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 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 2.5 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of oil, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 3 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
A pitot-static tube in a wind tunnel reads a difference of 600 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.
p0 − p = 600 Paρ = 1.2 kg/m³
v = m/s
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 150 mm, throat 100 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 370 mm. The discharge coefficient is 0.95. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².
d1 = 150 mmd2 = 100 mmΔh = 370 mmCd = 0.95ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 120 L/s through a pipe 800 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
Q = 120 L/sL = 800 mD = 300 mmf = 0.028g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 400 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.02) 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 120 L/s?
Q = 120 L/sL = 400 mD = 300 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 2
H = m
7.Pumps & pump powerMedium
A pump with an efficiency of 60% delivers 40 L/s of water (ρ = 1000 kg/m³) against a head of 32 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 40 L/sH = 32 mη = 60%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 80 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 6 m higher. The pipe is 200 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 5 (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 250 kPa. Find the head H the pump must add.
Q = 80 L/sΔz = 6 mL = 200 mD = 250 mmf = 0.02ΣK = 5ptank = 250 kPa
H = m
9.Open-channel flow (Manning)Medium
A rectangular brickwork channel is 1 m wide. Water flows uniformly 2.3 m deep. Manning n = 0.015 and the bed slope is S = 0.0005. Find the flow rate Q in m³/s.
b = 1 my = 2.3 mn = 0.015S = 0.0005
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular finished concrete channel 1 m wide must carry 2.8 m³/s in uniform flow at a depth of 1.1 m. Manning n = 0.012. Find the required bed slope S (in m/m).
Q = 2.8 m³/sb = 1 my = 1.1 mn = 0.012
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 312.7 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 = 3 + 2.5/2 = 4.25 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.25·7.5 = 312.7 kN
Answer
F = 312.7 kN
2.Hydrostatic force
Answer: F = 213.7 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 = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·5.5·4.5 = 213.7 kN
Answer
F = 213.7 kN
3.Continuity & Bernoulli
Answer: v = 31.62 m/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.
Bernoulli to the stagnation point
The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.
p0 − p = ½ρv²
v = √(2·600/1.2) = 31.62 m/s
Answer
v = 31.62 m/s
4.Continuity & Bernoulli
Answer: Q = 79.66 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.
Pressure difference from the manometer
Δp = (13600 − 1000)·9.81·0.37 = 45734 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 100/150 = 0.6667
v2 = √(2·45734/(1000·(1 − 0.1975))) = 10.68 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.95·0.007854·10.68 = 0.07966 m³/s = 79.66 L/s
Answer
Q = 79.66 L/s
5.Pipe friction & head loss
Answer: hf = 10.97 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·0.3²/4 = 0.07069 m²
V = Q/A = 0.12/0.07069 = 1.698 m/s
Velocity head
V²/(2g) = 1.698²/(2·9.81) = 0.1469 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(800/0.3)·1.698²/(2·9.81) = 10.97 m
Answer
hf = 10.97 m
6.Pipe friction & head loss
Answer: H = 4.608 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 = 0.12/0.07069 = 1.698 m/s
V²/(2g) = 0.1469 m
Loss coefficients
f·L/D = 0.02·400/0.3 = 26.67
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 + 26.67 + 3.7)·0.1469 = 4.608 m
Answer
H = 4.608 m
7.Pumps & pump power
Answer: Pshaft = 20.93 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 = 1000·9.81·0.04·32 = 12.56 kW
Shaft power
Pshaft = Ph/η = 12.56/0.6 = 20.93 kW
Answer
Pshaft = 20.93 kW
8.Pumps & pump power
Answer: H = 34.33 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 = 0.08/0.04909 = 1.63 m/s
V²/(2g) = 0.1354 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (16 + 5)·0.1354 = 2.843 m
Pressure head
p/(ρ·g) = 250000/(1000·9.81) = 25.48 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 6 + 25.48 + 2.843 = 34.33 m
Answer
H = 34.33 m
9.Open-channel flow (Manning)
Answer: Q = 1.894 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 = 1·2.3 = 2.3 m²
P = b + 2y = 1 + 2·2.3 = 5.6 m
R = A/P = 0.4107 m
Velocity
V = (1/0.015)·0.41072/3·0.00051/2 = 0.8237 m/s
Flow rate
Q = A·V = 2.3·0.8237 = 1.894 m³/s
Answer
Q = 1.894 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.003875 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·y = 1·1.1 = 1.1 m²
P = b + 2y = 1 + 2·1.1 = 3.2 m
R = A/P = 0.3438 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.012·2.8/(1.1·0.4907) = 0.06225
S = 0.06225² = 0.003875
Answer
S = 0.003875
That is a fall of 3.875 m per kilometre.
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