Environmental Hydraulics: Fluid Statics, Flow in Pipes, Flow Measurement and Channel Hydraulics
1. Fluid properties, pressure, hydrostatic force, buoyancy and flotation
A fluid deforms continuously under any shear stress, however small. Its properties are density ρ (water about 1000 kg/m³), specific weight γ = ρg (9.81 kN/m³ for water), dynamic viscosity μ, which relates shear stress to the velocity gradient in Newton’s law τ = μ du/dy (water about 1.0 × 10⁻³ Pa·s at 20 °C, falling as it warms), kinematic viscosity ν = μ/ρ, compressibility (bulk modulus), vapour pressure, and surface tension σ (about 0.073 N/m for water in air), which gives the capillary rise h = 4σ cos θ/(ρgd) in a tube of diameter d. The concepts of mechanics carried in from statics and dynamics are the free-body diagram, Newton’s second law, and the conservation of mass, momentum and energy — the three that become the continuity, momentum and energy equations of fluid flow.
In a fluid at rest pressure increases linearly with depth, p = p₀ + ρgh, and acts equally in all directions (Pascal’s law). Pressure is measured as absolute or as gauge (relative to the atmosphere) by a piezometer, a U-tube manometer (for a manometric liquid of specific gravity S_m reading a deflection x against water, the head difference is x(S_m − 1)), a differential or inclined manometer, or a Bourdon gauge. The hydrostatic force on a plane surface is the pressure at its centroid times its area, F = ρg h̄ A, and it acts at the centre of pressure, below the centroid by I_G/(A h̄): h_cp = h̄ + I_G/(A h̄). For a vertical rectangle of depth d with its top at the free surface, h_cp = 2d/3. On a curved surface the horizontal component is the force on its vertical projection and the vertical component is the weight of fluid above it.
Buoyancy is Archimedes’ principle: a body immersed in a fluid is pushed up by a force equal to the weight of the fluid it displaces, acting through the centre of buoyancy B, the centroid of the displaced volume. A floating body is in equilibrium when that force equals its weight. Its stability against rolling depends on the metacentre M: for a small heel the buoyancy acts through M, and the body is stable if M lies above the centre of gravity G. The metacentric height is GM = BM − BG, with BM = I/V, where I is the second moment of the waterline area about the tilting axis and V the displaced volume. For a rectangular pontoon of width b and draft d, BM = b²/(12d).
2. Kinematics of flow; the continuity, momentum and energy equations
Kinematics describes the motion without its causes. Flow is steady if conditions at a point do not change with time and uniform if they do not change along a streamline. A streamline is everywhere tangent to the velocity; a pathline is the track of one particle; a streakline joins all particles that passed one point; in steady flow the three coincide. The acceleration of a particle has a local part ∂V/∂t (zero in steady flow) and a convective part V ∂V/∂s (zero in uniform flow). A flow is irrotational if its vorticity, the curl of velocity, is zero; then a velocity potential φ exists with V = ∇φ. In two-dimensional incompressible flow a stream function ψ exists with u = ∂ψ/∂y, v = −∂ψ/∂x, and the difference of ψ between two streamlines is the discharge between them.
The continuity equation conserves mass: for an incompressible fluid in a pipe Q = A₁V₁ = A₂V₂, and in differential form ∂u/∂x + ∂v/∂y + ∂w/∂z = 0. The energy equation along a streamline for steady, incompressible, frictionless flow is Bernoulli’s equation, p/(ρg) + V²/(2g) + z = constant — pressure head, velocity head and datum head; with friction and machines it becomes p₁/ρg + V₁²/2g + z₁ + h_pump = p₂/ρg + V₂²/2g + z₂ + h_L. The momentum equation is Newton’s second law for a control volume: the net force equals the rate of momentum outflow minus inflow, ΣF = ρQ(V₂ − V₁), resolved in each direction. It gives the force on a pipe bend or a nozzle, the thrust of a jet on a plate, and — because it needs no knowledge of the energy lost — the hydraulic jump.
3. Laminar and turbulent flow, pipes, pipe networks and forces on immersed bodies
The Reynolds number Re = ρVD/μ = VD/ν compares inertia with viscosity. In a pipe, flow is laminar below about Re = 2000, turbulent above about 4000, and transitional between. Laminar pipe flow has a parabolic velocity profile with the maximum velocity twice the mean, and the Hagen–Poiseuille result for its head loss, h_f = 32μVL/(ρgD²). The general head-loss law for any regime is Darcy–Weisbach, h_f = fLV²/(2gD), where the friction factor is f = 64/Re in laminar flow and depends on Re and the relative roughness ε/D in turbulent flow (the Moody diagram; the Colebrook equation). Minor losses at entrances, bends, valves and expansions are written kV²/(2g); a sudden expansion loses (V₁ − V₂)²/(2g), and a pipe discharging into a tank loses its whole velocity head. The Hazen–Williams formula is the empirical alternative used for water mains.
Pipes in series carry the same flow and their head losses add; pipes in parallel share the same head loss and their flows add. For parallel pipes of equal length and f, h_f ∝ Q²/D⁵, so Q ∝ D^2.5: doubling the diameter multiplies the flow by 2^2.5 = 5.66. A pipe network — the looped grid of a distribution system — must satisfy continuity at every node and zero net head loss round every loop. The Hardy Cross method solves it iteratively: assume flows that satisfy continuity, compute the loop correction ΔQ = −Σh_f/(nΣ|h_f/Q|) (n = 2 for Darcy–Weisbach, 1.85 for Hazen–Williams), apply it to every pipe of the loop, and repeat until the corrections are small. A body in a flowing fluid experiences drag parallel to the flow and lift normal to it, F_D = C_D A ρV²/2 and F_L = C_L A ρV²/2. Drag is part skin friction and part pressure (form) drag; for a small sphere at low Re Stokes’ law gives F_D = 3πμVd, i.e. C_D = 24/Re — the law behind the settling of particles in Section 5.
| Quantity | Laminar (Re < about 2000) | Turbulent (Re > about 4000) |
|---|---|---|
| Velocity profile | Parabolic; V_max = 2V̄ | Flatter; V_max ≈ 1.2V̄ |
| Friction factor | f = 64/Re, independent of roughness | f(Re, ε/D); constant in fully rough flow |
| Head loss varies as | V (h_f ∝ V) | About V² (h_f ∝ V^1.75 to V²) |
4. Flow measurement in pipes and channels
In pipes, the venturimeter, orifice meter and flow nozzle all apply continuity and Bernoulli between a full section and a constriction and read the pressure drop: Q = C_d A₁A₂√(2gh)/√(A₁² − A₂²), where h is the difference in piezometric head. The venturimeter’s gradual diverging cone recovers most of the pressure, so C_d is about 0.95–0.99; the orifice plate is cheap but loses much more, with C_d about 0.6. The Pitot tube reads the stagnation head at a point, and V = √(2gΔh) gives the local velocity; electromagnetic and ultrasonic meters are used where no loss can be tolerated. In open channels discharge is measured by weirs and notches: a rectangular weir of length L gives Q = (2/3)C_d L √(2g) H^(3/2), and a triangular (V-) notch of angle θ gives Q = (8/15)C_d √(2g) tan(θ/2) H^(5/2) — more sensitive at low flows, which is why V-notches meter small effluent streams. The Parshall flume forces critical flow in a throat and gives Q from a single upstream depth, with little head loss and no pond for solids to settle in, which suits raw sewage; the current meter and area–velocity method gauge rivers.
5. Channel hydraulics: specific energy, critical flow, hydraulic jump, varied flow and lined channels
Uniform flow in an open channel, where the depth is constant and gravity balances friction, is described by Manning’s equation, V = (1/n)R^(2/3)S^(1/2), with R = A/P the hydraulic radius, S the bed slope and n the roughness (about 0.013–0.015 for concrete). The specific energy is the energy above the bed, E = y + V²/(2g) = y + q²/(2gy²) for a rectangular channel carrying q per unit width. For a given q, E has a minimum at the critical depth y_c = (q²/g)^(1/3), where E_min = 1.5y_c and the Froude number Fr = V/√(gy) = 1. Deeper, slower flow (Fr < 1) is subcritical; shallower, faster flow (Fr > 1) is supercritical or rapid. Two depths — alternate depths — share each E above the minimum, which is why a small hump in the bed can lower a subcritical surface and raise a supercritical one.
A hydraulic jump is the abrupt rise from supercritical to subcritical flow, as below a spillway or sluice gate, and it dissipates energy in intense turbulence — which is also why it is used as a rapid mixer for coagulants. The momentum equation gives the sequent depth, y₂ = (y₁/2)(√(1 + 8Fr₁²) − 1), and the energy lost is ΔE = (y₂ − y₁)³/(4y₁y₂). Rapidly varied flow — jumps, falls, flow over weirs and under gates — changes depth over a short distance; gradually varied flow changes it slowly under the balance of friction and slope, dy/dx = (S₀ − S_f)/(1 − Fr²), and its water-surface profiles (backwater behind a weir is the M1 curve on a mild slope) are classified by the slope (mild, steep, critical, horizontal, adverse) and by where the depth lies relative to the normal and critical depths. Lined channels are designed for a non-erodible boundary: the section is chosen for the most economical, or best hydraulic, section — the one of least wetted perimeter for a given area, which is half a hexagon for a trapezium and, for a rectangle, b = 2y with R = y/2 — and then checked for a velocity high enough not to deposit silt and a freeboard above the design depth.
| Regime | Froude number | Depth relative to y_c | Controlled from |
|---|---|---|---|
| Subcritical (tranquil) | Fr < 1 | y > y_c | Downstream |
| Critical | Fr = 1 | y = y_c; E is a minimum | The section itself (a control) |
| Supercritical (rapid) | Fr > 1 | y < y_c | Upstream |
Key takeaways
- τ = μ du/dy; capillary rise 4σ cos θ/(ρgd); F = ρg h̄ A at h_cp = h̄ + I_G/(A h̄); a floating body is stable if GM = I/V − BG > 0.
- Continuity Q = AV; Bernoulli p/ρg + V²/2g + z = constant with losses added; momentum ΣF = ρQ(V₂ − V₁), which works where energy is dissipated.
- Re < 2000 laminar, f = 64/Re; h_f = fLV²/(2gD); series pipes add losses, parallel pipes add flows with Q ∝ D^2.5; Hardy Cross balances loops; drag F_D = C_D A ρV²/2.
- Venturi Q = C_d A₁A₂√(2gh)/√(A₁² − A₂²); rectangular weir ∝ H^1.5, V-notch Q = (8/15)C_d√(2g) tan(θ/2) H^2.5; the Parshall flume suits raw sewage.
- Manning V = (1/n)R^(2/3)S^(1/2); y_c = (q²/g)^(1/3), E_min = 1.5y_c; jump y₂ = (y₁/2)(√(1 + 8Fr₁²) − 1), loss (y₂ − y₁)³/(4y₁y₂); best rectangle b = 2y.
Practice questions (14)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Water rises in a clean glass tube of 1 mm internal diameter. Taking σ = 0.073 N/m, a contact angle of zero, ρ = 1000 kg/m³ and g = 9.81 m/s², what is the capillary rise, in mm? Give the answer to one decimal place.
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Answer: 29.8
h = 4σ cos θ/(ρgd) = 4 × 0.073 × 1/(1000 × 9.81 × 0.001) = 0.292/9.81 = 0.0298 m = 29.8 mm. Using the radius in place of the diameter in this form gives 59.5 mm, twice the true rise.A vertical rectangular gate 2 m wide and 3 m high retains water, with its top edge at the water surface. What is the depth of the centre of pressure below the surface, in m?
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Answer: 2
h̄ = 1.5 m, A = 6 m², I_G = 2 × 3³/12 = 4.5 m⁴. h_cp = h̄ + I_G/(A h̄) = 1.5 + 4.5/9 = 2.0 m, which is 2d/3. The force is ρg h̄ A = 9.81 × 1.5 × 6 = 88.3 kN; placing it at the centroid, 1.5 m, is the usual slip.A rectangular pontoon 0.5 m wide floats with a draft of 0.28 m, and its centre of gravity is 0.20 m above its base. What is its metacentric height for rolling about its long axis, in mm? Give the answer to one decimal place.
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Answer: 14.4
BM = I/V = (Lb³/12)/(Lbd) = b²/(12d) = 0.25/3.36 = 0.0744 m. B is at d/2 = 0.14 m above the base, so BG = 0.20 − 0.14 = 0.06 m. GM = 0.0744 − 0.06 = 0.0144 m = 14.4 mm, positive, so the pontoon is stable. Measuring BG from the waterline instead of from B is the error to avoid.Which of the following statements about fluid kinematics and the flow equations are correct?
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Answer: A — In steady flow, streamlines, pathlines and streaklines coincide; B — A velocity potential exists only for irrotational flow; D — Convective acceleration can be non-zero in steady flow
(a) holds because the velocity field does not change with time. (b) V = ∇φ implies zero curl. (d) steady flow through a contraction accelerates the fluid in space even though nothing changes in time. (c) is false: a jump dissipates energy in turbulence, so it is analysed by momentum, with the loss found afterwards.Water flows at 1.5 m/s through a 300 mm pipe 1000 m long with a Darcy friction factor of 0.02. What is the head loss due to friction, in m? Give the answer to two decimal places. (g = 9.81 m/s²)
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Answer: 7.65
h_f = fLV²/(2gD) = 0.02 × 1000 × 1.5²/(2 × 9.81 × 0.3) = 45/5.886 = 7.65 m. Using the radius 0.15 m for D doubles it to 15.3 m.Two pipes of the same length and friction factor are laid in parallel between two reservoirs; one has twice the diameter of the other. What is the ratio of the flow in the larger pipe to that in the smaller? Give the answer to two decimal places.
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Answer: 5.66
Both see the same head loss h_f = fLV²/(2gD), so V ∝ √D and Q = (π/4)D²V ∝ D^2.5. The ratio is 2^2.5 = 5.66. Answers of 2 (proportional to D) or 4 (to area at equal velocity) forget that the larger pipe also flows faster.In the Hardy Cross analysis of a looped pipe network, which condition is enforced by the loop flow correction?
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Answer: A — The algebraic sum of head losses round each loop is zero
The initial flows are chosen to satisfy continuity at the nodes (option B), and adding the same ΔQ round a loop preserves it. The correction ΔQ = −Σh_f/(nΣ|h_f/Q|) is then iterated until the head losses round each loop sum to zero — the energy condition that makes the pressure at each node single-valued.Oil of kinematic viscosity 1 × 10⁻⁴ m²/s flows at 0.5 m/s in a 100 mm pipe. The flow is:
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Answer: A — laminar, with f = 0.128
Re = VD/ν = 0.5 × 0.1/10⁻⁴ = 500, well below 2000, so the flow is laminar and f = 64/Re = 64/500 = 0.128. Option D uses 16/Re, the Fanning factor, which is a quarter of the Darcy factor used in h_f = fLV²/(2gD).A venturimeter with a 300 mm inlet and a 150 mm throat is fitted in a water main. The difference in piezometric head between inlet and throat is 0.5 m of water and C_d = 0.98. What is the discharge, in m³/s? Give the answer to three decimal places. (g = 9.81 m/s²)
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Answer: 0.056
A₁ = π(0.3)²/4 = 0.07069 m², A₂ = π(0.15)²/4 = 0.01767 m². Q = C_d A₁A₂√(2gh)/√(A₁² − A₂²) = 0.98 × 0.001249 × 3.132/0.06844 = 0.056 m³/s (56 L/s). Omitting the approach term — Q = C_d A₂√(2gh) — gives 0.0542, which is the throat as if it drew from a reservoir.A 90° V-notch with C_d = 0.6 measures an effluent stream. What is the discharge when the head over the notch is 0.3 m, in m³/s? Give the answer to three decimal places. (g = 9.81 m/s²)
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Answer: 0.07
Q = (8/15)C_d√(2g) tan(θ/2) H^(5/2) = 0.5333 × 0.6 × 4.429 × tan 45° × 0.3^2.5 = 1.4174 × 0.04930 = 0.0699 ≈ 0.070 m³/s. Using the rectangular-weir exponent 3/2 gives a very different number; the V-notch’s H^2.5 is what makes it sensitive at low flows.A concrete-lined rectangular channel 4 m wide carries water at a uniform depth of 1 m on a slope of 0.001, with Manning’s n = 0.015. What is the discharge, in m³/s? Give the answer to two decimal places.
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Answer: 6.44
A = 4 × 1 = 4 m², P = 4 + 2 × 1 = 6 m, R = 0.667 m. V = (1/0.015) × 0.667^(2/3) × 0.001^(1/2) = 66.67 × 0.7631 × 0.03162 = 1.609 m/s, and Q = AV = 6.44 m³/s. Taking R as the depth (1 m) overestimates V by 31%.A wide rectangular channel carries 3 m³/s per metre of width. What is the critical depth, in m? Give the answer to two decimal places. (g = 9.81 m/s²)
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Answer: 0.97
y_c = (q²/g)^(1/3) = (9/9.81)^(1/3) = 0.9174^(1/3) = 0.97 m. The minimum specific energy is 1.5y_c = 1.46 m. Forgetting to square q gives (3/9.81)^(1/3) = 0.67 m.Supercritical flow at a depth of 0.5 m and a velocity of 7 m/s in a rectangular channel forms a hydraulic jump. What is the head lost in the jump, in m? Give the answer to two decimal places. (g = 9.81 m/s²)
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Answer: 0.84
Fr₁ = 7/√(9.81 × 0.5) = 3.161. y₂ = (0.5/2)(√(1 + 8 × 3.161²) − 1) = 0.25 × (8.995 − 1) = 1.999 m. ΔE = (y₂ − y₁)³/(4y₁y₂) = 1.499³/(4 × 0.5 × 1.999) = 3.368/3.998 = 0.84 m. Check by energy: E₁ = 0.5 + 49/19.62 = 2.997 m and E₂ = 1.999 + (3.5/1.999)²/19.62 = 2.155 m, a difference of 0.84 m.Which of the following statements about open-channel flow are correct?
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Answer: A — At critical depth the specific energy for a given discharge is a minimum; B — The best hydraulic rectangular section has a width equal to twice the depth; D — The backwater curve upstream of a weir on a mild slope is an M1 profile
(a) dE/dy = 0 gives Fr = 1. (b) minimising P = b + 2y for fixed A = by gives b = 2y and R = y/2. (d) on a mild slope the depth behind a weir exceeds the normal depth, which is zone 1 — the M1 curve. (c) is false: subcritical flow is controlled from downstream, because disturbances can travel upstream against it; it is supercritical flow that is controlled from upstream.