Answer:
D ≈ 8.45 m
L ≈ 100.02 m
Explanation:
Given
Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)
y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)
x = 100.00 m (distance between the upper and the lower channels)
We assume that:
We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):
P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂
Plugging the known values into the equation and simplifying we get
Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)
⇒ v₂ = 6.264 m/s
then we apply the formula
Q = v*A ⇒ A = Q/v ⇒ A = Q/v₂
⇒ A = (350 m³/s)/(6.264 m/s)
⇒ A = 55.873 m²
then, we get the diameter of the pipe as follows
A = π*D²/4 ⇒ D = 2*√(A/π)
⇒ D = 2*√(55.873 m²/π)
⇒ D = 8.434 m ≈ 8.45 m
Now, the length of the pipe can be obtained as follows
L² = x² + h²
⇒ L² = (100.00 m)² + (2.00 m)²
⇒ L ≈ 100.02 m
Answer: 2738.5 cubic metres
Explanation: Given that
Flow rate Q = 350m^3/s
Height h = 2m
Distance x = 100m
Using pythagorean theorem to find the length L of the pipe
L^2 = 100^2 + 2^2
L^2 = 10004
L = 100.02 m
Let assume that the pipe is uniform of same diameter at both ends and water flows through the pipe
Let also consider the atmospheric pressure at the upper channel
Using bernoulli equation
P1 = P2 + 1/2pV^2
Where P2 = phg
P1 = atmospheric pressure = 101325pa
V = velocity
p = density = 1000kg/m3
101325 = 1000×9.81×2 + (0.5×1000V^2)
101325 - 19620 = 500V^2
81705 = 500V^2
V^2 = 163.41
V = 12.8 m/s
Q = V × A
Where A = area of the pipe
A = Q/V
A = 350/12.8 = 27.4 square metre
The volume of the pipe = A × L
Volume = 27.4 × 100.02
Volume = 2738.5 cubic metres
The volume of the pipe determines how large a pipe is needed to carry the flow
