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 = 3 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
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 = 2.5 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows at 16 L/s through a horizontal pipe that narrows from 150 mm to 75 mm. The pressure in the narrow section is 270 kPa. Ignoring losses, find the pressure p1 in the wide section.
Q = 16 L/sd1 = 150 mmd2 = 75 mmp2 = 270 kPaρ = 1000 kg/m³
p1 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 25 mm while rising 5 m. At section 1 the pressure is 140 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 25 mmp1 = 140 kPav1 = 0.5 m/sΔz = 5 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 80 L/s through a pipe 500 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
Q = 80 L/sL = 500 mD = 250 mmf = 0.018g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 300 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.02) 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 = 5). 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 = 300 mD = 300 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5
H = m
7.Pumps & pump powerMedium
A pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 62 m. The pump efficiency is 60% and the electric motor driving it has an efficiency of 92%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 25 L/sH = 62 mηpump = 60%ηmotor = 92%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 75 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 400 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015; the minor-loss coefficients add up to ΣK = 3 (entrance and exit included). Use g = 9.81 m/s². The upper tank is closed and the air above its water is at a gauge pressure of 250 kPa. Find the head H the pump must add.
Q = 75 L/sΔz = 7 mL = 400 mD = 200 mmf = 0.015ΣK = 3ptank = 250 kPa
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal clean earth channel has a bottom width of 4 m and side slopes of 2:1 (horizontal:vertical). Water flows uniformly 2.1 m deep. Manning n = 0.022 and the bed slope is S = 0.0005. Find the flow rate Q in m³/s.
b = 4 my = 2.1 mz = 2n = 0.022S = 0.0005
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal finished concrete channel with a bottom width of 5 m and side slopes of 3:1 (horizontal:vertical) must carry 4.1 m³/s in uniform flow at a depth of 0.5 m. Manning n = 0.012. Find the required bed slope S (in m/m).
Q = 4.1 m³/sb = 5 my = 0.5 mz = 3n = 0.012
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 88.29 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 + 3/2 = 3 m
A = H·b = 3·1 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3·3 = 88.29 kN
Answer
F = 88.29 kN
2.Hydrostatic force
Answer: F = 138 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 + 2.5/2 = 3.75 m
A = H·b = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.75·3.75 = 138 kN
Answer
F = 138 kN
3.Continuity & Bernoulli
Answer: p1 = 276.1 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.
Velocities
v1 = Q/A1 = 0.016/0.01767 = 0.9054 m/s
v2 = Q/A2 = 0.016/0.004418 = 3.622 m/s
Bernoulli
p1 = p2 + ½ρ(v2² − v1²) = 270 + ½·1000·(3.622² − 0.9054²)/1000 = 270 + 6.148 = 276.1 kPa
Answer
p1 = 276.1 kPa
4.Continuity & Bernoulli
Answer: p2 = 59.08 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·(100/25)² = 8 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 8²)/1000 = −31.88 kPa
ρgΔz = 1000·9.81·5/1000 = 49.05 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 140 + (−31.88) − 49.05 = 59.08 kPa
Answer
p2 = 59.08 kPa
5.Pipe friction & head loss
Answer: hf = 4.874 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·0.25²/4 = 0.04909 m²
V = Q/A = 0.08/0.04909 = 1.63 m/s
Velocity head
V²/(2g) = 1.63²/(2·9.81) = 0.1354 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(500/0.25)·1.63²/(2·9.81) = 4.874 m
Answer
hf = 4.874 m
6.Pipe friction & head loss
Answer: H = 1.769 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 = 0.02·300/0.3 = 20
ΣK = 0.5 + 2·0.3 + 5 = 6.1
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 20 + 6.1)·0.06529 = 1.769 m
Answer
H = 1.769 m
7.Pumps & pump power
Answer: Pel = 27.55 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 = 1000·9.81·0.025·62 = 15.21 kW
Shaft power
Pshaft = 15.21/0.6 = 25.34 kW
Electrical power
Pel = 25.34/0.92 = 27.55 kW
Answer
Pel = 27.55 kW
8.Pumps & pump power
Answer: H = 42.07 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (30 + 3)·0.2905 = 9.586 m
Pressure head
p/(ρ·g) = 250000/(1000·9.81) = 25.48 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 7 + 25.48 + 9.586 = 42.07 m
Answer
H = 42.07 m
9.Open-channel flow (Manning)
Answer: Q = 20.7 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 = (4 + 2·2.1)·2.1 = 17.22 m²
P = b + 2y·√(1 + z²) = 4 + 2·2.1·√(1 + 2²) = 13.39 m
R = A/P = 1.286 m
Velocity
V = (1/0.022)·1.2862/3·0.00051/2 = 1.202 m/s
Flow rate
Q = A·V = 17.22·1.202 = 20.7 m³/s
Answer
Q = 20.7 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0007823 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 + 3·0.5)·0.5 = 3.25 m²
P = b + 2y·√(1 + z²) = 5 + 2·0.5·√(1 + 3²) = 8.162 m
R = A/P = 0.3982 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.012·4.1/(3.25·0.5412) = 0.02797
S = 0.02797² = 0.0007823
Answer
S = 0.0007823
That is a fall of 0.7823 m per kilometre.
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