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 1.5 m wide sits in the wall of a tank of water, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 2.5 mb = 1.5 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 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 3 mb = 1.5 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 20 mm. The water level is kept 4 m above the orifice and the discharge coefficient is 0.63. Find the flow rate Q. Use g = 9.81 m/s².
d = 20 mmh = 4 mCd = 0.63g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 150 mm, throat 75 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 150 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 = 75 mmΔh = 150 mmCd = 0.95ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.3 m/s through a pipe 1000 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.3 m/sL = 1000 mD = 250 mmf = 0.03g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022. The water surface of the upper reservoir is 30 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 4 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 = 30 mL = 150 mD = 200 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 85% delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 74 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 15 L/sH = 74 mη = 85%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 21 m higher. The pipe is 500 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.02; 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 = 75 L/sΔz = 21 mL = 500 mD = 200 mmf = 0.02ΣK = 8
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal grass-lined earth channel has a bottom width of 2 m and side slopes of 2:1 (horizontal:vertical). Water flows uniformly 2.4 m deep. Manning n = 0.03 and the bed slope is S = 0.0008. Find the flow rate Q in m³/s.
b = 2 my = 2.4 mz = 2n = 0.03S = 0.0008
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal finished concrete channel with a bottom width of 5.5 m and side slopes of 1.5:1 (horizontal:vertical) must carry 22 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 = 22 m³/sb = 5.5 my = 1.1 mz = 1.5n = 0.012
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 82.77 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 + 2.5/2 = 2.25 m
A = H·b = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.25·3.75 = 82.77 kN
Answer
F = 82.77 kN
2.Hydrostatic force
Answer: F = 88.29 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 = 0.5 + 3/2 = 2 m
A = H·b = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·4.5 = 88.29 kN
Answer
F = 88.29 kN
3.Continuity & Bernoulli
Answer: Q = 1.753 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) = √(2·9.81·4) = 8.859 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.63·0.0003142·8.859 = 0.001753 m³/s = 1.753 L/s
Answer
Q = 1.753 L/s
4.Continuity & Bernoulli
Answer: Q = 26.4 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)·9.81·0.15 = 18541 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 75/150 = 0.5
v2 = √(2·18541/(1000·(1 − 0.0625))) = 6.289 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.95·0.004418·6.289 = 0.0264 m³/s = 26.4 L/s
Answer
Q = 26.4 L/s
5.Pipe friction & head loss
Answer: hf = 32.35 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) = 2.3²/(2·9.81) = 0.2696 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.03·(1000/0.25)·2.3²/(2·9.81) = 32.35 m
Answer
hf = 32.35 m
6.Pipe friction & head loss
Answer: Q = 141 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 = 0.022·150/0.2 = 16.5
ΣK = 0.5 + 4·0.3 + 10 + 1.0 = 12.7
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
30 = (16.5 + 12.7)·V²/(2·9.81)
V = √(2·9.81·30/29.2) = 4.49 m/s
Flow rate
Q = V·π·D²/4 = 4.49·0.03142 = 0.141 m³/s = 141 L/s
Answer
Q = 141 L/s
7.Pumps & pump power
Answer: Pshaft = 12.81 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 = 1000·9.81·0.015·74 = 10.89 kW
Shaft power
Pshaft = Ph/η = 10.89/0.85 = 12.81 kW
Answer
Pshaft = 12.81 kW
8.Pumps & pump power
Answer: H = 37.85 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (50 + 8)·0.2905 = 16.85 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 21 + 16.85 = 37.85 m
Answer
H = 37.85 m
9.Open-channel flow (Manning)
Answer: Q = 18.16 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 + 2·2.4)·2.4 = 16.32 m²
P = b + 2y·√(1 + z²) = 2 + 2·2.4·√(1 + 2²) = 12.73 m
R = A/P = 1.282 m
Velocity
V = (1/0.03)·1.2822/3·0.00081/2 = 1.112 m/s
Flow rate
Q = A·V = 16.32·1.112 = 18.16 m³/s
Answer
Q = 18.16 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.001442 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 + 1.5·1.1)·1.1 = 7.865 m²
P = b + 2y·√(1 + z²) = 5.5 + 2·1.1·√(1 + 1.5²) = 9.466 m
R = A/P = 0.8309 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.012·22/(7.865·0.8838) = 0.03798
S = 0.03798² = 0.001442
Answer
S = 0.001442
That is a fall of 1.442 m per kilometre.
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