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

d = 2 mH = 2.5 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 2.5 mH = 1 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal venturi meter with an inlet diameter of 150 mm and a throat diameter of 100 mm. The pressure at the inlet is 50 kPa higher than at the throat. Ignoring losses, find the flow rate Q.

d₁ = 150 mmd₂ = 100 mmp₁ − p₂ = 50 kPaρ = 1000 kg/m³

Q = L/s

4.Continuity & BernoulliHard

A hose with an inner diameter of 32 mm siphons water from an open tank. The highest point C of the hose is 2 m above the water surface, and the outlet (open to the air) is 1 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.

d = 32 mmz_C = +2 mz₂ = −1 mg = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 1.5 m/s through a pipe 200 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss h_f.

V = 1.5 m/sL = 200 mD = 200 mmf = 0.022g = 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 = 0.018) 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 = 5). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 120 L/s?

Q = 120 L/sL = 400 mD = 300 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 5

H = m

7.Pumps & pump powerMedium

A pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 50 m. The pump efficiency is 85% 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 = 10 L/sH = 50 mη_pump = 85%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 6 m higher. The pipe is 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015; 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 = 6 mL = 100 mD = 200 mmf = 0.015ΣK = 6

H = m

9.Open-channel flow (Manning)Medium

A concrete storm sewer with an inside diameter of 900 mm flows half full. Manning n = 0.013 and the slope is S = 0.001. Find the flow rate Q in m³/s.

D = 900 mmn = 0.013S = 0.001

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal unfinished concrete channel with a bottom width of 6 m and side slopes of 2:1 (horizontal:vertical) must carry 64 m³/s in uniform flow at a depth of 2.3 m. Manning n = 0.014. Find the required bed slope S (in m/m).

Q = 64 m³/sb = 6 my = 2.3 mz = 2n = 0.014

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 159.4 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 + 2.5/2 = 3.25 m

A = H·b = 2.5·2 = 5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.25·5 = 159.4 kN

Answer

F = 159.4 kN

2.Hydrostatic force

Answer: y_cp = 3.028 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 = 2.5 + 1/2 = 3 m

A = H·b = 1·2 = 2 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3·2 = 58.86 kN

Centre of pressure

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

I_c = b·H³/12 = 2·1³/12 = 0.1667 m⁴

y_cp = y_c + I_c/(y_c·A) = 3 + 0.1667/(3·2) = 3.028 m

Answer

y_cp = 3.028 m

3.Continuity & Bernoulli

Answer: Q = 87.67 L/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.

Continuity

v₁ = v₂·(100/150)² = 0.4444·v₂

Bernoulli between inlet and throat

The pipe is horizontal, so only pressure and velocity change.

p₁ − p₂ = ½ρv₂²(1 − β⁴)

v₂ = √(2·50000/(1000·(1 − 0.1975))) = 11.16 m/s

v₁ = 4.961 m/s

Flow rate

Q = A₂·v₂ = 0.007854·11.16 = 0.08767 m³/s = 87.67 L/s

Answer

Q = 87.67 L/s

4.Continuity & Bernoulli

Answer: Q = 3.562 L/s

Velocity in the hose

Surface and outlet are both open to the air, so the pressure terms cancel.

v = √(2g·1) = √(2·9.81·1) = 4.429 m/s

Flow rate

Q = A·v = 0.0008042·4.429 = 0.003562 m³/s = 3.562 L/s

Answer

Q = 3.562 L/s

5.Pipe friction & head loss

Answer: h_f = 2.523 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) = 1.5²/(2·9.81) = 0.1147 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.022·(200/0.2)·1.5²/(2·9.81) = 2.523 m

Answer

h_f = 2.523 m

6.Pipe friction & head loss

Answer: H = 4.656 m

Velocity

V = Q/A = 0.12/0.07069 = 1.698 m/s

V²/(2g) = 0.1469 m

Loss coefficients

f·L/D = 0.018·400/0.3 = 24

ΣK = 0.5 + 4·0.3 + 5 = 6.7

Energy equation

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

H = (1 + 24 + 6.7)·0.1469 = 4.656 m

Answer

H = 4.656 m

7.Pumps & pump power

Answer: P_el = 6.074 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 = 1000·9.81·0.01·50 = 4.905 kW

Shaft power

P_shaft = 4.905/0.85 = 5.771 kW

Electrical power

P_el = 5.771/0.95 = 6.074 kW

Answer

P_el = 6.074 kW

8.Pumps & pump power

Answer: H = 10.46 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) = (7.5 + 6)·0.3305 = 4.462 m

Pump head

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

H = Δz + h_L = 6 + 4.462 = 10.46 m

Answer

H = 10.46 m

9.Open-channel flow (Manning)

Answer: Q = 0.2862 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 = 0.3181 m², R = D/4 = 0.225 m

Velocity

V = (1/0.013)·0.225^2/3·0.001^1/2 = 0.8999 m/s

Flow rate

Q = A·V = 0.3181·0.8999 = 0.2862 m³/s

Answer

Q = 0.2862 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.0007887 m/m

Geometry

A = (b + z·y)·y = (6 + 2·2.3)·2.3 = 24.38 m²

P = b + 2y·√(1 + z²) = 6 + 2·2.3·√(1 + 2²) = 16.29 m

R = A/P = 1.497 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.014·64/(24.38·1.309) = 0.02808

S = 0.02808² = 0.0007887

Answer

S = 0.0007887

That is a fall of 0.7887 m per kilometre.

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