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 `1.0m` and `0.04m` and the pressure drop across its length is `10N//m^(2)`.

Explanation

According to equation of continuity
`(v_(2))/(v_(1))=(A_(1))/(A_(2))=(r_(1)^(2))/(r_(2)^(2))=((0.1)^(2))/((0.04)^(2))=(25)/(4)`
and according to Bernoulli's equation for a horizontal tube,
`P_(1)+(1)/(2)rhov_(1)^(2)rhov_(1)^(2)=P_(2)+(1)/(2)rhov_(2)^(2)`
`v_(2)^(2)-v_(1)^(2)=2((P_(1)-P_(2)))/(rho)=16xx10^(-3)m^(2)//s^(2)`
but `v_(2)=(25)/(4)v_(1)=6.25v_(1)`
`:. [(6.25)^(2)-^(2)]v_(1)^(2)=16xx10^(-3)m^(2)//s^(2)`
or `v_(1)approx0.0205m//s`. the rate of volume flow
`=A_(1)v_(1)=pi(0.1)^(2)xx(0.02)=6.28xx10^(-4)m^(3)//s`
And the rate of mass flow is `(dm)/(dt)=rhoAv`
`=(1.25xx10^(3))xx(6.28xx10^(-4))=0.785kg//s`