Calculating Flow Velocity in a 500m Long Pipe

[Busybee12]

Water flows at the end of an aqueduct into the upper end of a 500m long pipe. The pipe has angle of 30 degrees to the horizontal and its low end exits into the bottom of a collection tank. The collection tank can hold a depth of 120m, above the centre line of the delivery pipe exit. Extra liquid can overflow the lip of the tank into a river.

The pipe tapers from a diameter of 2.0m at the top to 0.8m at the bottom. Assuming the head loss to be 8% of the potential head of the wateras it enters the pipe, determine the flow velocity of the water as it enters the collection tank.

I have worked out two CSA's for the pipe and are stuck. Should I use bernoulli's equation or the continuity equation to work out the flow velocity of the water?

 

[jvicens]

Hello,

Thank you for your question. I would recommend using both Bernoulli's equation and the continuity equation to determine the flow velocity of the water in this scenario.

First, let's use Bernoulli's equation to calculate the velocity at the top of the pipe:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Where:
P₁ = pressure at the top of the pipe
ρ = density of water
v₁ = velocity at the top of the pipe (unknown)
g = acceleration due to gravity
h₁ = height at the top of the pipe (unknown)
P₂ = pressure at the bottom of the pipe (unknown)
v₂ = velocity at the bottom of the pipe (unknown)
h₂ = height at the bottom of the pipe (120m)

We can rearrange this equation to solve for v₁:

v₁ = √(2gh₁ + (P₂ - P₁)/ρ)

To determine the pressure at the top and bottom of the pipe, we can use the hydrostatic equation:

P = ρgh

Where:
P = pressure
ρ = density of water
g = acceleration due to gravity
h = height of the water column

At the top of the pipe, the height of the water column is 500m, so the pressure is:

P₁ = ρgh₁ = 1000 kg/m³ * 9.8 m/s² * 500 m = 4,900,000 Pa

At the bottom of the pipe, the height of the water column is 120m, so the pressure is:

P₂ = ρgh₂ = 1000 kg/m³ * 9.8 m/s² * 120 m = 1,176,000 Pa

Substituting these values into the first equation, we get:

v₁ = √(2 * 9.8 m/s² * 500 m + (1,176,000 Pa - 4,900,000 Pa)/1000 kg/m³) = 22.1 m/s

Now, let's use the continuity equation to determine the velocity at the bottom of the pipe:

A₁v₁ = A₂v₂

Where:
A₁ = cross-sectional area of the pipe at the top (πr₁²)
A₂ = cross-sectional