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

d = 1.5 mH = 1 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 seawater, with its top edge 3 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 3 mH = 2 mb = 1 mρ = 1025 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

A pitot-static tube in a wind tunnel reads a difference of 1350 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.

p₀ − p = 1350 Paρ = 1.2 kg/m³

v = m/s

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 200 mm to 50 mm while rising 3 m. At section 1 the pressure is 160 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 200 mmd₂ = 50 mmp₁ = 160 kPav₁ = 0.75 m/sΔz = 3 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 2.4 m/s through a pipe 100 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the friction head loss h_f.

V = 2.4 m/sL = 100 mD = 300 mmf = 0.025g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

A pipe 400 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.02) 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 = 10). 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 = 400 mD = 300 mmf = 0.02K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 10

H = m

7.Pumps & pump powerMedium

A pump delivers 40 L/s of water (ρ = 1000 kg/m³) against a head of 50 m. The pump efficiency is 60% and the electric motor driving it has an efficiency of 92%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 40 L/sH = 50 mη_pump = 60%η_motor = 92%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 30 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 15 m higher. The pipe is 100 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.

Q = 30 L/sΔz = 15 mL = 100 mD = 150 mmf = 0.022ΣK = 8

H = m

9.Open-channel flow (Manning)Medium

A trapezoidal clean earth channel has a bottom width of 5.5 m and side slopes of 1:1 (horizontal:vertical). Water flows uniformly 0.7 m deep. Manning n = 0.022 and the bed slope is S = 0.0008. Find the flow rate Q in m³/s.

b = 5.5 my = 0.7 mz = 1n = 0.022S = 0.0008

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal clean earth channel with a bottom width of 2 m and side slopes of 3:1 (horizontal:vertical) must carry 15 m³/s in uniform flow at a depth of 1.9 m. Manning n = 0.022. Find the required bed slope S (in m/m).

Q = 15 m³/sb = 2 my = 1.9 mz = 3n = 0.022

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 29.43 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 = 1.5 + 1/2 = 2 m

A = H·b = 1·1.5 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2·1.5 = 29.43 kN

Answer

F = 29.43 kN

2.Hydrostatic force

Answer: y_cp = 4.083 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 = 3 + 2/2 = 4 m

A = H·b = 2·1 = 2 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·4·2 = 80.44 kN

Centre of pressure

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

I_c = b·H³/12 = 1·2³/12 = 0.6667 m⁴

y_cp = y_c + I_c/(y_c·A) = 4 + 0.6667/(4·2) = 4.083 m

Answer

y_cp = 4.083 m

3.Continuity & Bernoulli

Answer: v = 47.43 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 = √(2·1350/1.2) = 47.43 m/s

Answer

v = 47.43 m/s

4.Continuity & Bernoulli

Answer: p₂ = 58.85 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 0.75·(200/50)² = 12 m/s

½ρ(v₁² − v₂²) = ½·1000·(0.75² − 12²)/1000 = −71.72 kPa

ρgΔz = 1000·9.81·3/1000 = 29.43 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 160 + (−71.72) − 29.43 = 58.85 kPa

Answer

p₂ = 58.85 kPa

5.Pipe friction & head loss

Answer: h_f = 2.446 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) = 2.4²/(2·9.81) = 0.2936 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.025·(100/0.3)·2.4²/(2·9.81) = 2.446 m

Answer

h_f = 2.446 m

6.Pipe friction & head loss

Answer: H = 16.06 m

Velocity

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

V²/(2g) = 0.408 m

Loss coefficients

f·L/D = 0.02·400/0.3 = 26.67

ΣK = 0.5 + 4·0.3 + 10 = 11.7

Energy equation

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

H = (1 + 26.67 + 11.7)·0.408 = 16.06 m

Answer

H = 16.06 m

7.Pumps & pump power

Answer: P_el = 35.54 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 = 1000·9.81·0.04·50 = 19.62 kW

Shaft power

P_shaft = 19.62/0.6 = 32.7 kW

Electrical power

P_el = 32.7/0.92 = 35.54 kW

Answer

P_el = 35.54 kW

8.Pumps & pump power

Answer: H = 18.33 m

Velocity

V = Q/A = 0.03/0.01767 = 1.698 m/s

V²/(2g) = 0.1469 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (14.67 + 8)·0.1469 = 3.33 m

Pump head

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

H = Δz + h_L = 15 + 3.33 = 18.33 m

Answer

H = 18.33 m

9.Open-channel flow (Manning)

Answer: Q = 3.882 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 + 1·0.7)·0.7 = 4.34 m²

P = b + 2y·√(1 + z²) = 5.5 + 2·0.7·√(1 + 1²) = 7.48 m

R = A/P = 0.5802 m

Velocity

V = (1/0.022)·0.5802^2/3·0.0008^1/2 = 0.8944 m/s

Flow rate

Q = A·V = 4.34·0.8944 = 3.882 m³/s

Answer

Q = 3.882 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.0004806 m/m

Geometry

A = (b + z·y)·y = (2 + 3·1.9)·1.9 = 14.63 m²

P = b + 2y·√(1 + z²) = 2 + 2·1.9·√(1 + 3²) = 14.02 m

R = A/P = 1.044 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.022·15/(14.63·1.029) = 0.02192

S = 0.02192² = 0.0004806

Answer

S = 0.0004806

That is a fall of 0.4806 m per kilometre.

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