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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 2 mb = 1.5 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 oil, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 2.5 mb = 1 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
A pitot-static tube in a wind tunnel reads a difference of 100 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.
p0 − p = 100 Paρ = 1.2 kg/m³
v = m/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 75 mm to 25 mm while rising 3 m. At section 1 the pressure is 120 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 75 mmd2 = 25 mmp1 = 120 kPav1 = 1.25 m/sΔz = 3 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 1.1 m/s through a pipe 50 m long with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the friction head loss hf.
V = 1.1 m/sL = 50 mD = 75 mmf = 0.015g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 300 m long with an inside diameter of 250 mm (Darcy friction factor f = 0.022) 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), 2 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 60 L/s?
Q = 60 L/sL = 300 mD = 250 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 10
H = m
7.Pumps & pump powerMedium
A pump delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 22 m. The pump efficiency is 75% 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 = 20 L/sH = 22 mηpump = 75%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 9 m higher. The pipe is 500 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 80 L/sΔz = 9 mL = 500 mD = 250 mmf = 0.022ΣK = 6
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal finished concrete channel has a bottom width of 5 m and side slopes of 1:1 (horizontal:vertical). Water flows uniformly 1.1 m deep. Manning n = 0.012 and the bed slope is S = 0.005. Find the flow rate Q in m³/s.
b = 5 my = 1.1 mz = 1n = 0.012S = 0.005
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular clean earth channel 2 m wide must carry 2.5 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 = 2.5 m³/sb = 2 my = 0.7 mn = 0.022
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 73.58 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 = 2·1.5 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.5·3 = 73.58 kN
Answer
F = 73.58 kN
2.Hydrostatic force
Answer: F = 37.77 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 = 0.5 + 2.5/2 = 1.75 m
A = H·b = 2.5·1 = 2.5 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·1.75·2.5 = 37.77 kN
Answer
F = 37.77 kN
3.Continuity & Bernoulli
Answer: v = 12.91 m/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.
Bernoulli to the stagnation point
The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.
p0 − p = ½ρv²
v = √(2·100/1.2) = 12.91 m/s
Answer
v = 12.91 m/s
4.Continuity & Bernoulli
Answer: p2 = 28.07 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·(75/25)² = 11.25 m/s
½ρ(v1² − v2²) = ½·1000·(1.25² − 11.25²)/1000 = −62.5 kPa
ρgΔz = 1000·9.81·3/1000 = 29.43 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 120 + (−62.5) − 29.43 = 28.07 kPa
Answer
p2 = 28.07 kPa
5.Pipe friction & head loss
Answer: hf = 0.6167 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 head
V²/(2g) = 1.1²/(2·9.81) = 0.06167 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.015·(50/0.075)·1.1²/(2·9.81) = 0.6167 m
Answer
hf = 0.6167 m
6.Pipe friction & head loss
Answer: H = 2.932 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 = 0.06/0.04909 = 1.222 m/s
V²/(2g) = 0.07615 m
Loss coefficients
f·L/D = 0.022·300/0.25 = 26.4
ΣK = 0.5 + 2·0.3 + 10 = 11.1
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 26.4 + 11.1)·0.07615 = 2.932 m
Answer
H = 2.932 m
7.Pumps & pump power
Answer: Pel = 6.058 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 = 1000·9.81·0.02·22 = 4.316 kW
Shaft power
Pshaft = 4.316/0.75 = 5.755 kW
Electrical power
Pel = 5.755/0.95 = 6.058 kW
Answer
Pel = 6.058 kW
8.Pumps & pump power
Answer: H = 15.77 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 = 0.08/0.04909 = 1.63 m/s
V²/(2g) = 0.1354 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (44 + 6)·0.1354 = 6.769 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 9 + 6.769 = 15.77 m
Answer
H = 15.77 m
9.Open-channel flow (Manning)
Answer: Q = 34.84 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 = (5 + 1·1.1)·1.1 = 6.71 m²
P = b + 2y·√(1 + z²) = 5 + 2·1.1·√(1 + 1²) = 8.111 m
R = A/P = 0.8272 m
Velocity
V = (1/0.012)·0.82722/3·0.0051/2 = 5.193 m/s
Flow rate
Q = A·V = 6.71·5.193 = 34.84 m³/s
Answer
Q = 34.84 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.005038 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 = 2·0.7 = 1.4 m²
P = b + 2y = 2 + 2·0.7 = 3.4 m
R = A/P = 0.4118 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.022·2.5/(1.4·0.5535) = 0.07098
S = 0.07098² = 0.005038
Answer
S = 0.005038
That is a fall of 5.038 m per kilometre.
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