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

d = 1.5 mH = 1 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 3 m wide sits in the wall of a tank of oil, with its top edge 4 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 4 mH = 2.5 mb = 3 mρ = 880 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

A pitot-static tube in a water pipe reads a difference of 16 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.

p₀ − p = 16 kPaρ = 1000 kg/m³

v = m/s

4.Continuity & BernoulliHard

A closed water tank has compressed air above the water at a gauge pressure of 180 kPa. Water leaves through an opening with a diameter of 20 mm, 2.5 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².

p_air = 180 kPa (gauge)h = 2.5 md = 20 mmρ = 1000 kg/m³g = 9.81 m/s²

v = m/s

5.Pipe friction & head lossMedium

Water flows at 30 L/s through a pipe 400 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 30 L/sL = 400 mD = 250 mmf = 0.03g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.02. The minor losses are a square-edged entrance (K = 0.5), 2 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 200 L/s?

Q = 200 L/sL = 150 mD = 300 mmf = 0.02K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 5K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 46 m. The pump efficiency is 78% and the electric motor driving it has an efficiency of 90%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 25 L/sH = 46 mη_pump = 78%η_motor = 90%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 9 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 = 5 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 82%. Find the shaft power P_shaft in kW.

Q = 50 L/sΔz = 9 mL = 100 mD = 200 mmf = 0.02ΣK = 5η = 82%

P_shaft = kW

9.Open-channel flow (Manning)Medium

A rectangular gravel channel is 5 m wide. Water flows uniformly 1.9 m deep. Manning n = 0.025 and the bed slope is S = 0.005. Find the flow rate Q in m³/s.

b = 5 my = 1.9 mn = 0.025S = 0.005

Q = m³/s

10.Open-channel flow (Manning)Hard

A rectangular unfinished concrete channel 6 m wide must carry 15 m³/s in uniform flow at a depth of 0.7 m. Manning n = 0.014. Find the required bed slope S (in m/m).

Q = 15 m³/sb = 6 my = 0.7 mn = 0.014

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 19.62 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 + 1/2 = 2 m

A = H·b = 1·1 = 1 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2·1 = 19.62 kN

Answer

F = 19.62 kN

2.Hydrostatic force

Answer: y_cp = 5.349 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 = 4 + 2.5/2 = 5.25 m

A = H·b = 2.5·3 = 7.5 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·5.25·7.5 = 339.9 kN

Centre of pressure

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

I_c = b·H³/12 = 3·2.5³/12 = 3.906 m⁴

y_cp = y_c + I_c/(y_c·A) = 5.25 + 3.906/(5.25·7.5) = 5.349 m

Answer

y_cp = 5.349 m

3.Continuity & Bernoulli

Answer: v = 5.657 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 = √(2·16000/1000) = 5.657 m/s

Answer

v = 5.657 m/s

4.Continuity & Bernoulli

Answer: v = 20.22 m/s

Bernoulli from the surface to the jet

v = √(2·(180000/1000 + 9.81·2.5)) = 20.22 m/s

Answer

v = 20.22 m/s

5.Pipe friction & head loss

Answer: h_f = 0.9138 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·0.25²/4 = 0.04909 m²

V = Q/A = 0.03/0.04909 = 0.6112 m/s

Velocity head

V²/(2g) = 0.6112²/(2·9.81) = 0.01904 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.03·(400/0.25)·0.6112²/(2·9.81) = 0.9138 m

Answer

h_f = 0.9138 m

6.Pipe friction & head loss

Answer: Δz = 6.977 m

Velocity

V = Q/A = 0.2/0.07069 = 2.829 m/s

V²/(2g) = 0.408 m

Loss coefficients

f·L/D = 0.02·150/0.3 = 10

ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1

Total head loss

h_f = 10·0.408 = 4.08 m

h_m = 7.1·0.408 = 2.897 m

Δz = h_f + h_m = 6.977 m

Answer

Δz = 6.977 m

7.Pumps & pump power

Answer: P_el = 16.07 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 = 1000·9.81·0.025·46 = 11.28 kW

Shaft power

P_shaft = 11.28/0.78 = 14.46 kW

Electrical power

P_el = 14.46/0.9 = 16.07 kW

Answer

P_el = 16.07 kW

8.Pumps & pump power

Answer: P_shaft = 6.542 kW

Velocity

V = Q/A = 0.05/0.03142 = 1.592 m/s

V²/(2g) = 0.1291 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (10 + 5)·0.1291 = 1.937 m

Pump head

H = Δz + h_L = 9 + 1.937 = 10.94 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.05·10.94/0.82 = 6.542 kW

Answer

P_shaft = 6.542 kW

9.Open-channel flow (Manning)

Answer: Q = 28.28 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 = 5·1.9 = 9.5 m²

P = b + 2y = 5 + 2·1.9 = 8.8 m

R = A/P = 1.08 m

Velocity

V = (1/0.025)·1.08^2/3·0.005^1/2 = 2.976 m/s

Flow rate

Q = A·V = 9.5·2.976 = 28.28 m³/s

Answer

Q = 28.28 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.00532 m/m

Geometry

A = b·y = 6·0.7 = 4.2 m²

P = b + 2y = 6 + 2·0.7 = 7.4 m

R = A/P = 0.5676 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.014·15/(4.2·0.6855) = 0.07294

S = 0.07294² = 0.00532

Answer

S = 0.00532

That is a fall of 5.32 m per kilometre.

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