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 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 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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 1.5 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows at 26 L/s through a horizontal pipe that narrows from 150 mm to 100 mm. The pressure in the narrow section is 180 kPa. Ignoring losses, find the pressure p1 in the wide section.
Q = 26 L/sd1 = 150 mmd2 = 100 mmp2 = 180 kPaρ = 1000 kg/m³
p1 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 200 mm to 75 mm while rising 7 m. At section 1 the pressure is 290 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 200 mmd2 = 75 mmp1 = 290 kPav1 = 0.75 m/sΔz = 7 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 8 L/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.
Q = 8 L/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 100 m long with an inside diameter of 150 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 60 L/s?
Q = 60 L/sL = 100 mD = 150 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 68 m while its shaft takes 15.4 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 15 L/sH = 68 mPshaft = 15.4 kWg = 9.81 m/s²
η = %
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 600 m long with an inside diameter of 250 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 65%. Find the shaft power Pshaft in kW.
Q = 75 L/sΔz = 6 mL = 600 mD = 250 mmf = 0.015ΣK = 8η = 65%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A trapezoidal unfinished concrete channel has a bottom width of 2 m and side slopes of 1.5:1 (horizontal:vertical). Water flows uniformly 1.2 m deep. Manning n = 0.014 and the bed slope is S = 0.0015. Find the flow rate Q in m³/s.
b = 2 my = 1.2 mz = 1.5n = 0.014S = 0.0015
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 450 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 = 450 mmn = 0.013V = 1.2 m/s
S = m/m
3Solutions
1.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 = 0.5 + 3/2 = 2 m
A = H·b = 3·1 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·3 = 58.86 kN
Answer
F = 58.86 kN
2.Hydrostatic force
Answer: F = 143.5 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 + 1.5/2 = 3.25 m
A = H·b = 1.5·3 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.25·4.5 = 143.5 kN
Answer
F = 143.5 kN
3.Continuity & Bernoulli
Answer: p1 = 184.4 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.
Velocities
v1 = Q/A1 = 0.026/0.01767 = 1.471 m/s
v2 = Q/A2 = 0.026/0.007854 = 3.31 m/s
Bernoulli
p1 = p2 + ½ρ(v2² − v1²) = 180 + ½·1000·(3.31² − 1.471²)/1000 = 180 + 4.397 = 184.4 kPa
Answer
p1 = 184.4 kPa
4.Continuity & Bernoulli
Answer: p2 = 207.4 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 = 0.75·(200/75)² = 5.333 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 5.333²)/1000 = −13.94 kPa
ρgΔz = 1000·9.81·7/1000 = 68.67 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 290 + (−13.94) − 68.67 = 207.4 kPa
Answer
p2 = 207.4 kPa
5.Pipe friction & head loss
Answer: hf = 55.71 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.
Mean velocity
A = π·0.075²/4 = 0.004418 m²
V = Q/A = 0.008/0.004418 = 1.811 m/s
Velocity head
V²/(2g) = 1.811²/(2·9.81) = 0.1671 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.025·(1000/0.075)·1.811²/(2·9.81) = 55.71 m
Answer
hf = 55.71 m
6.Pipe friction & head loss
Answer: Δz = 8.285 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 = 0.06/0.01767 = 3.395 m/s
V²/(2g) = 0.5876 m
Loss coefficients
f·L/D = 0.015·100/0.15 = 10
ΣK = 0.5 + 2·0.3 + 2 + 1.0 = 4.1
Total head loss
hf = 10·0.5876 = 5.876 m
hm = 4.1·0.5876 = 2.409 m
Δz = hf + hm = 8.285 m
Answer
Δz = 8.285 m
7.Pumps & pump power
Answer: η = 64.98 %
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 = 1000·9.81·0.015·68 = 10.01 kW
Efficiency
η = Ph/Pshaft = 10.01/15.4 = 64.98%
Answer
η = 64.98%
8.Pumps & pump power
Answer: Pshaft = 12.72 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 = 0.075/0.04909 = 1.528 m/s
V²/(2g) = 0.119 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (36 + 8)·0.119 = 5.235 m
Pump head
H = Δz + hL = 6 + 5.235 = 11.24 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.075·11.24/0.65 = 12.72 kW
Answer
Pshaft = 12.72 kW
9.Open-channel flow (Manning)
Answer: Q = 10.14 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 + 1.5·1.2)·1.2 = 4.56 m²
P = b + 2y·√(1 + z²) = 2 + 2·1.2·√(1 + 1.5²) = 6.327 m
R = A/P = 0.7208 m
Velocity
V = (1/0.014)·0.72082/3·0.00151/2 = 2.224 m/s
Flow rate
Q = A·V = 4.56·2.224 = 10.14 m³/s
Answer
Q = 10.14 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.004481 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 = 0.45/4 = 0.1125 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·1.2/0.233)² = 0.004481
Answer
S = 0.004481
That is a fall of 4.481 m per kilometre.
ExamProblems · examproblems.com/courses/hydraulic-engineering