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 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 2 mb = 3 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 seawater, with its top edge 3 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 3 mH = 2.5 mb = 3 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 75 mm to 50 mm. In the wide section the pressure is 300 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 50 mmp1 = 300 kPav1 = 2.25 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 150 mm, throat 75 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 350 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 = 150 mmd2 = 75 mmΔh = 350 mmCd = 0.97ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.2 m/s through a pipe 200 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.2 m/sL = 200 mD = 50 mmf = 0.028g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022. The water surface of the upper reservoir is 21 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 = 5) 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 = 21 mL = 100 mD = 250 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 85% delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 66 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 80 L/sH = 66 mη = 85%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 6 m higher. The pipe is 150 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 80 L/sΔz = 6 mL = 150 mD = 250 mmf = 0.018ΣK = 8
H = m
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 750 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.0005. Find the flow rate Q in m³/s.
D = 750 mmn = 0.013S = 0.0005
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 250 mm (Manning n = 0.013) must reach a self-cleansing velocity of 1 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 250 mmn = 0.013V = 1 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 294.3 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 = 4 + 2/2 = 5 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·5·6 = 294.3 kN
Answer
F = 294.3 kN
2.Hydrostatic force
Answer: ycp = 4.373 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 = 3 + 2.5/2 = 4.25 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·4.25·7.5 = 320.5 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·2.5³/12 = 3.906 m⁴
ycp = yc + Ic/(yc·A) = 4.25 + 3.906/(4.25·7.5) = 4.373 m
Answer
ycp = 4.373 m
3.Continuity & Bernoulli
Answer: p2 = 289.7 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 = 2.25·(75/50)² = 5.063 m/s
½ρ(v1² − v2²) = ½·1000·(2.25² − 5.063²)/1000 = −10.28 kPa
p2 = p1 + ½ρ(v1² − v2²) = 300 + (−10.28) = 289.7 kPa
Answer
p2 = 289.7 kPa
4.Continuity & Bernoulli
Answer: Q = 41.17 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)·9.81·0.35 = 43262 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 75/150 = 0.5
v2 = √(2·43262/(1000·(1 − 0.0625))) = 9.607 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.97·0.004418·9.607 = 0.04117 m³/s = 41.17 L/s
Answer
Q = 41.17 L/s
5.Pipe friction & head loss
Answer: hf = 27.63 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) = 2.2²/(2·9.81) = 0.2467 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(200/0.05)·2.2²/(2·9.81) = 27.63 m
Answer
hf = 27.63 m
6.Pipe friction & head loss
Answer: Q = 249.9 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 = 0.022·100/0.25 = 8.8
ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
21 = (8.8 + 7.1)·V²/(2·9.81)
V = √(2·9.81·21/15.9) = 5.091 m/s
Flow rate
Q = V·π·D²/4 = 5.091·0.04909 = 0.2499 m³/s = 249.9 L/s
Answer
Q = 249.9 L/s
7.Pumps & pump power
Answer: Pshaft = 60.94 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 = 1000·9.81·0.08·66 = 51.8 kW
Shaft power
Pshaft = Ph/η = 51.8/0.85 = 60.94 kW
Answer
Pshaft = 60.94 kW
8.Pumps & pump power
Answer: H = 8.545 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 = 0.08/0.04909 = 1.63 m/s
V²/(2g) = 0.1354 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (10.8 + 8)·0.1354 = 2.545 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 6 + 2.545 = 8.545 m
Answer
H = 8.545 m
9.Open-channel flow (Manning)
Answer: Q = 0.2489 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 = 0.4418 m², R = D/4 = 0.1875 m
Velocity
V = (1/0.013)·0.18752/3·0.00051/2 = 0.5635 m/s
Flow rate
Q = A·V = 0.4418·0.5635 = 0.2489 m³/s
Answer
Q = 0.2489 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.006814 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 = 0.25/4 = 0.0625 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·1/0.1575)² = 0.006814
Answer
S = 0.006814
That is a fall of 6.814 m per kilometre.
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