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 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 = 1.5 mb = 1 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 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 = 3 mb = 1 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 40 mm. The water level is kept 4.5 m above the orifice and the discharge coefficient is 0.65. Find the flow rate Q. Use g = 9.81 m/s².
d = 40 mmh = 4.5 mCd = 0.65g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 50 mm while rising 8 m. At section 1 the pressure is 370 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 50 mmp1 = 370 kPav1 = 0.5 m/sΔz = 8 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 3 m/s through a horizontal pipe, 500 m long, with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 3 m/sL = 500 mD = 100 mmf = 0.015ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 100 m long with an inside diameter of 250 mm (Darcy friction factor f = 0.018) 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 = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 120 L/s?
Q = 120 L/sL = 100 mD = 250 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2
H = m
7.Pumps & pump powerMedium
A pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 52 m. The pump efficiency is 82% and the electric motor driving it has an efficiency of 90%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 25 L/sH = 52 mηpump = 82%ηmotor = 90%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 37 m higher. The pipe is 150 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 75%. Find the shaft power Pshaft in kW.
Q = 80 L/sΔz = 37 mL = 150 mD = 200 mmf = 0.022ΣK = 8η = 75%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 600 mm flows half full. Manning n = 0.013 and the slope is S = 0.004. Find the flow rate Q in m³/s.
D = 600 mmn = 0.013S = 0.004
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 600 mm (Manning n = 0.013) must reach a self-cleansing velocity of 0.75 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 600 mmn = 0.013V = 0.75 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 33.11 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 + 1.5/2 = 2.25 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.25·1.5 = 33.11 kN
Answer
F = 33.11 kN
2.Hydrostatic force
Answer: ycp = 2.8 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 + 3/2 = 2.5 m
A = H·b = 3·1 = 3 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·2.5·3 = 75.41 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1·3³/12 = 2.25 m⁴
ycp = yc + Ic/(yc·A) = 2.5 + 2.25/(2.5·3) = 2.8 m
Answer
ycp = 2.8 m
3.Continuity & Bernoulli
Answer: Q = 7.675 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.
Ideal jet velocity
√(2gh) = √(2·9.81·4.5) = 9.396 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.65·0.001257·9.396 = 0.007675 m³/s = 7.675 L/s
Answer
Q = 7.675 L/s
4.Continuity & Bernoulli
Answer: p2 = 289.6 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 = 0.5·(100/50)² = 2 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 2²)/1000 = −1.875 kPa
ρgΔz = 1000·9.81·8/1000 = 78.48 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 370 + (−1.875) − 78.48 = 289.6 kPa
Answer
p2 = 289.6 kPa
5.Pipe friction & head loss
Answer: Δp = 337.5 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) = 0.015·(500/0.1)·3²/(2·9.81) = 34.4 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 = 1000·9.81·34.4 = 337500 Pa = 337.5 kPa
Answer
Δp = 337.5 kPa
6.Pipe friction & head loss
Answer: H = 3.533 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Loss coefficients
f·L/D = 0.018·100/0.25 = 7.2
ΣK = 0.5 + 3·0.3 + 2 = 3.4
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 7.2 + 3.4)·0.3046 = 3.533 m
Answer
H = 3.533 m
7.Pumps & pump power
Answer: Pel = 17.28 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 = 1000·9.81·0.025·52 = 12.75 kW
Shaft power
Pshaft = 12.75/0.82 = 15.55 kW
Electrical power
Pel = 15.55/0.9 = 17.28 kW
Answer
Pel = 17.28 kW
8.Pumps & pump power
Answer: Pshaft = 47.19 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) = (16.5 + 8)·0.3305 = 8.097 m
Pump head
H = Δz + hL = 37 + 8.097 = 45.1 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.08·45.1/0.75 = 47.19 kW
Answer
Pshaft = 47.19 kW
9.Open-channel flow (Manning)
Answer: Q = 0.1942 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 = 0.1414 m², R = D/4 = 0.15 m
Velocity
V = (1/0.013)·0.152/3·0.0041/2 = 1.373 m/s
Flow rate
Q = A·V = 0.1414·1.373 = 0.1942 m³/s
Answer
Q = 0.1942 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.001193 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.
Hydraulic radius
R = D/4 = 0.6/4 = 0.15 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.75/0.2823)² = 0.001193
Answer
S = 0.001193
That is a fall of 1.193 m per kilometre.
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