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

d = 1.5 mH = 2 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 glycerin, 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 = 3 mb = 2 mρ = 1260 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

A pitot-static tube in a wind tunnel reads a difference of 150 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.

p₀ − p = 150 Paρ = 1.2 kg/m³

v = m/s

4.Continuity & BernoulliHard

A closed water tank has compressed air above the water at a gauge pressure of 100 kPa. Water leaves through an opening with a diameter of 50 mm, 1 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².

p_air = 100 kPa (gauge)h = 1 md = 50 mmρ = 1000 kg/m³g = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water flows at 25 L/s through a pipe 500 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 25 L/sL = 500 mD = 150 mmf = 0.02g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

A pipe 100 m long with an inside diameter of 250 mm (Darcy friction factor f = 0.022) 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 50 L/s?

Q = 50 L/sL = 100 mD = 250 mmf = 0.022K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 2

H = m

7.Pumps & pump powerMedium

A pump with an efficiency of 65% delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 78 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 25 L/sH = 78 mη = 65%g = 9.81 m/s²

P_shaft = 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 35 m higher. The pipe is 150 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.02; 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 250 kPa. Find the head H the pump must add.

Q = 120 L/sΔz = 35 mL = 150 mD = 250 mmf = 0.02ΣK = 8p_tank = 250 kPa

H = m

9.Open-channel flow (Manning)Medium

A trapezoidal finished concrete channel has a bottom width of 2 m and side slopes of 1.5:1 (horizontal:vertical). Water flows uniformly 1 m deep. Manning n = 0.012 and the bed slope is S = 0.001. Find the flow rate Q in m³/s.

b = 2 my = 1 mz = 1.5n = 0.012S = 0.001

Q = m³/s

10.Open-channel flow (Manning)Hard

A concrete sewer with an inside diameter of 300 mm (Manning n = 0.013) must reach a self-cleansing velocity of 0.75 m/s when flowing just full. Find the minimum slope S (in m/m).

D = 300 mmn = 0.013V = 0.75 m/s

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 98.1 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 + 2/2 = 2.5 m

A = H·b = 2·2 = 4 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2.5·4 = 98.1 kN

Answer

F = 98.1 kN

2.Hydrostatic force

Answer: y_cp = 4.188 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 + 3/2 = 4 m

A = H·b = 3·2 = 6 m²

Resultant force

F = ρ·g·y_c·A = 1260·9.81·4·6 = 296.7 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) = 4 + 4.5/(4·6) = 4.188 m

Answer

y_cp = 4.188 m

3.Continuity & Bernoulli

Answer: v = 15.81 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 = √(2·150/1.2) = 15.81 m/s

Answer

v = 15.81 m/s

4.Continuity & Bernoulli

Answer: Q = 29.1 L/s

Bernoulli from the surface to the jet

v = √(2·(100000/1000 + 9.81·1)) = 14.82 m/s

Flow rate

Q = A·v = 0.001963·14.82 = 0.0291 m³/s = 29.1 L/s

Answer

Q = 29.1 L/s

5.Pipe friction & head loss

Answer: h_f = 6.801 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 = 0.025/0.01767 = 1.415 m/s

Velocity head

V²/(2g) = 1.415²/(2·9.81) = 0.102 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.02·(500/0.15)·1.415²/(2·9.81) = 6.801 m

Answer

h_f = 6.801 m

6.Pipe friction & head loss

Answer: H = 0.7139 m

Velocity

V = Q/A = 0.05/0.04909 = 1.019 m/s

V²/(2g) = 0.05288 m

Loss coefficients

f·L/D = 0.022·100/0.25 = 8.8

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 + 8.8 + 3.7)·0.05288 = 0.7139 m

Answer

H = 0.7139 m

7.Pumps & pump power

Answer: P_shaft = 29.43 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 = 1000·9.81·0.025·78 = 19.13 kW

Shaft power

P_shaft = P_h/η = 19.13/0.65 = 29.43 kW

Answer

P_shaft = 29.43 kW

8.Pumps & pump power

Answer: H = 66.58 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) = (12 + 8)·0.3046 = 6.092 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 = 35 + 25.48 + 6.092 = 66.58 m

Answer

H = 66.58 m

9.Open-channel flow (Manning)

Answer: Q = 6.738 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 = (2 + 1.5·1)·1 = 3.5 m²

P = b + 2y·√(1 + z²) = 2 + 2·1·√(1 + 1.5²) = 5.606 m

R = A/P = 0.6244 m

Velocity

V = (1/0.012)·0.6244^2/3·0.001^1/2 = 1.925 m/s

Flow rate

Q = A·V = 3.5·1.925 = 6.738 m³/s

Answer

Q = 6.738 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.003006 m/m

Hydraulic radius

R = D/4 = 0.3/4 = 0.075 m

Solve Manning for S

S = (n·V/R^2/3)² = (0.013·0.75/0.1778)² = 0.003006

Answer

S = 0.003006

That is a fall of 3.006 m per kilometre.

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