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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 2 mb = 2 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 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 = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 200 mm and a throat diameter of 100 mm. The pressure at the inlet is 7 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 200 mmd2 = 100 mmp1 − p2 = 7 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 100 mm, throat 60 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 350 mm. The discharge coefficient is 0.99. 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 = 100 mmd2 = 60 mmΔh = 350 mmCd = 0.99ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 1.7 m/s through a pipe 1000 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 friction head loss hf.
V = 1.7 m/sL = 1000 mD = 75 mmf = 0.025g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022. The water surface of the upper reservoir is 23 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 = 10) 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 = 23 mL = 300 mD = 250 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 10Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 80% delivers 150 L/s of water (ρ = 1000 kg/m³) against a head of 54 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 150 L/sH = 54 mη = 80%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 30 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 5 m higher. The pipe is 500 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 70%. Find the shaft power Pshaft in kW.
Q = 30 L/sΔz = 5 mL = 500 mD = 150 mmf = 0.018ΣK = 5η = 70%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A trapezoidal gravel channel has a bottom width of 6 m and side slopes of 1:1 (horizontal:vertical). Water flows uniformly 1.8 m deep. Manning n = 0.025 and the bed slope is S = 0.003. Find the flow rate Q in m³/s.
b = 6 my = 1.8 mz = 1n = 0.025S = 0.003
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 300 mm (Manning n = 0.013) must reach a self-cleansing velocity of 1.2 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 300 mmn = 0.013V = 1.2 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 137.3 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 + 2/2 = 3.5 m
A = H·b = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.5·4 = 137.3 kN
Answer
F = 137.3 kN
2.Hydrostatic force
Answer: F = 29.43 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 = 1·1.5 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·1.5 = 29.43 kN
Answer
F = 29.43 kN
3.Continuity & Bernoulli
Answer: Q = 30.35 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.
Continuity
v1 = v2·(100/200)² = 0.25·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·7000/(1000·(1 − 0.0625))) = 3.864 m/s
v1 = 0.9661 m/s
Flow rate
Q = A2·v2 = 0.007854·3.864 = 0.03035 m³/s = 30.35 L/s
Answer
Q = 30.35 L/s
4.Continuity & Bernoulli
Answer: Q = 27.91 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)·9.81·0.35 = 43262 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 60/100 = 0.6
v2 = √(2·43262/(1000·(1 − 0.1296))) = 9.97 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.99·0.002827·9.97 = 0.02791 m³/s = 27.91 L/s
Answer
Q = 27.91 L/s
5.Pipe friction & head loss
Answer: hf = 49.1 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 head
V²/(2g) = 1.7²/(2·9.81) = 0.1473 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.025·(1000/0.075)·1.7²/(2·9.81) = 49.1 m
Answer
hf = 49.1 m
6.Pipe friction & head loss
Answer: Q = 168.1 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 = 0.022·300/0.25 = 26.4
ΣK = 0.5 + 2·0.3 + 10 + 1.0 = 12.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
23 = (26.4 + 12.1)·V²/(2·9.81)
V = √(2·9.81·23/38.5) = 3.424 m/s
Flow rate
Q = V·π·D²/4 = 3.424·0.04909 = 0.1681 m³/s = 168.1 L/s
Answer
Q = 168.1 L/s
7.Pumps & pump power
Answer: Pshaft = 99.33 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 = 1000·9.81·0.15·54 = 79.46 kW
Shaft power
Pshaft = Ph/η = 79.46/0.8 = 99.33 kW
Answer
Pshaft = 99.33 kW
8.Pumps & pump power
Answer: Pshaft = 6.116 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 = 0.03/0.01767 = 1.698 m/s
V²/(2g) = 0.1469 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (60 + 5)·0.1469 = 9.548 m
Pump head
H = Δz + hL = 5 + 9.548 = 14.55 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.03·14.55/0.7 = 6.116 kW
Answer
Pshaft = 6.116 kW
9.Open-channel flow (Manning)
Answer: Q = 36 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 = (6 + 1·1.8)·1.8 = 14.04 m²
P = b + 2y·√(1 + z²) = 6 + 2·1.8·√(1 + 1²) = 11.09 m
R = A/P = 1.266 m
Velocity
V = (1/0.025)·1.2662/3·0.0031/2 = 2.564 m/s
Flow rate
Q = A·V = 14.04·2.564 = 36 m³/s
Answer
Q = 36 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.007694 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.
Hydraulic radius
R = D/4 = 0.3/4 = 0.075 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·1.2/0.1778)² = 0.007694
Answer
S = 0.007694
That is a fall of 7.694 m per kilometre.
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