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 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 3 mb = 3 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 2 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 2 mH = 2.5 mb = 3 mρ = 1260 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 75 mm to 40 mm. In the wide section the pressure is 180 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 40 mmp1 = 180 kPav1 = 0.5 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A hose with an inner diameter of 25 mm siphons water from an open tank. The highest point C of the hose is 2.5 m above the water surface, and the outlet (open to the air) is 1 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.
d = 25 mmzC = +2.5 mz2 = −1 mg = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 3 L/s through a pipe 800 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
Q = 3 L/sL = 800 mD = 50 mmf = 0.028g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 150 m long with an inside diameter of 200 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), 3 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 40 L/s?
Q = 40 L/sL = 150 mD = 200 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 30 L/s of water (ρ = 1000 kg/m³) against a head of 26 m while its shaft takes 9 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 30 L/sH = 26 mPshaft = 9 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 8 m higher. The pipe is 200 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 2.5 (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 300 kPa. Find the head H the pump must add.
Q = 50 L/sΔz = 8 mL = 200 mD = 150 mmf = 0.018ΣK = 2.5ptank = 300 kPa
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal grass-lined earth channel has a bottom width of 4.5 m and side slopes of 3:1 (horizontal:vertical). Water flows uniformly 0.7 m deep. Manning n = 0.03 and the bed slope is S = 0.001. Find the flow rate Q in m³/s.
b = 4.5 my = 0.7 mz = 3n = 0.03S = 0.001
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal grass-lined earth channel with a bottom width of 5.5 m and side slopes of 2:1 (horizontal:vertical) must carry 18 m³/s in uniform flow at a depth of 1.3 m. Manning n = 0.03. Find the required bed slope S (in m/m).
Q = 18 m³/sb = 5.5 my = 1.3 mz = 2n = 0.03
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 176.6 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 = 3·3 = 9 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·9 = 176.6 kN
Answer
F = 176.6 kN
2.Hydrostatic force
Answer: ycp = 3.41 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 = 2 + 2.5/2 = 3.25 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·3.25·7.5 = 301.3 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·2.5³/12 = 3.906 m⁴
ycp = yc + Ic/(yc·A) = 3.25 + 3.906/(3.25·7.5) = 3.41 m
Answer
ycp = 3.41 m
3.Continuity & Bernoulli
Answer: p2 = 178.6 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 = 0.5·(75/40)² = 1.758 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 1.758²)/1000 = −1.42 kPa
p2 = p1 + ½ρ(v1² − v2²) = 180 + (−1.42) = 178.6 kPa
Answer
p2 = 178.6 kPa
4.Continuity & Bernoulli
Answer: Q = 2.174 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.
Velocity in the hose
Surface and outlet are both open to the air, so the pressure terms cancel.
v = √(2g·1) = √(2·9.81·1) = 4.429 m/s
Flow rate
Q = A·v = 0.0004909·4.429 = 0.002174 m³/s = 2.174 L/s
Answer
Q = 2.174 L/s
5.Pipe friction & head loss
Answer: hf = 53.3 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·0.05²/4 = 0.001963 m²
V = Q/A = 0.003/0.001963 = 1.528 m/s
Velocity head
V²/(2g) = 1.528²/(2·9.81) = 0.119 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(800/0.05)·1.528²/(2·9.81) = 53.3 m
Answer
hf = 53.3 m
6.Pipe friction & head loss
Answer: H = 2.14 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 = 0.04/0.03142 = 1.273 m/s
V²/(2g) = 0.08263 m
Loss coefficients
f·L/D = 0.018·150/0.2 = 13.5
ΣK = 0.5 + 3·0.3 + 10 = 11.4
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 13.5 + 11.4)·0.08263 = 2.14 m
Answer
H = 2.14 m
7.Pumps & pump power
Answer: η = 85.02 %
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 = 1000·9.81·0.03·26 = 7.652 kW
Efficiency
η = Ph/Pshaft = 7.652/9 = 85.02%
Answer
η = 85.02%
8.Pumps & pump power
Answer: H = 49.39 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 = 0.05/0.01767 = 2.829 m/s
V²/(2g) = 0.408 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (24 + 2.5)·0.408 = 10.81 m
Pressure head
p/(ρ·g) = 300000/(1000·9.81) = 30.58 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 8 + 30.58 + 10.81 = 49.39 m
Answer
H = 49.39 m
9.Open-channel flow (Manning)
Answer: Q = 3.139 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 = (4.5 + 3·0.7)·0.7 = 4.62 m²
P = b + 2y·√(1 + z²) = 4.5 + 2·0.7·√(1 + 3²) = 8.927 m
R = A/P = 0.5175 m
Velocity
V = (1/0.03)·0.51752/3·0.0011/2 = 0.6795 m/s
Flow rate
Q = A·V = 4.62·0.6795 = 3.139 m³/s
Answer
Q = 3.139 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.002894 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 = (5.5 + 2·1.3)·1.3 = 10.53 m²
P = b + 2y·√(1 + z²) = 5.5 + 2·1.3·√(1 + 2²) = 11.31 m
R = A/P = 0.9307 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.03·18/(10.53·0.9533) = 0.0538
S = 0.0538² = 0.002894
Answer
S = 0.002894
That is a fall of 2.894 m per kilometre.
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