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 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 1.5 mb = 3 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 water, with its top edge 0.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 0.5 mH = 2 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 20 mm. The water level is kept 5 m above the orifice and the discharge coefficient is 0.65. Find the flow rate Q. Use g = 9.81 m/s².
d = 20 mmh = 5 mCd = 0.65g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 50 mm and a gauge pressure of 140 kPa. The nozzle exit diameter is 20 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².
d1 = 50 mmp1 = 140 kPa (gauge)d2 = 20 mmρ = 1000 kg/m³g = 9.81 m/s²
H = m
5.Pipe friction & head lossMedium
Water flows at 25 L/s through a pipe 100 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
Q = 25 L/sL = 100 mD = 200 mmf = 0.022g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 100 m long with an inside diameter of 150 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), 3 bends (K = 0.3 each) and a valve (K = 5). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 25 L/s?
Q = 25 L/sL = 100 mD = 150 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 5
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 22 m while its shaft takes 28.8 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 80 L/sH = 22 mPshaft = 28.8 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 60 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 17 m higher. The pipe is 600 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015; 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 = 60 L/sΔz = 17 mL = 600 mD = 200 mmf = 0.015ΣK = 8
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal grass-lined earth channel has a bottom width of 2.5 m and side slopes of 1:1 (horizontal:vertical). Water flows uniformly 1.5 m deep. Manning n = 0.03 and the bed slope is S = 0.005. Find the flow rate Q in m³/s.
b = 2.5 my = 1.5 mz = 1n = 0.03S = 0.005
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal grass-lined earth channel with a bottom width of 2.5 m and side slopes of 3:1 (horizontal:vertical) must carry 22 m³/s in uniform flow at a depth of 2.1 m. Manning n = 0.03. Find the required bed slope S (in m/m).
Q = 22 m³/sb = 2.5 my = 2.1 mz = 3n = 0.03
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 209.7 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 + 1.5/2 = 4.75 m
A = H·b = 1.5·3 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.75·4.5 = 209.7 kN
Answer
F = 209.7 kN
2.Hydrostatic force
Answer: ycp = 1.722 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 = 0.5 + 2/2 = 1.5 m
A = H·b = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.5·4 = 58.86 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 2·2³/12 = 1.333 m⁴
ycp = yc + Ic/(yc·A) = 1.5 + 1.333/(1.5·4) = 1.722 m
Answer
ycp = 1.722 m
3.Continuity & Bernoulli
Answer: Q = 2.023 L/s
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.
Ideal jet velocity
√(2gh) = √(2·9.81·5) = 9.905 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.65·0.0003142·9.905 = 0.002023 m³/s = 2.023 L/s
Answer
Q = 2.023 L/s
4.Continuity & Bernoulli
Answer: H = 14.65 m
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.
Exit velocity
β = 20/50 = 0.4
v2 = √(2p1/(ρ(1 − β⁴))) = √(2·140000/(1000·0.9744)) = 16.95 m/s
Height of the jet
Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.
H = v2²/(2g) = 16.95²/(2·9.81) = 14.65 m
Answer
H = 14.65 m
5.Pipe friction & head loss
Answer: hf = 0.355 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 = 0.025/0.03142 = 0.7958 m/s
Velocity head
V²/(2g) = 0.7958²/(2·9.81) = 0.03228 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(100/0.2)·0.7958²/(2·9.81) = 0.355 m
Answer
hf = 0.355 m
6.Pipe friction & head loss
Answer: H = 2.251 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 = 0.025/0.01767 = 1.415 m/s
V²/(2g) = 0.102 m
Loss coefficients
f·L/D = 0.022·100/0.15 = 14.67
ΣK = 0.5 + 3·0.3 + 5 = 6.4
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 + 6.4)·0.102 = 2.251 m
Answer
H = 2.251 m
7.Pumps & pump power
Answer: η = 59.95 %
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 = 1000·9.81·0.08·22 = 17.27 kW
Efficiency
η = Ph/Pshaft = 17.27/28.8 = 59.95%
Answer
η = 59.95%
8.Pumps & pump power
Answer: H = 26.85 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 = 0.06/0.03142 = 1.91 m/s
V²/(2g) = 0.1859 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (45 + 8)·0.1859 = 9.853 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 17 + 9.853 = 26.85 m
Answer
H = 26.85 m
9.Open-channel flow (Manning)
Answer: Q = 13.08 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 = (2.5 + 1·1.5)·1.5 = 6 m²
P = b + 2y·√(1 + z²) = 2.5 + 2·1.5·√(1 + 1²) = 6.743 m
R = A/P = 0.8899 m
Velocity
V = (1/0.03)·0.88992/3·0.0051/2 = 2.181 m/s
Flow rate
Q = A·V = 6·2.181 = 13.08 m³/s
Answer
Q = 13.08 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.001033 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 = (2.5 + 3·2.1)·2.1 = 18.48 m²
P = b + 2y·√(1 + z²) = 2.5 + 2·2.1·√(1 + 3²) = 15.78 m
R = A/P = 1.171 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.03·22/(18.48·1.111) = 0.03215
S = 0.03215² = 0.001033
Answer
S = 0.001033
That is a fall of 1.033 m per kilometre.
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