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 1.5 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 = 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 seawater, with its top edge 1 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1 mH = 1 mb = 2 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 200 mm to 100 mm. In the wide section the pressure is 400 kPa and the velocity is 2.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 200 mmd2 = 100 mmp1 = 400 kPav1 = 2.5 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 75 mm to 50 mm while rising 9 m. At section 1 the pressure is 280 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 75 mmd2 = 50 mmp1 = 280 kPav1 = 2.25 m/sΔz = 9 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 60 L/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 hf.
Q = 60 L/sL = 1000 mD = 200 mmf = 0.028g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 200 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.025) 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), 3 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 = 200 mD = 300 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10
H = m
7.Pumps & pump powerMedium
A pump delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 34 m. The pump efficiency is 65% and the electric motor driving it has an efficiency of 95%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 15 L/sH = 34 mηpump = 65%ηmotor = 95%g = 9.81 m/s²
Pel = 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 31 m higher. The pipe is 300 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 70%. Find the shaft power Pshaft in kW.
Q = 80 L/sΔz = 31 mL = 300 mD = 200 mmf = 0.022ΣK = 8η = 70%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 450 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.005. Find the flow rate Q in m³/s.
D = 450 mmn = 0.013S = 0.005
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular clean earth channel 6 m wide must carry 15 m³/s in uniform flow at a depth of 1.7 m. Manning n = 0.022. Find the required bed slope S (in m/m).
Q = 15 m³/sb = 6 my = 1.7 mn = 0.022
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 119.6 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 = 2 + 2.5/2 = 3.25 m
A = H·b = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.25·3.75 = 119.6 kN
Answer
F = 119.6 kN
2.Hydrostatic force
Answer: ycp = 1.556 m
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 + 1/2 = 1.5 m
A = H·b = 1·2 = 2 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·1.5·2 = 30.17 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 2·1³/12 = 0.1667 m⁴
ycp = yc + Ic/(yc·A) = 1.5 + 0.1667/(1.5·2) = 1.556 m
Answer
ycp = 1.556 m
3.Continuity & Bernoulli
Answer: p2 = 353.1 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.
Continuity and Bernoulli
Work in Pa, then divide by 1000 to get kPa.
v2 = v1·(d1/d2)² = 2.5·(200/100)² = 10 m/s
½ρ(v1² − v2²) = ½·1000·(2.5² − 10²)/1000 = −46.88 kPa
p2 = p1 + ½ρ(v1² − v2²) = 400 + (−46.88) = 353.1 kPa
Answer
p2 = 353.1 kPa
4.Continuity & Bernoulli
Answer: p2 = 181.4 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 = 2.25·(75/50)² = 5.063 m/s
½ρ(v1² − v2²) = ½·1000·(2.25² − 5.063²)/1000 = −10.28 kPa
ρgΔz = 1000·9.81·9/1000 = 88.29 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 280 + (−10.28) − 88.29 = 181.4 kPa
Answer
p2 = 181.4 kPa
5.Pipe friction & head loss
Answer: hf = 26.03 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.
Mean velocity
A = π·0.2²/4 = 0.03142 m²
V = Q/A = 0.06/0.03142 = 1.91 m/s
Velocity head
V²/(2g) = 1.91²/(2·9.81) = 0.1859 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(1000/0.2)·1.91²/(2·9.81) = 26.03 m
Answer
hf = 26.03 m
6.Pipe friction & head loss
Answer: H = 1.898 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 = 0.025·200/0.3 = 16.67
ΣK = 0.5 + 3·0.3 + 10 = 11.4
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 16.67 + 11.4)·0.06529 = 1.898 m
Answer
H = 1.898 m
7.Pumps & pump power
Answer: Pel = 8.102 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 = 1000·9.81·0.015·34 = 5.003 kW
Shaft power
Pshaft = 5.003/0.65 = 7.697 kW
Electrical power
Pel = 7.697/0.95 = 8.102 kW
Answer
Pel = 8.102 kW
8.Pumps & pump power
Answer: Pshaft = 49.95 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.
Velocity
V = Q/A = 0.08/0.03142 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (33 + 8)·0.3305 = 13.55 m
Pump head
H = Δz + hL = 31 + 13.55 = 44.55 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.08·44.55/0.7 = 49.95 kW
Answer
Pshaft = 49.95 kW
9.Open-channel flow (Manning)
Answer: Q = 0.2016 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 = 0.159 m², R = D/4 = 0.1125 m
Velocity
V = (1/0.013)·0.11252/3·0.0051/2 = 1.268 m/s
Flow rate
Q = A·V = 0.159·1.268 = 0.2016 m³/s
Answer
Q = 0.2016 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0009387 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·y = 6·1.7 = 10.2 m²
P = b + 2y = 6 + 2·1.7 = 9.4 m
R = A/P = 1.085 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.022·15/(10.2·1.056) = 0.03064
S = 0.03064² = 0.0009387
Answer
S = 0.0009387
That is a fall of 0.9387 m per kilometre.
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