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 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1 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 seawater, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 1 mb = 1.5 mρ = 1025 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 6 m above the orifice and the discharge coefficient is 0.63. Find the flow rate Q. Use g = 9.81 m/s².
d = 50 mmh = 6 mCd = 0.63g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 200 mm to 100 mm while rising 6 m. At section 1 the pressure is 330 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 200 mmd2 = 100 mmp1 = 330 kPav1 = 2 m/sΔz = 6 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 25 L/s through a pipe 150 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
Q = 25 L/sL = 150 mD = 200 mmf = 0.028g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 200 m long with an inside diameter of 250 mm (Darcy friction factor f = 0.022) 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 = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 60 L/s?
Q = 60 L/sL = 200 mD = 250 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2
H = m
7.Pumps & pump powerMedium
A pump with an efficiency of 60% delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 44 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 60 L/sH = 44 mη = 60%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 80 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 200 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 3 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 80 L/sΔz = 8 mL = 200 mD = 200 mmf = 0.022ΣK = 3
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal brickwork channel has a bottom width of 4 m and side slopes of 1.5:1 (horizontal:vertical). Water flows uniformly 1 m deep. Manning n = 0.015 and the bed slope is S = 0.003. Find the flow rate Q in m³/s.
b = 4 my = 1 mz = 1.5n = 0.015S = 0.003
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular gravel channel 2.5 m wide must carry 3.3 m³/s in uniform flow at a depth of 1.5 m. Manning n = 0.025. Find the required bed slope S (in m/m).
Q = 3.3 m³/sb = 2.5 my = 1.5 mn = 0.025
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 44.15 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 + 1/2 = 1.5 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.5·3 = 44.15 kN
Answer
F = 44.15 kN
2.Hydrostatic force
Answer: F = 45.25 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 + 1/2 = 3 m
A = H·b = 1·1.5 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·3·1.5 = 45.25 kN
Answer
F = 45.25 kN
3.Continuity & Bernoulli
Answer: Q = 13.42 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) = √(2·9.81·6) = 10.85 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.63·0.001963·10.85 = 0.01342 m³/s = 13.42 L/s
Answer
Q = 13.42 L/s
4.Continuity & Bernoulli
Answer: p2 = 241.1 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·(200/100)² = 8 m/s
½ρ(v1² − v2²) = ½·1000·(2² − 8²)/1000 = −30 kPa
ρgΔz = 1000·9.81·6/1000 = 58.86 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 330 + (−30) − 58.86 = 241.1 kPa
Answer
p2 = 241.1 kPa
5.Pipe friction & head loss
Answer: hf = 0.6778 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·0.2²/4 = 0.03142 m²
V = Q/A = 0.025/0.03142 = 0.7958 m/s
Velocity head
V²/(2g) = 0.7958²/(2·9.81) = 0.03228 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(150/0.2)·0.7958²/(2·9.81) = 0.6778 m
Answer
hf = 0.6778 m
6.Pipe friction & head loss
Answer: H = 1.675 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 = 0.06/0.04909 = 1.222 m/s
V²/(2g) = 0.07615 m
Loss coefficients
f·L/D = 0.022·200/0.25 = 17.6
ΣK = 0.5 + 3·0.3 + 2 = 3.4
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 17.6 + 3.4)·0.07615 = 1.675 m
Answer
H = 1.675 m
7.Pumps & pump power
Answer: Pshaft = 43.16 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 = 1000·9.81·0.06·44 = 25.9 kW
Shaft power
Pshaft = Ph/η = 25.9/0.6 = 43.16 kW
Answer
Pshaft = 43.16 kW
8.Pumps & pump power
Answer: H = 16.26 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 = 0.08/0.03142 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (22 + 3)·0.3305 = 8.263 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 8 + 8.263 = 16.26 m
Answer
H = 16.26 m
9.Open-channel flow (Manning)
Answer: Q = 16.18 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 + 1.5·1)·1 = 5.5 m²
P = b + 2y·√(1 + z²) = 4 + 2·1·√(1 + 1.5²) = 7.606 m
R = A/P = 0.7232 m
Velocity
V = (1/0.015)·0.72322/3·0.0031/2 = 2.942 m/s
Flow rate
Q = A·V = 5.5·2.942 = 16.18 m³/s
Answer
Q = 16.18 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0008065 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·y = 2.5·1.5 = 3.75 m²
P = b + 2y = 2.5 + 2·1.5 = 5.5 m
R = A/P = 0.6818 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.025·3.3/(3.75·0.7747) = 0.0284
S = 0.0284² = 0.0008065
Answer
S = 0.0008065
That is a fall of 0.8065 m per kilometre.
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