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 = 1 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1 m wide sits in the wall of a tank of seawater, with its top edge 1 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1 mH = 1 mb = 1 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 100 mm to 50 mm. In the wide section the pressure is 140 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 100 mmd2 = 50 mmp1 = 140 kPav1 = 1.25 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 75 mm while rising 8 m. At section 1 the pressure is 320 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 75 mmp1 = 320 kPav1 = 1.25 m/sΔz = 8 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 1 m/s through a horizontal pipe, 400 m long, with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 1 m/sL = 400 mD = 100 mmf = 0.018ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 4 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 75 L/s?
Q = 75 L/sL = 200 mD = 200 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 82% delivers 40 L/s of water (ρ = 1000 kg/m³) against a head of 58 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 40 L/sH = 58 mη = 82%g = 9.81 m/s²
Pshaft = kW
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 12 m higher. The pipe is 100 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 70%. Find the shaft power Pshaft in kW.
Q = 20 L/sΔz = 12 mL = 100 mD = 100 mmf = 0.022ΣK = 5η = 70%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A rectangular finished concrete channel is 2 m wide. Water flows uniformly 0.7 m deep. Manning n = 0.012 and the bed slope is S = 0.0008. Find the flow rate Q in m³/s.
b = 2 my = 0.7 mn = 0.012S = 0.0008
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular grass-lined earth channel 1 m wide must carry 1.3 m³/s in uniform flow at a depth of 2.2 m. Manning n = 0.03. Find the required bed slope S (in m/m).
Q = 1.3 m³/sb = 1 my = 2.2 mn = 0.03
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 29.43 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 + 1/2 = 1 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1·3 = 29.43 kN
Answer
F = 29.43 kN
2.Hydrostatic force
Answer: ycp = 1.556 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 = 1 + 1/2 = 1.5 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·1.5·1 = 15.08 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1·1³/12 = 0.08333 m⁴
ycp = yc + Ic/(yc·A) = 1.5 + 0.08333/(1.5·1) = 1.556 m
Answer
ycp = 1.556 m
3.Continuity & Bernoulli
Answer: p2 = 128.3 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 = 1.25·(100/50)² = 5 m/s
½ρ(v1² − v2²) = ½·1000·(1.25² − 5²)/1000 = −11.72 kPa
p2 = p1 + ½ρ(v1² − v2²) = 140 + (−11.72) = 128.3 kPa
Answer
p2 = 128.3 kPa
4.Continuity & Bernoulli
Answer: p2 = 239.8 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 = 1.25·(100/75)² = 2.222 m/s
½ρ(v1² − v2²) = ½·1000·(1.25² − 2.222²)/1000 = −1.688 kPa
ρgΔz = 1000·9.81·8/1000 = 78.48 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 320 + (−1.688) − 78.48 = 239.8 kPa
Answer
p2 = 239.8 kPa
5.Pipe friction & head loss
Answer: Δp = 36 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) = 0.018·(400/0.1)·1²/(2·9.81) = 3.67 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 = 1000·9.81·3.67 = 36000 Pa = 36 kPa
Answer
Δp = 36 kPa
6.Pipe friction & head loss
Answer: Δz = 8.046 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Loss coefficients
f·L/D = 0.015·200/0.2 = 15
ΣK = 0.5 + 4·0.3 + 10 + 1.0 = 12.7
Total head loss
hf = 15·0.2905 = 4.357 m
hm = 12.7·0.2905 = 3.689 m
Δz = hf + hm = 8.046 m
Answer
Δz = 8.046 m
7.Pumps & pump power
Answer: Pshaft = 27.76 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 = 1000·9.81·0.04·58 = 22.76 kW
Shaft power
Pshaft = Ph/η = 22.76/0.82 = 27.76 kW
Answer
Pshaft = 27.76 kW
8.Pumps & pump power
Answer: Pshaft = 5.865 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) = (22 + 5)·0.3305 = 8.924 m
Pump head
H = Δz + hL = 12 + 8.924 = 20.92 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.02·20.92/0.7 = 5.865 kW
Answer
Pshaft = 5.865 kW
9.Open-channel flow (Manning)
Answer: Q = 1.826 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 = 2·0.7 = 1.4 m²
P = b + 2y = 2 + 2·0.7 = 3.4 m
R = A/P = 0.4118 m
Velocity
V = (1/0.012)·0.41182/3·0.00081/2 = 1.305 m/s
Flow rate
Q = A·V = 1.4·1.305 = 1.826 m³/s
Answer
Q = 1.826 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.001041 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 = 1·2.2 = 2.2 m²
P = b + 2y = 1 + 2·2.2 = 5.4 m
R = A/P = 0.4074 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.03·1.3/(2.2·0.5496) = 0.03226
S = 0.03226² = 0.001041
Answer
S = 0.001041
That is a fall of 1.041 m per kilometre.
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