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 1 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 1.5 mb = 1 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 oil, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 2 mb = 3 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 100 mm to 50 mm. In the wide section the pressure is 380 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 100 mmd2 = 50 mmp1 = 380 kPav1 = 1.25 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 40 mm while rising 1 m. At section 1 the pressure is 180 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 40 mmp1 = 180 kPav1 = 0.75 m/sΔz = 1 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 20 L/s through a pipe 50 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss hf.
Q = 20 L/sL = 50 mD = 150 mmf = 0.03g = 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 250 mm and a Darcy friction factor f = 0.015. The water surface of the upper reservoir is 7 m above that of the lower one. 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. Find the flow rate Q in L/s.
Δz = 7 mL = 150 mD = 250 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 10Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 82% delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 68 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 15 L/sH = 68 mη = 82%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 75 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 300 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 = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 80%. Find the shaft power Pshaft in kW.
Q = 75 L/sΔz = 12 mL = 300 mD = 200 mmf = 0.022ΣK = 6η = 80%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 1200 mm flows half full. Manning n = 0.013 and the slope is S = 0.004. Find the flow rate Q in m³/s.
D = 1200 mmn = 0.013S = 0.004
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular brickwork channel 5.5 m wide must carry 17 m³/s in uniform flow at a depth of 1.5 m. Manning n = 0.015. Find the required bed slope S (in m/m).
Q = 17 m³/sb = 5.5 my = 1.5 mn = 0.015
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 47.82 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 + 1.5/2 = 3.25 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.25·1.5 = 47.82 kN
Answer
F = 47.82 kN
2.Hydrostatic force
Answer: F = 77.7 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 + 2/2 = 1.5 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·1.5·6 = 77.7 kN
Answer
F = 77.7 kN
3.Continuity & Bernoulli
Answer: p2 = 368.3 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) = 380 + (−11.72) = 368.3 kPa
Answer
p2 = 368.3 kPa
4.Continuity & Bernoulli
Answer: p2 = 114.9 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 = 0.75·(150/40)² = 10.55 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 10.55²)/1000 = −55.34 kPa
ρgΔz = 1000·9.81·1/1000 = 9.81 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 180 + (−55.34) − 9.81 = 114.9 kPa
Answer
p2 = 114.9 kPa
5.Pipe friction & head loss
Answer: hf = 0.6529 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·0.15²/4 = 0.01767 m²
V = Q/A = 0.02/0.01767 = 1.132 m/s
Velocity head
V²/(2g) = 1.132²/(2·9.81) = 0.06529 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.03·(50/0.15)·1.132²/(2·9.81) = 0.6529 m
Answer
hf = 0.6529 m
6.Pipe friction & head loss
Answer: Q = 125.2 L/s
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.
Loss coefficients
f·L/D = 0.015·150/0.25 = 9
ΣK = 0.5 + 2·0.3 + 10 + 1.0 = 12.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
7 = (9 + 12.1)·V²/(2·9.81)
V = √(2·9.81·7/21.1) = 2.551 m/s
Flow rate
Q = V·π·D²/4 = 2.551·0.04909 = 0.1252 m³/s = 125.2 L/s
Answer
Q = 125.2 L/s
7.Pumps & pump power
Answer: Pshaft = 12.2 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 = 1000·9.81·0.015·68 = 10.01 kW
Shaft power
Pshaft = Ph/η = 10.01/0.82 = 12.2 kW
Answer
Pshaft = 12.2 kW
8.Pumps & pump power
Answer: Pshaft = 21.46 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (33 + 6)·0.2905 = 11.33 m
Pump head
H = Δz + hL = 12 + 11.33 = 23.33 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.075·23.33/0.8 = 21.46 kW
Answer
Pshaft = 21.46 kW
9.Open-channel flow (Manning)
Answer: Q = 1.233 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 = 0.5655 m², R = D/4 = 0.3 m
Velocity
V = (1/0.013)·0.32/3·0.0041/2 = 2.18 m/s
Flow rate
Q = A·V = 0.5655·2.18 = 1.233 m³/s
Answer
Q = 1.233 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0009942 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 = 5.5·1.5 = 8.25 m²
P = b + 2y = 5.5 + 2·1.5 = 8.5 m
R = A/P = 0.9706 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.015·17/(8.25·0.9803) = 0.03153
S = 0.03153² = 0.0009942
Answer
S = 0.0009942
That is a fall of 0.9942 m per kilometre.
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