Hydraulic Engineering: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Hydrostatic force

Pressure at depth: p = ρgh
Force on a vertical wall: F = ρg·h_c·A
Centre of pressure: y_p = y_c + I_c/(y_cA)
Rectangle: I_c = bH³/12

Continuity & Bernoulli

Continuity: A₁v₁ = A₂v₂
Bernoulli: p₁ + ½ρv₁² + ρgz₁ = p₂ + ½ρv₂² + ρgz₂
Flow rate: Q = A·v
Circle area: A = πd²/4

Pipe friction & head loss

Reynolds number: Re = V·D/ν = ρ·V·D/μ
Flow regime: laminar below Re ≈ 2000, turbulent above Re ≈ 4000
Mean velocity: V = Q/A, A = π·D²/4
Darcy–Weisbach: h_f = f·(L/D)·V²/(2g)
Minor losses: h_m = ΣK·V²/(2g)
Typical K values: entrance 0.5, exit 1.0, 90° bend ≈ 0.3
Pressure drop: Δp = ρ·g·h_f
Two open reservoirs: Δz = (f·L/D + ΣK)·V²/(2g)
Free outlet: H = (1 + f·L/D + ΣK)·V²/(2g)

Pumps & pump power

Hydraulic power: P_h = ρ·g·Q·H = Q·Δp
Shaft power: P_shaft = P_h/η_pump
Electrical power: P_el = P_shaft/η_motor
Pump head: H = Δz + Δp/(ρ·g) + h_L
System losses: h_L = (f·L/D + ΣK)·V²/(2g)

Open-channel flow (Manning)

Manning equation: V = (1/n)·R^2/3·S^1/2
Flow rate: Q = A·V
Hydraulic radius: R = A/P (P = wetted perimeter, free surface excluded)
Rectangle: A = b·y, P = b + 2y
Trapezoid (side slope z:1): A = (b + z·y)·y, P = b + 2y·√(1 + z²)
Circular pipe, full or half full: R = D/4
Required slope: S = (n·Q/(A·R^2/3))²

2Practice problems

1.Hydrostatic forceMedium

A vertical rectangular gate 1.5 m wide sits in the wall of a tank of water, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 0.5 mH = 1.5 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 1 m wide sits in the wall of a tank of water, with its top edge 1.5 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 1.5 mH = 2.5 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

A pitot-static tube in a water pipe reads a difference of 20 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.

p₀ − p = 20 kPaρ = 1000 kg/m³

v = m/s

4.Continuity & BernoulliHard

A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 50 mm and a gauge pressure of 130 kPa. The nozzle exit diameter is 25 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².

d₁ = 50 mmp₁ = 130 kPa (gauge)d₂ = 25 mmρ = 1000 kg/m³g = 9.81 m/s²

H = m

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 0.9 m/s through a horizontal pipe, 400 m long, with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 0.9 m/sL = 400 mD = 250 mmf = 0.022ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.02. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 10) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 12 L/s?

Q = 12 L/sL = 100 mD = 100 mmf = 0.02K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 10K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 12 m. The pump efficiency is 75% and the electric motor driving it has an efficiency of 90%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 50 L/sH = 12 mη_pump = 75%η_motor = 90%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 80 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 22 m higher. The pipe is 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.

Q = 80 L/sΔz = 22 mL = 150 mD = 200 mmf = 0.02ΣK = 6

H = m

9.Open-channel flow (Manning)Medium

A concrete storm sewer with an inside diameter of 600 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.004. Find the flow rate Q in m³/s.

D = 600 mmn = 0.013S = 0.004

Q = m³/s

10.Open-channel flow (Manning)Hard

A rectangular brickwork channel 1.5 m wide must carry 7.9 m³/s in uniform flow at a depth of 1.9 m. Manning n = 0.015. Find the required bed slope S (in m/m).

Q = 7.9 m³/sb = 1.5 my = 1.9 mn = 0.015

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 27.59 kN

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 0.5 + 1.5/2 = 1.25 m

A = H·b = 1.5·1.5 = 2.25 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·1.25·2.25 = 27.59 kN

Answer

F = 27.59 kN

2.Hydrostatic force

Answer: y_cp = 2.939 m

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 1.5 + 2.5/2 = 2.75 m

A = H·b = 2.5·1 = 2.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2.75·2.5 = 67.44 kN

Centre of pressure

Pressure grows with depth, so the resultant acts below the centroid.

I_c = b·H³/12 = 1·2.5³/12 = 1.302 m⁴

y_cp = y_c + I_c/(y_c·A) = 2.75 + 1.302/(2.75·2.5) = 2.939 m

Answer

y_cp = 2.939 m

3.Continuity & Bernoulli

Answer: v = 6.325 m/s

Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.

Bernoulli to the stagnation point

The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.

p₀ − p = ½ρv²

v = √(2·20000/1000) = 6.325 m/s

Answer

v = 6.325 m/s

4.Continuity & Bernoulli

Answer: H = 14.14 m

Exit velocity

β = 25/50 = 0.5

v₂ = √(2p₁/(ρ(1 − β⁴))) = √(2·130000/(1000·0.9375)) = 16.65 m/s

Height of the jet

Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.

H = v₂²/(2g) = 16.65²/(2·9.81) = 14.14 m

Answer

H = 14.14 m

5.Pipe friction & head loss

Answer: Δp = 14.26 kPa

Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss h_f = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses h_m = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = h_f + h_m, with the exit loss K = 1 included.

Head loss

h_f = f·(L/D)·V²/(2g) = 0.022·(400/0.25)·0.9²/(2·9.81) = 1.453 m

Pressure drop

The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·h_f.

Δp = ρ·g·h_f = 1000·9.81·1.453 = 14256 Pa = 14.26 kPa

Answer

Δp = 14.26 kPa

6.Pipe friction & head loss

Answer: Δz = 3.855 m

Velocity

V = Q/A = 0.012/0.007854 = 1.528 m/s

V²/(2g) = 0.119 m

Loss coefficients

f·L/D = 0.02·100/0.1 = 20

ΣK = 0.5 + 3·0.3 + 10 + 1.0 = 12.4

Total head loss

h_f = 20·0.119 = 2.38 m

h_m = 12.4·0.119 = 1.475 m

Δz = h_f + h_m = 3.855 m

Answer

Δz = 3.855 m

7.Pumps & pump power

Answer: P_el = 8.72 kW

Pump power: the power given to the water is the hydraulic power P_h = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: P_shaft = P_h/η_pump, and the motor draws P_el = P_shaft/η_motor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.

Hydraulic power

P_h = ρ·g·Q·H = 1000·9.81·0.05·12 = 5.886 kW

Shaft power

P_shaft = 5.886/0.75 = 7.848 kW

Electrical power

P_el = 7.848/0.9 = 8.72 kW

Answer

P_el = 8.72 kW

8.Pumps & pump power

Answer: H = 28.94 m

Velocity

V = Q/A = 0.08/0.03142 = 2.546 m/s

V²/(2g) = 0.3305 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (15 + 6)·0.3305 = 6.941 m

Pump head

Both surfaces are open and practically at rest, so pressure and velocity terms drop out.

H = Δz + h_L = 22 + 6.941 = 28.94 m

Answer

H = 28.94 m

9.Open-channel flow (Manning)

Answer: Q = 0.3883 m³/s

Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R^2/3·S^1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.

Geometry

A = 0.2827 m², R = D/4 = 0.15 m

Velocity

V = (1/0.013)·0.15^2/3·0.004^1/2 = 1.373 m/s

Flow rate

Q = A·V = 0.2827·1.373 = 0.3883 m³/s

Answer

Q = 0.3883 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.003954 m/m

Geometry

A = b·y = 1.5·1.9 = 2.85 m²

P = b + 2y = 1.5 + 2·1.9 = 5.3 m

R = A/P = 0.5377 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.015·7.9/(2.85·0.6613) = 0.06288

S = 0.06288² = 0.003954

Answer

S = 0.003954

That is a fall of 3.954 m per kilometre.

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