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

d = 1.5 mH = 2.5 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 3 m wide sits in the wall of a tank of water, 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 = 2 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

A pitot-static tube in a water pipe reads a difference of 38 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.

p₀ − p = 38 kPaρ = 1000 kg/m³

v = m/s

4.Continuity & BernoulliHard

A horizontal venturi meter (inlet 100 mm, throat 60 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 60 mm. The discharge coefficient is 0.99. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².

d₁ = 100 mmd₂ = 60 mmΔh = 60 mmC_d = 0.99ρ_Hg = 13 600 kg/m³g = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water flows at 20 L/s through a pipe 800 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 20 L/sL = 800 mD = 150 mmf = 0.025g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

A pipe 200 m long with an inside diameter of 150 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 40 L/s?

Q = 40 L/sL = 200 mD = 150 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 56 m. The pump efficiency is 70% 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 = 56 mη_pump = 70%η_motor = 92%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 120 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 13 m higher. The pipe is 400 m long with an inside diameter of 250 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. The upper tank is closed and the air above its water is at a gauge pressure of 400 kPa. Find the head H the pump must add.

Q = 120 L/sΔz = 13 mL = 400 mD = 250 mmf = 0.022ΣK = 8p_tank = 400 kPa

H = m

9.Open-channel flow (Manning)Medium

A trapezoidal unfinished concrete channel has a bottom width of 3.5 m and side slopes of 2:1 (horizontal:vertical). Water flows uniformly 1.7 m deep. Manning n = 0.014 and the bed slope is S = 0.0015. Find the flow rate Q in m³/s.

b = 3.5 my = 1.7 mz = 2n = 0.014S = 0.0015

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal grass-lined earth channel with a bottom width of 3 m and side slopes of 2:1 (horizontal:vertical) must carry 4.8 m³/s in uniform flow at a depth of 1.3 m. Manning n = 0.03. Find the required bed slope S (in m/m).

Q = 4.8 m³/sb = 3 my = 1.3 mz = 2n = 0.03

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 67.44 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 + 2.5/2 = 2.75 m

A = H·b = 2.5·1 = 2.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2.75·2.5 = 67.44 kN

Answer

F = 67.44 kN

2.Hydrostatic force

Answer: y_cp = 3.111 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 + 2/2 = 3 m

A = H·b = 2·3 = 6 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3·6 = 176.6 kN

Centre of pressure

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

I_c = b·H³/12 = 3·2³/12 = 2 m⁴

y_cp = y_c + I_c/(y_c·A) = 3 + 2/(3·6) = 3.111 m

Answer

y_cp = 3.111 m

3.Continuity & Bernoulli

Answer: v = 8.718 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 = √(2·38000/1000) = 8.718 m/s

Answer

v = 8.718 m/s

4.Continuity & Bernoulli

Answer: Q = 11.56 L/s

Pressure difference from the manometer

Δp = (13600 − 1000)·9.81·0.06 = 7416 Pa

Throat velocity

Continuity and Bernoulli together give the ideal throat velocity.

β = 60/100 = 0.6

v₂ = √(2·7416/(1000·(1 − 0.1296))) = 4.128 m/s

Actual flow rate

Q = C_d·A₂·v₂ = 0.99·0.002827·4.128 = 0.01156 m³/s = 11.56 L/s

Answer

Q = 11.56 L/s

5.Pipe friction & head loss

Answer: h_f = 8.705 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·0.15²/4 = 0.01767 m²

V = Q/A = 0.02/0.01767 = 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) = 0.025·(800/0.15)·1.132²/(2·9.81) = 8.705 m

Answer

h_f = 8.705 m

6.Pipe friction & head loss

Answer: H = 7.495 m

Velocity

V = Q/A = 0.04/0.01767 = 2.264 m/s

V²/(2g) = 0.2611 m

Loss coefficients

f·L/D = 0.018·200/0.15 = 24

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 + 24 + 3.7)·0.2611 = 7.495 m

Answer

H = 7.495 m

7.Pumps & pump power

Answer: P_el = 8.53 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 = 1000·9.81·0.01·56 = 5.494 kW

Shaft power

P_shaft = 5.494/0.7 = 7.848 kW

Electrical power

P_el = 7.848/0.92 = 8.53 kW

Answer

P_el = 8.53 kW

8.Pumps & pump power

Answer: H = 66.93 m

Velocity

V = Q/A = 0.12/0.04909 = 2.445 m/s

V²/(2g) = 0.3046 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (35.2 + 8)·0.3046 = 13.16 m

Pressure head

p/(ρ·g) = 400000/(1000·9.81) = 40.77 m

Pump head

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

H = Δz + p/(ρ·g) + h_L = 13 + 40.77 + 13.16 = 66.93 m

Answer

H = 66.93 m

9.Open-channel flow (Manning)

Answer: Q = 33.66 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 = (3.5 + 2·1.7)·1.7 = 11.73 m²

P = b + 2y·√(1 + z²) = 3.5 + 2·1.7·√(1 + 2²) = 11.1 m

R = A/P = 1.057 m

Velocity

V = (1/0.014)·1.057^2/3·0.0015^1/2 = 2.87 m/s

Flow rate

Q = A·V = 11.73·2.87 = 33.66 m³/s

Answer

Q = 33.66 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.0005049 m/m

Geometry

A = (b + z·y)·y = (3 + 2·1.3)·1.3 = 7.28 m²

P = b + 2y·√(1 + z²) = 3 + 2·1.3·√(1 + 2²) = 8.814 m

R = A/P = 0.826 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.03·4.8/(7.28·0.8803) = 0.02247

S = 0.02247² = 0.0005049

Answer

S = 0.0005049

That is a fall of 0.5049 m per kilometre.

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