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 = 1.5 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 3 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

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

y_cp = m

3.Continuity & BernoulliMedium

A pitot-static tube in a wind tunnel reads a difference of 1300 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.

p₀ − p = 1300 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 30 kPa. Water leaves through an opening with a diameter of 20 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 = 30 kPa (gauge)h = 1 md = 20 mmρ = 1000 kg/m³g = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 1.5 m/s through a pipe 800 m long with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss h_f.

V = 1.5 m/sL = 800 mD = 100 mmf = 0.02g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

A pipe 150 m long with an inside diameter of 100 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 = 5). 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 = 0.015K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 5

H = m

7.Pumps & pump powerMedium

A pump delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 70 m. The pump efficiency is 72% and the electric motor driving it has an efficiency of 95%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 50 L/sH = 70 mη_pump = 72%η_motor = 95%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 15 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 5 m higher. The pipe is 400 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 82%. Find the shaft power P_shaft in kW.

Q = 15 L/sΔz = 5 mL = 400 mD = 100 mmf = 0.018ΣK = 8η = 82%

P_shaft = kW

9.Open-channel flow (Manning)Medium

A rectangular clean earth channel is 4.5 m wide. Water flows uniformly 1.2 m deep. Manning n = 0.022 and the bed slope is S = 0.0008. Find the flow rate Q in m³/s.

b = 4.5 my = 1.2 mn = 0.022S = 0.0008

Q = m³/s

10.Open-channel flow (Manning)Hard

A rectangular grass-lined earth channel 2 m wide must carry 3.7 m³/s in uniform flow at a depth of 1.9 m. Manning n = 0.03. Find the required bed slope S (in m/m).

Q = 3.7 m³/sb = 2 my = 1.9 mn = 0.03

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 55.18 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 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.75·1.5 = 55.18 kN

Answer

F = 55.18 kN

2.Hydrostatic force

Answer: y_cp = 4.667 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 = 3 + 3/2 = 4.5 m

A = H·b = 3·2 = 6 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·4.5·6 = 233.1 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) = 4.5 + 4.5/(4.5·6) = 4.667 m

Answer

y_cp = 4.667 m

3.Continuity & Bernoulli

Answer: v = 46.55 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 = √(2·1300/1.2) = 46.55 m/s

Answer

v = 46.55 m/s

4.Continuity & Bernoulli

Answer: Q = 2.803 L/s

Bernoulli from the surface to the jet

v = √(2·(30000/1000 + 9.81·1)) = 8.923 m/s

Flow rate

Q = A·v = 0.0003142·8.923 = 0.002803 m³/s = 2.803 L/s

Answer

Q = 2.803 L/s

5.Pipe friction & head loss

Answer: h_f = 18.35 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.

Velocity head

V²/(2g) = 1.5²/(2·9.81) = 0.1147 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.02·(800/0.1)·1.5²/(2·9.81) = 18.35 m

Answer

h_f = 18.35 m

6.Pipe friction & head loss

Answer: H = 9.882 m

Velocity

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

V²/(2g) = 0.3305 m

Loss coefficients

f·L/D = 0.015·150/0.1 = 22.5

ΣK = 0.5 + 3·0.3 + 5 = 6.4

Energy equation

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

H = (1 + 22.5 + 6.4)·0.3305 = 9.882 m

Answer

H = 9.882 m

7.Pumps & pump power

Answer: P_el = 50.2 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 = 1000·9.81·0.05·70 = 34.34 kW

Shaft power

P_shaft = 34.34/0.72 = 47.69 kW

Electrical power

P_el = 47.69/0.95 = 50.2 kW

Answer

P_el = 50.2 kW

8.Pumps & pump power

Answer: P_shaft = 3.566 kW

Velocity

V = Q/A = 0.015/0.007854 = 1.91 m/s

V²/(2g) = 0.1859 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (72 + 8)·0.1859 = 14.87 m

Pump head

H = Δz + h_L = 5 + 14.87 = 19.87 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.015·19.87/0.82 = 3.566 kW

Answer

P_shaft = 3.566 kW

9.Open-channel flow (Manning)

Answer: Q = 5.896 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 = 4.5·1.2 = 5.4 m²

P = b + 2y = 4.5 + 2·1.2 = 6.9 m

R = A/P = 0.7826 m

Velocity

V = (1/0.022)·0.7826^2/3·0.0008^1/2 = 1.092 m/s

Flow rate

Q = A·V = 5.4·1.092 = 5.896 m³/s

Answer

Q = 5.896 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.001499 m/m

Geometry

A = b·y = 2·1.9 = 3.8 m²

P = b + 2y = 2 + 2·1.9 = 5.8 m

R = A/P = 0.6552 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.03·3.7/(3.8·0.7543) = 0.03872

S = 0.03872² = 0.001499

Answer

S = 0.001499

That is a fall of 1.499 m per kilometre.

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