Question

A pipe is running full of water. At a certain point `A` it tapers from `60 cm` diameter to `2.0 cm` diameter at `B`. The pressure difference between `A` and `B` is `100 cm` of water column. Find the rate of flow of water through the pipe.

Answer 1

Diameter of tube at `A=60 cm`

Radius at `A =r_(A)=30 cm`, radius at `B=r_(B)=10cm`
area of cross section at `A=pi(30)^(2)cm^(2)=A_(1)`
area of cross section at `B=pi(10)^(2)=A_(2)`
`P_(A)-P_(B)=100 cm` of water of `100xx1xx980dyn//cm^(2)`
Using the formula `Q=A_(1)A_(2)sqrt((2(P_(A)-P_(B)))/((A_(1)^(2)-A_(2)^(2))d))`
Substitute the value
`Q=(900pi)(100pi)sqrt((2xx(100xx1xx980))/((900^(2)pi^(2)-100^(2)pi^(2)))0`
`implies Q=1.4xx10^(5)cm^(3)//s=140 1//s`