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 2 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 = 3 mb = 2 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 water, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 1 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 25 mm. The water level is kept 5 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².
d = 25 mmh = 5 mCd = 0.62g = 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 75 mm and a gauge pressure of 130 kPa. The nozzle exit diameter is 30 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 = 75 mmp1 = 130 kPa (gauge)d2 = 30 mmρ = 1000 kg/m³g = 9.81 m/s²
H = m
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 1.8 m/s through a horizontal pipe, 800 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 pressure drop Δp over the pipe in kPa.
V = 1.8 m/sL = 800 mD = 50 mmf = 0.022ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 300 m long with an inside diameter of 250 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 = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 80 L/s?
Q = 80 L/sL = 300 mD = 250 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 2
H = m
7.Pumps & pump powerMedium
A pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 12 m. The pump efficiency is 72% and the electric motor driving it has an efficiency of 88%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 10 L/sH = 12 mηpump = 72%ηmotor = 88%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 120 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 27 m higher. The pipe is 200 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 = 8 (entrance and exit included). Use g = 9.81 m/s². The upper tank is closed and the air above its water is at a gauge pressure of 150 kPa. Find the head H the pump must add.
Q = 120 L/sΔz = 27 mL = 200 mD = 250 mmf = 0.015ΣK = 8ptank = 150 kPa
H = m
9.Open-channel flow (Manning)Medium
A rectangular clean earth channel is 5.5 m wide. Water flows uniformly 0.8 m deep. Manning n = 0.022 and the bed slope is S = 0.0005. Find the flow rate Q in m³/s.
b = 5.5 my = 0.8 mn = 0.022S = 0.0005
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal clean earth channel with a bottom width of 5.5 m and side slopes of 1:1 (horizontal:vertical) must carry 11 m³/s in uniform flow at a depth of 1 m. Manning n = 0.022. Find the required bed slope S (in m/m).
Q = 11 m³/sb = 5.5 my = 1 mz = 1n = 0.022
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 323.7 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 + 3/2 = 5.5 m
A = H·b = 3·2 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·5.5·6 = 323.7 kN
Answer
F = 323.7 kN
2.Hydrostatic force
Answer: F = 24.52 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 + 1/2 = 2.5 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.5·1 = 24.52 kN
Answer
F = 24.52 kN
3.Continuity & Bernoulli
Answer: Q = 3.014 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) = 0.62·0.0004909·9.905 = 0.003014 m³/s = 3.014 L/s
Answer
Q = 3.014 L/s
4.Continuity & Bernoulli
Answer: H = 13.6 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 = 30/75 = 0.4
v2 = √(2p1/(ρ(1 − β⁴))) = √(2·130000/(1000·0.9744)) = 16.33 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) = 16.33²/(2·9.81) = 13.6 m
Answer
H = 13.6 m
5.Pipe friction & head loss
Answer: Δp = 570.2 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) = 0.022·(800/0.05)·1.8²/(2·9.81) = 58.13 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 = 1000·9.81·58.13 = 570240 Pa = 570.2 kPa
Answer
Δp = 570.2 kPa
6.Pipe friction & head loss
Answer: H = 4.21 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 = 0.08/0.04909 = 1.63 m/s
V²/(2g) = 0.1354 m
Loss coefficients
f·L/D = 0.022·300/0.25 = 26.4
ΣK = 0.5 + 4·0.3 + 2 = 3.7
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 26.4 + 3.7)·0.1354 = 4.21 m
Answer
H = 4.21 m
7.Pumps & pump power
Answer: Pel = 1.858 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 = 1000·9.81·0.01·12 = 1.177 kW
Shaft power
Pshaft = 1.177/0.72 = 1.635 kW
Electrical power
Pel = 1.635/0.88 = 1.858 kW
Answer
Pel = 1.858 kW
8.Pumps & pump power
Answer: H = 48.38 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (12 + 8)·0.3046 = 6.092 m
Pressure head
p/(ρ·g) = 150000/(1000·9.81) = 15.29 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 27 + 15.29 + 6.092 = 48.38 m
Answer
H = 48.38 m
9.Open-channel flow (Manning)
Answer: Q = 3.251 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 = 5.5·0.8 = 4.4 m²
P = b + 2y = 5.5 + 2·0.8 = 7.1 m
R = A/P = 0.6197 m
Velocity
V = (1/0.022)·0.61972/3·0.00051/2 = 0.7388 m/s
Flow rate
Q = A·V = 4.4·0.7388 = 3.251 m³/s
Answer
Q = 3.251 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.001929 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 = (5.5 + 1·1)·1 = 6.5 m²
P = b + 2y·√(1 + z²) = 5.5 + 2·1·√(1 + 1²) = 8.328 m
R = A/P = 0.7805 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.022·11/(6.5·0.8477) = 0.04392
S = 0.04392² = 0.001929
Answer
S = 0.001929
That is a fall of 1.929 m per kilometre.
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