Question

The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The height difference from the upper to the lower channels is about 2 m and the distance between them is about 100 m. How large a pipe is needed to carry the flow? State any assumptions you make

Answer 1

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

Answer 2

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:

• the upper and the lower channels are at the same pressure (the atmospheric pressure).
• the velocity of water in the upper channel is zero (v₁ = 0 m/s).
• y₁ = 2.00 m  (height of the upper channel)
• y₂ = 0.00 m  (height of the lower channel)
• g = 9.81 m/s²
• ρ = 1000 Kg/m³ (density of water)

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