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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 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 seawater, with its top edge 4 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 4 mH = 2 mb = 1.5 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 100 mm and a throat diameter of 60 mm. The pressure at the inlet is 48 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 100 mmd2 = 60 mmp1 − p2 = 48 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 300 mm, throat 150 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 170 mm. The discharge coefficient is 0.98. 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 = 300 mmd2 = 150 mmΔh = 170 mmCd = 0.98ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.9 m/s through a horizontal pipe, 400 m long, with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.9 m/sL = 400 mD = 250 mmf = 0.018ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each), a valve (K = 2) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 120 L/s?
Q = 120 L/sL = 150 mD = 300 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 82% delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 48 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 25 L/sH = 48 mη = 82%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 15 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 10 m higher. The pipe is 150 m long with an inside diameter of 100 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 pump efficiency is 70%. Find the shaft power Pshaft in kW.
Q = 15 L/sΔz = 10 mL = 150 mD = 100 mmf = 0.015ΣK = 8η = 70%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 600 mm flows half full. Manning n = 0.013 and the slope is S = 0.004. Find the flow rate Q in m³/s.
D = 600 mmn = 0.013S = 0.004
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular clean earth channel 5 m wide must carry 3.5 m³/s in uniform flow at a depth of 0.5 m. Manning n = 0.022. Find the required bed slope S (in m/m).
Q = 3.5 m³/sb = 5 my = 0.5 mn = 0.022
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 101.2 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 + 2.5/2 = 2.75 m
A = H·b = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.75·3.75 = 101.2 kN
Answer
F = 101.2 kN
2.Hydrostatic force
Answer: ycp = 5.067 m
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 + 2/2 = 5 m
A = H·b = 2·1.5 = 3 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·5·3 = 150.8 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1.5·2³/12 = 1 m⁴
ycp = yc + Ic/(yc·A) = 5 + 1/(5·3) = 5.067 m
Answer
ycp = 5.067 m
3.Continuity & Bernoulli
Answer: Q = 29.69 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·(60/100)² = 0.36·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·48000/(1000·(1 − 0.1296))) = 10.5 m/s
v1 = 3.781 m/s
Flow rate
Q = A2·v2 = 0.002827·10.5 = 0.02969 m³/s = 29.69 L/s
Answer
Q = 29.69 L/s
4.Continuity & Bernoulli
Answer: Q = 116 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)·9.81·0.17 = 21013 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 150/300 = 0.5
v2 = √(2·21013/(1000·(1 − 0.0625))) = 6.695 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.98·0.01767·6.695 = 0.116 m³/s = 116 L/s
Answer
Q = 116 L/s
5.Pipe friction & head loss
Answer: Δp = 121.1 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) = 0.018·(400/0.25)·2.9²/(2·9.81) = 12.34 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 = 1000·9.81·12.34 = 121104 Pa = 121.1 kPa
Answer
Δp = 121.1 kPa
6.Pipe friction & head loss
Answer: Δz = 1.704 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 = 0.015·150/0.3 = 7.5
ΣK = 0.5 + 2·0.3 + 2 + 1.0 = 4.1
Total head loss
hf = 7.5·0.1469 = 1.102 m
hm = 4.1·0.1469 = 0.6023 m
Δz = hf + hm = 1.704 m
Answer
Δz = 1.704 m
7.Pumps & pump power
Answer: Pshaft = 14.36 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 = 1000·9.81·0.025·48 = 11.77 kW
Shaft power
Pshaft = Ph/η = 11.77/0.82 = 14.36 kW
Answer
Pshaft = 14.36 kW
8.Pumps & pump power
Answer: Pshaft = 3.294 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 = 0.015/0.007854 = 1.91 m/s
V²/(2g) = 0.1859 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (22.5 + 8)·0.1859 = 5.67 m
Pump head
H = Δz + hL = 10 + 5.67 = 15.67 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.015·15.67/0.7 = 3.294 kW
Answer
Pshaft = 3.294 kW
9.Open-channel flow (Manning)
Answer: Q = 0.1942 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 = 0.1414 m², R = D/4 = 0.15 m
Velocity
V = (1/0.013)·0.152/3·0.0041/2 = 1.373 m/s
Flow rate
Q = A·V = 0.1414·1.373 = 0.1942 m³/s
Answer
Q = 0.1942 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.003048 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 = 5·0.5 = 2.5 m²
P = b + 2y = 5 + 2·0.5 = 6 m
R = A/P = 0.4167 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.022·3.5/(2.5·0.5579) = 0.05521
S = 0.05521² = 0.003048
Answer
S = 0.003048
That is a fall of 3.048 m per kilometre.
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