A sewer pipe 8 inches in diameter can carry 0.662 ft3/s when flowing at a depth of 4 inches.

a.) What flow rate can the pipe carry when flowing full?

b.) What would be the velocity when flowing at a depth of 4 inches?

a) the flow under full capacity is q₂ = 1.334 ft³/s

b) the velocity would be v = 3.793 ft/s

Explanation:

a) Since the pipe has 8 inches in diameter but 4 are covered with water flow (half of a circle in area=A₁), q₁=0.662 ft³/s then

q₁=A₁*v

then for the same velocity v but area A₂=2*A₁

flow under full capacity = q₂ = A₂*v = 2*A₁*v = 2*q₁=2*0.662 ft³/s = 1.334 ft³/s

b) when flowing at a depth of 4 inches

A₁ = (1/2) * (π*D²/4) = π * (1/8) * (8 in) ² = 8π in² * (1 ft² / 144 in²) = π/18 ft² = 0.1745 ft²

then

v=q₁/A₁ = 0.662 ft³/s/0.1745 ft² = 3.793 ft/s

v = 3.793 ft/s