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 3 m wide sits in the wall of a tank of glycerin, with its top edge 3 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 3 mH = 3 mb = 3 mρ = 1260 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 150 mm and a throat diameter of 75 mm. The pressure at the inlet is 34 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 150 mmd2 = 75 mmp1 − p2 = 34 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 75 mm to 50 mm while rising 4 m. At section 1 the pressure is 290 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 75 mmd2 = 50 mmp1 = 290 kPav1 = 2 m/sΔz = 4 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 4 L/s through a pipe 150 m long with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
Q = 4 L/sL = 150 mD = 75 mmf = 0.022g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 500 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.015. The water surface of the upper reservoir is 16 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each), a valve (K = 2) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². Find the flow rate Q in L/s.
Δz = 16 mL = 500 mD = 300 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 72% delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 58 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 50 L/sH = 58 mη = 72%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 20 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 33 m higher. The pipe is 200 m long with an inside diameter of 100 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 78%. Find the shaft power Pshaft in kW.
Q = 20 L/sΔz = 33 mL = 200 mD = 100 mmf = 0.022ΣK = 8η = 78%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 750 mm flows half full. Manning n = 0.013 and the slope is S = 0.001. Find the flow rate Q in m³/s.
D = 750 mmn = 0.013S = 0.001
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 450 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 = 450 mmn = 0.013V = 0.75 m/s
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: ycp = 4.667 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 = 3 + 3/2 = 4.5 m
A = H·b = 3·3 = 9 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·4.5·9 = 500.6 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·3³/12 = 6.75 m⁴
ycp = yc + Ic/(yc·A) = 4.5 + 6.75/(4.5·9) = 4.667 m
Answer
ycp = 4.667 m
3.Continuity & Bernoulli
Answer: Q = 37.63 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.
Continuity
v1 = v2·(75/150)² = 0.25·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·34000/(1000·(1 − 0.0625))) = 8.517 m/s
v1 = 2.129 m/s
Flow rate
Q = A2·v2 = 0.004418·8.517 = 0.03763 m³/s = 37.63 L/s
Answer
Q = 37.63 L/s
4.Continuity & Bernoulli
Answer: p2 = 242.6 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 = 2·(75/50)² = 4.5 m/s
½ρ(v1² − v2²) = ½·1000·(2² − 4.5²)/1000 = −8.125 kPa
ρgΔz = 1000·9.81·4/1000 = 39.24 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 290 + (−8.125) − 39.24 = 242.6 kPa
Answer
p2 = 242.6 kPa
5.Pipe friction & head loss
Answer: hf = 1.838 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·0.075²/4 = 0.004418 m²
V = Q/A = 0.004/0.004418 = 0.9054 m/s
Velocity head
V²/(2g) = 0.9054²/(2·9.81) = 0.04178 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(150/0.075)·0.9054²/(2·9.81) = 1.838 m
Answer
hf = 1.838 m
6.Pipe friction & head loss
Answer: Q = 232.2 L/s
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.
Loss coefficients
f·L/D = 0.015·500/0.3 = 25
ΣK = 0.5 + 2·0.3 + 2 + 1.0 = 4.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
16 = (25 + 4.1)·V²/(2·9.81)
V = √(2·9.81·16/29.1) = 3.284 m/s
Flow rate
Q = V·π·D²/4 = 3.284·0.07069 = 0.2322 m³/s = 232.2 L/s
Answer
Q = 232.2 L/s
7.Pumps & pump power
Answer: Pshaft = 39.51 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 = 1000·9.81·0.05·58 = 28.45 kW
Shaft power
Pshaft = Ph/η = 28.45/0.72 = 39.51 kW
Answer
Pshaft = 39.51 kW
8.Pumps & pump power
Answer: Pshaft = 12.62 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 = 0.02/0.007854 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (44 + 8)·0.3305 = 17.19 m
Pump head
H = Δz + hL = 33 + 17.19 = 50.19 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.02·50.19/0.78 = 12.62 kW
Answer
Pshaft = 12.62 kW
9.Open-channel flow (Manning)
Answer: Q = 0.176 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 = 0.2209 m², R = D/4 = 0.1875 m
Velocity
V = (1/0.013)·0.18752/3·0.0011/2 = 0.7969 m/s
Flow rate
Q = A·V = 0.2209·0.7969 = 0.176 m³/s
Answer
Q = 0.176 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.00175 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 = 0.45/4 = 0.1125 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.75/0.233)² = 0.00175
Answer
S = 0.00175
That is a fall of 1.75 m per kilometre.
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