Calculate the rate of flow of glycerine of density `1.25xx10^(3) kgm^(-3)` through the conical section of a horizontal 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

A, C, D

Explanation

From continuity equation,
`A_(1)v_(1)=A_(2)v_(2)`
or, `(v_(1))/(v_(2))=(A_(2))/(A_(1))-(pi r_(2)^(2))/(pi r_(1)^(2))=((r_2)/(r_1))^(2)`
`=((0.04)/(0.1))^(2)=(4)/(25)` …(i)
From Bernouli's equation,

`p_(1)+(1)/(2)rho v_(1)^(2)=p_(2)+(1)/(2)rho v_(2)^(2)` or `v_(2)^(2)-v_(1)^(2)=(2(p_(1)-p_(2)))/(rho)`
or `v_(2)^(2)-v_(1)^(2)=(2xx10)/(1.25xx10^(3))=1.6xx10^(-2) m^(2)//s^(2)` ...(ii)
Solving Eqs, (i) ans (ii), we get `v_(2)~~0.128 m//s`
`:.` Rate of volume flow through the tube
`Q=A_(2)v_(2)=(pi r_(2)^(2))v_(2)`
`=pi(0.04^(2))(0.128)`
`=6.43xx10^(-4) m^(3)//s`.