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 m wide sits in the wall of a tank of water, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 1 mH = 2 mb = 1 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 oil, with its top edge 2 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 2 mH = 3 mb = 2 mρ = 880 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 200 mm to 100 mm. In the wide section the pressure is 290 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 200 mmd₂ = 100 mmp₁ = 290 kPav₁ = 1.25 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

A closed water tank has compressed air above the water at a gauge pressure of 140 kPa. Water leaves through an opening with a diameter of 40 mm, 3.5 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².

p_air = 140 kPa (gauge)h = 3.5 md = 40 mmρ = 1000 kg/m³g = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 1.1 m/s through a horizontal pipe, 50 m long, with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 1.1 m/sL = 50 mD = 200 mmf = 0.02ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 150 m long with an inside diameter of 100 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 = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 20 L/s?

Q = 20 L/sL = 150 mD = 100 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 2

H = m

7.Pumps & pump powerMedium

A pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 74 m. The pump efficiency is 72% 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 = 10 L/sH = 74 mη_pump = 72%η_motor = 92%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 75 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 150 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 = 5 (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 250 kPa. Find the head H the pump must add.

Q = 75 L/sΔz = 8 mL = 150 mD = 200 mmf = 0.015ΣK = 5p_tank = 250 kPa

H = m

9.Open-channel flow (Manning)Medium

A concrete storm sewer with an inside diameter of 750 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.002. Find the flow rate Q in m³/s.

D = 750 mmn = 0.013S = 0.002

Q = m³/s

10.Open-channel flow (Manning)Hard

A rectangular gravel channel 5 m wide must carry 23 m³/s in uniform flow at a depth of 1.8 m. Manning n = 0.025. Find the required bed slope S (in m/m).

Q = 23 m³/sb = 5 my = 1.8 mn = 0.025

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 39.24 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 + 2/2 = 2 m

A = H·b = 2·1 = 2 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2·2 = 39.24 kN

Answer

F = 39.24 kN

2.Hydrostatic force

Answer: y_cp = 3.714 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 + 3/2 = 3.5 m

A = H·b = 3·2 = 6 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·3.5·6 = 181.3 kN

Centre of pressure

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

I_c = b·H³/12 = 2·3³/12 = 4.5 m⁴

y_cp = y_c + I_c/(y_c·A) = 3.5 + 4.5/(3.5·6) = 3.714 m

Answer

y_cp = 3.714 m

3.Continuity & Bernoulli

Answer: p₂ = 278.3 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 = 1.25·(200/100)² = 5 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.25² − 5²)/1000 = −11.72 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 290 + (−11.72) = 278.3 kPa

Answer

p₂ = 278.3 kPa

4.Continuity & Bernoulli

Answer: Q = 23.46 L/s

Bernoulli from the surface to the jet

v = √(2·(140000/1000 + 9.81·3.5)) = 18.67 m/s

Flow rate

Q = A·v = 0.001257·18.67 = 0.02346 m³/s = 23.46 L/s

Answer

Q = 23.46 L/s

5.Pipe friction & head loss

Answer: Δp = 3.025 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) = 0.02·(50/0.2)·1.1²/(2·9.81) = 0.3084 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 = 1000·9.81·0.3084 = 3025 Pa = 3.025 kPa

Answer

Δp = 3.025 kPa

6.Pipe friction & head loss

Answer: H = 10.48 m

Velocity

V = Q/A = 0.02/0.007854 = 2.546 m/s

V²/(2g) = 0.3305 m

Loss coefficients

f·L/D = 0.018·150/0.1 = 27

ΣK = 0.5 + 4·0.3 + 2 = 3.7

Energy equation

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

H = (1 + 27 + 3.7)·0.3305 = 10.48 m

Answer

H = 10.48 m

7.Pumps & pump power

Answer: P_el = 10.96 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 = 1000·9.81·0.01·74 = 7.259 kW

Shaft power

P_shaft = 7.259/0.72 = 10.08 kW

Electrical power

P_el = 10.08/0.92 = 10.96 kW

Answer

P_el = 10.96 kW

8.Pumps & pump power

Answer: H = 38.2 m

Velocity

V = Q/A = 0.075/0.03142 = 2.387 m/s

V²/(2g) = 0.2905 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (11.25 + 5)·0.2905 = 4.72 m

Pressure head

p/(ρ·g) = 250000/(1000·9.81) = 25.48 m

Pump head

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

H = Δz + p/(ρ·g) + h_L = 8 + 25.48 + 4.72 = 38.2 m

Answer

H = 38.2 m

9.Open-channel flow (Manning)

Answer: Q = 0.4979 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 = 0.4418 m², R = D/4 = 0.1875 m

Velocity

V = (1/0.013)·0.1875^2/3·0.002^1/2 = 1.127 m/s

Flow rate

Q = A·V = 0.4418·1.127 = 0.4979 m³/s

Answer

Q = 0.4979 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.003842 m/m

Geometry

A = b·y = 5·1.8 = 9 m²

P = b + 2y = 5 + 2·1.8 = 8.6 m

R = A/P = 1.047 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.025·23/(9·1.031) = 0.06198

S = 0.06198² = 0.003842

Answer

S = 0.003842

That is a fall of 3.842 m per kilometre.

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