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

d = 3 mH = 3 mb = 1 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 3 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 3 mH = 1.5 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²

F = kN

3.Continuity & BernoulliMedium

A pitot-static tube in a water pipe reads a difference of 5 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.

p₀ − p = 5 kPaρ = 1000 kg/m³

v = m/s

4.Continuity & BernoulliHard

A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 50 mm and a gauge pressure of 60 kPa. The nozzle exit diameter is 20 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 = 50 mmp₁ = 60 kPa (gauge)d₂ = 20 mmρ = 1000 kg/m³g = 9.81 m/s²

H = m

5.Pipe friction & head lossMedium

Water flows at 60 L/s through a pipe 100 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 = 60 L/sL = 100 mD = 300 mmf = 0.015g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 10) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 20 L/s?

Q = 20 L/sL = 100 mD = 150 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 10K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump with an efficiency of 82% delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 74 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 50 L/sH = 74 mη = 82%g = 9.81 m/s²

P_shaft = 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 12 m higher. The pipe is 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 4 (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 100 kPa. Find the head H the pump must add.

Q = 80 L/sΔz = 12 mL = 150 mD = 200 mmf = 0.025ΣK = 4p_tank = 100 kPa

H = m

9.Open-channel flow (Manning)Medium

A rectangular unfinished concrete channel is 6 m wide. Water flows uniformly 1.8 m deep. Manning n = 0.014 and the bed slope is S = 0.0008. Find the flow rate Q in m³/s.

b = 6 my = 1.8 mn = 0.014S = 0.0008

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal grass-lined earth channel with a bottom width of 6 m and side slopes of 1.5:1 (horizontal:vertical) must carry 17 m³/s in uniform flow at a depth of 1.7 m. Manning n = 0.03. Find the required bed slope S (in m/m).

Q = 17 m³/sb = 6 my = 1.7 mz = 1.5n = 0.03

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 132.4 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 = 3 + 3/2 = 4.5 m

A = H·b = 3·1 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4.5·3 = 132.4 kN

Answer

F = 132.4 kN

2.Hydrostatic force

Answer: F = 72.84 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 = 3 + 1.5/2 = 3.75 m

A = H·b = 1.5·1.5 = 2.25 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·3.75·2.25 = 72.84 kN

Answer

F = 72.84 kN

3.Continuity & Bernoulli

Answer: v = 3.162 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 = √(2·5000/1000) = 3.162 m/s

Answer

v = 3.162 m/s

4.Continuity & Bernoulli

Answer: H = 6.277 m

Exit velocity

β = 20/50 = 0.4

v₂ = √(2p₁/(ρ(1 − β⁴))) = √(2·60000/(1000·0.9744)) = 11.1 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) = 11.1²/(2·9.81) = 6.277 m

Answer

H = 6.277 m

5.Pipe friction & head loss

Answer: h_f = 0.1836 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 = 0.06/0.07069 = 0.8488 m/s

Velocity head

V²/(2g) = 0.8488²/(2·9.81) = 0.03672 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.015·(100/0.3)·0.8488²/(2·9.81) = 0.1836 m

Answer

h_f = 0.1836 m

6.Pipe friction & head loss

Answer: Δz = 1.462 m

Velocity

V = Q/A = 0.02/0.01767 = 1.132 m/s

V²/(2g) = 0.06529 m

Loss coefficients

f·L/D = 0.015·100/0.15 = 10

ΣK = 0.5 + 3·0.3 + 10 + 1.0 = 12.4

Total head loss

h_f = 10·0.06529 = 0.6529 m

h_m = 12.4·0.06529 = 0.8095 m

Δz = h_f + h_m = 1.462 m

Answer

Δz = 1.462 m

7.Pumps & pump power

Answer: P_shaft = 44.26 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 = 1000·9.81·0.05·74 = 36.3 kW

Shaft power

P_shaft = P_h/η = 36.3/0.82 = 44.26 kW

Answer

P_shaft = 44.26 kW

8.Pumps & pump power

Answer: H = 29.71 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) = (18.75 + 4)·0.3305 = 7.519 m

Pressure head

p/(ρ·g) = 100000/(1000·9.81) = 10.19 m

Pump head

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

H = Δz + p/(ρ·g) + h_L = 12 + 10.19 + 7.519 = 29.71 m

Answer

H = 29.71 m

9.Open-channel flow (Manning)

Answer: Q = 23.6 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 = 6·1.8 = 10.8 m²

P = b + 2y = 6 + 2·1.8 = 9.6 m

R = A/P = 1.125 m

Velocity

V = (1/0.014)·1.125^2/3·0.0008^1/2 = 2.185 m/s

Flow rate

Q = A·V = 10.8·2.185 = 23.6 m³/s

Answer

Q = 23.6 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.0009673 m/m

Geometry

A = (b + z·y)·y = (6 + 1.5·1.7)·1.7 = 14.54 m²

P = b + 2y·√(1 + z²) = 6 + 2·1.7·√(1 + 1.5²) = 12.13 m

R = A/P = 1.198 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.03·17/(14.54·1.128) = 0.0311

S = 0.0311² = 0.0009673

Answer

S = 0.0009673

That is a fall of 0.9673 m per kilometre.

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