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 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 mb = 2 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 1.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1.5 mH = 3 mb = 3 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 100 mm. In the wide section the pressure is 400 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 100 mmp1 = 400 kPav1 = 3 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 10 m. At section 1 the pressure is 300 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 75 mmp1 = 300 kPav1 = 1.5 m/sΔz = 10 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 5 L/s through a pipe 50 m long with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss hf.
Q = 5 L/sL = 50 mD = 75 mmf = 0.03g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 2) 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 60 L/s?
Q = 60 L/sL = 200 mD = 250 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 72% delivers 30 L/s of water (ρ = 1000 kg/m³) against a head of 46 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 30 L/sH = 46 mη = 72%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 11 m higher. The pipe is 200 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 = 4 (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 400 kPa. Find the head H the pump must add.
Q = 75 L/sΔz = 11 mL = 200 mD = 200 mmf = 0.022ΣK = 4ptank = 400 kPa
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal finished concrete channel has a bottom width of 3 m and side slopes of 1:1 (horizontal:vertical). Water flows uniformly 0.8 m deep. Manning n = 0.012 and the bed slope is S = 0.004. Find the flow rate Q in m³/s.
b = 3 my = 0.8 mz = 1n = 0.012S = 0.004
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 1.2 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 300 mmn = 0.013V = 1.2 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 88.29 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 + 1/2 = 4.5 m
A = H·b = 1·2 = 2 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.5·2 = 88.29 kN
Answer
F = 88.29 kN
2.Hydrostatic force
Answer: ycp = 3.25 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 = 1.5 + 3/2 = 3 m
A = H·b = 3·3 = 9 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·3·9 = 271.5 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·3³/12 = 6.75 m⁴
ycp = yc + Ic/(yc·A) = 3 + 6.75/(3·9) = 3.25 m
Answer
ycp = 3.25 m
3.Continuity & Bernoulli
Answer: p2 = 381.7 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 = 3·(150/100)² = 6.75 m/s
½ρ(v1² − v2²) = ½·1000·(3² − 6.75²)/1000 = −18.28 kPa
p2 = p1 + ½ρ(v1² − v2²) = 400 + (−18.28) = 381.7 kPa
Answer
p2 = 381.7 kPa
4.Continuity & Bernoulli
Answer: p2 = 185 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 = 1.5·(150/75)² = 6 m/s
½ρ(v1² − v2²) = ½·1000·(1.5² − 6²)/1000 = −16.88 kPa
ρgΔz = 1000·9.81·10/1000 = 98.1 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 300 + (−16.88) − 98.1 = 185 kPa
Answer
p2 = 185 kPa
5.Pipe friction & head loss
Answer: hf = 1.306 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·0.075²/4 = 0.004418 m²
V = Q/A = 0.005/0.004418 = 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·(50/0.075)·1.132²/(2·9.81) = 1.306 m
Answer
hf = 1.306 m
6.Pipe friction & head loss
Answer: Δz = 1.675 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 = 0.06/0.04909 = 1.222 m/s
V²/(2g) = 0.07615 m
Loss coefficients
f·L/D = 0.022·200/0.25 = 17.6
ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4
Total head loss
hf = 17.6·0.07615 = 1.34 m
hm = 4.4·0.07615 = 0.3351 m
Δz = hf + hm = 1.675 m
Answer
Δz = 1.675 m
7.Pumps & pump power
Answer: Pshaft = 18.8 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 = 1000·9.81·0.03·46 = 13.54 kW
Shaft power
Pshaft = Ph/η = 13.54/0.72 = 18.8 kW
Answer
Pshaft = 18.8 kW
8.Pumps & pump power
Answer: H = 59.33 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (22 + 4)·0.2905 = 7.553 m
Pressure head
p/(ρ·g) = 400000/(1000·9.81) = 40.77 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 11 + 40.77 + 7.553 = 59.33 m
Answer
H = 59.33 m
9.Open-channel flow (Manning)
Answer: Q = 11.11 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 + z·y)·y = (3 + 1·0.8)·0.8 = 3.04 m²
P = b + 2y·√(1 + z²) = 3 + 2·0.8·√(1 + 1²) = 5.263 m
R = A/P = 0.5776 m
Velocity
V = (1/0.012)·0.57762/3·0.0041/2 = 3.656 m/s
Flow rate
Q = A·V = 3.04·3.656 = 11.11 m³/s
Answer
Q = 11.11 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.007694 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 = (0.013·1.2/0.1778)² = 0.007694
Answer
S = 0.007694
That is a fall of 7.694 m per kilometre.
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