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 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 1.5 mb = 3 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 water, with its top edge 4 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 4 mH = 1.5 mb = 1.5 mρ = 1000 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 360 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 40 mmp1 = 360 kPav1 = 2 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 300 mm, throat 150 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 400 mm. The discharge coefficient is 0.95. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².
d1 = 300 mmd2 = 150 mmΔh = 400 mmCd = 0.95ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.6 m/s through a pipe 300 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.6 m/sL = 300 mD = 50 mmf = 0.022g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 150 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.02) 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), 2 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 120 L/s?
Q = 120 L/sL = 150 mD = 300 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 10
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 22 m while its shaft takes 36 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 100 L/sH = 22 mPshaft = 36 kWg = 9.81 m/s²
η = %
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 5 m higher. The pipe is 200 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.015; the minor-loss coefficients add up to ΣK = 2.5 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 80%. Find the shaft power Pshaft in kW.
Q = 20 L/sΔz = 5 mL = 200 mD = 100 mmf = 0.015ΣK = 2.5η = 80%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A trapezoidal finished concrete 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 = 0.012 and the bed slope is S = 0.005. Find the flow rate Q in m³/s.
b = 4.5 my = 0.7 mz = 3n = 0.012S = 0.005
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 300 mm (Manning n = 0.013) must reach a self-cleansing velocity of 1.2 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 300 mmn = 0.013V = 1.2 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 121.4 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 = 2 + 1.5/2 = 2.75 m
A = H·b = 1.5·3 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.75·4.5 = 121.4 kN
Answer
F = 121.4 kN
2.Hydrostatic force
Answer: ycp = 4.789 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 = 4 + 1.5/2 = 4.75 m
A = H·b = 1.5·1.5 = 2.25 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.75·2.25 = 104.8 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) = 4.75 + 0.4219/(4.75·2.25) = 4.789 m
Answer
ycp = 4.789 m
3.Continuity & Bernoulli
Answer: p2 = 337.3 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·(75/40)² = 7.031 m/s
½ρ(v1² − v2²) = ½·1000·(2² − 7.031²)/1000 = −22.72 kPa
p2 = p1 + ½ρ(v1² − v2²) = 360 + (−22.72) = 337.3 kPa
Answer
p2 = 337.3 kPa
4.Continuity & Bernoulli
Answer: Q = 172.4 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.
Pressure difference from the manometer
Δp = (13600 − 1000)·9.81·0.4 = 49442 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 150/300 = 0.5
v2 = √(2·49442/(1000·(1 − 0.0625))) = 10.27 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.95·0.01767·10.27 = 0.1724 m³/s = 172.4 L/s
Answer
Q = 172.4 L/s
5.Pipe friction & head loss
Answer: hf = 45.48 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) = 2.6²/(2·9.81) = 0.3445 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(300/0.05)·2.6²/(2·9.81) = 45.48 m
Answer
hf = 45.48 m
6.Pipe friction & head loss
Answer: H = 3.246 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 = 0.12/0.07069 = 1.698 m/s
V²/(2g) = 0.1469 m
Loss coefficients
f·L/D = 0.02·150/0.3 = 10
ΣK = 0.5 + 2·0.3 + 10 = 11.1
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 10 + 11.1)·0.1469 = 3.246 m
Answer
H = 3.246 m
7.Pumps & pump power
Answer: η = 59.95 %
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 = 1000·9.81·0.1·22 = 21.58 kW
Efficiency
η = Ph/Pshaft = 21.58/36 = 59.95%
Answer
η = 59.95%
8.Pumps & pump power
Answer: Pshaft = 3.861 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) = (30 + 2.5)·0.3305 = 10.74 m
Pump head
H = Δz + hL = 5 + 10.74 = 15.74 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.02·15.74/0.8 = 3.861 kW
Answer
Pshaft = 3.861 kW
9.Open-channel flow (Manning)
Answer: Q = 17.55 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 = (1/0.012)·0.51752/3·0.0051/2 = 3.798 m/s
Flow rate
Q = A·V = 4.62·3.798 = 17.55 m³/s
Answer
Q = 17.55 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.007694 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 = 0.3/4 = 0.075 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·1.2/0.1778)² = 0.007694
Answer
S = 0.007694
That is a fall of 7.694 m per kilometre.
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