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 2 m wide sits in the wall of a tank of water, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1.5 mb = 2 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 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 2 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 100 mm. In the wide section the pressure is 180 kPa and the velocity is 2.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 100 mmp1 = 180 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 40 mm while rising 9 m. At section 1 the pressure is 330 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 75 mmd2 = 40 mmp1 = 330 kPav1 = 3 m/sΔz = 9 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 1.2 m/s through a horizontal pipe, 300 m long, with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 1.2 m/sL = 300 mD = 75 mmf = 0.03ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.025. The minor losses are a square-edged entrance (K = 0.5), 4 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 50 L/s?
Q = 50 L/sL = 100 mD = 250 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 5Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 78 m. The pump efficiency is 78% 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 = 78 mηpump = 78%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 21 m higher. The pipe is 400 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 = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 65%. Find the shaft power Pshaft in kW.
Q = 80 L/sΔz = 21 mL = 400 mD = 200 mmf = 0.02ΣK = 6η = 65%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A trapezoidal finished concrete channel has a bottom width of 1.5 m and side slopes of 3:1 (horizontal:vertical). Water flows uniformly 2 m deep. Manning n = 0.012 and the bed slope is S = 0.002. Find the flow rate Q in m³/s.
b = 1.5 my = 2 mz = 3n = 0.012S = 0.002
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal clean earth channel with a bottom width of 5 m and side slopes of 3:1 (horizontal:vertical) must carry 8.1 m³/s in uniform flow at a depth of 0.7 m. Manning n = 0.022. Find the required bed slope S (in m/m).
Q = 8.1 m³/sb = 5 my = 0.7 mz = 3n = 0.022
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 51.5 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 = 1 + 1.5/2 = 1.75 m
A = H·b = 1.5·2 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.75·3 = 51.5 kN
Answer
F = 51.5 kN
2.Hydrostatic force
Answer: F = 176.6 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 + 2/2 = 3 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3·6 = 176.6 kN
Answer
F = 176.6 kN
3.Continuity & Bernoulli
Answer: p2 = 167.3 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·(150/100)² = 5.625 m/s
½ρ(v1² − v2²) = ½·1000·(2.5² − 5.625²)/1000 = −12.7 kPa
p2 = p1 + ½ρ(v1² − v2²) = 180 + (−12.7) = 167.3 kPa
Answer
p2 = 167.3 kPa
4.Continuity & Bernoulli
Answer: p2 = 190.6 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 = 3·(75/40)² = 10.55 m/s
½ρ(v1² − v2²) = ½·1000·(3² − 10.55²)/1000 = −51.12 kPa
ρgΔz = 1000·9.81·9/1000 = 88.29 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 330 + (−51.12) − 88.29 = 190.6 kPa
Answer
p2 = 190.6 kPa
5.Pipe friction & head loss
Answer: Δp = 86.4 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) = 0.03·(300/0.075)·1.2²/(2·9.81) = 8.807 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 = 1000·9.81·8.807 = 86400 Pa = 86.4 kPa
Answer
Δp = 86.4 kPa
6.Pipe friction & head loss
Answer: Δz = 0.936 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 = 0.05/0.04909 = 1.019 m/s
V²/(2g) = 0.05288 m
Loss coefficients
f·L/D = 0.025·100/0.25 = 10
ΣK = 0.5 + 4·0.3 + 5 + 1.0 = 7.7
Total head loss
hf = 10·0.05288 = 0.5288 m
hm = 7.7·0.05288 = 0.4072 m
Δz = hf + hm = 0.936 m
Answer
Δz = 0.936 m
7.Pumps & pump power
Answer: Pel = 15.49 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 = 1000·9.81·0.015·78 = 11.48 kW
Shaft power
Pshaft = 11.48/0.78 = 14.71 kW
Electrical power
Pel = 14.71/0.95 = 15.49 kW
Answer
Pel = 15.49 kW
8.Pumps & pump power
Answer: Pshaft = 43.71 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) = (40 + 6)·0.3305 = 15.2 m
Pump head
H = Δz + hL = 21 + 15.2 = 36.2 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.08·36.2/0.65 = 43.71 kW
Answer
Pshaft = 43.71 kW
9.Open-channel flow (Manning)
Answer: Q = 58.12 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 = (b + z·y)·y = (1.5 + 3·2)·2 = 15 m²
P = b + 2y·√(1 + z²) = 1.5 + 2·2·√(1 + 3²) = 14.15 m
R = A/P = 1.06 m
Velocity
V = (1/0.012)·1.062/3·0.0021/2 = 3.875 m/s
Flow rate
Q = A·V = 15·3.875 = 58.12 m³/s
Answer
Q = 58.12 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.003019 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 = (5 + 3·0.7)·0.7 = 4.97 m²
P = b + 2y·√(1 + z²) = 5 + 2·0.7·√(1 + 3²) = 9.427 m
R = A/P = 0.5272 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.022·8.1/(4.97·0.6526) = 0.05494
S = 0.05494² = 0.003019
Answer
S = 0.003019
That is a fall of 3.019 m per kilometre.
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