Given:
Pipe 1
d=1 in (diameter)
v= 2ft/s (average water flow speed)

The equation sets up like the following.
Q=Volume flow rate=(average velocity)*(Cross-sectional area of flow)
So I get
(2ft/s)*(pi/4)(1/12 ft)^2 which would give me a flow rate in terms of (ft/s)^3

Is this set up right? And why/what is pi/4 accounting for?

Question

Given:
Pipe 1 
d=1 in (diameter)
v= 2ft/s (average water flow speed)

The equation sets up like the following.
Q=Volume flow rate=(average velocity)*(Cross-sectional area of flow)
So I get 
(2ft/s)*(pi/4)(1/12 ft)^2 which would give me a flow rate in terms of (ft/s)^3

Is this set up right? And why/what is pi/4 accounting for?

anonymous · Answer

$$2\frac{ ft }{s} * \frac{ \pi }{ 4 } * (1 inch)^2= \frac{ ft^3 }{ s }$$

anonymous · Answer

lol any ideas?