Question

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 56.3 cm ( 0.563 m) and the flow speed of the petroleum is 10.9 m/s. At the refinery, the petroleum flows at 6.77 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery

Answer 1

a) 2.71 m³/s

b) 0.71 m

Step-by-step explanation:

The flow rate, Q = πr²v

This is also referred to as the amount per second that flows through the pipe, so

Q = 3.142 * 0.2815² * 10.9

Q = 34.2478 * 0.07924

Q = 2.71 m³/s

Assuming that oil has not been added, subtracted or even compressed, then the flow rate Q, is the same everywhere. Thus,

πD²/4 * 6.77 = Q, knowing the value for our Q, we substitute

π/4 * D² * 6.77 = 2.71

D² * 6.77 = (2.71 * 4) / π

D² = 3.45 / 6.77

D² = 0.5096

D = √0.5096

D = 0.714 m

The diameter of the pipe is 0.71 m

Answer 2

Volume flow rate: 2.71 cubic meters per second

Diameter at refinery: 0.714 m

Step-by-step explanation:

The volume flow rate would be the product of pipe crossectional area and the flow speed:

`\dot{V} = Av = \pi(d/2)^2v = \pi (0.563/2)^2 10.9 =2.71 m^3/s`

Assume this is steady flow, so the volume rate would be constant. So the volume rate at the refinery would be the same. Knowing its flow speed we can also calculate the cross-sectional area at this point:

`A_2 = \dot{V}/v_2 = 2.71 / 6.77 = 0.4 m^2`

So the radius and diameter at the refinery is

`\pi r_2^2 = A_2 = 0.4`

`r_2^2 = A_2 / \pi = 0.128`

`r_2 = √(0.128) = 0.357 m`

`d_2 = r_2*2 = 0.357*2 = 0.714 m`