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 3 m wide sits in the wall of a tank of water, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 1.5 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 2 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 = 2 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 50 mm. The water level is kept 2 m above the orifice and the discharge coefficient is 0.63. Find the flow rate Q. Use g = 9.81 m/s².
d = 50 mmh = 2 mCd = 0.63g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 75 mm while rising 2 m. At section 1 the pressure is 350 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 75 mmp1 = 350 kPav1 = 3 m/sΔz = 2 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.7 m/s through a pipe 1000 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 = 2.7 m/sL = 1000 mD = 75 mmf = 0.015g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.02. The water surface of the upper reservoir is 12 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 4 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 = 12 mL = 200 mD = 200 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 20 m. The pump efficiency is 85% and the electric motor driving it has an efficiency of 90%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 100 L/sH = 20 mηpump = 85%ηmotor = 90%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 120 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 500 m long with an inside diameter of 250 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 70%. Find the shaft power Pshaft in kW.
Q = 120 L/sΔz = 7 mL = 500 mD = 250 mmf = 0.02ΣK = 6η = 70%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 300 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.001. Find the flow rate Q in m³/s.
D = 300 mmn = 0.013S = 0.001
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular finished concrete channel 3.5 m wide must carry 15 m³/s in uniform flow at a depth of 1.2 m. Manning n = 0.012. Find the required bed slope S (in m/m).
Q = 15 m³/sb = 3.5 my = 1.2 mn = 0.012
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 165.5 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 = 3 + 1.5/2 = 3.75 m
A = H·b = 1.5·3 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.75·4.5 = 165.5 kN
Answer
F = 165.5 kN
2.Hydrostatic force
Answer: F = 183.9 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 = 2.5·2 = 5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.75·5 = 183.9 kN
Answer
F = 183.9 kN
3.Continuity & Bernoulli
Answer: Q = 7.749 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.
Ideal jet velocity
√(2gh) = √(2·9.81·2) = 6.264 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.63·0.001963·6.264 = 0.007749 m³/s = 7.749 L/s
Answer
Q = 7.749 L/s
4.Continuity & Bernoulli
Answer: p2 = 320.7 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·(100/75)² = 5.333 m/s
½ρ(v1² − v2²) = ½·1000·(3² − 5.333²)/1000 = −9.722 kPa
ρgΔz = 1000·9.81·2/1000 = 19.62 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 350 + (−9.722) − 19.62 = 320.7 kPa
Answer
p2 = 320.7 kPa
5.Pipe friction & head loss
Answer: hf = 74.31 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) = 2.7²/(2·9.81) = 0.3716 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.015·(1000/0.075)·2.7²/(2·9.81) = 74.31 m
Answer
hf = 74.31 m
6.Pipe friction & head loss
Answer: Q = 96.99 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 = 0.02·200/0.2 = 20
ΣK = 0.5 + 4·0.3 + 2 + 1.0 = 4.7
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
12 = (20 + 4.7)·V²/(2·9.81)
V = √(2·9.81·12/24.7) = 3.087 m/s
Flow rate
Q = V·π·D²/4 = 3.087·0.03142 = 0.09699 m³/s = 96.99 L/s
Answer
Q = 96.99 L/s
7.Pumps & pump power
Answer: Pel = 25.65 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 = 1000·9.81·0.1·20 = 19.62 kW
Shaft power
Pshaft = 19.62/0.85 = 23.08 kW
Electrical power
Pel = 23.08/0.9 = 25.65 kW
Answer
Pel = 25.65 kW
8.Pumps & pump power
Answer: Pshaft = 35.34 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (40 + 6)·0.3046 = 14.01 m
Pump head
H = Δz + hL = 7 + 14.01 = 21.01 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.12·21.01/0.7 = 35.34 kW
Answer
Pshaft = 35.34 kW
9.Open-channel flow (Manning)
Answer: Q = 0.03058 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 = 0.07069 m², R = D/4 = 0.075 m
Velocity
V = (1/0.013)·0.0752/3·0.0011/2 = 0.4326 m/s
Flow rate
Q = A·V = 0.07069·0.4326 = 0.03058 m³/s
Answer
Q = 0.03058 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.00289 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 = 3.5·1.2 = 4.2 m²
P = b + 2y = 3.5 + 2·1.2 = 5.9 m
R = A/P = 0.7119 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.012·15/(4.2·0.7973) = 0.05376
S = 0.05376² = 0.00289
Answer
S = 0.00289
That is a fall of 2.89 m per kilometre.
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