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

d = 2 mH = 2.5 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 1 m wide sits in the wall of a tank of water, 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 = 1.5 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows at 18 L/s through a horizontal pipe that narrows from 150 mm to 100 mm. The pressure in the narrow section is 290 kPa. Ignoring losses, find the pressure p₁ in the wide section.

Q = 18 L/sd₁ = 150 mmd₂ = 100 mmp₂ = 290 kPaρ = 1000 kg/m³

p₁ = 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 2 m above the water surface, and the outlet (open to the air) is 2.5 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 = +2 mz₂ = −2.5 mg = 9.81 m/s²p_atm = 101.3 kPa

p_C = kPa

5.Pipe friction & head lossMedium

Water flows at 50 L/s through a pipe 500 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 50 L/sL = 500 mD = 300 mmf = 0.022g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.02. The minor losses are a square-edged entrance (K = 0.5), 4 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 100 L/s?

Q = 100 L/sL = 300 mD = 200 mmf = 0.02K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 10K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 18 m. The pump efficiency is 75% and the electric motor driving it has an efficiency of 94%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 60 L/sH = 18 mη_pump = 75%η_motor = 94%g = 9.81 m/s²

P_el = 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 24 m higher. The pipe is 100 m long with an inside diameter of 200 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. Find the head H the pump must add.

Q = 80 L/sΔz = 24 mL = 100 mD = 200 mmf = 0.02ΣK = 8

H = m

9.Open-channel flow (Manning)Medium

A rectangular grass-lined earth channel is 1.5 m wide. Water flows uniformly 0.8 m deep. Manning n = 0.03 and the bed slope is S = 0.003. Find the flow rate Q in m³/s.

b = 1.5 my = 0.8 mn = 0.03S = 0.003

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal finished concrete channel with a bottom width of 2 m and side slopes of 3:1 (horizontal:vertical) must carry 7.2 m³/s in uniform flow at a depth of 0.6 m. Manning n = 0.012. Find the required bed slope S (in m/m).

Q = 7.2 m³/sb = 2 my = 0.6 mz = 3n = 0.012

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 159.4 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 = 2 + 2.5/2 = 3.25 m

A = H·b = 2.5·2 = 5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.25·5 = 159.4 kN

Answer

F = 159.4 kN

2.Hydrostatic force

Answer: y_cp = 1.4 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 + 1.5/2 = 1.25 m

A = H·b = 1.5·1 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·1.25·1.5 = 18.39 kN

Centre of pressure

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

I_c = b·H³/12 = 1·1.5³/12 = 0.2813 m⁴

y_cp = y_c + I_c/(y_c·A) = 1.25 + 0.2813/(1.25·1.5) = 1.4 m

Answer

y_cp = 1.4 m

3.Continuity & Bernoulli

Answer: p₁ = 292.1 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.

Velocities

v₁ = Q/A₁ = 0.018/0.01767 = 1.019 m/s

v₂ = Q/A₂ = 0.018/0.007854 = 2.292 m/s

Bernoulli

p₁ = p₂ + ½ρ(v₂² − v₁²) = 290 + ½·1000·(2.292² − 1.019²)/1000 = 290 + 2.107 = 292.1 kPa

Answer

p₁ = 292.1 kPa

4.Continuity & Bernoulli

Answer: p_C = 57.15 kPa

Velocity in the hose

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

v = √(2g·2.5) = √(2·9.81·2.5) = 7.004 m/s

Bernoulli from the surface to C

p_atm = p_C + ½ρv² + ρgz_C

p_C = 101.3 − 1000·(7.004²/2 + 9.81·2)/1000 = 101.3 − 44.15 = 57.15 kPa

Answer

p_C = 57.15 kPa

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

5.Pipe friction & head loss

Answer: h_f = 0.9351 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 = 0.05/0.07069 = 0.7074 m/s

Velocity head

V²/(2g) = 0.7074²/(2·9.81) = 0.0255 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.022·(500/0.3)·0.7074²/(2·9.81) = 0.9351 m

Answer

h_f = 0.9351 m

6.Pipe friction & head loss

Answer: Δz = 22.05 m

Velocity

V = Q/A = 0.1/0.03142 = 3.183 m/s

V²/(2g) = 0.5164 m

Loss coefficients

f·L/D = 0.02·300/0.2 = 30

ΣK = 0.5 + 4·0.3 + 10 + 1.0 = 12.7

Total head loss

h_f = 30·0.5164 = 15.49 m

h_m = 12.7·0.5164 = 6.559 m

Δz = h_f + h_m = 22.05 m

Answer

Δz = 22.05 m

7.Pumps & pump power

Answer: P_el = 15.03 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 = 1000·9.81·0.06·18 = 10.59 kW

Shaft power

P_shaft = 10.59/0.75 = 14.13 kW

Electrical power

P_el = 14.13/0.94 = 15.03 kW

Answer

P_el = 15.03 kW

8.Pumps & pump power

Answer: H = 29.95 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) = (10 + 8)·0.3305 = 5.949 m

Pump head

Both surfaces are open and practically at rest, so pressure and velocity terms drop out.

H = Δz + h_L = 24 + 5.949 = 29.95 m

Answer

H = 29.95 m

9.Open-channel flow (Manning)

Answer: Q = 1.164 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 = 1.5·0.8 = 1.2 m²

P = b + 2y = 1.5 + 2·0.8 = 3.1 m

R = A/P = 0.3871 m

Velocity

V = (1/0.03)·0.3871^2/3·0.003^1/2 = 0.9697 m/s

Flow rate

Q = A·V = 1.2·0.9697 = 1.164 m³/s

Answer

Q = 1.164 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.004981 m/m

Geometry

A = (b + z·y)·y = (2 + 3·0.6)·0.6 = 2.28 m²

P = b + 2y·√(1 + z²) = 2 + 2·0.6·√(1 + 3²) = 5.795 m

R = A/P = 0.3935 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.012·7.2/(2.28·0.537) = 0.07057

S = 0.07057² = 0.004981

Answer

S = 0.004981

That is a fall of 4.981 m per kilometre.

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