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 2 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 = 1.5 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 1.5 m wide sits in the wall of a tank of seawater, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 2.5 mH = 3 mb = 1.5 mρ = 1025 kg/m³g = 9.81 m/s²

F = kN

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 75 mm to 50 mm. In the wide section the pressure is 180 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 75 mmd₂ = 50 mmp₁ = 180 kPav₁ = 1.75 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 100 mm to 50 mm while rising 10 m. At section 1 the pressure is 240 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 100 mmd₂ = 50 mmp₁ = 240 kPav₁ = 1 m/sΔz = 10 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 2.3 m/s through a pipe 150 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss h_f.

V = 2.3 m/sL = 150 mD = 50 mmf = 0.018g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.022. The water surface of the upper reservoir is 23 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 2) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². Find the flow rate Q in L/s.

Δz = 23 mL = 300 mD = 300 mmf = 0.022K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 2K_exit = 1.0

Q = L/s

7.Pumps & pump powerMedium

A pump with an efficiency of 80% delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 14 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 15 L/sH = 14 mη = 80%g = 9.81 m/s²

P_shaft = kW

8.Pumps & pump powerHard

A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 8 m higher. The pipe is 400 m long with an inside diameter of 200 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 upper tank is closed and the air above its water is at a gauge pressure of 200 kPa. Find the head H the pump must add.

Q = 50 L/sΔz = 8 mL = 400 mD = 200 mmf = 0.025ΣK = 8p_tank = 200 kPa

H = m

9.Open-channel flow (Manning)Medium

A trapezoidal unfinished concrete channel has a bottom width of 5.5 m and side slopes of 2:1 (horizontal:vertical). Water flows uniformly 1.8 m deep. Manning n = 0.014 and the bed slope is S = 0.002. Find the flow rate Q in m³/s.

b = 5.5 my = 1.8 mz = 2n = 0.014S = 0.002

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal finished concrete channel with a bottom width of 5.5 m and side slopes of 2:1 (horizontal:vertical) must carry 28 m³/s in uniform flow at a depth of 1.2 m. Manning n = 0.012. Find the required bed slope S (in m/m).

Q = 28 m³/sb = 5.5 my = 1.2 mz = 2n = 0.012

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 139.8 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 + 1.5/2 = 4.75 m

A = H·b = 1.5·2 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4.75·3 = 139.8 kN

Answer

F = 139.8 kN

2.Hydrostatic force

Answer: F = 181 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 + 3/2 = 4 m

A = H·b = 3·1.5 = 4.5 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·4·4.5 = 181 kN

Answer

F = 181 kN

3.Continuity & Bernoulli

Answer: p₂ = 173.8 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 = 1.75·(75/50)² = 3.938 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.75² − 3.938²)/1000 = −6.221 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 180 + (−6.221) = 173.8 kPa

Answer

p₂ = 173.8 kPa

4.Continuity & Bernoulli

Answer: p₂ = 134.4 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 1·(100/50)² = 4 m/s

½ρ(v₁² − v₂²) = ½·1000·(1² − 4²)/1000 = −7.5 kPa

ρgΔz = 1000·9.81·10/1000 = 98.1 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 240 + (−7.5) − 98.1 = 134.4 kPa

Answer

p₂ = 134.4 kPa

5.Pipe friction & head loss

Answer: h_f = 14.56 m

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.

Velocity head

V²/(2g) = 2.3²/(2·9.81) = 0.2696 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.018·(150/0.05)·2.3²/(2·9.81) = 14.56 m

Answer

h_f = 14.56 m

6.Pipe friction & head loss

Answer: Q = 292.2 L/s

Loss coefficients

f·L/D = 0.022·300/0.3 = 22

ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4

Energy equation

Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.

23 = (22 + 4.4)·V²/(2·9.81)

V = √(2·9.81·23/26.4) = 4.134 m/s

Flow rate

Q = V·π·D²/4 = 4.134·0.07069 = 0.2922 m³/s = 292.2 L/s

Answer

Q = 292.2 L/s

7.Pumps & pump power

Answer: P_shaft = 2.575 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 = 1000·9.81·0.015·14 = 2.06 kW

Shaft power

P_shaft = P_h/η = 2.06/0.8 = 2.575 kW

Answer

P_shaft = 2.575 kW

8.Pumps & pump power

Answer: H = 35.88 m

Velocity

V = Q/A = 0.05/0.03142 = 1.592 m/s

V²/(2g) = 0.1291 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (50 + 8)·0.1291 = 7.488 m

Pressure head

p/(ρ·g) = 200000/(1000·9.81) = 20.39 m

Pump head

Both surfaces are practically at rest, so there is no velocity head term.

H = Δz + p/(ρ·g) + h_L = 8 + 20.39 + 7.488 = 35.88 m

Answer

H = 35.88 m

9.Open-channel flow (Manning)

Answer: Q = 59.38 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 = (5.5 + 2·1.8)·1.8 = 16.38 m²

P = b + 2y·√(1 + z²) = 5.5 + 2·1.8·√(1 + 2²) = 13.55 m

R = A/P = 1.209 m

Velocity

V = (1/0.014)·1.209^2/3·0.002^1/2 = 3.625 m/s

Flow rate

Q = A·V = 16.38·3.625 = 59.38 m³/s

Answer

Q = 59.38 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.001507 m/m

Geometry

A = (b + z·y)·y = (5.5 + 2·1.2)·1.2 = 9.48 m²

P = b + 2y·√(1 + z²) = 5.5 + 2·1.2·√(1 + 2²) = 10.87 m

R = A/P = 0.8724 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.012·28/(9.48·0.913) = 0.03882

S = 0.03882² = 0.001507

Answer

S = 0.001507

That is a fall of 1.507 m per kilometre.

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