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 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1.5 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 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 pipe that narrows from 200 mm to 75 mm. In the wide section the pressure is 210 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 200 mmd2 = 75 mmp1 = 210 kPav1 = 1.75 m/sρ = 1000 kg/m³
p2 = kPa
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 260 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 = 150 mmd2 = 75 mmΔh = 260 mmCd = 0.99ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 1.6 m/s through a horizontal pipe, 400 m long, with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 1.6 m/sL = 400 mD = 100 mmf = 0.028ρ = 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 = 0.025. The water surface of the upper reservoir is 34 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 3 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 = 34 mL = 100 mD = 200 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 62 m. The pump efficiency is 72% and the electric motor driving it has an efficiency of 92%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 100 L/sH = 62 mηpump = 72%ηmotor = 92%g = 9.81 m/s²
Pel = 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 6 m higher. The pipe is 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 6 (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 150 kPa. Find the head H the pump must add.
Q = 75 L/sΔz = 6 mL = 150 mD = 200 mmf = 0.025ΣK = 6ptank = 150 kPa
H = m
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 900 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.004. Find the flow rate Q in m³/s.
D = 900 mmn = 0.013S = 0.004
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 375 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 = 375 mmn = 0.013V = 1.2 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 77.25 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 + 1.5/2 = 1.75 m
A = H·b = 1.5·3 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.75·4.5 = 77.25 kN
Answer
F = 77.25 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: p2 = 134.1 kPa
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 and Bernoulli
Work in Pa, then divide by 1000 to get kPa.
v2 = v1·(d1/d2)² = 1.75·(200/75)² = 12.44 m/s
½ρ(v1² − v2²) = ½·1000·(1.75² − 12.44²)/1000 = −75.9 kPa
p2 = p1 + ½ρ(v1² − v2²) = 210 + (−75.9) = 134.1 kPa
Answer
p2 = 134.1 kPa
4.Continuity & Bernoulli
Answer: Q = 36.21 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)·9.81·0.26 = 32138 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 75/150 = 0.5
v2 = √(2·32138/(1000·(1 − 0.0625))) = 8.28 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.99·0.004418·8.28 = 0.03621 m³/s = 36.21 L/s
Answer
Q = 36.21 L/s
5.Pipe friction & head loss
Answer: Δp = 143.4 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) = 0.028·(400/0.1)·1.6²/(2·9.81) = 14.61 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 = 1000·9.81·14.61 = 143360 Pa = 143.4 kPa
Answer
Δp = 143.4 kPa
6.Pipe friction & head loss
Answer: Q = 162.6 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 = 0.025·100/0.2 = 12.5
ΣK = 0.5 + 3·0.3 + 10 + 1.0 = 12.4
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
34 = (12.5 + 12.4)·V²/(2·9.81)
V = √(2·9.81·34/24.9) = 5.176 m/s
Flow rate
Q = V·π·D²/4 = 5.176·0.03142 = 0.1626 m³/s = 162.6 L/s
Answer
Q = 162.6 L/s
7.Pumps & pump power
Answer: Pel = 91.82 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 = 1000·9.81·0.1·62 = 60.82 kW
Shaft power
Pshaft = 60.82/0.72 = 84.48 kW
Electrical power
Pel = 84.48/0.92 = 91.82 kW
Answer
Pel = 91.82 kW
8.Pumps & pump power
Answer: H = 28.48 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) = (18.75 + 6)·0.2905 = 7.19 m
Pressure head
p/(ρ·g) = 150000/(1000·9.81) = 15.29 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 6 + 15.29 + 7.19 = 28.48 m
Answer
H = 28.48 m
9.Open-channel flow (Manning)
Answer: Q = 1.145 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 = 0.6362 m², R = D/4 = 0.225 m
Velocity
V = (1/0.013)·0.2252/3·0.0041/2 = 1.8 m/s
Flow rate
Q = A·V = 0.6362·1.8 = 1.145 m³/s
Answer
Q = 1.145 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.005714 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 = 0.375/4 = 0.09375 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·1.2/0.2064)² = 0.005714
Answer
S = 0.005714
That is a fall of 5.714 m per kilometre.
ExamProblems · examproblems.com/courses/hydraulic-engineering