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 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 3 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 water, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 2 mb = 3 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 4.5 m above the orifice and the discharge coefficient is 0.61. Find the flow rate Q. Use g = 9.81 m/s².
d = 50 mmh = 4.5 mCd = 0.61g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 75 mm to 40 mm while rising 10 m. At section 1 the pressure is 200 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 75 mmd2 = 40 mmp1 = 200 kPav1 = 0.75 m/sΔz = 10 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.1 m/s through a horizontal pipe, 50 m long, with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.1 m/sL = 50 mD = 200 mmf = 0.015ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 100 m long with an inside diameter of 150 mm (Darcy friction factor f = 0.018) 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), 4 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 20 L/s?
Q = 20 L/sL = 100 mD = 150 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10
H = m
7.Pumps & pump powerMedium
A pump with an efficiency of 78% delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 12 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 20 L/sH = 12 mη = 78%g = 9.81 m/s²
Pshaft = 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 29 m higher. The pipe is 500 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 8 (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 = 29 mL = 500 mD = 200 mmf = 0.018ΣK = 8ptank = 250 kPa
H = m
9.Open-channel flow (Manning)Medium
A rectangular clean earth channel is 6 m wide. Water flows uniformly 2.5 m deep. Manning n = 0.022 and the bed slope is S = 0.003. Find the flow rate Q in m³/s.
b = 6 my = 2.5 mn = 0.022S = 0.003
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal clean earth channel with a bottom width of 3.5 m and side slopes of 3:1 (horizontal:vertical) must carry 24 m³/s in uniform flow at a depth of 1.8 m. Manning n = 0.022. Find the required bed slope S (in m/m).
Q = 24 m³/sb = 3.5 my = 1.8 mz = 3n = 0.022
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 = 0.5 + 3/2 = 2 m
A = H·b = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·4.5 = 88.29 kN
Answer
F = 88.29 kN
2.Hydrostatic force
Answer: F = 235.4 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 + 2/2 = 4 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4·6 = 235.4 kN
Answer
F = 235.4 kN
3.Continuity & Bernoulli
Answer: Q = 11.25 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) = √(2·9.81·4.5) = 9.396 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.61·0.001963·9.396 = 0.01125 m³/s = 11.25 L/s
Answer
Q = 11.25 L/s
4.Continuity & Bernoulli
Answer: p2 = 98.71 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 = 0.75·(75/40)² = 2.637 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 2.637²)/1000 = −3.195 kPa
ρgΔz = 1000·9.81·10/1000 = 98.1 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 200 + (−3.195) − 98.1 = 98.71 kPa
Answer
p2 = 98.71 kPa
5.Pipe friction & head loss
Answer: Δp = 8.269 kPa
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.
Head loss
hf = f·(L/D)·V²/(2g) = 0.015·(50/0.2)·2.1²/(2·9.81) = 0.8429 m
Pressure drop
The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·hf.
Δp = ρ·g·hf = 1000·9.81·0.8429 = 8269 Pa = 8.269 kPa
Answer
Δp = 8.269 kPa
6.Pipe friction & head loss
Answer: H = 1.613 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 = 0.02/0.01767 = 1.132 m/s
V²/(2g) = 0.06529 m
Loss coefficients
f·L/D = 0.018·100/0.15 = 12
ΣK = 0.5 + 4·0.3 + 10 = 11.7
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 12 + 11.7)·0.06529 = 1.613 m
Answer
H = 1.613 m
7.Pumps & pump power
Answer: Pshaft = 3.018 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 = 1000·9.81·0.02·12 = 2.354 kW
Shaft power
Pshaft = Ph/η = 2.354/0.78 = 3.018 kW
Answer
Pshaft = 3.018 kW
8.Pumps & pump power
Answer: H = 69.88 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) = (45 + 8)·0.2905 = 15.4 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 = 29 + 25.48 + 15.4 = 69.88 m
Answer
H = 69.88 m
9.Open-channel flow (Manning)
Answer: Q = 45.92 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·y = 6·2.5 = 15 m²
P = b + 2y = 6 + 2·2.5 = 11 m
R = A/P = 1.364 m
Velocity
V = (1/0.022)·1.3642/3·0.0031/2 = 3.062 m/s
Flow rate
Q = A·V = 15·3.062 = 45.92 m³/s
Answer
Q = 45.92 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0009848 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 = (3.5 + 3·1.8)·1.8 = 16.02 m²
P = b + 2y·√(1 + z²) = 3.5 + 2·1.8·√(1 + 3²) = 14.88 m
R = A/P = 1.076 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.022·24/(16.02·1.05) = 0.03138
S = 0.03138² = 0.0009848
Answer
S = 0.0009848
That is a fall of 0.9848 m per kilometre.
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