{\partial} t = 0\) (steady) or \(\ne 0\) (unsteady) — velocity constant at a point over time |
| Uniform / Non-uniform | \(\partial \dfrac{V}{\partial} s = 0\) (uniform) or \(\ne 0\) — velocity constant along a streamline at a given instant |
| Laminar / Turbulent | Re < 2000 (laminar, viscous dominates) or Re > 4000 (turbulent, inertia dominates) |
| Compressible / Incompressible | Density ρ varies significantly, or is essentially constant |
A streamline is tangent at every point to the velocity direction at that instant, with no flow crossing it. A pathline is the actual trajectory traced by a single fluid particle over time. A streakline is the locus of all particles that have passed through a fixed point (what dye injection visualises). A streamtube is a bundle of streamlines through which no fluid crosses the walls. For steady flow, all three (streamline, pathline, streakline) coincide exactly; for unsteady flow, they are generally all different.
The continuity equation expresses conservation of mass. In its most general (3D, compressible) form it is a partial differential equation; for steady, incompressible flow it reduces to the divergence of velocity being zero; and for one-dimensional flow through a streamtube it reduces to the simple statement that volumetric flow rate is the same at every cross-section (or mass flow rate, for compressible flow).
The stream function ψ exists for any 2D incompressible flow (it automatically satisfies continuity) — lines of constant ψ are streamlines, and the difference in ψ between two streamlines equals the flow rate between them. The velocity potential φ exists only for irrotational flow, and satisfies Laplace's equation; lines of constant φ (equipotential lines) are always perpendicular to streamlines, forming an orthogonal flow net.
Vorticity is the curl of the velocity field, physically equal to twice the local angular velocity of a fluid element. Circulation is the line integral of velocity around a closed contour, and by Stokes' theorem equals the surface integral of vorticity over any surface bounded by that contour. In irrotational flow, vorticity is zero everywhere, so circulation around any contour not enclosing a singular vortex is also zero.
General (3D, compressible): \( \dfrac{\partial\rho}{\partial t}+\dfrac{\partial(\rho u)}{\partial x}+\dfrac{\partial(\rho v)}{\partial y}+\dfrac{\partial(\rho w)}{\partial z}=0 \)
Steady, incompressible: \( \dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0 \)
1D steady (streamtube): \( A_1V_1=A_2V_2=Q \)
Compressible: \( \rho_1A_1V_1=\rho_2A_2V_2=\dot m \)
\( u = \partial\dfrac{\psi}{\partial} y \), \( v = -\partial\dfrac{\psi}{\partial} x \)
ψ = constant along a streamline; Δψ between streamlines = flow rate per unit depth
\( u=-\partial\dfrac{\phi}{\partial} x \), \( v=-\partial\dfrac{\phi}{\partial} y \), \( w=-\partial\dfrac{\phi}{\partial} z \)
Satisfies Laplace's equation: \( \nabla^2\phi=0 \)
Exists only for irrotational flow
Rotation: \( \omega_z=\tfrac12\left(\dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial y}\right) \); irrotational if \(\omega=0\)
Vorticity: \( \vec\omega=\nabla\times\vec V \)
Circulation: \( \Gamma=\oint\vec V\cdot d\vec s \)
Stokes' theorem: \( \Gamma=\iint(\nabla\times\vec V)\cdot d\vec A \)
Given: Water flows through a pipe that reduces from 300 mm to 150 mm diameter. If the velocity in the larger section is 2 m/s, find the velocity in the smaller section.
Solution:
\( A_1V_1=A_2V_2 \Rightarrow V_2 = V_1(\dfrac{D_1}{D_2})^2 = 2\times(\dfrac{300}{150})^2 = 2\times4=8\ \text{m/s} \)
Answer: \(V_2=8\ \text{m/s}\).
Given: A 2D flow has stream function \(\psi = 3x^2y - y^3\). Find the velocity components \(u\) and \(v\) at the point (1, 2).
Solution:
\( u = \partial\dfrac{\psi}{\partial} y = 3x^2 - 3y^2 \); at (1,2): \( u=3(1)-3(4)=3-12=-9 \)
\( v = -\partial\dfrac{\psi}{\partial} x = -6xy \); at (1,2): \( v=-6(1)(2)=-12 \)
Answer: \(u=-9\), \(v=-12\) (units per the problem's length/time units).
Given: A flow field has \(u = 2x\), \(v = -2y\). Check whether the flow is irrotational.
Solution:
\( \partial \dfrac{v}{\partial} x = 0 \); \( \partial \dfrac{u}{\partial} y = 0 \)
\( \omega_z = \tfrac12(0-0) = 0 \)
Answer: The flow is irrotational (\(\omega_z=0\) everywhere) — a velocity potential φ exists for it.
Fig. 5.1 — Flow net: streamlines (constant ψ) and equipotential lines (constant φ) intersect at right angles for irrotational flow.
Euler's equation of motion, integrated along a streamline for steady, incompressible, inviscid, irrotational flow, gives Bernoulli's equation: the sum of pressure head, velocity head and datum (elevation) head remains constant along a streamline — a statement of conservation of mechanical energy per unit weight of fluid. In real (viscous) systems, a modified form accounts for head losses and any pump head added between two points.
| Device | Type | \(C_d\) Typical |
|---|---|---|
| Venturimeter | Differential pressure (pipe); smooth converging-diverging section | 0.96–0.99 |
| Orifice meter | Differential pressure (pipe); simple plate with a hole | 0.60–0.65 |
| Pitot tube | Point velocity measurement from stagnation pressure | \(C_v\approx0.98\text{–}1.0\) |
| Rotameter | Variable area (float rises in a tapered tube) | — |
| Notch / Weir | Open channel measurement (see Ch. 13) | — |
A Venturimeter's smooth, gradual contraction and expansion give it a much higher discharge coefficient than the simple orifice meter, at the cost of greater length and expense — this trade-off between accuracy/efficiency and simplicity/cost is a recurring theme across flow-measuring devices.
Torricelli's theorem gives the theoretical jet velocity from an orifice under head h. The actual velocity is always slightly less, captured by the coefficient of velocity \(C_v\) (friction losses). The jet also contracts to a minimum cross-section (the vena contracta) downstream of a sharp-edged orifice, captured by the coefficient of contraction \(C_c\). The overall discharge coefficient combines both effects, and its value depends strongly on orifice edge geometry — sharp-edged, bell-mouth, or re-entrant.
The momentum equation (derived from Newton's second law applied to a control volume via the Reynolds Transport Theorem) relates the net force on the fluid to the rate of change of momentum flux across the control surface. Applied to a pipe bend, it gives the force components needed to hold the bend in place — by Newton's third law, the fluid exerts an equal and opposite force on the pipe, which is what anchor blocks and thrust restraints must resist.
Euler (along streamline): \( \dfrac{dp}{\rho g}+\dfrac{V\,dV}{g}+dz=0 \)
Bernoulli: \( \dfrac{p}{\rho g}+\dfrac{V^2}{2g}+z = H = \text{const} \)
Modified (with losses \(h_L\), pump head \(H_p\)): \( H_1+H_p = H_2+h_L \)
Venturimeter: \( Q = C_d\dfrac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh} \)
Orifice meter: \( Q = C_d A_o\sqrt{2gh} \)
Pitot tube: \( V = C_v\sqrt{2g\,\Delta h} \)
Torricelli (theoretical): \( V_t = \sqrt{2gh} \)
Actual velocity: \( V_a = C_v V_t \), \(C_v\approx0.98\)
Contraction: \( C_c = A_{vena}/A_{orifice}\approx0.64 \)
Discharge: \( Q = C_dA\sqrt{2gh} \), \(C_d=C_vC_c\approx0.61\) (sharp-edged); 0.82 (bell-mouth); 0.51 (re-entrant)
Submerged orifice: \( Q = C_dA\sqrt{2g(h_1-h_2)} \)
\( \sum\vec F = \rho Q(\vec V_2-\vec V_1) \)
Pipe bend: \( F_x=\rho Q(V_2\cos\theta_2-V_1\cos\theta_1)+(p_2A_2\cos\theta_2-p_1A_1\cos\theta_1) \)
\( F_y=\rho Q(V_2\sin\theta_2-V_1\sin\theta_1)+(p_2A_2\sin\theta_2-p_1A_1\sin\theta_1) \)
Resultant: \( F_R=\sqrt{F_x^2+F_y^2} \)
Given: A Venturimeter has \(A_1=0.05\ \text{m}^2\), \(A_2=0.02\ \text{m}^2\), \(C_d=0.98\), manometer reading gives \(h=0.3\ \text{m}\) of water. Find the discharge.
Solution:
\( Q = 0.98\times\dfrac{0.05\times0.02}{\sqrt{0.05^2-0.02^2}}\times\sqrt{2\times9.81\times0.3} \)
\( = 0.98\times\dfrac{0.001}{0.0458}\times2.425 = 0.98\times0.02184\times2.425 \approx 0.0519\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q\approx0.052\ \dfrac{\text{m}^3}{\text{s}} = 52\ \text{L/s}\).
Given: An orifice is located 2.5 m below the free surface of a tank. Find the theoretical jet velocity and, if \(C_v=0.98\), the actual velocity.
Solution:
\( V_t = \sqrt{2\times9.81\times2.5} = \sqrt{49.05} \approx 7.00\ \text{m/s} \)
\( V_a = 0.98\times7.00 = 6.86\ \text{m/s} \)
Answer: \(V_t\approx7.00\ \text{m/s}\), \(V_a\approx6.86\ \text{m/s}\).
Given: A 90° pipe bend has \(Q=0.15\ \dfrac{\text{m}^3}{\text{s}}\), inlet velocity \(V_1=3\ \text{m/s}\) (along x), exit velocity \(V_2=3\ \text{m/s}\) (along y, since area is constant). Neglect pressure forces. Find the resultant force on the fluid.
Solution:
\( F_x = \rho Q(V_2\cos90° - V_1\cos0°) = 1000\times0.15\times(0-3) = -450\ \text{N} \)
\( F_y = \rho Q(V_2\sin90° - V_1\sin0°) = 1000\times0.15\times(3-0) = 450\ \text{N} \)
\( F_R = \sqrt{450^2+450^2} = 450\sqrt2 \approx 636.4\ \text{N} \)
Answer: \(F_R\approx636\ \text{N}\), directed at 45° to the inlet axis.
Fig. 6.1 — Venturimeter: the smooth converging-diverging section creates a pressure differential between inlet and throat, from which discharge Q is computed via Bernoulli's equation.
For fully developed laminar flow in a circular pipe, viscous shear balances the pressure gradient exactly, giving an exact parabolic velocity profile with maximum velocity at the centreline. Integrating this profile over the cross-section gives the Hagen-Poiseuille law for volumetric flow rate, directly proportional to the fourth power of diameter — meaning a small change in pipe diameter has a dramatic effect on capacity. The head loss predicted by this exact solution corresponds to the Darcy-Weisbach friction factor \(f=\dfrac{64}{Re}\), which is why the Darcy-Weisbach equation applies equally to laminar and turbulent flow.
Flow between two flat plates has two limiting cases superimposed in the general Couette-Poiseuille profile: pure Couette flow (linear velocity profile, driven purely by one plate moving relative to the other, zero pressure gradient) and pure Poiseuille flow (parabolic velocity profile, driven purely by a pressure gradient with both plates fixed). Real bearing and lubrication flows are usually a combination of both mechanisms.
The Reynolds number is the ratio of inertia forces to viscous forces in a flow, and is the single most important dimensionless parameter for classifying pipe flow. Below the lower critical value, viscous forces dominate and flow remains laminar (smooth, layered); above the upper value, inertia dominates and flow becomes turbulent (chaotic, well-mixed); between the two lies an unstable transition zone that may intermittently switch between the two regimes.
Velocity profile: \( u(r) = \dfrac{(R^2-r^2)}{4\mu}\left(-\dfrac{dp}{dx}\right) \) (parabolic)
Max. velocity: \( u_{max} = \dfrac{R^2}{4\mu}\left(-\dfrac{dp}{dx}\right) \)
Mean velocity: \( V_{mean} = \dfrac{u_{max}}{2} \)
Flow rate: \( Q = \dfrac{\pi R^4}{8\mu}\left(-\dfrac{dp}{dx}\right) = \dfrac{\pi D^4\Delta P}{128\mu L} \)
Head loss: \( h_f = \dfrac{32\mu LV}{\rho gD^2} = \dfrac{128\mu LQ}{\pi\rho gD^4} \)
Correction factors (laminar): \(\alpha=2.0\) (kinetic energy); \(\beta=\dfrac{4}{3}\) (momentum)
Couette-Poiseuille: \( u(y) = \dfrac{Uy}{h}+\dfrac{1}{2\mu}\left(-\dfrac{dp}{dx}\right)y(h-y) \)
Pure Couette: \( u = \dfrac{Uy}{h} \) (linear)
Pure Poiseuille: \( u = \dfrac{h^2-y^2}{4\mu}\left(-\dfrac{dp}{dx}\right) \) (parabolic)
Mean velocity (Poiseuille): \( V = \dfrac{h^2}{12\mu}\left(-\dfrac{dp}{dx}\right) \)
\( Re = \dfrac{\rho VD}{\mu} = \dfrac{VD}{\nu} \)
\(Re<2000\): laminar; \(2000
Critical Re in a pipe: 2300 (theoretical), 2000 (engineering, conservative)
Given: Oil (\(\mu=0.1\ \text{Pa·s}\)) flows through a 20 mm diameter pipe, 5 m long, with a pressure drop of 4000 Pa. Find the discharge, assuming laminar flow.
Solution:
\( Q = \dfrac{\pi\times(0.02)^4\times4000}{128\times0.1\times5} = \dfrac{\pi\times1.6\times10^{-7}\times4000}{64} \)
\( Q = \dfrac{2.011\times10^{-3}}{64} \approx 3.14\times10^{-5}\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q\approx31.4\ \text{mL/s}\).
Given: Water (\(\nu=1.0\times10^{-6}\ \dfrac{\text{m}^2}{\text{s}}\)) flows at 0.5 m/s through a 10 mm diameter tube. Determine the flow regime.
Solution:
\( Re = \dfrac{VD}{\nu} = \dfrac{0.5\times0.01}{1.0\times10^{-6}} = 5000 \)
Answer: \(Re=5000>4000\) — the flow is turbulent.
Given: Laminar flow in a pipe has a mean velocity of 0.8 m/s. Find the maximum (centreline) velocity.
Solution:
\( u_{max} = 2\times V_{mean} = 2\times0.8 = 1.6\ \text{m/s} \)
Answer: \(u_{max}=1.6\ \text{m/s}\).
Fig. 7.1 — Parabolic (Hagen-Poiseuille) velocity profile: maximum velocity at the centreline is exactly twice the mean (average) velocity across the pipe.
The Darcy-Weisbach equation is the universal formula for friction (major) head loss in pipe flow, applicable to both laminar and turbulent flow provided the correct friction factor \(f\) is used. For laminar flow, \(f=\dfrac{64}{Re}\) exactly; for turbulent flow, \(f\) depends on both Reynolds number and relative pipe roughness, given approximately by the Blasius formula for smooth pipes at moderate Re, or more generally (and implicitly) by the Colebrook-White equation across the full turbulent range, visualised graphically in the Moody diagram.
Minor losses occur at pipe fittings, bends, valves, and sudden changes in cross-section — despite the name, they can be significant in short pipe systems with many fittings. Each loss is expressed as a multiple \(K\) of the velocity head; sudden expansion losses can be derived theoretically (Borda-Carnot), while most other loss coefficients are determined experimentally.
Pipes in series carry the same flow rate through every pipe, and head losses add up along the path (just like resistors in series). Pipes in parallel share the same head loss across each parallel branch, and the total flow rate is the sum of the flow through each branch (just like resistors in parallel, but for flow rather than current). An equivalent single pipe, matching the same total head loss for the same total flow, can be computed for either arrangement.
Real distribution networks (water supply grids) form closed loops rather than simple series/parallel combinations. The Hardy-Cross method solves such networks iteratively: an initial guess of flow in each pipe is corrected loop-by-loop until both continuity at every junction and zero net head loss around every loop are satisfied simultaneously.
Sudden valve closure in a pipeline abruptly decelerates the flow, converting kinetic energy into a pressure surge that propagates as a wave at the speed of sound in the fluid-pipe system (the wave speed is reduced from the value in an infinite fluid by the elasticity of the pipe wall). If the valve closes faster than the time for the pressure wave to travel to the reservoir and back (the critical closure time), the full Joukowsky pressure rise develops; slower closure allows partial pressure relief, reducing the severity of the surge.
\( h_f = \dfrac{fLV^2}{2gD} = \dfrac{fLQ^2}{3.03D^5} \)
Chezy: \( V = C\sqrt{RS} \), \(R=\dfrac{D}{4}\) (full pipe); relation: \(f=\dfrac{8g}{C^2}\)
Laminar: \(f=\dfrac{64}{Re}\)
Blasius (smooth, \(4000 Colebrook-White: \(\dfrac{1}{\sqrt f} = -2\log\left(\dfrac{\epsilon}{3.7D}+\dfrac{2.51}{Re\sqrt f}\right)\)
\( h_m = \dfrac{KV^2}{2g} \)
Entry (sharp): \(K=0.5\); Entry (bell-mouth): \(K\approx0.04\); Exit: \(K=1.0\)
Sudden expansion: \( h_e = (V_1-V_2)\dfrac{^2}{2g} \), \(K=(1-A_1/A_2)^2\)
Sudden contraction: \(K=0.5(1-A_2/A_1)\)
Bend (90°): \(K=0.3\text{–}1.5\); Gate valve (open): \(K=0.2\); Globe valve: \(K=6\text{–}10\)
Series: \(Q_1=Q_2=Q_3=Q\); \(h_{f,total}=h_{f1}+h_{f2}+h_{f3}\)
Parallel: \(h_{f1}=h_{f2}=h_{f3}\); \(Q=Q_1+Q_2+Q_3\)
Equivalent pipe (series): \(\dfrac{L_{eq}}{D_{eq}^5}=\sum(\dfrac{L_i}{D_i^5})\)
Equivalent pipe (parallel): \(\dfrac{L_{eq}}{D_{eq}^5} = \dfrac{L}{(\sum D_i^{5/2})^2}\)
Hardy-Cross correction: \( \Delta Q = -\dfrac{\sum h_f}{n\sum|h_f/Q|} \) (\(n=2\) Darcy-Weisbach; \(n=1.85\) Hazen-Williams)
Joukowsky: \( \Delta P = \rho c\,\Delta V \)
Celerity: \( c = \dfrac{1}{\sqrt{\rho(1/K+D/(eE))}} \); rigid pipe: \(c=\sqrt{\dfrac{K}{\rho}}\approx1480\ \text{m/s}\) (water)
Critical closure time: \( t_c = \dfrac{2L}{c} \)
Given: Water flows through a 200 mm diameter, 500 m long pipe at 1.5 m/s, with \(f=0.02\). Find the friction head loss.
Solution:
\( h_f = \dfrac{0.02\times500\times1.5^2}{2\times9.81\times0.2} = \dfrac{0.02\times500\times2.25}{3.924} = \dfrac{22.5}{3.924} \approx 5.73\ \text{m} \)
Answer: \(h_f\approx5.73\ \text{m}\).
Given: Two pipes in series: Pipe 1 (\(L_1=200\ \text{m}, D_1=0.3\ \text{m}\)), Pipe 2 (\(L_2=150\ \text{m}, D_2=0.2\ \text{m}\)). Find the equivalent length for an equivalent pipe of diameter 0.25 m.
Solution:
\( \sum(\dfrac{L_i}{D_i^5}) = \dfrac{200}{0.3^5}+\dfrac{150}{0.2^5} = \dfrac{200}{0.00243}+\dfrac{150}{0.00032} = 82{,}305+468{,}750 = 551{,}055 \)
\( L_{eq} = 551{,}055\times D_{eq}^5 = 551{,}055\times0.25^5 = 551{,}055\times0.000977 \approx 538.4\ \text{m} \)
Answer: \(L_{eq}\approx538\ \text{m}\) of 0.25 m diameter pipe.
Given: A rigid pipe carries water at \(V_0=2.5\ \text{m/s}\); a valve closes suddenly (\(t < t_c\)). Find the pressure rise.
Solution:
\( c = \sqrt{\dfrac{K}{\rho}} = \sqrt{2.18\dfrac{\times10^9}{1000}} \approx 1476\ \text{m/s} \)
\( \Delta P = \rho c\Delta V = 1000\times1476\times2.5 = 3.69\times10^6\ \text{Pa} \)
Answer: \(\Delta P\approx3.69\ \text{MPa}\) — illustrating why rapid valve closure is dangerous in long pipelines.
Fig. 8.1 — Pipes in series (same discharge, head losses add) vs. parallel (same head loss across each branch, discharges add).
Drag is the force component parallel to the flow direction (resisting motion), while lift is the force component perpendicular to it. Both are expressed via empirical or theoretical dimensionless coefficients \(C_D\) and \(C_L\), which depend on body shape, orientation, and Reynolds number, and are typically obtained from wind-tunnel or towing-tank experiments.
| Body Shape | Re Range | \(C_D\) | Notes |
|---|---|---|---|
| Sphere | Re < 0.5 | 24/Re (Stokes) | Creeping flow; \(F_D=3\pi\mu DV\) |
| Sphere | 0.5–1000 | Intermediate; decreasing | Oseen/intermediate range |
| Sphere | 10³–3×10⁵ | ≈ 0.4 | Newton's drag regime |
| Sphere | > 3×10⁵ | ≈ 0.1–0.2 | Drag crisis; turbulent BL (dimples on golf ball) |
| Long circular cylinder | 10³–10⁵ | ≈ 1.0–1.2 | High form drag; separated flow |
| Flat plate (normal) | Turbulent | ≈ 1.9 | Full pressure drag; wake ≈ plate size |
| Flat plate (parallel) | Laminar/Turbulent | ≈ 0.001–0.005 | Friction drag only |
| Aerofoil (thin) | Large Re | 0.01–0.05 | Streamlined; mostly friction drag |
For very low Reynolds number (Re < 1), viscous forces completely dominate inertia, and Stokes derived an exact expression for the drag force on a sphere. Setting this drag equal to the net gravity/buoyancy force on a settling particle gives the terminal (settling) velocity — the constant velocity a particle reaches once these forces balance. This principle underlies the design of sedimentation tanks, centrifuges, and particle-size analysis by sedimentation.
\( F_D = C_D\times\tfrac12\rho V^2A \)
\( F_L = C_L\times\tfrac12\rho V^2A \)
A = frontal area (drag) or planform area (lift).
Stokes drag (Re < 1): \( F_D = 3\pi\mu DV \)
\( C_D = \dfrac{24}{Re} \)
Terminal velocity: \( V_t = \dfrac{(\rho_s-\rho_f)gD^2}{18\mu} \)
Given: A sphere of diameter 0.1 m moves through air (\(\rho=1.2\ \text{kg/m}^3\)) at 20 m/s; \(C_D=0.4\). Find the drag force.
Solution:
\( A = \dfrac{\pi}{4}\times(0.1)^2 = 0.00785\ \text{m}^2 \)
\( F_D = 0.4\times0.5\times1.2\times20^2\times0.00785 = 0.4\times0.5\times1.2\times400\times0.00785 \)
Answer: \(F_D\approx0.754\ \text{N}\).
Given: A sand particle, \(D=0.1\ \text{mm}\), \(\rho_s=2650\ \text{kg/m}^3\), settles in water (\(\rho_f=1000\ \text{kg/m}^3\), \(\mu=1.0\times10^{-3}\ \text{Pa·s}\)). Find the terminal velocity (assuming Stokes' law applies).
Solution:
\( V_t = \dfrac{(2650-1000)\times9.81\times(1\times10^{-4})^2}{18\times1\times10^{-3}} = \dfrac{1650\times9.81\times1\times10^{-8}}{0.018} \)
\( V_t = \dfrac{1.619\times10^{-4}}{0.018} \approx 8.99\times10^{-3}\ \text{m/s} \)
Answer: \(V_t\approx9\ \text{mm/s}\).
Given: A sphere of diameter 2 mm settles in oil (\(\mu=0.5\ \text{Pa·s}\)) at 0.01 m/s. Find the Stokes drag force.
Solution:
\( F_D = 3\pi\times0.5\times0.002\times0.01 = 3\pi\times10^{-5} \approx 9.42\times10^{-5}\ \text{N} \)
Answer: \(F_D\approx9.42\times10^{-5}\ \text{N} = 94.2\ \mu\text{N}\).
Fig. 9.1 — Drag coefficient of a sphere vs. Reynolds number: Stokes regime, roughly constant C_D≈0.4, then a sharp "drag crisis" drop as the boundary layer transitions to turbulent.
Prandtl (1904) introduced the boundary layer concept: for high-Reynolds-number flows over a body, viscous effects are confined to a thin layer near the surface — the boundary layer. Outside this layer, the flow is essentially inviscid (potential flow), which is why potential flow theory (accurate everywhere except this thin layer) can still predict pressure distributions well, even though it completely ignores viscosity.
Blasius solved the boundary layer equations exactly for laminar flow over a flat plate with zero pressure gradient, giving closed-form expressions for boundary layer thickness, displacement thickness, momentum thickness, and skin friction, all as functions of the local Reynolds number \(Re_x\) (based on distance from the leading edge). Transition to turbulence on a smooth flat plate typically occurs around \(Re_x \approx 5\times10^5\).
Once the boundary layer transitions to turbulence, it grows faster (a 1/5 power law rather than the laminar 1/2 power law) and has higher skin friction, since turbulent mixing continuously replenishes momentum near the wall. The turbulent velocity profile near a wall follows the universal law of the wall: a linear viscous sublayer very close to the surface, transitioning to a logarithmic profile further out.
Separation occurs when the wall shear stress falls to zero, caused by an adverse pressure gradient (pressure increasing in the flow direction) that decelerates the near-wall flow faster than the free stream. Common triggers include bluff bodies, sharp corners, diverging channels, and airfoils at high angle of attack; separation causes a large wake, increased pressure drag, and (for a wing) loss of lift known as stall. Prevention strategies include streamlining the body, boundary-layer suction, deliberately tripping the boundary layer turbulent (since a turbulent BL resists separation better), or vortex generators.
The von Kármán momentum integral equation is an approximate, integrated form of the boundary layer equations, relating wall shear stress to the growth rate of momentum thickness along the surface. It underlies most approximate (non-exact) boundary layer solution methods, including flows with a non-zero pressure gradient where an exact solution (like Blasius') is not available.
Thickness δ: distance where velocity = \(0.99U\)
Displacement thickness: \( \delta^* = \int_0^\delta(1-\dfrac{u}{U})\,dy \)
Momentum thickness: \( \theta = \int_0^\delta \dfrac{u}{U}(1-\dfrac{u}{U})\,dy \)
Shape factor: \( H = \delta^*/\theta \)
\( \dfrac{\delta}{x} = \dfrac{5.0}{\sqrt{Re_x}} \)
\( \delta^*/x = \dfrac{1.72}{\sqrt{Re_x}} \), \( \dfrac{\theta}{x} = \dfrac{0.664}{\sqrt{Re_x}} \)
Local skin friction: \( C_f = \dfrac{0.664}{\sqrt{Re_x}} \)
Average drag: \( C_D = \dfrac{1.328}{\sqrt{Re_L}} \)
Transition: \( Re_{x,cr}\approx5\times10^5 \)
\( \dfrac{\delta}{x} = \dfrac{0.37}{Re_x^{0.2}} \) (1/5 power law)
Local skin friction: \( C_f = \dfrac{0.0576}{Re_x^{0.2}} \)
Average drag: \( C_D = \dfrac{0.074}{Re_L^{0.2}} \) (\(Re<10^7\))
Mixed (laminar+turbulent): \( C_D = \dfrac{0.074}{Re_L^{0.2}} - \dfrac{A}{Re_L} \), \(A=1742\) for \(Re_{cr}=5\times10^5\)
Log-law: \( u^+ = 2.5\ln(y^+)+5.0 \), \(\kappa=0.41\), \(u^+=\dfrac{u}{u^*}\), \(y^+=yu^*/\nu\)
Separation: \( \left.\dfrac{\partial u}{\partial y}\right|_{y=0}=0 \) (adverse \(\partial \dfrac{p}{\partial} x>0\))
\( \dfrac{\tau_w}{\rho U^2} = \dfrac{d\theta}{dx}+(2+H)\dfrac{\theta}{U}\dfrac{dU}{dx} \)
Flat plate (zero pressure gradient): \( \tau_w = \rho U^2\,d\dfrac{\theta}{dx} \)
Given: Air (\(\nu=1.5\times10^{-5}\ \dfrac{\text{m}^2}{\text{s}}\)) flows at 5 m/s over a flat plate. Find the boundary layer thickness at \(x=0.5\ \text{m}\) from the leading edge.
Solution:
\( Re_x = \dfrac{Ux}{\nu} = 5\dfrac{\times0.5}{1.5}\times10^{-5} = 1.667\times10^5 \)
\( \delta = 5.0\dfrac{\times0.5}{\sqrt{1.667\times10^5}} = \dfrac{2.5}{408.3} \approx 0.00612\ \text{m} \)
Answer: \(\delta\approx6.1\ \text{mm}\) (still laminar since \(Re_x<5\times10^5\)).
Given: For a flat plate of length 1 m in the same flow as Example 1, find the average drag coefficient at the trailing edge.
Solution:
\( Re_L = 5\dfrac{\times1}{1.5}\times10^{-5} = 3.33\times10^5 \)
\( C_D = \dfrac{1.328}{\sqrt{3.33\times10^5}} = \dfrac{1.328}{577.4} \approx 0.0023 \)
Answer: \(C_D\approx0.0023\) (still within the laminar range, marginally below transition).
Given: Water flows over a flat plate at high Re such that the boundary layer is turbulent, with \(Re_x=2\times10^6\) at \(x=2\ \text{m}\). Find δ.
Solution:
\( \delta = 0.37\dfrac{\times2}{(2\times10^6)^{0.2}} \)
\( (2\times10^6)^{0.2} \approx 18.9 \)
\( \delta = \dfrac{0.74}{18.9} \approx 0.0392\ \text{m} \)
Answer: \(\delta\approx39.2\ \text{mm}\) — notably thicker than an equivalent laminar boundary layer at the same distance.
Fig. 10.1 — Boundary layer growth over a flat plate: laminar layer thickens as \(\sqrt x\), transitions near \(Re_x\approx5\times10^5\), then grows faster (\(\approx x^{0.8}\)) as a turbulent layer.
Turbulent flow is characterised by chaotic, irregular velocity fluctuations in space and time. It involves mixing, vortex stretching, an energy cascade from large eddies down to progressively smaller eddies, and ultimate dissipation as heat at the smallest (Kolmogorov) scale. Reynolds decomposition splits the instantaneous velocity into a time-averaged mean component and a fluctuating component, which by definition averages to zero over time but still transports momentum — giving rise to an additional apparent stress (Reynolds stress) beyond ordinary viscous shear.
Prandtl's mixing length theory models turbulent shear stress using an empirical length scale that grows linearly with distance from the wall. The resulting velocity profile in a turbulent pipe is much flatter than the parabolic laminar profile, commonly approximated by a power law whose exponent depends on Reynolds number; the ratio of mean to maximum velocity is consequently much closer to 1.0 than the laminar value of 0.5, reflecting the more uniform velocity distribution that turbulent mixing produces.
Because the velocity profile is flatter in turbulent flow, the kinetic energy and momentum correction factors (α and β) are much closer to 1.0 than their laminar counterparts (2.0 and 4/3). The Moody diagram summarises the Darcy friction factor as a function of both Reynolds number and relative pipe roughness across the entire laminar-to-fully-rough-turbulent range, with the Blasius formula and Colebrook-White equation as its two key underlying correlations.
\( u = \bar u + u' \) (mean + fluctuating); \( \bar{u'} = 0 \)
Reynolds (turbulent) stress: \( \tau_t = -\rho\,\overline{u'v'} \)
Total shear: \( \tau_{total} = \mu\dfrac{du}{dy} - \rho\overline{u'v'} \)
\( \tau_t = \rho l^2\left|\dfrac{du}{dy}\right|\dfrac{du}{dy} \), \(l\approx\kappa y\) (\(\kappa=0.41\))
Power law: \( \dfrac{u}{u_{max}} = (\dfrac{y}{R})^{1/n} \), \(n\approx7\) at \(Re\approx10^5\)
\( \dfrac{V_{mean}}{u_{max}} = \dfrac{2n^2}{(n+1)(2n+1)} \approx 0.817 \) (for \(n=7\))
Friction velocity: \( u^* = \sqrt{\dfrac{\tau_0}{\rho}} \)
Smooth pipe: \( \dfrac{u}{u^*} = 5.75\log(yu^*/\nu)+5.5 \)
Rough pipe: \( \dfrac{u}{u^*} = 5.75\log(\dfrac{y}{\epsilon})+8.5 \)
Turbulent: \(\alpha\approx1.05\text{–}1.10\); \(\beta\approx1.02\text{–}1.05\)
Laminar (for comparison): \(\alpha=2.0\); \(\beta=\dfrac{4}{3}\)
Smooth (Blasius): \( f=\dfrac{0.316}{Re^{0.25}} \) (\(4000 Fully rough: \( \dfrac{1}{\sqrt f} = 2\log(\dfrac{3.7D}{\varepsilon})+\text{const} \) (f independent of Re)
Given: Turbulent pipe flow follows the 1/7th power law (\(n=7\)). If the centreline (maximum) velocity is 4 m/s, find the mean velocity.
Solution:
\( V_{mean} = 0.817\times u_{max} = 0.817\times4 = 3.27\ \text{m/s} \)
Answer: \(V_{mean}\approx3.27\ \text{m/s}\).
Given: Water flows in a pipe with wall shear stress \(\tau_0=5\ \text{Pa}\). Find the friction velocity.
Solution:
\( u^* = \sqrt{\dfrac{5}{1000}} = \sqrt{0.005} \approx 0.0707\ \text{m/s} \)
Answer: \(u^*\approx70.7\ \text{mm/s}\).
Given: Water flows in a smooth pipe at \(Re=20{,}000\). Find the Darcy friction factor using the Blasius formula.
Solution:
\( f = \dfrac{0.316}{Re^{0.25}} = \dfrac{0.316}{20000^{0.25}} = \dfrac{0.316}{11.89} \approx 0.0266 \)
Answer: \(f\approx0.0266\).
Fig. 11.1 — Laminar (parabolic) vs. turbulent (flatter, fuller) velocity profile in a pipe: turbulent mixing redistributes momentum, giving V_mean/u_max ≈ 0.817 vs. 0.5 for laminar flow.
The Buckingham π theorem states that if a physical phenomenon can be described by \(n\) variables involving \(m\) fundamental dimensions, it can be reduced to a relationship between \((n-m)\) independent dimensionless groups (π groups). This dramatically reduces the number of experiments needed to characterise a phenomenon — instead of varying every variable independently, only the dimensionless groups need be varied.
| Number | Symbol | Formula | Significance | Model Similarity |
|---|---|---|---|---|
| Reynolds | Re | \(\rho \dfrac{VL}{\mu}\) | Inertia / Viscous; pipe flow, boundary layer | Viscous models (pipe, BL) |
| Froude | Fr | \(\dfrac{V}{\sqrt{gL}}\) | Inertia / Gravity; free-surface flow | Open channel, ship, spillway |
| Euler | Eu | \(\dfrac{p}{\rho V^2}\) | Pressure / Inertia; cavitation | Pressure-dominated flows |
| Weber | We | \(\rho \dfrac{V^2L}{\sigma}\) | Inertia / Surface tension; droplets | Capillary flow, atomisation |
| Mach | Ma | \(\dfrac{V}{c}\) | Inertia / Compressibility; sound waves | Compressible (aerodynamics) |
| Strouhal | St | \(\dfrac{fL}{V}\) | Vortex shedding frequency; oscillatory flows | Vortex-induced vibration |
| Prandtl | Pr | \(\mu \dfrac{c_p}{k}=\dfrac{\nu}{\alpha}\) | Momentum / Thermal diffusivity | Heat transfer similarity |
Complete similarity between a model and its prototype requires geometric similarity (same shape), kinematic similarity (same velocity pattern, scaled), and dynamic similarity (same force ratios). Since it is generally impossible to match every dimensionless number simultaneously, the dominant force ratio for the phenomenon being studied is matched: Froude similarity for free-surface flows (open channels, ships, spillways, where gravity dominates), Reynolds similarity for fully-enclosed viscous flows (pipes, submarines), and Mach similarity for compressible/high-speed aerodynamic flows.
Froude and Reynolds similarity generally cannot both be satisfied in the same model using the same fluid — this is the fundamental model-scale problem in hydraulic engineering, and is why ship model tests (which must satisfy Froude similarity for wave-making) accept some Reynolds-number mismatch, correcting for it separately.
Number of π groups \( = n - m \)
Procedure: (1) list variables and dimensions; (2) choose \(m\) repeating variables spanning all dimensions; (3) form \((n-m)\) groups; (4) solve for exponents by dimensional homogeneity.
Reynolds: \( Re = \rho \dfrac{VL}{\mu} = \dfrac{VL}{\nu} \)
Froude: \( Fr = \dfrac{V}{\sqrt{gL}} \)
Weber: \( We = \rho \dfrac{V^2L}{\sigma} \)
Mach: \( Ma = \dfrac{V}{c} \)
\( Fr_p = Fr_m \)
\( V_r = \sqrt{L_r} \), \( T_r = \sqrt{L_r} \)
\( Q_r = L_r^{5/2} \), \( F_r = L_r^3 \)
Reynolds: \( Re_p = Re_m \) → \( V_r = \dfrac{\nu_r}{L_r} \); \( Q_r = \nu_r L_r \)
Mach: \( Ma_p=Ma_m \) → \( V_r = c_r = \sqrt{\dfrac{K_r}{\rho_r}} \)
Given: A phenomenon involves 6 variables and 3 fundamental dimensions (M, L, T). Find the number of dimensionless π groups.
Solution:
\( n-m = 6-3 = 3 \)
Answer: 3 independent dimensionless π groups.
Given: A spillway model is built at a length scale ratio \(L_r=25\) (prototype:model). Find the velocity and discharge scale ratios (Froude similarity).
Solution:
\( V_r = \sqrt{25} = 5 \)
\( Q_r = L_r^{5/2} = 25^{2.5} = 25^2\times\sqrt{25} = 625\times5 = 3125 \)
Answer: \(V_r=5\), \(Q_r=3125\) — a model discharge of 1 L/s corresponds to a prototype discharge of 3125 L/s.
Given: A submarine model at \(L_r=20\) is tested in the same fluid as the prototype (\(\nu_r=1\)). Find the required velocity scale ratio for Reynolds similarity.
Solution:
\( V_r = \dfrac{\nu_r}{L_r} = \dfrac{1}{20} = 0.05 \)
Answer: The model velocity must be \(\dfrac{1}{0.05}=20\) times the prototype velocity — illustrating why Reynolds-similarity models often require very high model velocities or a different, less viscous fluid.
Fig. 12.1 — Model-prototype similarity: geometric (shape scaled by L_r), kinematic (velocity pattern scaled), and dynamic (matching the dominant dimensionless force ratio — Froude, Reynolds, or Mach).
A rectangular sharp-crested weir gives discharge proportional to \(H^{3/2}\), derived by integrating the velocity (from Torricelli's theorem) over the flow depth above the crest. Real weirs deviate from this theoretical value through the discharge coefficient. When the weir does not span the full channel width, end contractions reduce the effective crest length (the Francis formula), and when the approach velocity is significant, a velocity-of-approach correction adds the upstream kinetic energy head to the effective head.
A V-notch gives discharge proportional to \(H^{5/2}\) — a steeper power than the rectangular weir's \(H^{3/2}\) — because both the head and the effective flow width increase together as the water rises in the V shape. This makes the V-notch far more sensitive at low flows (a small change in Q produces a larger, more measurable change in H) and self-cleaning (no stagnant zone at low flow, unlike a rectangular notch), which is why it is preferred for measuring small or highly variable discharges.
The Cipolletti weir uses side slopes of 1:4 (horizontal:vertical), chosen specifically so that the increase in flow area from the sloping sides exactly compensates for the reduction in discharge caused by end contractions — allowing the same simple formula as a suppressed (no end contraction) rectangular weir to be used directly, without a separate contraction correction.
A broad-crested weir is wide enough (relative to head) that the flow becomes critical on the crest itself, giving a discharge formula derived from critical-flow theory rather than the sharp-crested weir's orifice-like formula. An ogee (overflow spillway) profile is shaped to exactly follow the underside of the nappe from a sharp-crested weir at its design head, minimising negative pressure on the spillway face and making it the most hydraulically efficient shape for large dam spillways.
When the downstream (tailwater) level rises above the weir crest, the discharge is reduced compared to free (unsubmerged) flow — the Villemonte equation quantifies this reduction as a function of the submergence ratio (downstream head over upstream head). Only mild submergence (ratio below about 0.67) causes a small reduction; near-complete submergence causes a severe reduction in discharge capacity for the same upstream head.
Theoretical: \( Q_{th} = \tfrac23 L\sqrt{2g}\,H^{3/2} \)
Actual: \( Q = C_d\tfrac23 L\sqrt{2g}\,H^{3/2} \), \(C_d\approx0.611\text{–}0.62\)
Francis (end contractions, \(n\)=no. of contractions): \( Q = C_d[L-0.1nH]\tfrac23\sqrt{2g}\,H^{3/2} \)
Velocity of approach: \( H_{eff}=H+\dfrac{V_a^2}{2g} \)
\( Q = \tfrac{8}{15}C_d\tan(\dfrac{\theta}{2})\sqrt{2g}\,H^{5/2} \)
For 90° V-notch (\(C_d\approx0.62\)): \( Q\approx1.417\,H^{5/2} \)
Cipolletti (1:4 side slope): \( Q = \tfrac23C_dL\sqrt{2g}\,H^{3/2} \) (no contraction correction needed)
Broad-crested: \( Q \approx 1.705\,C_d\,L\,H^{3/2} \), \(C_d\approx0.848\)
Ogee spillway: \(C_d\approx0.74\) (at design head)
\( \dfrac{Q_{subm}}{Q_{free}} = \left[1-(H_2/H_1)^{1.5}\right]^{0.385} \)
\(S<0.67\) → <10% reduction; \(S>0.9\) → severe reduction.
Given: A suppressed rectangular weir, \(L=2\ \text{m}\), head \(H=0.4\ \text{m}\), \(C_d=0.62\). Find the discharge.
Solution:
\( Q = 0.62\times\tfrac23\times2\times\sqrt{2\times9.81}\times0.4^{1.5} \)
\( = 0.62\times1.333\times2\times4.429\times0.2530 \approx 1.851\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q\approx1.85\ \dfrac{\text{m}^3}{\text{s}}\).
Given: A 90° V-notch weir has head \(H=0.25\ \text{m}\). Find the discharge using the standard coefficient.
Solution:
\( Q = 1.417\times H^{2.5} = 1.417\times0.25^{2.5} = 1.417\times0.03125 \approx 0.0443\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q\approx44.3\ \text{L/s}\).
Given: A rectangular weir, \(L=1.5\ \text{m}\), \(H=0.3\ \text{m}\), with two end contractions (\(n=2\)), \(C_d=0.62\). Find the discharge.
Solution:
\( L_{eff} = 1.5-0.1\times2\times0.3 = 1.5-0.06=1.44\ \text{m} \)
\( Q = 0.62\times1.44\times\tfrac23\times\sqrt{2\times9.81}\times0.3^{1.5} \)
\( = 0.62\times1.44\times0.667\times4.429\times0.1643 \approx 0.4297\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q\approx0.43\ \dfrac{\text{m}^3}{\text{s}}\).
Fig. 13.1 — Rectangular weir (Q ∝ H^1.5) vs. V-notch weir (Q ∝ H^2.5): the V-notch's growing effective width with head makes it more sensitive at low flows.
A jet striking a stationary flat plate perpendicular to its path exerts a force but does no work (the plate doesn't move, so power = force × 0 = 0). A jet striking a series of curved vanes moving at velocity \(u\) does useful work, since momentum is exchanged between the fluid and the moving vane — this is the fundamental principle behind impulse turbines. Work done is maximised at a specific vane-to-jet velocity ratio, and the theoretical maximum efficiency approaches 100% only for a semicircular vane deflecting the jet through a full 180° with no friction.
A Pelton wheel is a pure impulse turbine: a high-velocity jet (accelerated through a nozzle from a high head) strikes a series of curved buckets on a wheel, converting the jet's kinetic energy into mechanical work at essentially atmospheric pressure throughout the runner. Maximum hydraulic efficiency occurs when the bucket speed is exactly half the jet velocity — beyond this point, the bucket "runs away" from the jet too fast for effective energy transfer; below it, the bucket doesn't move fast enough to fully utilise the jet's relative velocity.
Unlike the Pelton wheel, a Francis turbine is a reaction turbine — pressure drops continuously as water flows through the runner, not just in a nozzle beforehand. Work done is captured by the Euler turbine equation, relating the change in whirl (tangential) velocity component to the work extracted. A draft tube below the runner allows the turbine to be positioned above tailwater level (easing maintenance access) while still recovering the kinetic energy leaving the runner as useful pressure head, rather than wasting it.
A Kaplan turbine is an axial-flow reaction turbine suited to low head, high flow-rate sites (run-of-river hydropower); its adjustable blades (vs. the fixed blades of a simple propeller turbine) let it maintain high efficiency across a wide range of flow conditions, which is essential where river flow varies substantially with season.
| Turbine | Head Range | Specific Speed \(N_s\) | Type |
|---|---|---|---|
| Pelton | > 300 m | 4–70 | Impulse; 1–2 jets |
| Francis | 30–500 m | 60–300 | Reaction; radial/mixed inflow |
| Kaplan | < 50 m | 300–900 | Reaction; axial flow, adjustable blades |
Stationary flat plate: \( F = \rho aV^2 \)
Curved vane (moving series, steady state): \( F_x = \rho aV(V-u)(1-\cos\beta) \)
Work done/second: \( = F_x\times u = \rho aV(V-u)^2(1-\cos\beta) \)
Optimum: \( u=\dfrac{V}{2} \); Max. efficiency: \( \eta_{max}=\dfrac{1-\cos\beta}{2} \) (\(\beta=180°\Rightarrow\eta=1.0\) ideal)
Jet velocity: \( V_1 = C_v\sqrt{2gH} \), \(C_v\approx0.96\text{–}0.99\)
Work/kg: \( W = (V_1-u)u(1+k\cos\beta) \), \(k\approx0.85\text{–}0.95\), \(\beta\approx165°\text{–}170°\)
Hydraulic efficiency: \( \eta_h = \dfrac{2u(V_1-u)(1+k\cos\beta)}{V_1^2} \)
Max. at \(u=\dfrac{V_1}{2}\): \( \eta_{h,max}=\dfrac{1+k\cos\beta}{2}\approx90\text{–}95\% \)
Speed ratio: \( \phi = \dfrac{u}{\sqrt{2gH}} = 0.43\text{–}0.47 \)
Euler equation: \( W = (V_{w1}u_1-V_{w2}u_2)/g \)
Draft tube efficiency: \( \eta_{dt} = \dfrac{(V_3^2-V_4^2)/(2g)}{H_s+(V_3^2-V_4^2)/(2g)} \)
\(\eta_{max}\approx90\text{–}92\%\)
\( N_s = N\dfrac{\sqrt{P}}{H^{5/4}} \) (\(P\) in kW, \(N\) in rpm, \(H\) in m)
Given: A Pelton wheel has \(H=250\ \text{m}\), \(C_v=0.98\), bucket angle from the reversed jet direction \(\beta=15°\) (i.e. the jet is deflected through 165°), \(k=0.9\), running at optimum speed ratio. Find the maximum hydraulic efficiency.
Solution:
\( \eta_{h,max} = \dfrac{1+k\cos\beta}{2} = \dfrac{1+0.9\times\cos15°}{2} = \dfrac{1+0.9\times0.966}{2} = \dfrac{1+0.869}{2} \approx 0.935 \)
Answer: \(\eta_{h,max}\approx93.5\%\).
Given: A Pelton wheel operates under a net head of 400 m with \(C_v=0.98\). Find the jet velocity and the optimum bucket velocity.
Solution:
\( V_1 = 0.98\sqrt{2\times9.81\times400} = 0.98\times88.59 \approx 86.8\ \text{m/s} \)
\( u_{opt} = \dfrac{V_1}{2} \approx 43.4\ \text{m/s} \)
Answer: \(V_1\approx86.8\ \text{m/s}\), \(u_{opt}\approx43.4\ \text{m/s}\).
Given: A turbine develops 5000 kW at 300 rpm under a head of 400 m. Find \(N_s\) and identify the likely turbine type.
Solution:
\( N_s = \dfrac{300\times\sqrt{5000}}{400^{1.25}} = \dfrac{300\times70.71}{400^{1.25}} \)
\( 400^{1.25} = 400\times400^{0.25} \approx 400\times4.47 = 1788 \)
\( N_s = \dfrac{21{,}213}{1788} \approx 11.9 \)
Answer: \(N_s\approx12\) — falls in the Pelton range (4–70), consistent with the very high head of 400 m.
Fig. 14.1 — Jet striking a Pelton bucket: the split, double-curved bucket shape reverses the jet direction by nearly 180°, maximising momentum transfer; efficiency peaks when bucket speed u = V₁/2.
| Type | Mechanism | \(N_s\) | Application |
|---|---|---|---|
| Centrifugal (radial) | Centrifugal action via impeller; most common | 20–80 | High head, moderate flow; water supply |
| Mixed flow | Combines centrifugal + axial | 80–160 | Medium head and flow |
| Axial flow (propeller) | Axial lift; low head; high flow | 160–400 | Irrigation, drainage; very low head |
| Reciprocating (positive displacement) | Piston in cylinder; fixed volume per stroke | — | Dosing, very high pressure, viscous fluids |
| Gear / Rotary | Intermeshing gears or rotors; positive displacement | — | Lubrication, hydraulic systems |
Euler's pump equation (the mirror image of the turbine equation) gives the theoretical head developed by a centrifugal pump impeller in terms of the change in whirl velocity across it — for the common case of purely radial entry (no pre-swirl), the equation simplifies considerably. Net Positive Suction Head (NPSH) is the margin between the actual pressure at the pump inlet and the fluid's vapour pressure; if the available NPSH falls below the pump's required NPSH (a design characteristic), local pressure drops below vapour pressure somewhere inside the pump and cavitation occurs, which is why pump suction lift is practically limited to about 6–8 m for water.
Specific speed is a dimensionless (or semi-dimensionless, depending on unit convention) parameter that characterises the "shape" of a pump's performance and is the standard basis for pump classification. The affinity laws predict how a pump's performance changes with rotational speed (for the same pump) or with impeller diameter (for geometrically similar pumps): flow scales linearly, head scales with the square, and power scales with the cube — the cubic power relationship means even small speed increases substantially raise motor power requirements.
Pumps in series pass the same flow through each pump, with heads adding — used when a single pump cannot generate sufficient head. Pumps in parallel operate at the same head with flows adding — used when a single pump cannot deliver sufficient flow. The actual operating point of any pump-system combination is the intersection of the pump's head-flow characteristic curve with the system's resistance curve (static lift plus friction losses, which grow with the square of flow rate).
A reciprocating pump displaces a fixed volume per stroke (piston area × stroke length), giving theoretical discharge directly proportional to speed — a double-acting design (pumping on both the forward and return stroke) roughly doubles output for the same piston. Slip is the shortfall between theoretical and actual discharge (due to leakage past valves and the piston); an air vessel near the pump smooths the inherently pulsating flow of a reciprocating pump and reduces the acceleration head that would otherwise stress the suction/delivery pipework.
\( H_m = (V_{w2}u_2-V_{w1}u_1)/g \); radial entry (\(V_{w1}=0\)): \( H_m=V_{w2}u_2/g \)
\( NPSH_a = \dfrac{p_s}{\rho g}+\dfrac{V_s^2}{2g}-\dfrac{p_v}{\rho g} \)
Cavitation if \(NPSH_a Suction limit: \( H_s\le\dfrac{p_{atm}}{\rho g}-\dfrac{p_v}{\rho g}-NPSH_r-h_{fs} \); practical: \(H_s\le6\text{–}8\ \text{m}\)
\( N_s = N\sqrt Q/H^{3/4} \)
Same pump, varying speed: \( Q\propto N \); \( H\propto N^2 \); \( P\propto N^3 \)
Similar pumps, different size: \( Q\propto D^3 \); \( H\propto D^2 \); \( P\propto D^5 \)
Combined: \( Q\propto ND^3 \); \( H\propto N^2D^2 \); \( P\propto\rho N^3D^5 \)
Series: \( Q_{total}=Q_{each} \); \( H_{total}=\sum H_i \)
Parallel: \( H_{total}=H_{each} \); \( Q_{total}=\sum Q_i \)
System curve: \( H_{system}=H_{static}+kQ^2 \)
Single-acting: \( Q_{mean} = \dfrac{ALN}{60} \)
Double-acting: \( Q_{mean} = (2A-a)\dfrac{LN}{60} \)
Slip: \( S = Q_{th}-Q_{act} \); \( C_d = \dfrac{Q_{act}}{Q_{th}} = 1-\dfrac{S}{Q_{th}} \)
Given: A pump delivers \(Q=100\ \text{L/s}\), \(H=20\ \text{m}\), \(P=25\ \text{kW}\) at 1450 rpm. Find the performance at 1750 rpm.
Solution:
\( Q_2 = 100\times(\dfrac{1750}{1450}) = 120.7\ \text{L/s} \)
\( H_2 = 20\times(\dfrac{1750}{1450})^2 = 20\times1.456 = 29.1\ \text{m} \)
\( P_2 = 25\times(\dfrac{1750}{1450})^3 = 25\times1.757 = 43.9\ \text{kW} \)
Answer: \(Q_2\approx120.7\ \text{L/s}\), \(H_2\approx29.1\ \text{m}\), \(P_2\approx43.9\ \text{kW}\).
Given: A pump delivers \(Q=0.5\ \dfrac{\text{m}^3}{\text{s}}\) against \(H=15\ \text{m}\) at 750 rpm. Find the specific speed.
Solution:
\( N_s = \dfrac{750\times\sqrt{0.5}}{15^{0.75}} = \dfrac{750\times0.707}{7.62} = \dfrac{530.3}{7.62} \approx 69.6 \)
Answer: \(N_s\approx70\) — in the centrifugal (radial) pump range.
Given: A single-acting reciprocating pump, piston diameter 150 mm, stroke 300 mm, running at 60 rpm. Find the theoretical discharge.
Solution:
\( A = \dfrac{\pi}{4}\times0.15^2 = 0.01767\ \text{m}^2 \)
\( Q_{mean} = \dfrac{ALN}{60} = 0.01767\times0.3\dfrac{\times60}{60} = 0.00530\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q_{mean}\approx5.3\ \text{L/s}\).
Fig. 15.1 — Pump operating point: intersection of the pump's head-discharge characteristic curve with the system's resistance curve (static lift + friction losses).
Open channel flow (a free surface exposed to atmosphere) is governed by Chezy's and Manning's formulas, both relating mean velocity to the hydraulic radius (area over wetted perimeter) and the channel's energy slope. Manning's formula is by far the most widely used in practice, since Manning's roughness coefficient \(n\) is well-tabulated for a huge range of natural and constructed channel surfaces.
| Channel Type | Manning's n |
|---|---|
| Smooth concrete | 0.011–0.013 |
| Unfinished concrete | 0.014–0.017 |
| Brick lined | 0.013–0.016 |
| Clean earth channel | 0.020–0.025 |
| Grassed earth channel | 0.030–0.040 |
| Natural (clean, winding) | 0.025–0.040 |
| Natural (weedy, deep pools) | 0.050–0.100 |
| Flood plain (cultivated areas) | 0.030–0.040 |
For a given cross-sectional area, slope, and roughness, discharge is maximised by minimising the wetted perimeter (which maximises the hydraulic radius) — this defines the "most efficient" or "best hydraulic" section for each shape. Among all possible cross-sectional shapes, the semicircle is theoretically the most efficient of all; among practically constructible shapes, the best rectangle has width equal to twice the depth, and the best trapezoid has 60° side walls (half of an equilateral triangle).
Specific energy is the total mechanical energy measured relative to the channel bed (rather than an absolute horizontal datum) — the sum of flow depth and velocity head. For a given discharge, specific energy has a minimum value at a particular depth, the critical depth, at which the Froude number equals exactly 1. Below critical depth, flow is supercritical (fast, shallow, "shooting", controlled from upstream); above critical depth, flow is subcritical (slow, deep, "tranquil", controlled from downstream) — this upstream/downstream control distinction is fundamental to understanding how disturbances propagate in open channels.
A hydraulic jump is the abrupt, turbulent transition from supercritical to subcritical flow, dissipating substantial kinetic energy as turbulence and heat (rather than smoothly, as in a gradual transition) — it is deliberately used at the base of spillways and stilling basins to dissipate excess energy before water re-enters a natural channel, protecting the channel bed from erosion. The ratio of downstream to upstream depth (the sequent or conjugate depth ratio) depends only on the upstream Froude number, and jumps are classified by strength (and energy dissipation effectiveness) according to that Froude number.
When the channel bed slope, roughness, or cross-section changes gradually (rather than abruptly, as in a hydraulic jump), the flow depth adjusts gradually along the channel length according to the GVF equation, relating the rate of change of depth to the mismatch between the actual bed slope and the friction slope required to sustain uniform flow. Depending on how the actual depth compares to the normal depth (uniform-flow depth) and critical depth, twelve standard water-surface profile types are classified (M1-M3 for mild slopes, S1-S3 for steep slopes, and further categories for critical, horizontal and adverse slopes).
Trapezoidal channels (the most common constructed cross-section) and partially-filled circular pipes (sewers, culverts) both have closed-form expressions for area, wetted perimeter and hydraulic radius as functions of depth, allowing direct application of Manning's formula. For a circular section, maximum discharge and maximum velocity occur at slightly different partial-depth ratios (94% and 81% of full diameter respectively) — a subtlety in sewer design that is frequently tested.
Chezy: \( V = C\sqrt{RS} \)
Manning: \( V = \dfrac1n R^{2/3}S^{1/2} \); \( Q = \dfrac An R^{2/3}S^{1/2} \)
\( R=\dfrac{A}{P} \); relation: \( C = \dfrac{R^{1/6}}{n} \)
Rectangle: \( B=2y \), \( R=\dfrac{y}{2} \)
Trapezoid: \(\theta=60°\), side slope \(1:\sqrt3\), \( R=\dfrac{y}{2} \)
Circle: \( Q_{max} \) at \( d=0.94D \); \( V_{max} \) at \( d=0.81D \)
\( E = y+\dfrac{V^2}{2g} = y+\dfrac{Q^2}{2gA^2} \)
Critical (rectangular): \( y_c = (\dfrac{Q^2}{gB^2})^{1/3} \); \( V_c=\sqrt{gy_c} \); \( E_c=\tfrac32y_c \)
\( Fr = \dfrac{V}{\sqrt{gy}} \); \(Fr<1\) subcritical, \(Fr=1\) critical, \(Fr>1\) supercritical
Sequent depth: \( \dfrac{y_2}{y_1} = \tfrac12\left[-1+\sqrt{1+8Fr_1^2}\right] \)
Energy loss: \( \Delta E = (y_2-y_1)\dfrac{^3}{4y_1y_2} \)
Jump length: \( L_j\approx5\text{–}7\times y_2 \) (empirical)
Classification: \(Fr_1=1\text{–}1.7\) undular; 1.7–2.5 weak; 2.5–4.5 oscillating; 4.5–9.0 steady (best dissipation); >9.0 strong
\( \dfrac{dy}{dx} = \dfrac{S_0-S_f}{1-Fr^2} \), \( S_f = \dfrac{n^2V^2}{R^{4/3}} \)
Profile types: M1,M2,M3 (mild); S1,S2,S3 (steep); C1,C2 (critical); H1-3 (horizontal); A1,A2 (adverse)
Trapezoid: \( A=(B+my)y \); \( P=B+2y\sqrt{1+m^2} \); \( R=\dfrac{A}{P} \)
Circular (partial depth \(d\)): \(\theta=2\arccos(1-\dfrac{2d}{D})\); \( A=(\dfrac{D^2}{8})(\theta-\sin\theta) \); \( P=D\dfrac{\theta}{2} \)
Given: A rectangular channel, \(B=4\ \text{m}\), depth \(y=1.5\ \text{m}\), bed slope \(S=0.0004\), \(n=0.015\). Find the discharge.
Solution:
\( A = 4\times1.5=6\ \text{m}^2 \); \( P = 4+2\times1.5=7\ \text{m} \); \( R=\dfrac{6}{7}=0.857\ \text{m} \)
\( Q = \dfrac{6}{0.015}\times0.857^{2/3}\times0.0004^{1/2} = 400\times0.902\times0.02 \approx 7.22\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q\approx7.22\ \dfrac{\text{m}^3}{\text{s}}\).
Given: A rectangular channel, \(B=3\ \text{m}\), carries \(Q=9\ \dfrac{\text{m}^3}{\text{s}}\) at an upstream depth \(y_1=0.5\ \text{m}\) (supercritical). Find \(y_c\), \(Fr_1\), and the sequent depth \(y_2\).
Solution:
\( y_c = (\dfrac{9^2}{9.81\times3^2})^{1/3} = (\dfrac{81}{88.29})^{1/3} = (0.9174)^{1/3} \approx 0.972\ \text{m} \)
\( V_1 = \dfrac{Q}{By_1} = \dfrac{9}{3\times0.5} = 6\ \text{m/s} \); \( Fr_1 = \dfrac{6}{\sqrt{9.81\times0.5}} = \dfrac{6}{2.215} \approx 2.71 \)
\( \dfrac{y_2}{y_1} = \tfrac12[-1+\sqrt{1+8\times2.71^2}] = \tfrac12[-1+\sqrt{1+58.75}]=\tfrac12[-1+7.727]=3.36 \)
\( y_2 = 3.36\times0.5 = 1.68\ \text{m} \)
Answer: \(y_c\approx0.97\ \text{m}\), \(Fr_1\approx2.71\), \(y_2\approx1.68\ \text{m}\) (oscillating-type jump, since \(2.5
Given: Design the most efficient rectangular channel section to carry \(Q=5\ \dfrac{\text{m}^3}{\text{s}}\) at \(S=0.001\), \(n=0.014\).
Solution:
For the best rectangle: \(B=2y\), so \(A=2y^2\), \(P=4y\), \(R=\dfrac{A}{P}=\dfrac{y}{2}\)
\( Q = \dfrac{2y^2}{n}\left(\dfrac y2\right)^{2/3}\sqrt S = \dfrac{2^{1/3}}{n}\sqrt S\;y^{8/3} \)
\( 5 = \dfrac{2^{1/3}}{0.014}\times\sqrt{0.001}\times y^{8/3} = 2.846\,y^{8/3} \)
\( y^{8/3} = 1.757 \ \Rightarrow\ y = 1.757^{3/8} \approx 1.235\ \text{m} \)
Answer: \(y\approx1.24\ \text{m}\), \(B=2y\approx2.47\ \text{m}\).
Fig. 16.1 — (Left) Specific energy diagram: E_min at y_c, with two possible depths for the same E (alternate depths). (Right) Hydraulic jump: supercritical → subcritical, with sequent depth \(\dfrac{y_2}{y_1}=\tfrac12(\sqrt{1+8Fr_1^2}-1)\).
\( \tau=\mu\,\dfrac{du}{dy} \); \(\nu=\dfrac{\mu}{\rho}\); \(S_{Hg}=13.6\); \(\gamma_w=9.81\ \text{kN/m}^3\)
Droplet: \(4\dfrac{\sigma}{d}\); Bubble: \(8\dfrac{\sigma}{d}\); Jet: \(2\dfrac{\sigma}{d}\); Capillary rise: \(h=4\sigma\cos\theta/(\rho gd)\)
\( p=\rho gh \); \( F=\rho g\bar yA \); \(y_{cp}=\bar y+\dfrac{I_G}{\bar yA}\)
\(F_B=\rho_fgV_{sub}\); fraction submerged \(=\dfrac{\rho_b}{\rho_f}\); \(BM=\dfrac{I_{WL}}{V_{sub}}\); \(GM=BM-BG\)
\(A_1V_1=A_2V_2=Q\); \(u=\partial\dfrac{\psi}{\partial} y\), \(v=-\partial\dfrac{\psi}{\partial} x\)
Bernoulli: \(\dfrac{p}{\rho} g+\dfrac{V^2}{2g}+z=H\); \(Q=C_dA\sqrt{2gh}\), \(C_d=C_vC_c\)
\(Q=\pi D^4\Delta \dfrac{P}{128\mu L}\); \(u_{max}=2V_{mean}\); \(f=\dfrac{64}{Re}\) (laminar)
\(h_f=\dfrac{fLV^2}{2gD}\); \(h_m=\dfrac{KV^2}{2g}\); water hammer \(\Delta P=\rho c\Delta V\)
\(F_D=C_D\times\tfrac12\rho V^2A\); Stokes: \(F_D=3\pi\mu DV\), \(V_t=(\rho_s-\rho_f)\dfrac{gD^2}{18\mu}\)
Laminar BL: \(\delta=\dfrac{5x}{\sqrt{Re_x}}\); Turbulent BL: \(\delta=\dfrac{0.37x}{Re_x^{0.2}}\)
π groups \(=n-m\); \(Fr=\dfrac{V}{\sqrt{gL}}\); Froude: \(Q_r=L_r^{5/2}\)
Rectangle weir: \(Q=C_d\tfrac23L\sqrt{2g}H^{3/2}\); Manning: \(V=\tfrac1nR^{2/3}S^{1/2}\); jump: \(\dfrac{y_2}{y_1}=\tfrac12(\sqrt{1+8Fr_1^2}-1)\)
| Parameter | Value | Context |
|---|---|---|
| S_mercury | 13.6 | Specific gravity; manometry |
| C_d (orifice) | ≈ 0.61 | Sharp-edged; 0.96–0.99 for Venturi |
| C_v | ≈ 0.98 | Velocity coefficient |
| C_c | ≈ 0.64 | Contraction coefficient |
| Laminar Re | < 2000 | Pipe flow transition |
| Turbulent Re | > 4000 | Pipe flow |
| f (laminar) | 64/Re | Darcy friction factor |
| α (laminar / turbulent) | 2.0 / ~1.05 | Kinetic energy correction |
| Exit loss K | 1.0 | Pipe discharges to reservoir |
| Entry loss K | 0.5 | Sharp-edged entry |
| Re (BL transition) | 5 × 10⁵ | Flat plate |
| Pelton speed ratio φ | 0.43–0.47 | \(\dfrac{u}{\sqrt{2gH}}\) at best efficiency |
| C_d weir (sharp) | 0.62 | Rectangular and V-notch |
| E_c / y_c | 1.5 | Critical specific energy |
| Best rectangle | B = 2y | Most efficient open channel |
| Topic | GATE Focus | ESE Focus | SSC JE Focus |
|---|---|---|---|
| Fluid Properties | Newton's law; viscosity types; capillarity formulae; surface tension ΔP | Non-Newtonian fluids; Andrade equation; compressibility; cavitation number | Definitions; units; S_Hg=13.6; μ increases/decreases with T |
| Manometry | U-tube calculation; differential manometer p_A−p_B | Inverted U-tube; micro-manometer; inclined manometer amplification | Simple U-tube reading; manometer types; Pascal's law |
| Hydrostatics | F=ρgȳA; y_cp formula; curved surface F_H and F_V | Full curved surface analysis; dam overturning; lock gate forces | F=ρgȳA; CP below CG; dam pressure triangle |
| Buoyancy | BM=I_WL/V_sub; GM; experimental GM; fraction submerged | Stability conditions; metacentric height calc; layered fluids | Archimedes; fraction submerged; float vs sink |
| Kinematics | Continuity A₁V₁=A₂V₂; stream function; irrotationality | Vorticity; circulation; flow net; potential functions | Continuity concept; streamline definition |
| Bernoulli / Flow Measurement | Venturimeter Q; C_d values; Pitot tube; momentum on bend | Modified Bernoulli; Euler derivation; impact force on plate | Bernoulli equation; Torricelli \(V=\sqrt{2gh}\) |
| Pipe Flow | Darcy-Weisbach; f=64/Re; minor losses; series/parallel | Hardy-Cross; water hammer; Moody diagram; Colebrook-White | Darcy-Weisbach formula; minor loss names |
| Boundary Layer | BL thickness formulae; C_D laminar/turbulent; transition Re | Von Kármán integral; separation; Blasius solution | BL concept; laminar vs turbulent BL |
| Dim. Analysis | π theorem (n−m); Froude Q_r=L_r^5/2; Re similarity | Full model scale law derivations | Dimensionless number names/meanings |
| Weirs | Rectangle and V-notch Q formulae; Francis correction | Broad-crested weir; ogee spillway; Villemonte submerged | Rectangular weir Q; V-notch 90°; C_d≈0.62 |
| Turbines | Pelton speed ratio u=V/2; Euler equation; N_s ranges | Full turbine design; velocity triangles; cavitation | Impulse vs reaction; N_s comparison |
| Pumps | Affinity laws; series vs parallel; NPSH | Pump characteristics; system curve; reciprocating pump | Centrifugal pump concept; affinity basics |
| Open Channel | Manning; y_c; hydraulic jump; E_c=1.5y_c | GVF profiles; best hydraulic section | Manning's formula; critical flow concept |
1. A capillary tube of 0.5 mm diameter is dipped in mercury (σ=0.51 N/m, θ=140°). Find the capillary depression.
2. A vertical circular gate of diameter 1.5 m has its centre 2 m below the water surface. Find the total hydrostatic force.
3. Water flows through a 150 mm diameter pipe at Re=8000. Find the Darcy friction factor using the Blasius formula.
4. A Pelton wheel operates under H=300 m with Cv=0.98. Find the jet velocity and optimum bucket speed.
5. A rectangular channel 5 m wide carries 12 m³/s. Find the critical depth.