Fluid Mechanics and Hydraulics

Syllabus

• Fluid mechanics: fluid properties, fluid statics, continuity, momentum and energy equations and their applications, potential flow, laminar and turbulent flow, flow in pipes and pipe networks, boundary layers, dimensional analysis and similitude.
• Hydraulics: forces on immersed bodies, flow measurement, open channel flow (specific energy, critical flow, hydraulic jump, uniform flow, gradually varied flow), channel design.

Key concepts and formulas

Hydrostatics. p = ρgh. The force on a plane surface is F = ρgAh̄ (h̄ = depth of the centroid), acting at the centre of pressure h_cp = h̄ + I_G/(Ah̄) for a vertical surface.

Bernoulli's equation along a streamline (steady, incompressible, frictionless): p/(ρg) + V²/(2g) + z = constant. Continuity: Q = A₁V₁ = A₂V₂.

Pipe flow.

• Reynolds number Re = ρVD/μ = VD/ν. Flow is laminar for Re < 2000 in a pipe.
• Darcy-Weisbach head loss: h_f = fLV²/(2gD). For laminar flow, f = 64/Re.
• Hagen-Poiseuille (laminar): the velocity profile is parabolic and V_max = 2V_avg.
• Pipes in series: the same Q and added head losses. Pipes in parallel: the same head loss and added Q.
• Power of a jet or pump: P = ρgQH.

Dimensional analysis. Buckingham π theorem: n variables with m fundamental dimensions give n − m dimensionless groups. Froude similarity (open channels, spillways) gives a velocity scale √L_r, a discharge scale L_r^5/2 and a time scale √L_r.

Open channel flow.

• Manning's equation: V = (1/n) R^2/3 S^1/2, with hydraulic radius R = A/P.
• Froude number Fr = V/√(gD) (D = hydraulic depth); Fr < 1 subcritical, = 1 critical, > 1 supercritical.
• Rectangular channel, discharge per unit width q: critical depth y_c = (q²/g)^1/3, minimum specific energy E_min = 1.5y_c.
• Hydraulic jump in a rectangular channel: y₂/y₁ = ½(−1 + √(1 + 8Fr₁²)), with energy loss ΔE = (y₂ − y₁)³/(4y₁y₂).
• Most economical rectangular section: width = 2 × depth (R = y/2); trapezoidal: half a hexagon.

Worked example

A rectangular channel 4 m wide carries 12 m³/s. Find the critical depth and minimum specific energy. A hydraulic jump forms where the depth is 0.4 m: find the sequent depth and the energy loss.

import math

g, Q, b = 9.81, 12.0, 4.0
q = Q / b
yc = (q * q / g) ** (1 / 3)
print(f"q = {q} m2/s, yc = {yc:.4f} m, Emin = {1.5 * yc:.4f} m")
y1 = 0.4
Fr1 = (q / y1) / math.sqrt(g * y1)
y2 = 0.5 * y1 * (-1 + math.sqrt(1 + 8 * Fr1 ** 2))
loss = (y2 - y1) ** 3 / (4 * y1 * y2)
print(f"Fr1 = {Fr1:.3f}, y2 = {y2:.4f} m, energy loss = {loss:.4f} m")

# Laminar pipe flow check and head loss
V, D, nu, L = 0.01, 0.1, 1e-6, 100
Re = V * D / nu
f = 64 / Re
print(f"Re = {Re:.0f} ({'laminar' if Re < 2000 else 'not laminar'}), f = {f:.4f}, hf = {f * L * V * V / (2 * g * D) * 1000:.4f} mm")

q = 3.0 m2/s, yc = 0.9717 m, Emin = 1.4575 m
Fr1 = 3.786, y2 = 1.9511 m, energy loss = 1.1954 m
Re = 1000 (laminar), f = 0.0640, hf = 0.3262 mm

Practice questions