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 1.5 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 = 2 mb = 1.5 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 = 2.5 mb = 2 mρ = 1260 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 150 mm to 75 mm. In the wide section the pressure is 110 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 150 mmd₂ = 75 mmp₁ = 110 kPav₁ = 1 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 100 mm to 40 mm while rising 1 m. At section 1 the pressure is 260 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 100 mmd₂ = 40 mmp₁ = 260 kPav₁ = 2 m/sΔz = 1 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 2 m/s through a pipe 1000 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss h_f.

V = 2 m/sL = 1000 mD = 200 mmf = 0.028g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

A pipe 100 m long with an inside diameter of 150 mm (Darcy friction factor f = 0.02) 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), 2 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 60 L/s?

Q = 60 L/sL = 100 mD = 150 mmf = 0.02K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 2

H = m

7.Pumps & pump powerMedium

A pump with an efficiency of 85% delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 58 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 80 L/sH = 58 mη = 85%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 14 m higher. The pipe is 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.015; 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 150 kPa. Find the head H the pump must add.

Q = 120 L/sΔz = 14 mL = 300 mD = 250 mmf = 0.015ΣK = 8p_tank = 150 kPa

H = m

9.Open-channel flow (Manning)Medium

A rectangular grass-lined earth channel is 5.5 m wide. Water flows uniformly 1.5 m deep. Manning n = 0.03 and the bed slope is S = 0.003. Find the flow rate Q in m³/s.

b = 5.5 my = 1.5 mn = 0.03S = 0.003

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 1.2 m/s when flowing just full. Find the minimum slope S (in m/m).

D = 300 mmn = 0.013V = 1.2 m/s

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 117.7 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 + 2/2 = 4 m

A = H·b = 2·1.5 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4·3 = 117.7 kN

Answer

F = 117.7 kN

2.Hydrostatic force

Answer: y_cp = 3.889 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 + 2.5/2 = 3.75 m

A = H·b = 2.5·2 = 5 m²

Resultant force

F = ρ·g·y_c·A = 1260·9.81·3.75·5 = 231.8 kN

Centre of pressure

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

I_c = b·H³/12 = 2·2.5³/12 = 2.604 m⁴

y_cp = y_c + I_c/(y_c·A) = 3.75 + 2.604/(3.75·5) = 3.889 m

Answer

y_cp = 3.889 m

3.Continuity & Bernoulli

Answer: p₂ = 102.5 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·(150/75)² = 4 m/s

½ρ(v₁² − v₂²) = ½·1000·(1² − 4²)/1000 = −7.5 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 110 + (−7.5) = 102.5 kPa

Answer

p₂ = 102.5 kPa

4.Continuity & Bernoulli

Answer: p₂ = 174.1 kPa

Continuity and Bernoulli

Work in Pa, then divide by 1000 to get kPa.

v₂ = v₁·(d₁/d₂)² = 2·(100/40)² = 12.5 m/s

½ρ(v₁² − v₂²) = ½·1000·(2² − 12.5²)/1000 = −76.13 kPa

ρgΔz = 1000·9.81·1/1000 = 9.81 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 260 + (−76.13) − 9.81 = 174.1 kPa

Answer

p₂ = 174.1 kPa

5.Pipe friction & head loss

Answer: h_f = 28.54 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) = 2²/(2·9.81) = 0.2039 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.028·(1000/0.2)·2²/(2·9.81) = 28.54 m

Answer

h_f = 28.54 m

6.Pipe friction & head loss

Answer: H = 10.24 m

Velocity

V = Q/A = 0.06/0.01767 = 3.395 m/s

V²/(2g) = 0.5876 m

Loss coefficients

f·L/D = 0.02·100/0.15 = 13.33

ΣK = 0.5 + 2·0.3 + 2 = 3.1

Energy equation

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

H = (1 + 13.33 + 3.1)·0.5876 = 10.24 m

Answer

H = 10.24 m

7.Pumps & pump power

Answer: P_shaft = 53.55 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 = 1000·9.81·0.08·58 = 45.52 kW

Shaft power

P_shaft = P_h/η = 45.52/0.85 = 53.55 kW

Answer

P_shaft = 53.55 kW

8.Pumps & pump power

Answer: H = 37.21 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) = (18 + 8)·0.3046 = 7.919 m

Pressure head

p/(ρ·g) = 150000/(1000·9.81) = 15.29 m

Pump head

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

H = Δz + p/(ρ·g) + h_L = 14 + 15.29 + 7.919 = 37.21 m

Answer

H = 37.21 m

9.Open-channel flow (Manning)

Answer: Q = 14.77 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 = 5.5·1.5 = 8.25 m²

P = b + 2y = 5.5 + 2·1.5 = 8.5 m

R = A/P = 0.9706 m

Velocity

V = (1/0.03)·0.9706^2/3·0.003^1/2 = 1.79 m/s

Flow rate

Q = A·V = 8.25·1.79 = 14.77 m³/s

Answer

Q = 14.77 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.007694 m/m

Hydraulic radius

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

Solve Manning for S

S = (n·V/R^2/3)² = (0.013·1.2/0.1778)² = 0.007694

Answer

S = 0.007694

That is a fall of 7.694 m per kilometre.

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