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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 3 mb = 1 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 glycerin, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 3 mb = 1.5 mρ = 1260 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 100 mm and a throat diameter of 50 mm. The pressure at the inlet is 56 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 100 mmd2 = 50 mmp1 − p2 = 56 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
A hose with an inner diameter of 50 mm siphons water from an open tank. The highest point C of the hose is 1.5 m above the water surface, and the outlet (open to the air) is 1.5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.
d = 50 mmzC = +1.5 mz2 = −1.5 mg = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 0.9 m/s through a horizontal pipe, 100 m long, with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 0.9 m/sL = 100 mD = 150 mmf = 0.025ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 150 m long with an inside diameter of 250 mm (Darcy friction factor f = 0.025) runs from an open reservoir and discharges as a free jet into the air. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each) and a valve (K = 0.2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 120 L/s?
Q = 120 L/sL = 150 mD = 250 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 0.2
H = m
7.Pumps & pump powerMedium
A pump with an efficiency of 82% delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 34 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 100 L/sH = 34 mη = 82%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 19 m higher. The pipe is 200 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 75%. Find the shaft power Pshaft in kW.
Q = 50 L/sΔz = 19 mL = 200 mD = 150 mmf = 0.02ΣK = 6η = 75%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A trapezoidal brickwork channel has a bottom width of 2.5 m and side slopes of 2:1 (horizontal:vertical). Water flows uniformly 1.2 m deep. Manning n = 0.015 and the bed slope is S = 0.0005. Find the flow rate Q in m³/s.
b = 2.5 my = 1.2 mz = 2n = 0.015S = 0.0005
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal unfinished concrete channel with a bottom width of 6 m and side slopes of 2:1 (horizontal:vertical) must carry 51 m³/s in uniform flow at a depth of 1.6 m. Manning n = 0.014. Find the required bed slope S (in m/m).
Q = 51 m³/sb = 6 my = 1.6 mz = 2n = 0.014
S = m/m
3Solutions
1.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 = 1.5 + 3/2 = 3 m
A = H·b = 3·1 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3·3 = 88.29 kN
Answer
F = 88.29 kN
2.Hydrostatic force
Answer: F = 194.7 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 + 3/2 = 3.5 m
A = H·b = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·3.5·4.5 = 194.7 kN
Answer
F = 194.7 kN
3.Continuity & Bernoulli
Answer: Q = 21.46 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·(50/100)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 = √(2·56000/(1000·(1 − 0.0625))) = 10.93 m/s
v1 = 2.733 m/s
Flow rate
Q = A2·v2 = 0.001963·10.93 = 0.02146 m³/s = 21.46 L/s
Answer
Q = 21.46 L/s
4.Continuity & Bernoulli
Answer: Q = 10.65 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.
Velocity in the hose
Surface and outlet are both open to the air, so the pressure terms cancel.
v = √(2g·1.5) = √(2·9.81·1.5) = 5.425 m/s
Flow rate
Q = A·v = 0.001963·5.425 = 0.01065 m³/s = 10.65 L/s
Answer
Q = 10.65 L/s
5.Pipe friction & head loss
Answer: Δp = 6.75 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) = 0.025·(100/0.15)·0.9²/(2·9.81) = 0.6881 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 = 1000·9.81·0.6881 = 6750 Pa = 6.75 kPa
Answer
Δp = 6.75 kPa
6.Pipe friction & head loss
Answer: H = 5.452 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Loss coefficients
f·L/D = 0.025·150/0.25 = 15
ΣK = 0.5 + 4·0.3 + 0.2 = 1.9
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 15 + 1.9)·0.3046 = 5.452 m
Answer
H = 5.452 m
7.Pumps & pump power
Answer: Pshaft = 40.68 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 = 1000·9.81·0.1·34 = 33.35 kW
Shaft power
Pshaft = Ph/η = 33.35/0.82 = 40.68 kW
Answer
Pshaft = 40.68 kW
8.Pumps & pump power
Answer: Pshaft = 21.14 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 = 0.05/0.01767 = 2.829 m/s
V²/(2g) = 0.408 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (26.67 + 6)·0.408 = 13.33 m
Pump head
H = Δz + hL = 19 + 13.33 = 32.33 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.05·32.33/0.75 = 21.14 kW
Answer
Pshaft = 21.14 kW
9.Open-channel flow (Manning)
Answer: Q = 7.219 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 = (2.5 + 2·1.2)·1.2 = 5.88 m²
P = b + 2y·√(1 + z²) = 2.5 + 2·1.2·√(1 + 2²) = 7.867 m
R = A/P = 0.7475 m
Velocity
V = (1/0.015)·0.74752/3·0.00051/2 = 1.228 m/s
Flow rate
Q = A·V = 5.88·1.228 = 7.219 m³/s
Answer
Q = 7.219 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.002025 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 = (6 + 2·1.6)·1.6 = 14.72 m²
P = b + 2y·√(1 + z²) = 6 + 2·1.6·√(1 + 2²) = 13.16 m
R = A/P = 1.119 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.014·51/(14.72·1.078) = 0.045
S = 0.045² = 0.002025
Answer
S = 0.002025
That is a fall of 2.025 m per kilometre.
ExamProblems · examproblems.com/courses/hydraulic-engineering