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 4 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 4 mH = 3 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 glycerin, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 2.5 mH = 2.5 mb = 1 mρ = 1260 kg/m³g = 9.81 m/s²

F = kN

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 75 mm to 40 mm. In the wide section the pressure is 300 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 75 mmd₂ = 40 mmp₁ = 300 kPav₁ = 1.5 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 10 m. At section 1 the pressure is 230 kPa and the velocity is 2.75 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 150 mmd₂ = 75 mmp₁ = 230 kPav₁ = 2.75 m/sΔz = 10 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 2.3 m/s through a horizontal pipe, 150 m long, with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 2.3 m/sL = 150 mD = 300 mmf = 0.02ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 100 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.025) runs from an open reservoir and discharges as a free jet into the air. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each) and a valve (K = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 200 L/s?

Q = 200 L/sL = 100 mD = 300 mmf = 0.025K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 2

H = m

7.Pumps & pump powerMedium

A pump delivers 150 L/s of water (ρ = 1000 kg/m³) against a head of 38 m. The pump efficiency is 82% and the electric motor driving it has an efficiency of 94%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 150 L/sH = 38 mη_pump = 82%η_motor = 94%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 120 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 36 m higher. The pipe is 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 70%. Find the shaft power P_shaft in kW.

Q = 120 L/sΔz = 36 mL = 300 mD = 250 mmf = 0.025ΣK = 8η = 70%

P_shaft = kW

9.Open-channel flow (Manning)Medium

A trapezoidal brickwork channel has a bottom width of 3.5 m and side slopes of 1:1 (horizontal:vertical). Water flows uniformly 1.8 m deep. Manning n = 0.015 and the bed slope is S = 0.002. Find the flow rate Q in m³/s.

b = 3.5 my = 1.8 mz = 1n = 0.015S = 0.002

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal brickwork channel with a bottom width of 3 m and side slopes of 1.5:1 (horizontal:vertical) must carry 5 m³/s in uniform flow at a depth of 0.6 m. Manning n = 0.015. Find the required bed slope S (in m/m).

Q = 5 m³/sb = 3 my = 0.6 mz = 1.5n = 0.015

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 242.8 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 = 4 + 3/2 = 5.5 m

A = H·b = 3·1.5 = 4.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·5.5·4.5 = 242.8 kN

Answer

F = 242.8 kN

2.Hydrostatic force

Answer: F = 115.9 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 = 2.5 + 2.5/2 = 3.75 m

A = H·b = 2.5·1 = 2.5 m²

Resultant force

F = ρ·g·y_c·A = 1260·9.81·3.75·2.5 = 115.9 kN

Answer

F = 115.9 kN

3.Continuity & Bernoulli

Answer: p₂ = 287.2 kPa

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.

Continuity and Bernoulli

Work in Pa, then divide by 1000 to get kPa.

v₂ = v₁·(d₁/d₂)² = 1.5·(75/40)² = 5.273 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.5² − 5.273²)/1000 = −12.78 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 300 + (−12.78) = 287.2 kPa

Answer

p₂ = 287.2 kPa

4.Continuity & Bernoulli

Answer: p₂ = 75.18 kPa

Continuity and Bernoulli

Work in Pa, then divide by 1000 to get kPa.

v₂ = v₁·(d₁/d₂)² = 2.75·(150/75)² = 11 m/s

½ρ(v₁² − v₂²) = ½·1000·(2.75² − 11²)/1000 = −56.72 kPa

ρgΔz = 1000·9.81·10/1000 = 98.1 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 230 + (−56.72) − 98.1 = 75.18 kPa

Answer

p₂ = 75.18 kPa

5.Pipe friction & head loss

Answer: Δp = 26.45 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) = 0.02·(150/0.3)·2.3²/(2·9.81) = 2.696 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 = 1000·9.81·2.696 = 26450 Pa = 26.45 kPa

Answer

Δp = 26.45 kPa

6.Pipe friction & head loss

Answer: H = 5.318 m

Velocity

V = Q/A = 0.2/0.07069 = 2.829 m/s

V²/(2g) = 0.408 m

Loss coefficients

f·L/D = 0.025·100/0.3 = 8.333

ΣK = 0.5 + 4·0.3 + 2 = 3.7

Energy equation

The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.

H = (1 + 8.333 + 3.7)·0.408 = 5.318 m

Answer

H = 5.318 m

7.Pumps & pump power

Answer: P_el = 72.54 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 = 1000·9.81·0.15·38 = 55.92 kW

Shaft power

P_shaft = 55.92/0.82 = 68.19 kW

Electrical power

P_el = 68.19/0.94 = 72.54 kW

Answer

P_el = 72.54 kW

8.Pumps & pump power

Answer: P_shaft = 80.01 kW

Velocity

V = Q/A = 0.12/0.04909 = 2.445 m/s

V²/(2g) = 0.3046 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (30 + 8)·0.3046 = 11.57 m

Pump head

H = Δz + h_L = 36 + 11.57 = 47.57 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.12·47.57/0.7 = 80.01 kW

Answer

P_shaft = 80.01 kW

9.Open-channel flow (Manning)

Answer: Q = 30.5 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 = (b + z·y)·y = (3.5 + 1·1.8)·1.8 = 9.54 m²

P = b + 2y·√(1 + z²) = 3.5 + 2·1.8·√(1 + 1²) = 8.591 m

R = A/P = 1.11 m

Velocity

V = (1/0.015)·1.11^2/3·0.002^1/2 = 3.197 m/s

Flow rate

Q = A·V = 9.54·3.197 = 30.5 m³/s

Answer

Q = 30.5 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.002951 m/m

Geometry

A = (b + z·y)·y = (3 + 1.5·0.6)·0.6 = 2.34 m²

P = b + 2y·√(1 + z²) = 3 + 2·0.6·√(1 + 1.5²) = 5.163 m

R = A/P = 0.4532 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.015·5/(2.34·0.59) = 0.05432

S = 0.05432² = 0.002951

Answer

S = 0.002951

That is a fall of 2.951 m per kilometre.

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