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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 3 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 water, with its top edge 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 1 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
A pitot-static tube in a water pipe reads a difference of 40 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.
p0 − p = 40 kPaρ = 1000 kg/m³
v = m/s
4.Continuity & BernoulliHard
A hose with an inner diameter of 32 mm siphons water from an open tank. The highest point C of the hose is 1 m above the water surface, and the outlet (open to the air) is 3.5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.
d = 32 mmzC = +1 mz2 = −3.5 mg = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2 m/s through a horizontal pipe, 500 m long, with an inside diameter of 75 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 m/sL = 500 mD = 75 mmf = 0.025ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015. The water surface of the upper reservoir is 10 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each), a valve (K = 0.2) 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 = 10 mL = 100 mD = 200 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 0.2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 16 m while its shaft takes 3.7 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 20 L/sH = 16 mPshaft = 3.7 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 20 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 12 m higher. The pipe is 200 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 4 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 78%. Find the shaft power Pshaft in kW.
Q = 20 L/sΔz = 12 mL = 200 mD = 100 mmf = 0.018ΣK = 4η = 78%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A trapezoidal gravel channel has a bottom width of 4 m and side slopes of 1.5:1 (horizontal:vertical). Water flows uniformly 1.5 m deep. Manning n = 0.025 and the bed slope is S = 0.003. Find the flow rate Q in m³/s.
b = 4 my = 1.5 mz = 1.5n = 0.025S = 0.003
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal grass-lined earth channel with a bottom width of 1 m and side slopes of 1:1 (horizontal:vertical) must carry 3.3 m³/s in uniform flow at a depth of 1.4 m. Manning n = 0.03. Find the required bed slope S (in m/m).
Q = 3.3 m³/sb = 1 my = 1.4 mz = 1n = 0.03
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 264.9 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 = 1.5 + 3/2 = 3 m
A = H·b = 3·3 = 9 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3·9 = 264.9 kN
Answer
F = 264.9 kN
2.Hydrostatic force
Answer: F = 58.86 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 + 1/2 = 2 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·3 = 58.86 kN
Answer
F = 58.86 kN
3.Continuity & Bernoulli
Answer: v = 8.944 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 = √(2·40000/1000) = 8.944 m/s
Answer
v = 8.944 m/s
4.Continuity & Bernoulli
Answer: Q = 6.665 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.
Velocity in the hose
Surface and outlet are both open to the air, so the pressure terms cancel.
v = √(2g·3.5) = √(2·9.81·3.5) = 8.287 m/s
Flow rate
Q = A·v = 0.0008042·8.287 = 0.006665 m³/s = 6.665 L/s
Answer
Q = 6.665 L/s
5.Pipe friction & head loss
Answer: Δp = 333.3 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·(500/0.075)·2²/(2·9.81) = 33.98 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 = 1000·9.81·33.98 = 333333 Pa = 333.3 kPa
Answer
Δp = 333.3 kPa
6.Pipe friction & head loss
Answer: Q = 140.6 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 = 0.015·100/0.2 = 7.5
ΣK = 0.5 + 2·0.3 + 0.2 + 1.0 = 2.3
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
10 = (7.5 + 2.3)·V²/(2·9.81)
V = √(2·9.81·10/9.8) = 4.474 m/s
Flow rate
Q = V·π·D²/4 = 4.474·0.03142 = 0.1406 m³/s = 140.6 L/s
Answer
Q = 140.6 L/s
7.Pumps & pump power
Answer: η = 84.84 %
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 = 1000·9.81·0.02·16 = 3.139 kW
Efficiency
η = Ph/Pshaft = 3.139/3.7 = 84.84%
Answer
η = 84.84%
8.Pumps & pump power
Answer: Pshaft = 6.344 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.
Velocity
V = Q/A = 0.02/0.007854 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (36 + 4)·0.3305 = 13.22 m
Pump head
H = Δz + hL = 12 + 13.22 = 25.22 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.02·25.22/0.78 = 6.344 kW
Answer
Pshaft = 6.344 kW
9.Open-channel flow (Manning)
Answer: Q = 20.49 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 = (4 + 1.5·1.5)·1.5 = 9.375 m²
P = b + 2y·√(1 + z²) = 4 + 2·1.5·√(1 + 1.5²) = 9.408 m
R = A/P = 0.9965 m
Velocity
V = (1/0.025)·0.99652/3·0.0031/2 = 2.186 m/s
Flow rate
Q = A·V = 9.375·2.186 = 20.49 m³/s
Answer
Q = 20.49 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.001459 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 + 1·1.4)·1.4 = 3.36 m²
P = b + 2y·√(1 + z²) = 1 + 2·1.4·√(1 + 1²) = 4.96 m
R = A/P = 0.6774 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.03·3.3/(3.36·0.7713) = 0.0382
S = 0.0382² = 0.001459
Answer
S = 0.001459
That is a fall of 1.459 m per kilometre.
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