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 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 2.5 mb = 1 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 glycerin, with its top edge 1 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1 mH = 3 mb = 3 mρ = 1260 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 75 mm to 25 mm. In the wide section the pressure is 340 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 25 mmp1 = 340 kPav1 = 1.25 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 200 mm to 75 mm while rising 1 m. At section 1 the pressure is 260 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 200 mmd2 = 75 mmp1 = 260 kPav1 = 0.5 m/sΔz = 1 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 80 L/s through a pipe 300 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
Q = 80 L/sL = 300 mD = 200 mmf = 0.018g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 200 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.022) runs from an open reservoir and discharges as a free jet into the air. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each) and a valve (K = 10). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 150 L/s?
Q = 150 L/sL = 200 mD = 300 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 150 L/s of water (ρ = 1000 kg/m³) against a head of 58 m while its shaft takes 142.2 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 150 L/sH = 58 mPshaft = 142.2 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 15 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 150 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 = 6 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 15 L/sΔz = 8 mL = 150 mD = 100 mmf = 0.022ΣK = 6
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal clean earth channel has a bottom width of 3.5 m and side slopes of 1:1 (horizontal:vertical). Water flows uniformly 1.4 m deep. Manning n = 0.022 and the bed slope is S = 0.0008. Find the flow rate Q in m³/s.
b = 3.5 my = 1.4 mz = 1n = 0.022S = 0.0008
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 0.9 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 200 mmn = 0.013V = 0.9 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 42.92 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 + 2.5/2 = 1.75 m
A = H·b = 2.5·1 = 2.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.75·2.5 = 42.92 kN
Answer
F = 42.92 kN
2.Hydrostatic force
Answer: ycp = 2.8 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 = 1 + 3/2 = 2.5 m
A = H·b = 3·3 = 9 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·2.5·9 = 278.1 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) = 2.5 + 6.75/(2.5·9) = 2.8 m
Answer
ycp = 2.8 m
3.Continuity & Bernoulli
Answer: p2 = 277.5 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 = 1.25·(75/25)² = 11.25 m/s
½ρ(v1² − v2²) = ½·1000·(1.25² − 11.25²)/1000 = −62.5 kPa
p2 = p1 + ½ρ(v1² − v2²) = 340 + (−62.5) = 277.5 kPa
Answer
p2 = 277.5 kPa
4.Continuity & Bernoulli
Answer: p2 = 244 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 = 0.5·(200/75)² = 3.556 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 3.556²)/1000 = −6.196 kPa
ρgΔz = 1000·9.81·1/1000 = 9.81 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 260 + (−6.196) − 9.81 = 244 kPa
Answer
p2 = 244 kPa
5.Pipe friction & head loss
Answer: hf = 8.924 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·0.2²/4 = 0.03142 m²
V = Q/A = 0.08/0.03142 = 2.546 m/s
Velocity head
V²/(2g) = 2.546²/(2·9.81) = 0.3305 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(300/0.2)·2.546²/(2·9.81) = 8.924 m
Answer
hf = 8.924 m
6.Pipe friction & head loss
Answer: H = 6.281 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 = 0.15/0.07069 = 2.122 m/s
V²/(2g) = 0.2295 m
Loss coefficients
f·L/D = 0.022·200/0.3 = 14.67
ΣK = 0.5 + 4·0.3 + 10 = 11.7
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 14.67 + 11.7)·0.2295 = 6.281 m
Answer
H = 6.281 m
7.Pumps & pump power
Answer: η = 60.02 %
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 = 1000·9.81·0.15·58 = 85.35 kW
Efficiency
η = Ph/Pshaft = 85.35/142.2 = 60.02%
Answer
η = 60.02%
8.Pumps & pump power
Answer: H = 15.25 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 = 0.015/0.007854 = 1.91 m/s
V²/(2g) = 0.1859 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (33 + 6)·0.1859 = 7.251 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 8 + 7.251 = 15.25 m
Answer
H = 15.25 m
9.Open-channel flow (Manning)
Answer: Q = 8.34 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 = (3.5 + 1·1.4)·1.4 = 6.86 m²
P = b + 2y·√(1 + z²) = 3.5 + 2·1.4·√(1 + 1²) = 7.46 m
R = A/P = 0.9196 m
Velocity
V = (1/0.022)·0.91962/3·0.00081/2 = 1.216 m/s
Flow rate
Q = A·V = 6.86·1.216 = 8.34 m³/s
Answer
Q = 8.34 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.007432 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 = (0.013·0.9/0.1357)² = 0.007432
Answer
S = 0.007432
That is a fall of 7.432 m per kilometre.
ExamProblems · examproblems.com/courses/hydraulic-engineering