Answer:
Flow rate 2.34 m3/s
Diameter 0.754 m
Explanation:
Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.
The area at the well head is
[tex]A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2[/tex]
So the volume flow rate along the pipe is
[tex]\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s[/tex]
We can use the similar logic to find the cross-section area at the refinery
[tex]A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2[/tex]
The radius of the pipe at the refinery is:
[tex]A_r = \pi r^2[/tex]
[tex]r^2 =A_r/\pi = 0.446/\pi = 0.141[/tex]
[tex]r = \sqrt{0.141} = 0.377m[/tex]
So the diameter is twice the radius = 0.38*2 = 0.754m
