Calculate the rate of flow of glycerine of density `1.25 xx 10^(3) kg//m^(3)` through the conical section of a pipe if the radii of its ends are 0.1m and 0.04 m and the pressure drop across its lengths 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)=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)`
`therefore [(6.25)^(2)-1^(2)]v_(1)^(2)=16xx10^(-3)m^(2)//s^(2)`
`or v_(1)~~0.0205m//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`