Hydrostatic force
- Pressure at depth
- p = ρgh
- Force on a vertical wall
- F = ρg·hc·A
- Centre of pressure
- yp = yc + Ic/(ycA)
- Rectangle
- Ic = bH³/12
Hydraulic Engineering: cram pack
Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
1Formulas and methods
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
- hf = f·(L/D)·V²/(2g)
- Minor losses
- hm = ΣK·V²/(2g)
- Typical K values
- entrance 0.5, exit 1.0, 90° bend ≈ 0.3
- Pressure drop
- Δp = ρ·g·hf
- 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
- Ph = ρ·g·Q·H = Q·Δp
- Shaft power
- Pshaft = Ph/ηpump
- Electrical power
- Pel = Pshaft/ηmotor
- Pump head
- H = Δz + Δp/(ρ·g) + hL
- System losses
- hL = (f·L/D + ΣK)·V²/(2g)
Open-channel flow (Manning)
- Manning equation
- V = (1/n)·R2/3·S1/2
- 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·R2/3))²
2Practice problems
1.Hydrostatic forceMedium
A vertical rectangular gate 3 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 = 2 mb = 3 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 water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 2.5 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 250 mm and a throat diameter of 125 mm. The pressure at the inlet is 22 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 250 mmd2 = 125 mmp1 − p2 = 22 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 200 mm to 75 mm while rising 1 m. At section 1 the pressure is 250 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 200 mmd2 = 75 mmp1 = 250 kPav1 = 1 m/sΔz = 1 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.5 m/s through a horizontal pipe, 200 m long, with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.5 m/sL = 200 mD = 200 mmf = 0.018ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 300 m long with an inside diameter of 300 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), 4 bends (K = 0.3 each) and a valve (K = 10). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 80 L/s?
Q = 80 L/sL = 300 mD = 300 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10
H = m
7.Pumps & pump powerMedium
A pump with an efficiency of 82% delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 30 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 50 L/sH = 30 mη = 82%g = 9.81 m/s²
Pshaft = 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 35 m higher. The pipe is 200 m long with an inside diameter of 250 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. Find the head H the pump must add.
Q = 120 L/sΔz = 35 mL = 200 mD = 250 mmf = 0.018ΣK = 8
H = m
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 750 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.0008. Find the flow rate Q in m³/s.
D = 750 mmn = 0.013S = 0.0008
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal brickwork channel with a bottom width of 3.5 m and side slopes of 2:1 (horizontal:vertical) must carry 1.8 m³/s in uniform flow at a depth of 0.5 m. Manning n = 0.015. Find the required bed slope S (in m/m).
Q = 1.8 m³/sb = 3.5 my = 0.5 mz = 2n = 0.015
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 147.2 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 1.5 + 2/2 = 2.5 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.5·6 = 147.2 kN
Answer
F = 147.2 kN
2.Hydrostatic force
Answer: F = 257.5 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 4 + 2.5/2 = 5.25 m
A = H·b = 2.5·2 = 5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·5.25·5 = 257.5 kN
Answer
F = 257.5 kN
3.Continuity & Bernoulli
Answer: Q = 84.07 L/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.
Continuity
v1 = v2·(125/250)² = 0.25·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·22000/(1000·(1 − 0.0625))) = 6.851 m/s
v1 = 1.713 m/s
Flow rate
Q = A2·v2 = 0.01227·6.851 = 0.08407 m³/s = 84.07 L/s
Answer
Q = 84.07 L/s
4.Continuity & Bernoulli
Answer: p2 = 215.4 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·(200/75)² = 7.111 m/s
½ρ(v1² − v2²) = ½·1000·(1² − 7.111²)/1000 = −24.78 kPa
ρgΔz = 1000·9.81·1/1000 = 9.81 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 250 + (−24.78) − 9.81 = 215.4 kPa
Answer
p2 = 215.4 kPa
5.Pipe friction & head loss
Answer: Δp = 56.25 kPa
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 hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Head loss
hf = f·(L/D)·V²/(2g) = 0.018·(200/0.2)·2.5²/(2·9.81) = 5.734 m
Pressure drop
The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·hf.
Δp = ρ·g·hf = 1000·9.81·5.734 = 56250 Pa = 56.25 kPa
Answer
Δp = 56.25 kPa
6.Pipe friction & head loss
Answer: H = 2.135 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 hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Velocity
V = Q/A = 0.08/0.07069 = 1.132 m/s
V²/(2g) = 0.06529 m
Loss coefficients
f·L/D = 0.02·300/0.3 = 20
ΣK = 0.5 + 4·0.3 + 10 = 11.7
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 20 + 11.7)·0.06529 = 2.135 m
Answer
H = 2.135 m
7.Pumps & pump power
Answer: Pshaft = 17.95 kW
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Hydraulic power
Ph = ρ·g·Q·H = 1000·9.81·0.05·30 = 14.71 kW
Shaft power
Pshaft = Ph/η = 14.71/0.82 = 17.95 kW
Answer
Pshaft = 17.95 kW
8.Pumps & pump power
Answer: H = 41.82 m
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Velocity
V = Q/A = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (14.4 + 8)·0.3046 = 6.823 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 35 + 6.823 = 41.82 m
Answer
H = 41.82 m
9.Open-channel flow (Manning)
Answer: Q = 0.3149 m³/s
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 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 = 0.4418 m², R = D/4 = 0.1875 m
Velocity
V = (1/0.013)·0.18752/3·0.00081/2 = 0.7127 m/s
Flow rate
Q = A·V = 0.4418·0.7127 = 0.3149 m³/s
Answer
Q = 0.3149 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0005015 m/m
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 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 = (3.5 + 2·0.5)·0.5 = 2.25 m²
P = b + 2y·√(1 + z²) = 3.5 + 2·0.5·√(1 + 2²) = 5.736 m
R = A/P = 0.3923 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.015·1.8/(2.25·0.5359) = 0.02239
S = 0.02239² = 0.0005015
Answer
S = 0.0005015
That is a fall of 0.5015 m per kilometre.
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