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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 2.5 mb = 3 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 resultant hydrostatic force F on the gate.
d = 1.5 mH = 3 mb = 3 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 100 mm. In the wide section the pressure is 350 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 100 mmp1 = 350 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 8 m. At section 1 the pressure is 240 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 75 mmp1 = 240 kPav1 = 1.75 m/sΔz = 8 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, 100 m long, with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.9 m/sL = 100 mD = 50 mmf = 0.025ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022. The water surface of the upper reservoir is 9 m above that of the lower one. 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. Find the flow rate Q in L/s.
Δz = 9 mL = 200 mD = 200 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 16 m while its shaft takes 9.8 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 50 L/sH = 16 mPshaft = 9.8 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 9 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 = 5 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 50 L/sΔz = 9 mL = 200 mD = 200 mmf = 0.022ΣK = 5
H = m
9.Open-channel flow (Manning)Medium
A rectangular gravel channel is 1 m wide. Water flows uniformly 0.6 m deep. Manning n = 0.025 and the bed slope is S = 0.004. Find the flow rate Q in m³/s.
b = 1 my = 0.6 mn = 0.025S = 0.004
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal gravel channel with a bottom width of 1.5 m and side slopes of 1.5:1 (horizontal:vertical) must carry 17 m³/s in uniform flow at a depth of 2.5 m. Manning n = 0.025. Find the required bed slope S (in m/m).
Q = 17 m³/sb = 1.5 my = 2.5 mz = 1.5n = 0.025
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 275.9 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 + 2.5/2 = 3.75 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.75·7.5 = 275.9 kN
Answer
F = 275.9 kN
2.Hydrostatic force
Answer: F = 271.5 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 + 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
Answer
F = 271.5 kN
3.Continuity & Bernoulli
Answer: p2 = 331.7 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) = 350 + (−18.28) = 331.7 kPa
Answer
p2 = 331.7 kPa
4.Continuity & Bernoulli
Answer: p2 = 138.6 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.
Continuity and Bernoulli
Work in Pa, then divide by 1000 to get kPa.
v2 = v1·(d1/d2)² = 1.75·(150/75)² = 7 m/s
½ρ(v1² − v2²) = ½·1000·(1.75² − 7²)/1000 = −22.97 kPa
ρgΔz = 1000·9.81·8/1000 = 78.48 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 240 + (−22.97) − 78.48 = 138.6 kPa
Answer
p2 = 138.6 kPa
5.Pipe friction & head loss
Answer: Δp = 210.3 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) = 0.025·(100/0.05)·2.9²/(2·9.81) = 21.43 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 = 1000·9.81·21.43 = 210250 Pa = 210.3 kPa
Answer
Δp = 210.3 kPa
6.Pipe friction & head loss
Answer: Q = 81.25 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 = 0.022·200/0.2 = 22
ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
9 = (22 + 4.4)·V²/(2·9.81)
V = √(2·9.81·9/26.4) = 2.586 m/s
Flow rate
Q = V·π·D²/4 = 2.586·0.03142 = 0.08125 m³/s = 81.25 L/s
Answer
Q = 81.25 L/s
7.Pumps & pump power
Answer: η = 80.08 %
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 = 1000·9.81·0.05·16 = 7.848 kW
Efficiency
η = Ph/Pshaft = 7.848/9.8 = 80.08%
Answer
η = 80.08%
8.Pumps & pump power
Answer: H = 12.49 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 = 0.05/0.03142 = 1.592 m/s
V²/(2g) = 0.1291 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (22 + 5)·0.1291 = 3.486 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 9 + 3.486 = 12.49 m
Answer
H = 12.49 m
9.Open-channel flow (Manning)
Answer: Q = 0.6384 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 = 1·0.6 = 0.6 m²
P = b + 2y = 1 + 2·0.6 = 2.2 m
R = A/P = 0.2727 m
Velocity
V = (1/0.025)·0.27272/3·0.0041/2 = 1.064 m/s
Flow rate
Q = A·V = 0.6·1.064 = 0.6384 m³/s
Answer
Q = 0.6384 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0007801 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 = (1.5 + 1.5·2.5)·2.5 = 13.13 m²
P = b + 2y·√(1 + z²) = 1.5 + 2·2.5·√(1 + 1.5²) = 10.51 m
R = A/P = 1.248 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.025·17/(13.13·1.159) = 0.02793
S = 0.02793² = 0.0007801
Answer
S = 0.0007801
That is a fall of 0.7801 m per kilometre.
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