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

d = 1 mH = 1 mb = 2 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 0.5 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 0.5 mH = 2.5 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 100 mm to 75 mm. In the wide section the pressure is 370 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 100 mmd₂ = 75 mmp₁ = 370 kPav₁ = 1.25 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

A hose with an inner diameter of 25 mm siphons water from an open tank. The highest point C of the hose is 1.5 m above the water surface, and the outlet (open to the air) is 1 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the absolute pressure p_C at point C. The atmospheric pressure is 101.3 kPa.

d = 25 mmz_C = +1.5 mz₂ = −1 mg = 9.81 m/s²p_atm = 101.3 kPa

p_C = kPa

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 2.5 m/s through a pipe 800 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss h_f.

V = 2.5 m/sL = 800 mD = 200 mmf = 0.02g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.022. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 5) 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 60 L/s?

Q = 60 L/sL = 200 mD = 150 mmf = 0.022K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 5K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

In a test, a pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 56 m while its shaft takes 17.2 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.

Q = 25 L/sH = 56 mP_shaft = 17.2 kWg = 9.81 m/s²

η = %

8.Pumps & pump powerHard

A pump moves 40 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 500 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 85%. Find the shaft power P_shaft in kW.

Q = 40 L/sΔz = 5 mL = 500 mD = 150 mmf = 0.022ΣK = 6η = 85%

P_shaft = kW

9.Open-channel flow (Manning)Medium

A trapezoidal grass-lined earth channel has a bottom width of 4.5 m and side slopes of 3:1 (horizontal:vertical). Water flows uniformly 2.3 m deep. Manning n = 0.03 and the bed slope is S = 0.0005. Find the flow rate Q in m³/s.

b = 4.5 my = 2.3 mz = 3n = 0.03S = 0.0005

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal clean earth channel with a bottom width of 5 m and side slopes of 3:1 (horizontal:vertical) must carry 12 m³/s in uniform flow at a depth of 1.2 m. Manning n = 0.022. Find the required bed slope S (in m/m).

Q = 12 m³/sb = 5 my = 1.2 mz = 3n = 0.022

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 29.43 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 + 1/2 = 1.5 m

A = H·b = 1·2 = 2 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·1.5·2 = 29.43 kN

Answer

F = 29.43 kN

2.Hydrostatic force

Answer: y_cp = 2.048 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 = 0.5 + 2.5/2 = 1.75 m

A = H·b = 2.5·1.5 = 3.75 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·1.75·3.75 = 56.65 kN

Centre of pressure

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

I_c = b·H³/12 = 1.5·2.5³/12 = 1.953 m⁴

y_cp = y_c + I_c/(y_c·A) = 1.75 + 1.953/(1.75·3.75) = 2.048 m

Answer

y_cp = 2.048 m

3.Continuity & Bernoulli

Answer: p₂ = 368.3 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 = 1.25·(100/75)² = 2.222 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.25² − 2.222²)/1000 = −1.688 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 370 + (−1.688) = 368.3 kPa

Answer

p₂ = 368.3 kPa

4.Continuity & Bernoulli

Answer: p_C = 76.78 kPa

Velocity in the hose

Surface and outlet are both open to the air, so the pressure terms cancel.

v = √(2g·1) = √(2·9.81·1) = 4.429 m/s

Bernoulli from the surface to C

p_atm = p_C + ½ρv² + ρgz_C

p_C = 101.3 − 1000·(4.429²/2 + 9.81·1.5)/1000 = 101.3 − 24.53 = 76.78 kPa

Answer

p_C = 76.78 kPa

If this pressure falls to the vapour pressure of the water, the siphon stops working.

5.Pipe friction & head loss

Answer: h_f = 25.48 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) = 2.5²/(2·9.81) = 0.3186 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.02·(800/0.2)·2.5²/(2·9.81) = 25.48 m

Answer

h_f = 25.48 m

6.Pipe friction & head loss

Answer: Δz = 21.76 m

Velocity

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

V²/(2g) = 0.5876 m

Loss coefficients

f·L/D = 0.022·200/0.15 = 29.33

ΣK = 0.5 + 4·0.3 + 5 + 1.0 = 7.7

Total head loss

h_f = 29.33·0.5876 = 17.24 m

h_m = 7.7·0.5876 = 4.524 m

Δz = h_f + h_m = 21.76 m

Answer

Δz = 21.76 m

7.Pumps & pump power

Answer: η = 79.85 %

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 = 1000·9.81·0.025·56 = 13.73 kW

Efficiency

η = P_h/P_shaft = 13.73/17.2 = 79.85%

Answer

η = 79.85%

8.Pumps & pump power

Answer: P_shaft = 11.87 kW

Velocity

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

V²/(2g) = 0.2611 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (73.33 + 6)·0.2611 = 20.72 m

Pump head

H = Δz + h_L = 5 + 20.72 = 25.72 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.04·25.72/0.85 = 11.87 kW

Answer

P_shaft = 11.87 kW

9.Open-channel flow (Manning)

Answer: Q = 24.18 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 = (4.5 + 3·2.3)·2.3 = 26.22 m²

P = b + 2y·√(1 + z²) = 4.5 + 2·2.3·√(1 + 3²) = 19.05 m

R = A/P = 1.377 m

Velocity

V = (1/0.03)·1.377^2/3·0.0005^1/2 = 0.9224 m/s

Flow rate

Q = A·V = 26.22·0.9224 = 24.18 m³/s

Answer

Q = 24.18 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.000853 m/m

Geometry

A = (b + z·y)·y = (5 + 3·1.2)·1.2 = 10.32 m²

P = b + 2y·√(1 + z²) = 5 + 2·1.2·√(1 + 3²) = 12.59 m

R = A/P = 0.8197 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.022·12/(10.32·0.8759) = 0.02921

S = 0.02921² = 0.000853

Answer

S = 0.000853

That is a fall of 0.853 m per kilometre.

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