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

y_cp = m

3.Continuity & BernoulliMedium

Water flows at 38 L/s through a horizontal pipe that narrows from 150 mm to 100 mm. The pressure in the narrow section is 260 kPa. Ignoring losses, find the pressure p₁ in the wide section.

Q = 38 L/sd₁ = 150 mmd₂ = 100 mmp₂ = 260 kPaρ = 1000 kg/m³

p₁ = kPa

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 100 mm to 75 mm while rising 2 m. At section 1 the pressure is 140 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 100 mmd₂ = 75 mmp₁ = 140 kPav₁ = 2 m/sΔz = 2 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 1.8 m/s through a horizontal pipe, 300 m long, with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 1.8 m/sL = 300 mD = 300 mmf = 0.028ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 150 m long with an inside diameter of 250 mm (Darcy friction factor f = 0.015) 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), 3 bends (K = 0.3 each) and a valve (K = 5). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 50 L/s?

Q = 50 L/sL = 150 mD = 250 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 5

H = m

7.Pumps & pump powerMedium

A pump delivers 150 L/s of water (ρ = 1000 kg/m³) against a head of 26 m. The pump efficiency is 65% and the electric motor driving it has an efficiency of 95%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 150 L/sH = 26 mη_pump = 65%η_motor = 95%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 17 m higher. The pipe is 400 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018; 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 = 17 mL = 400 mD = 200 mmf = 0.018ΣK = 6

H = m

9.Open-channel flow (Manning)Medium

A trapezoidal grass-lined earth channel has a bottom width of 3 m and side slopes of 1.5:1 (horizontal:vertical). Water flows uniformly 0.5 m deep. Manning n = 0.03 and the bed slope is S = 0.0008. Find the flow rate Q in m³/s.

b = 3 my = 0.5 mz = 1.5n = 0.03S = 0.0008

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal gravel channel with a bottom width of 6 m and side slopes of 3:1 (horizontal:vertical) must carry 58 m³/s in uniform flow at a depth of 1.7 m. Manning n = 0.025. Find the required bed slope S (in m/m).

Q = 58 m³/sb = 6 my = 1.7 mz = 3n = 0.025

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 = 2.333 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 + 1.5/2 = 2.25 m

A = H·b = 1.5·1 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2.25·1.5 = 33.11 kN

Centre of pressure

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

I_c = b·H³/12 = 1·1.5³/12 = 0.2813 m⁴

y_cp = y_c + I_c/(y_c·A) = 2.25 + 0.2813/(2.25·1.5) = 2.333 m

Answer

y_cp = 2.333 m

3.Continuity & Bernoulli

Answer: p₁ = 269.4 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.

Velocities

v₁ = Q/A₁ = 0.038/0.01767 = 2.15 m/s

v₂ = Q/A₂ = 0.038/0.007854 = 4.838 m/s

Bernoulli

p₁ = p₂ + ½ρ(v₂² − v₁²) = 260 + ½·1000·(4.838² − 2.15²)/1000 = 260 + 9.393 = 269.4 kPa

Answer

p₁ = 269.4 kPa

4.Continuity & Bernoulli

Answer: p₂ = 116.1 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 2·(100/75)² = 3.556 m/s

½ρ(v₁² − v₂²) = ½·1000·(2² − 3.556²)/1000 = −4.321 kPa

ρgΔz = 1000·9.81·2/1000 = 19.62 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 140 + (−4.321) − 19.62 = 116.1 kPa

Answer

p₂ = 116.1 kPa

5.Pipe friction & head loss

Answer: Δp = 45.36 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) = 0.028·(300/0.3)·1.8²/(2·9.81) = 4.624 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 = 1000·9.81·4.624 = 45360 Pa = 45.36 kPa

Answer

Δp = 45.36 kPa

6.Pipe friction & head loss

Answer: H = 0.8673 m

Velocity

V = Q/A = 0.05/0.04909 = 1.019 m/s

V²/(2g) = 0.05288 m

Loss coefficients

f·L/D = 0.015·150/0.25 = 9

ΣK = 0.5 + 3·0.3 + 5 = 6.4

Energy equation

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

H = (1 + 9 + 6.4)·0.05288 = 0.8673 m

Answer

H = 0.8673 m

7.Pumps & pump power

Answer: P_el = 61.96 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 = 1000·9.81·0.15·26 = 38.26 kW

Shaft power

P_shaft = 38.26/0.65 = 58.86 kW

Electrical power

P_el = 58.86/0.95 = 61.96 kW

Answer

P_el = 61.96 kW

8.Pumps & pump power

Answer: H = 30.88 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) = (36 + 6)·0.3305 = 13.88 m

Pump head

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

H = Δz + h_L = 17 + 13.88 = 30.88 m

Answer

H = 30.88 m

9.Open-channel flow (Manning)

Answer: Q = 0.9443 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 + 1.5·0.5)·0.5 = 1.875 m²

P = b + 2y·√(1 + z²) = 3 + 2·0.5·√(1 + 1.5²) = 4.803 m

R = A/P = 0.3904 m

Velocity

V = (1/0.03)·0.3904^2/3·0.0008^1/2 = 0.5036 m/s

Flow rate

Q = A·V = 1.875·0.5036 = 0.9443 m³/s

Answer

Q = 0.9443 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.005038 m/m

Geometry

A = (b + z·y)·y = (6 + 3·1.7)·1.7 = 18.87 m²

P = b + 2y·√(1 + z²) = 6 + 2·1.7·√(1 + 3²) = 16.75 m

R = A/P = 1.126 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.025·58/(18.87·1.083) = 0.07098

S = 0.07098² = 0.005038

Answer

S = 0.005038

That is a fall of 5.038 m per kilometre.

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