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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 2 mb = 1.5 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 water, with its top edge 1.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1.5 mH = 2 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows at 27 L/s through a horizontal pipe that narrows from 200 mm to 150 mm. The pressure in the narrow section is 220 kPa. Ignoring losses, find the pressure p1 in the wide section.
Q = 27 L/sd1 = 200 mmd2 = 150 mmp2 = 220 kPaρ = 1000 kg/m³
p1 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 3 m. At section 1 the pressure is 320 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 75 mmp1 = 320 kPav1 = 1.25 m/sΔz = 3 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.7 m/s through a horizontal pipe, 1000 m long, with an inside diameter of 150 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 = 2.7 m/sL = 1000 mD = 150 mmf = 0.018ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 100 m long with an inside diameter of 100 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 = 5). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 10 L/s?
Q = 10 L/sL = 100 mD = 100 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 5
H = m
7.Pumps & pump powerMedium
A pump delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 28 m. The pump efficiency is 78% and the electric motor driving it has an efficiency of 94%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 60 L/sH = 28 mηpump = 78%ηmotor = 94%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 40 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 7 m higher. The pipe is 150 m long with an inside diameter of 200 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 = 40 L/sΔz = 7 mL = 150 mD = 200 mmf = 0.018ΣK = 8
H = m
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 900 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.003. Find the flow rate Q in m³/s.
D = 900 mmn = 0.013S = 0.003
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular unfinished concrete channel 3.5 m wide must carry 10 m³/s in uniform flow at a depth of 1.3 m. Manning n = 0.014. Find the required bed slope S (in m/m).
Q = 10 m³/sb = 3.5 my = 1.3 mn = 0.014
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 103 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 = 2.5 + 2/2 = 3.5 m
A = H·b = 2·1.5 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.5·3 = 103 kN
Answer
F = 103 kN
2.Hydrostatic force
Answer: ycp = 2.633 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 = 1.5 + 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
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·2³/12 = 2 m⁴
ycp = yc + Ic/(yc·A) = 2.5 + 2/(2.5·6) = 2.633 m
Answer
ycp = 2.633 m
3.Continuity & Bernoulli
Answer: p1 = 220.8 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 = 0.027/0.03142 = 0.8594 m/s
v2 = Q/A2 = 0.027/0.01767 = 1.528 m/s
Bernoulli
p1 = p2 + ½ρ(v2² − v1²) = 220 + ½·1000·(1.528² − 0.8594²)/1000 = 220 + 0.7979 = 220.8 kPa
Answer
p1 = 220.8 kPa
4.Continuity & Bernoulli
Answer: p2 = 278.9 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 = 1.25·(150/75)² = 5 m/s
½ρ(v1² − v2²) = ½·1000·(1.25² − 5²)/1000 = −11.72 kPa
ρgΔz = 1000·9.81·3/1000 = 29.43 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 320 + (−11.72) − 29.43 = 278.9 kPa
Answer
p2 = 278.9 kPa
5.Pipe friction & head loss
Answer: Δp = 437.4 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·(1000/0.15)·2.7²/(2·9.81) = 44.59 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 = 1000·9.81·44.59 = 437400 Pa = 437.4 kPa
Answer
Δp = 437.4 kPa
6.Pipe friction & head loss
Answer: H = 2.099 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 = 0.01/0.007854 = 1.273 m/s
V²/(2g) = 0.08263 m
Loss coefficients
f·L/D = 0.018·100/0.1 = 18
ΣK = 0.5 + 3·0.3 + 5 = 6.4
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 18 + 6.4)·0.08263 = 2.099 m
Answer
H = 2.099 m
7.Pumps & pump power
Answer: Pel = 22.48 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 = 1000·9.81·0.06·28 = 16.48 kW
Shaft power
Pshaft = 16.48/0.78 = 21.13 kW
Electrical power
Pel = 21.13/0.94 = 22.48 kW
Answer
Pel = 22.48 kW
8.Pumps & pump power
Answer: H = 8.776 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 = 0.04/0.03142 = 1.273 m/s
V²/(2g) = 0.08263 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (13.5 + 8)·0.08263 = 1.776 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 7 + 1.776 = 8.776 m
Answer
H = 8.776 m
9.Open-channel flow (Manning)
Answer: Q = 0.9915 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 = 0.6362 m², R = D/4 = 0.225 m
Velocity
V = (1/0.013)·0.2252/3·0.0031/2 = 1.559 m/s
Flow rate
Q = A·V = 0.6362·1.559 = 0.9915 m³/s
Answer
Q = 0.9915 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0014 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 = 3.5·1.3 = 4.55 m²
P = b + 2y = 3.5 + 2·1.3 = 6.1 m
R = A/P = 0.7459 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.014·10/(4.55·0.8225) = 0.03741
S = 0.03741² = 0.0014
Answer
S = 0.0014
That is a fall of 1.4 m per kilometre.
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