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 = 3 mb = 3 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 water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 1.5 mb = 2 mρ = 1000 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 20 mm. The water level is kept 1 m above the orifice and the discharge coefficient is 0.63. Find the flow rate Q. Use g = 9.81 m/s².
d = 20 mmh = 1 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 1 m. At section 1 the pressure is 210 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 200 mmd2 = 100 mmp1 = 210 kPav1 = 1.75 m/sΔz = 1 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 75 L/s through a pipe 400 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
Q = 75 L/sL = 400 mD = 250 mmf = 0.022g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.025. The minor losses are a square-edged entrance (K = 0.5), 3 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 100 L/s?
Q = 100 L/sL = 300 mD = 250 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 68 m while its shaft takes 22.2 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 25 L/sH = 68 mPshaft = 22.2 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 120 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 11 m higher. The pipe is 600 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The upper tank is closed and the air above its water is at a gauge pressure of 100 kPa. Find the head H the pump must add.
Q = 120 L/sΔz = 11 mL = 600 mD = 250 mmf = 0.022ΣK = 8ptank = 100 kPa
H = m
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 450 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.002. Find the flow rate Q in m³/s.
D = 450 mmn = 0.013S = 0.002
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 300 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 = 300 mmn = 0.013V = 0.75 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 176.6 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 + 3/2 = 2 m
A = H·b = 3·3 = 9 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·9 = 176.6 kN
Answer
F = 176.6 kN
2.Hydrostatic force
Answer: F = 139.8 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 + 1.5/2 = 4.75 m
A = H·b = 1.5·2 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.75·3 = 139.8 kN
Answer
F = 139.8 kN
3.Continuity & Bernoulli
Answer: Q = 0.8767 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) = √(2·9.81·1) = 4.429 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.63·0.0003142·4.429 = 0.0008767 m³/s = 0.8767 L/s
Answer
Q = 0.8767 L/s
4.Continuity & Bernoulli
Answer: p2 = 177.2 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 = 1.75·(200/100)² = 7 m/s
½ρ(v1² − v2²) = ½·1000·(1.75² − 7²)/1000 = −22.97 kPa
ρgΔz = 1000·9.81·1/1000 = 9.81 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 210 + (−22.97) − 9.81 = 177.2 kPa
Answer
p2 = 177.2 kPa
5.Pipe friction & head loss
Answer: hf = 4.188 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 = 0.075/0.04909 = 1.528 m/s
Velocity head
V²/(2g) = 1.528²/(2·9.81) = 0.119 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(400/0.25)·1.528²/(2·9.81) = 4.188 m
Answer
hf = 4.188 m
6.Pipe friction & head loss
Answer: Δz = 8.969 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 = 0.1/0.04909 = 2.037 m/s
V²/(2g) = 0.2115 m
Loss coefficients
f·L/D = 0.025·300/0.25 = 30
ΣK = 0.5 + 3·0.3 + 10 + 1.0 = 12.4
Total head loss
hf = 30·0.2115 = 6.346 m
hm = 12.4·0.2115 = 2.623 m
Δz = hf + hm = 8.969 m
Answer
Δz = 8.969 m
7.Pumps & pump power
Answer: η = 75.12 %
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 = 1000·9.81·0.025·68 = 16.68 kW
Efficiency
η = Ph/Pshaft = 16.68/22.2 = 75.12%
Answer
η = 75.12%
8.Pumps & pump power
Answer: H = 39.71 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (52.8 + 8)·0.3046 = 18.52 m
Pressure head
p/(ρ·g) = 100000/(1000·9.81) = 10.19 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 11 + 10.19 + 18.52 = 39.71 m
Answer
H = 39.71 m
9.Open-channel flow (Manning)
Answer: Q = 0.1275 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 = 0.159 m², R = D/4 = 0.1125 m
Velocity
V = (1/0.013)·0.11252/3·0.0021/2 = 0.8017 m/s
Flow rate
Q = A·V = 0.159·0.8017 = 0.1275 m³/s
Answer
Q = 0.1275 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.003006 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 = 0.3/4 = 0.075 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.75/0.1778)² = 0.003006
Answer
S = 0.003006
That is a fall of 3.006 m per kilometre.
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