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 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 2 mb = 2 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 glycerin, with its top edge 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 2.5 mb = 2 mρ = 1260 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 0.6. Find the flow rate Q. Use g = 9.81 m/s².
d = 50 mmh = 4.5 mCd = 0.6g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 250 mm, throat 125 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 310 mm. The discharge coefficient is 0.97. 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 = 250 mmd2 = 125 mmΔh = 310 mmCd = 0.97ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 120 L/s through a pipe 800 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
Q = 120 L/sL = 800 mD = 250 mmf = 0.028g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.022. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each), a valve (K = 10) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 30 L/s?
Q = 30 L/sL = 150 mD = 150 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 10Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 65% delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 78 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 15 L/sH = 78 mη = 65%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 40 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 12 m higher. The pipe is 500 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 72%. Find the shaft power Pshaft in kW.
Q = 40 L/sΔz = 12 mL = 500 mD = 150 mmf = 0.025ΣK = 8η = 72%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A rectangular finished concrete channel is 3.5 m wide. Water flows uniformly 0.8 m deep. Manning n = 0.012 and the bed slope is S = 0.004. Find the flow rate Q in m³/s.
b = 3.5 my = 0.8 mn = 0.012S = 0.004
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 450 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 = 450 mmn = 0.013V = 0.75 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 78.48 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 = 1 + 2/2 = 2 m
A = H·b = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·4 = 78.48 kN
Answer
F = 78.48 kN
2.Hydrostatic force
Answer: F = 170 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 = 1.5 + 2.5/2 = 2.75 m
A = H·b = 2.5·2 = 5 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·2.75·5 = 170 kN
Answer
F = 170 kN
3.Continuity & Bernoulli
Answer: Q = 11.07 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) = 0.6·0.001963·9.396 = 0.01107 m³/s = 11.07 L/s
Answer
Q = 11.07 L/s
4.Continuity & Bernoulli
Answer: Q = 107.6 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)·9.81·0.31 = 38318 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 125/250 = 0.5
v2 = √(2·38318/(1000·(1 − 0.0625))) = 9.041 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.97·0.01227·9.041 = 0.1076 m³/s = 107.6 L/s
Answer
Q = 107.6 L/s
5.Pipe friction & head loss
Answer: hf = 27.29 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·0.25²/4 = 0.04909 m²
V = Q/A = 0.12/0.04909 = 2.445 m/s
Velocity head
V²/(2g) = 2.445²/(2·9.81) = 0.3046 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(800/0.25)·2.445²/(2·9.81) = 27.29 m
Answer
hf = 27.29 m
6.Pipe friction & head loss
Answer: Δz = 5.009 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 = 0.03/0.01767 = 1.698 m/s
V²/(2g) = 0.1469 m
Loss coefficients
f·L/D = 0.022·150/0.15 = 22
ΣK = 0.5 + 2·0.3 + 10 + 1.0 = 12.1
Total head loss
hf = 22·0.1469 = 3.232 m
hm = 12.1·0.1469 = 1.777 m
Δz = hf + hm = 5.009 m
Answer
Δz = 5.009 m
7.Pumps & pump power
Answer: Pshaft = 17.66 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 = 1000·9.81·0.015·78 = 11.48 kW
Shaft power
Pshaft = Ph/η = 11.48/0.65 = 17.66 kW
Answer
Pshaft = 17.66 kW
8.Pumps & pump power
Answer: Pshaft = 19.54 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 = 0.04/0.01767 = 2.264 m/s
V²/(2g) = 0.2611 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (83.33 + 8)·0.2611 = 23.85 m
Pump head
H = Δz + hL = 12 + 23.85 = 35.85 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.04·35.85/0.72 = 19.54 kW
Answer
Pshaft = 19.54 kW
9.Open-channel flow (Manning)
Answer: Q = 9.895 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 = 3.5·0.8 = 2.8 m²
P = b + 2y = 3.5 + 2·0.8 = 5.1 m
R = A/P = 0.549 m
Velocity
V = (1/0.012)·0.5492/3·0.0041/2 = 3.534 m/s
Flow rate
Q = A·V = 2.8·3.534 = 9.895 m³/s
Answer
Q = 9.895 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.00175 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 = 0.45/4 = 0.1125 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.75/0.233)² = 0.00175
Answer
S = 0.00175
That is a fall of 1.75 m per kilometre.
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