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

Options

A. `62.8 xx 10^(-4)`
B. `6.28 xx 10^(-4)`
C. `2 xx 10^(-2)`
D. `0.2 xx 10^(-2)`

Answer

B

Explanation

According to continuity equation, `(v_1)/(v_2) = (A_1)/(A_2) = (pi xx (0.1)^2)/(pi xx (0.04)^2) = 25/4 " " (i)`
and according to Bernoulli.s equation for horizontal tube,

`p_1 = 1/2 rho v_1^2 = p_2 + 1/2 rho v_2^2` , ie `v_2^2 - v_1^2 = 2 (p_1 - p_2) //rho`
ie. `v_2^2 - v_1^2 = (2 xx 10)//(1.25 xx 10^3) = 16 xx 10^(-3) " " (ii)`
Substituting the value of `v_2` from equ. (i) and (ii),
`(6.25v_1)^2 -v_1^2 = 16 xx 10^(-3)`, ie, `v_1 ~= 0.0205 m//s`
So rate of flow through the tube
`R = A_1 v_1 (=A_2v_2) = pi xx (0.1)^2 xx 0.02 = 6.28 xx 10^(-4) m^3//s`