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 1.5 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 = 1.5 mb = 1.5 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 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 = 1.5 mb = 2 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows at 21 L/s through a horizontal pipe that narrows from 100 mm to 50 mm. The pressure in the narrow section is 140 kPa. Ignoring losses, find the pressure p1 in the wide section.
Q = 21 L/sd1 = 100 mmd2 = 50 mmp2 = 140 kPaρ = 1000 kg/m³
p1 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 75 mm while rising 5 m. At section 1 the pressure is 310 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 75 mmp1 = 310 kPav1 = 0.75 m/sΔz = 5 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.9 m/s through a horizontal pipe, 400 m long, with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.9 m/sL = 400 mD = 200 mmf = 0.028ρ = 1000 kg/m³
Δp = kPa
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 = 0.022. The water surface of the upper reservoir is 31 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 4 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. Find the flow rate Q in L/s.
Δz = 31 mL = 300 mD = 250 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 70% delivers 30 L/s of water (ρ = 1000 kg/m³) against a head of 20 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 30 L/sH = 20 mη = 70%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 100 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 8 m higher. The pipe is 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.015; the minor-loss coefficients add up to ΣK = 4 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 80%. Find the shaft power Pshaft in kW.
Q = 100 L/sΔz = 8 mL = 300 mD = 250 mmf = 0.015ΣK = 4η = 80%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A rectangular unfinished concrete channel is 1.5 m wide. Water flows uniformly 0.6 m deep. Manning n = 0.014 and the bed slope is S = 0.001. Find the flow rate Q in m³/s.
b = 1.5 my = 0.6 mn = 0.014S = 0.001
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular brickwork channel 6 m wide must carry 14 m³/s in uniform flow at a depth of 1.2 m. Manning n = 0.015. Find the required bed slope S (in m/m).
Q = 14 m³/sb = 6 my = 1.2 mn = 0.015
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 49.66 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 + 1.5/2 = 2.25 m
A = H·b = 1.5·1.5 = 2.25 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.25·2.25 = 49.66 kN
Answer
F = 49.66 kN
2.Hydrostatic force
Answer: ycp = 3.8 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 + 1.5/2 = 3.75 m
A = H·b = 1.5·2 = 3 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·3.75·3 = 113.1 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 2·1.5³/12 = 0.5625 m⁴
ycp = yc + Ic/(yc·A) = 3.75 + 0.5625/(3.75·3) = 3.8 m
Answer
ycp = 3.8 m
3.Continuity & Bernoulli
Answer: p1 = 193.6 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.
Velocities
v1 = Q/A1 = 0.021/0.007854 = 2.674 m/s
v2 = Q/A2 = 0.021/0.001963 = 10.7 m/s
Bernoulli
p1 = p2 + ½ρ(v2² − v1²) = 140 + ½·1000·(10.7² − 2.674²)/1000 = 140 + 53.62 = 193.6 kPa
Answer
p1 = 193.6 kPa
4.Continuity & Bernoulli
Answer: p2 = 260.3 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 = 0.75·(100/75)² = 1.333 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 1.333²)/1000 = −0.6076 kPa
ρgΔz = 1000·9.81·5/1000 = 49.05 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 310 + (−0.6076) − 49.05 = 260.3 kPa
Answer
p2 = 260.3 kPa
5.Pipe friction & head loss
Answer: Δp = 235.5 kPa
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.
Head loss
hf = f·(L/D)·V²/(2g) = 0.028·(400/0.2)·2.9²/(2·9.81) = 24 m
Pressure drop
The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·hf.
Δp = ρ·g·hf = 1000·9.81·24 = 235480 Pa = 235.5 kPa
Answer
Δp = 235.5 kPa
6.Pipe friction & head loss
Answer: Q = 217.1 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 = 0.022·300/0.25 = 26.4
ΣK = 0.5 + 4·0.3 + 2 + 1.0 = 4.7
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
31 = (26.4 + 4.7)·V²/(2·9.81)
V = √(2·9.81·31/31.1) = 4.422 m/s
Flow rate
Q = V·π·D²/4 = 4.422·0.04909 = 0.2171 m³/s = 217.1 L/s
Answer
Q = 217.1 L/s
7.Pumps & pump power
Answer: Pshaft = 8.409 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 = 1000·9.81·0.03·20 = 5.886 kW
Shaft power
Pshaft = Ph/η = 5.886/0.7 = 8.409 kW
Answer
Pshaft = 8.409 kW
8.Pumps & pump power
Answer: Pshaft = 15.52 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 = 0.1/0.04909 = 2.037 m/s
V²/(2g) = 0.2115 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (18 + 4)·0.2115 = 4.654 m
Pump head
H = Δz + hL = 8 + 4.654 = 12.65 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.1·12.65/0.8 = 15.52 kW
Answer
Pshaft = 15.52 kW
9.Open-channel flow (Manning)
Answer: Q = 0.9773 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·y = 1.5·0.6 = 0.9 m²
P = b + 2y = 1.5 + 2·0.6 = 2.7 m
R = A/P = 0.3333 m
Velocity
V = (1/0.014)·0.33332/3·0.0011/2 = 1.086 m/s
Flow rate
Q = A·V = 0.9·1.086 = 0.9773 m³/s
Answer
Q = 0.9773 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.001045 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 = 6·1.2 = 7.2 m²
P = b + 2y = 6 + 2·1.2 = 8.4 m
R = A/P = 0.8571 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.015·14/(7.2·0.9023) = 0.03232
S = 0.03232² = 0.001045
Answer
S = 0.001045
That is a fall of 1.045 m per kilometre.
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