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 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 = 2.5 mb = 1 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 oil, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 1 mb = 1 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows at 21 L/s through a horizontal pipe that narrows from 200 mm to 100 mm. The pressure in the narrow section is 120 kPa. Ignoring losses, find the pressure p1 in the wide section.
Q = 21 L/sd1 = 200 mmd2 = 100 mmp2 = 120 kPaρ = 1000 kg/m³
p1 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 75 mm to 40 mm while rising 6 m. At section 1 the pressure is 330 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 75 mmd2 = 40 mmp1 = 330 kPav1 = 2 m/sΔz = 6 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.9 m/s through a pipe 50 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.9 m/sL = 50 mD = 50 mmf = 0.022g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 400 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.018. 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. What difference in water level Δz between the reservoirs gives a flow rate of 150 L/s?
Q = 150 L/sL = 400 mD = 250 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 70% delivers 150 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 = 150 L/sH = 66 mη = 70%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 400 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.02; 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 = 20 L/sΔz = 21 mL = 400 mD = 100 mmf = 0.02ΣK = 8
H = m
9.Open-channel flow (Manning)Medium
A rectangular grass-lined earth channel is 3.5 m wide. Water flows uniformly 2 m deep. Manning n = 0.03 and the bed slope is S = 0.0005. Find the flow rate Q in m³/s.
b = 3.5 my = 2 mn = 0.03S = 0.0005
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 200 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 = 200 mmn = 0.013V = 0.75 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 67.44 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 + 2.5/2 = 2.75 m
A = H·b = 2.5·1 = 2.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.75·2.5 = 67.44 kN
Answer
F = 67.44 kN
2.Hydrostatic force
Answer: F = 25.9 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 = 2.5 + 1/2 = 3 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·3·1 = 25.9 kN
Answer
F = 25.9 kN
3.Continuity & Bernoulli
Answer: p1 = 123.4 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.
Velocities
v1 = Q/A1 = 0.021/0.03142 = 0.6685 m/s
v2 = Q/A2 = 0.021/0.007854 = 2.674 m/s
Bernoulli
p1 = p2 + ½ρ(v2² − v1²) = 120 + ½·1000·(2.674² − 0.6685²)/1000 = 120 + 3.351 = 123.4 kPa
Answer
p1 = 123.4 kPa
4.Continuity & Bernoulli
Answer: p2 = 248.4 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 = 2·(75/40)² = 7.031 m/s
½ρ(v1² − v2²) = ½·1000·(2² − 7.031²)/1000 = −22.72 kPa
ρgΔz = 1000·9.81·6/1000 = 58.86 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 330 + (−22.72) − 58.86 = 248.4 kPa
Answer
p2 = 248.4 kPa
5.Pipe friction & head loss
Answer: hf = 9.43 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) = 2.9²/(2·9.81) = 0.4286 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(50/0.05)·2.9²/(2·9.81) = 9.43 m
Answer
hf = 9.43 m
6.Pipe friction & head loss
Answer: Δz = 17.09 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 = 0.15/0.04909 = 3.056 m/s
V²/(2g) = 0.4759 m
Loss coefficients
f·L/D = 0.018·400/0.25 = 28.8
ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1
Total head loss
hf = 28.8·0.4759 = 13.71 m
hm = 7.1·0.4759 = 3.379 m
Δz = hf + hm = 17.09 m
Answer
Δz = 17.09 m
7.Pumps & pump power
Answer: Pshaft = 138.7 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 = 1000·9.81·0.15·66 = 97.12 kW
Shaft power
Pshaft = Ph/η = 97.12/0.7 = 138.7 kW
Answer
Pshaft = 138.7 kW
8.Pumps & pump power
Answer: H = 50.08 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) = (80 + 8)·0.3305 = 29.08 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 21 + 29.08 = 50.08 m
Answer
H = 50.08 m
9.Open-channel flow (Manning)
Answer: Q = 4.983 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 = 3.5·2 = 7 m²
P = b + 2y = 3.5 + 2·2 = 7.5 m
R = A/P = 0.9333 m
Velocity
V = (1/0.03)·0.93332/3·0.00051/2 = 0.7118 m/s
Flow rate
Q = A·V = 7·0.7118 = 4.983 m³/s
Answer
Q = 4.983 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.005161 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 = 0.2/4 = 0.05 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.75/0.1357)² = 0.005161
Answer
S = 0.005161
That is a fall of 5.161 m per kilometre.
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