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)`.

Options

A. `6.43xx10^(-3)m^3s^(-1)`
B. `6.43xx10^(-4)m^3s^(-1)`
C. `6.43xx10^(-4)m^2s^(-1)`
D. `6.43xx10^(3)m^3s^(-1)`

Answer

B

Explanation

According to continuity equation ,
`A_1v_1 = A_2v_2`
or `v_1/v_2=A_2/A_1 = (pir_2^2)/(pir_1^2)`
`=(r_2/r_1)^2=((0.04)/(0.1))^2 = (4)/(25) " "…(i)`
From bernoulli's equation
`p_1+1/2rhov_1^2 = p_2 + 1/2 rhov_2^2 " " (As h_1=h_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^2s^(-2) " "...(ii)`
Solving eqs. (i) and (ii) , we get
`v_2-0.128ms^(-1)`
`therefore` Rate of volume flow through the tube
`Q=A_2v_2=(pir_2^2)v_2=pi(0.04)^2(0.128)`
`=6.43xx10^(-4)m^3s^(-1)`