Question

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?

Answer 1

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

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

Step-by-step 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