Answer:
The dept of the water needed is 26.11m
Explanation:
Given the following parameters:
Length of the pipe = [tex]150m[/tex]
Diameter of the pipe = [tex]75mm = 0.75m[/tex]
Volumetric flow rate [tex]1m^{3}/s[/tex]
Knowing that:
Volumetric flowrate [tex]= area (A) * Velocity(V)[/tex]
==> [tex]0.1 = A*V[/tex] *Knowing that [tex]V = \sqrt{2gh}[/tex]
==> [tex]0.1 = \frac{\pi }{4}* (0.075)^{2} * \sqrt{2gh}[/tex]
==> [tex]h = 26.11m[/tex]
Hence, the dept of water needed to produce a volumetric flow rate of [tex]1m^{3}/s[/tex] is 26.11m.
