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)`.

Explanation

From countinuity equation ,
`A_(1)v_(1) = A_(2)v_(2)`
or `(v_(1))/(v_(2)) = (A_(2))/(A_(1)) = (pir_(2)^(2))/(pir_(1)^(2)) = ((r_(2))/(r_(1))) = ((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) = (2 xx 10)/(1.25 xx 10^(3)) = 1.6 xx 10^(-2)m^(2) //s`
Solving Eqs. `(i)` and `(ii)`, we get
`v_(2) approx 0.128 m//s`
`:.` Rate of volume flow through the tube
`Q = A_(2)v_(2) = (pir_(2)^(2))v_(2)`
`= pi(0.04)^(2)(0.128)`
`= 6.43 xx 10^(-4)m^(3)//s`