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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 1.5 mb = 2 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 oil, with its top edge 1.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1.5 mH = 1.5 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 100 mm. In the wide section the pressure is 140 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 100 mmp1 = 140 kPav1 = 1.5 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 75 mm and a gauge pressure of 130 kPa. The nozzle exit diameter is 20 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².
d1 = 75 mmp1 = 130 kPa (gauge)d2 = 20 mmρ = 1000 kg/m³g = 9.81 m/s²
H = m
5.Pipe friction & head lossMedium
Water flows at 6 L/s through a pipe 100 m long with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
Q = 6 L/sL = 100 mD = 75 mmf = 0.018g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.025. The water surface of the upper reservoir is 7 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 0.2) 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 = 7 mL = 100 mD = 300 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 0.2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump delivers 30 L/s of water (ρ = 1000 kg/m³) against a head of 30 m. The pump efficiency is 85% 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 = 30 L/sH = 30 mηpump = 85%ηmotor = 95%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 11 m higher. The pipe is 500 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 4 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 120 L/sΔz = 11 mL = 500 mD = 250 mmf = 0.018ΣK = 4
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal brickwork channel has a bottom width of 2.5 m and side slopes of 3:1 (horizontal:vertical). Water flows uniformly 1.4 m deep. Manning n = 0.015 and the bed slope is S = 0.001. Find the flow rate Q in m³/s.
b = 2.5 my = 1.4 mz = 3n = 0.015S = 0.001
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 200 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 = 200 mmn = 0.013V = 0.75 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 95.65 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 + 1.5/2 = 3.25 m
A = H·b = 1.5·2 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.25·3 = 95.65 kN
Answer
F = 95.65 kN
2.Hydrostatic force
Answer: ycp = 2.333 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 = 1.5 + 1.5/2 = 2.25 m
A = H·b = 1.5·1.5 = 2.25 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·2.25·2.25 = 43.7 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1.5·1.5³/12 = 0.4219 m⁴
ycp = yc + Ic/(yc·A) = 2.25 + 0.4219/(2.25·2.25) = 2.333 m
Answer
ycp = 2.333 m
3.Continuity & Bernoulli
Answer: p2 = 135.4 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 = 1.5·(150/100)² = 3.375 m/s
½ρ(v1² − v2²) = ½·1000·(1.5² − 3.375²)/1000 = −4.57 kPa
p2 = p1 + ½ρ(v1² − v2²) = 140 + (−4.57) = 135.4 kPa
Answer
p2 = 135.4 kPa
4.Continuity & Bernoulli
Answer: H = 13.32 m
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.
Exit velocity
β = 20/75 = 0.2667
v2 = √(2p1/(ρ(1 − β⁴))) = √(2·130000/(1000·0.9949)) = 16.17 m/s
Height of the jet
Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.
H = v2²/(2g) = 16.17²/(2·9.81) = 13.32 m
Answer
H = 13.32 m
5.Pipe friction & head loss
Answer: hf = 2.256 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 = 0.006/0.004418 = 1.358 m/s
Velocity head
V²/(2g) = 1.358²/(2·9.81) = 0.09401 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(100/0.075)·1.358²/(2·9.81) = 2.256 m
Answer
hf = 2.256 m
6.Pipe friction & head loss
Answer: Q = 250.5 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 = 0.025·100/0.3 = 8.333
ΣK = 0.5 + 3·0.3 + 0.2 + 1.0 = 2.6
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
7 = (8.333 + 2.6)·V²/(2·9.81)
V = √(2·9.81·7/10.93) = 3.544 m/s
Flow rate
Q = V·π·D²/4 = 3.544·0.07069 = 0.2505 m³/s = 250.5 L/s
Answer
Q = 250.5 L/s
7.Pumps & pump power
Answer: Pel = 10.93 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 = 1000·9.81·0.03·30 = 8.829 kW
Shaft power
Pshaft = 8.829/0.85 = 10.39 kW
Electrical power
Pel = 10.39/0.95 = 10.93 kW
Answer
Pel = 10.93 kW
8.Pumps & pump power
Answer: H = 23.18 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (36 + 4)·0.3046 = 12.18 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 11 + 12.18 = 23.18 m
Answer
H = 23.18 m
9.Open-channel flow (Manning)
Answer: Q = 17.41 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 = (2.5 + 3·1.4)·1.4 = 9.38 m²
P = b + 2y·√(1 + z²) = 2.5 + 2·1.4·√(1 + 3²) = 11.35 m
R = A/P = 0.8261 m
Velocity
V = (1/0.015)·0.82612/3·0.0011/2 = 1.856 m/s
Flow rate
Q = A·V = 9.38·1.856 = 17.41 m³/s
Answer
Q = 17.41 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.005161 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 = 0.2/4 = 0.05 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.75/0.1357)² = 0.005161
Answer
S = 0.005161
That is a fall of 5.161 m per kilometre.
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