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)`.

Explanation

According to the question, we draw the following diagram.

From continuity equation,
`A_(1)v_(1)=A_(2)v_(2)`
or `(v_(1))/(v_(2))=(A_(2))/(A_(1))=(pir_(2)^(2))/(pi r_(1)^(2))`
`=((r_(2))/(r_(1)))^(2)=((0.04)/(0.1))^(2)=(4)/(25)`
From Bernoulli's equation,
`p_(1)+(1)/(2)rhov_(1)^(2)=p_(2)+(1)/(2)rhov_(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)`
Solving Eqs. (i) and (ii), we get
`v_(2) ~~0.128 ms^(-1)`
`therefore` 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^(-1)`