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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 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 glycerin, with its top edge 2 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 2 mH = 3 mb = 1 mρ = 1260 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
A pitot-static tube in a wind tunnel reads a difference of 950 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.
p0 − p = 950 Paρ = 1.2 kg/m³
v = m/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 75 mm to 25 mm while rising 10 m. At section 1 the pressure is 160 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 75 mmd2 = 25 mmp1 = 160 kPav1 = 0.75 m/sΔz = 10 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 80 L/s through a pipe 100 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
Q = 80 L/sL = 100 mD = 300 mmf = 0.028g = 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 200 mm and a Darcy friction factor f = 0.025. The minor losses are a square-edged entrance (K = 0.5), 4 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. What difference in water level Δz between the reservoirs gives a flow rate of 50 L/s?
Q = 50 L/sL = 150 mD = 200 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 5Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 65% delivers 30 L/s of water (ρ = 1000 kg/m³) against a head of 48 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 30 L/sH = 48 mη = 65%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 21 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. Find the head H the pump must add.
Q = 20 L/sΔz = 21 mL = 100 mD = 100 mmf = 0.022ΣK = 5
H = m
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 750 mm flows half full. Manning n = 0.013 and the slope is S = 0.001. Find the flow rate Q in m³/s.
D = 750 mmn = 0.013S = 0.001
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal gravel channel with a bottom width of 2.5 m and side slopes of 3:1 (horizontal:vertical) must carry 31 m³/s in uniform flow at a depth of 1.6 m. Manning n = 0.025. Find the required bed slope S (in m/m).
Q = 31 m³/sb = 2.5 my = 1.6 mz = 3n = 0.025
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 58.86 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 = 1.5 + 1/2 = 2 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·3 = 58.86 kN
Answer
F = 58.86 kN
2.Hydrostatic force
Answer: ycp = 3.714 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 = 2 + 3/2 = 3.5 m
A = H·b = 3·1 = 3 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·3.5·3 = 129.8 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1·3³/12 = 2.25 m⁴
ycp = yc + Ic/(yc·A) = 3.5 + 2.25/(3.5·3) = 3.714 m
Answer
ycp = 3.714 m
3.Continuity & Bernoulli
Answer: v = 39.79 m/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.
Bernoulli to the stagnation point
The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.
p0 − p = ½ρv²
v = √(2·950/1.2) = 39.79 m/s
Answer
v = 39.79 m/s
4.Continuity & Bernoulli
Answer: p2 = 39.4 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·(75/25)² = 6.75 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 6.75²)/1000 = −22.5 kPa
ρgΔz = 1000·9.81·10/1000 = 98.1 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 160 + (−22.5) − 98.1 = 39.4 kPa
Answer
p2 = 39.4 kPa
5.Pipe friction & head loss
Answer: hf = 0.6093 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·0.3²/4 = 0.07069 m²
V = Q/A = 0.08/0.07069 = 1.132 m/s
Velocity head
V²/(2g) = 1.132²/(2·9.81) = 0.06529 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(100/0.3)·1.132²/(2·9.81) = 0.6093 m
Answer
hf = 0.6093 m
6.Pipe friction & head loss
Answer: Δz = 3.415 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 = 0.05/0.03142 = 1.592 m/s
V²/(2g) = 0.1291 m
Loss coefficients
f·L/D = 0.025·150/0.2 = 18.75
ΣK = 0.5 + 4·0.3 + 5 + 1.0 = 7.7
Total head loss
hf = 18.75·0.1291 = 2.421 m
hm = 7.7·0.1291 = 0.9941 m
Δz = hf + hm = 3.415 m
Answer
Δz = 3.415 m
7.Pumps & pump power
Answer: Pshaft = 21.73 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 = 1000·9.81·0.03·48 = 14.13 kW
Shaft power
Pshaft = Ph/η = 14.13/0.65 = 21.73 kW
Answer
Pshaft = 21.73 kW
8.Pumps & pump power
Answer: H = 29.92 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 = 0.02/0.007854 = 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
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 21 + 8.924 = 29.92 m
Answer
H = 29.92 m
9.Open-channel flow (Manning)
Answer: Q = 0.176 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 = 0.2209 m², R = D/4 = 0.1875 m
Velocity
V = (1/0.013)·0.18752/3·0.0011/2 = 0.7969 m/s
Flow rate
Q = A·V = 0.2209·0.7969 = 0.176 m³/s
Answer
Q = 0.176 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.004881 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 + z·y)·y = (2.5 + 3·1.6)·1.6 = 11.68 m²
P = b + 2y·√(1 + z²) = 2.5 + 2·1.6·√(1 + 3²) = 12.62 m
R = A/P = 0.9256 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.025·31/(11.68·0.9497) = 0.06986
S = 0.06986² = 0.004881
Answer
S = 0.004881
That is a fall of 4.881 m per kilometre.
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