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)`.

Options

A. `5.28 xx 10^(-4) m^(3)//s`
B. `6.28 xx 10^(-4) m^(3)//s`
C. `7.28 xx 10^(-4) m^(3)//s`
D. `8.28 xx 10^(-4) m^(3)//s`

Answer

B

Explanation

According to equation of continuity
`(v_(2))/(v_(1)) = (A_(1))/(A_(2)) = (pi xx (0.1)^(2))/(pi (0.04)^(2)) = (25)/(4)` …(i)
According to Bernoulli.s equation for horizontal tube,
`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) = ((2 xx 10))/((1.25 xx 10^(3))) = 16 xx 10^(-3)`...(ii)
Substituting the value of `v_(2)` from equation (i) in (ii), we get
`(6.25 v_(1))^(2) - v_(1)^(2) = 16 xx 10^(-3) or v_(1) = 0.02 m//s`
So rate of flow through the tube `= A_(1) v_(1) = (A_(2) v_(2)) = pi xx (0.1)^(2) xx 0.02 = 6.28 xx 10^(-4) m^(3)//s`