Calculate the rate of flow of glycerine of density `1.25xx10^(3) kg m^(-3)` through the conical section of a pipe, if the radii of its ends are`0.1` m and `0.04` m and the pressure drop across its length is `10 Nm^(-2)`.

Answer

`6.43 xx 10^(-4) m^(3)//s`

Explanation

`A_(1)v_(1) = A_(2)v_(2)` ……(i)
`P_(1) + (1)/(2)rhov_(1)^(2) ÷ 0 = P_(2) + (1)/(2) rhov_(2)^(2) + 0` ……..(ii)
Given, `P_(1) - P_(2) = 10N//m^(2)`
`rho = 1.25 xx 10^(3)`
`r_(1) = 0.1 m`
`r_(2) = 0.04 m`
Solving, rate of flow of glycerin `= A_(1)V_(1) = A_(2)V_(2) = 6.43 xx 10^(-4) m^(3)//sec`.