Answer:
[tex]d=3.1\ ft[/tex]
Explanation:
Discharge (Hydrology)
The continuity equation states that for an incompressible fluid (like water) the discharge Q equals to the product of the stream's cross-sectional area (A) by its average velocity (v):
[tex]Q=v.A[/tex]
The pipe described in the problem carries [tex]Q=75\ ft^3/s[/tex] of a liquid at an average velocity of v=10 ft/s. Let's compute the cross-sectional area by solving for A
[tex]\displaystyle A=\frac{Q}{v}[/tex]
[tex]\displaystyle A=\frac{75}{10}=7.5\ ft^2[/tex]
The pipe has a cross-section with the shape of a circle. Let's set d as the diameter of the circle, it's area is computed as
[tex]\displaystyle A=\frac{\pi d^2}{4}[/tex]
Solving for d
[tex]\displaystyle d=2\sqrt{\frac{A}{\pi}}[/tex]
Replacing the value of A
[tex]\displaystyle d=2\sqrt{\frac{7.5}{\pi}}[/tex]
[tex]\boxed{d=3.1\ ft}[/tex]
