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

d = 4 mH = 1.5 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²

F = kN

3.Continuity & BernoulliMedium

Water flows at 18 L/s through a horizontal pipe that narrows from 200 mm to 100 mm. The pressure in the narrow section is 280 kPa. Ignoring losses, find the pressure p₁ in the wide section.

Q = 18 L/sd₁ = 200 mmd₂ = 100 mmp₂ = 280 kPaρ = 1000 kg/m³

p₁ = kPa

4.Continuity & BernoulliHard

A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 100 mm and a gauge pressure of 20 kPa. The nozzle exit diameter is 40 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².

d₁ = 100 mmp₁ = 20 kPa (gauge)d₂ = 40 mmρ = 1000 kg/m³g = 9.81 m/s²

H = m

5.Pipe friction & head lossMedium

Water flows at 80 L/s through a pipe 150 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 80 L/sL = 150 mD = 300 mmf = 0.015g = 9.81 m/s²

h_f = m

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 = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 75 L/s?

Q = 75 L/sL = 150 mD = 250 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 2

H = m

7.Pumps & pump powerMedium

A pump delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 20 m. The pump efficiency is 70% and the electric motor driving it has an efficiency of 90%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 15 L/sH = 20 mη_pump = 70%η_motor = 90%g = 9.81 m/s²

P_el = 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 11 m higher. The pipe is 600 m long with an inside diameter of 150 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. The upper tank is closed and the air above its water is at a gauge pressure of 350 kPa. Find the head H the pump must add.

Q = 50 L/sΔz = 11 mL = 600 mD = 150 mmf = 0.018ΣK = 6p_tank = 350 kPa

H = m

9.Open-channel flow (Manning)Medium

A rectangular grass-lined earth channel is 3 m wide. Water flows uniformly 1.8 m deep. Manning n = 0.03 and the bed slope is S = 0.004. Find the flow rate Q in m³/s.

b = 3 my = 1.8 mn = 0.03S = 0.004

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal finished concrete channel with a bottom width of 2.5 m and side slopes of 1:1 (horizontal:vertical) must carry 9.7 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 = 9.7 m³/sb = 2.5 my = 1.2 mz = 1n = 0.012

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 323.7 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 = 3·2 = 6 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·5.5·6 = 323.7 kN

Answer

F = 323.7 kN

2.Hydrostatic force

Answer: F = 92.26 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 = 1.5·1.5 = 2.25 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·4.75·2.25 = 92.26 kN

Answer

F = 92.26 kN

3.Continuity & Bernoulli

Answer: p₁ = 282.5 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 = 0.018/0.03142 = 0.573 m/s

v₂ = Q/A₂ = 0.018/0.007854 = 2.292 m/s

Bernoulli

p₁ = p₂ + ½ρ(v₂² − v₁²) = 280 + ½·1000·(2.292² − 0.573²)/1000 = 280 + 2.462 = 282.5 kPa

Answer

p₁ = 282.5 kPa

4.Continuity & Bernoulli

Answer: H = 2.092 m

Exit velocity

β = 40/100 = 0.4

v₂ = √(2p₁/(ρ(1 − β⁴))) = √(2·20000/(1000·0.9744)) = 6.407 m/s

Height of the jet

Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.

H = v₂²/(2g) = 6.407²/(2·9.81) = 2.092 m

Answer

H = 2.092 m

5.Pipe friction & head loss

Answer: h_f = 0.4896 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.

Mean velocity

A = π·0.3²/4 = 0.07069 m²

V = Q/A = 0.08/0.07069 = 1.132 m/s

Velocity head

V²/(2g) = 1.132²/(2·9.81) = 0.06529 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.015·(150/0.3)·1.132²/(2·9.81) = 0.4896 m

Answer

h_f = 0.4896 m

6.Pipe friction & head loss

Answer: H = 1.594 m

Velocity

V = Q/A = 0.075/0.04909 = 1.528 m/s

V²/(2g) = 0.119 m

Loss coefficients

f·L/D = 0.015·150/0.25 = 9

ΣK = 0.5 + 3·0.3 + 2 = 3.4

Energy equation

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

H = (1 + 9 + 3.4)·0.119 = 1.594 m

Answer

H = 1.594 m

7.Pumps & pump power

Answer: P_el = 4.671 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 = 1000·9.81·0.015·20 = 2.943 kW

Shaft power

P_shaft = 2.943/0.7 = 4.204 kW

Electrical power

P_el = 4.204/0.9 = 4.671 kW

Answer

P_el = 4.671 kW

8.Pumps & pump power

Answer: H = 78.5 m

Velocity

V = Q/A = 0.05/0.01767 = 2.829 m/s

V²/(2g) = 0.408 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (72 + 6)·0.408 = 31.83 m

Pressure head

p/(ρ·g) = 350000/(1000·9.81) = 35.68 m

Pump head

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

H = Δz + p/(ρ·g) + h_L = 11 + 35.68 + 31.83 = 78.5 m

Answer

H = 78.5 m

9.Open-channel flow (Manning)

Answer: Q = 9.959 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·y = 3·1.8 = 5.4 m²

P = b + 2y = 3 + 2·1.8 = 6.6 m

R = A/P = 0.8182 m

Velocity

V = (1/0.03)·0.8182^2/3·0.004^1/2 = 1.844 m/s

Flow rate

Q = A·V = 5.4·1.844 = 9.959 m³/s

Answer

Q = 9.959 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.001003 m/m

Geometry

A = (b + z·y)·y = (2.5 + 1·1.2)·1.2 = 4.44 m²

P = b + 2y·√(1 + z²) = 2.5 + 2·1.2·√(1 + 1²) = 5.894 m

R = A/P = 0.7533 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.012·9.7/(4.44·0.8279) = 0.03167

S = 0.03167² = 0.001003

Answer

S = 0.001003

That is a fall of 1.003 m per kilometre.

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