Question

A water pipe tapers down from an initial radius of R1 = 0.21 m to a final radius of R2 = 0.11 m. The water flows at a velocity v1 = 0.84 m/s in the larger section of pipe. 1) What is the volume flow rate of the water?

Answer 1

`0.116 m^3/s`

Step-by-step explanation:

The volume flow rate of a fluid in a pipe is given by:

`Q=Av`

where

A is the cross-sectional area of the pipe

v is the speed of the fluid

In this problem, at the initial point we have

v = 0.84 m/s is the speed of the water

r = 0.21 m is the radius of the pipe, so the cross-sectional area is

`A=\pi r^2 = \pi (0.21 m)^2 =0.138 m^2`

So, the volume flow rate is

`Q=(0.138 m^2)(0.84 m/s)=0.116 m^3/s`