Calculate rate of flow of glycerin of density 1.25xx10^(3)kg//m^(3) through the conical section of a horizontal pipe, if the radii of its ends are 0.1m and 0.04m and pressure drop across its length is 10N//m^(2).

Calculate rate of flow of glycerin of density 1.25xx10^(3)kg//m^(3) th

1.	Calculate rate of flow of glycerin of density 1.25xx10^(3)kg//m^(3) through the conical section of a horizontal pipe, if the radii of its ends are 0.1m and 0.04m and pressure drop across its length is 10N//m^(2).
Answer» Solution :According to CONTINUITY EQUATION, `(V_(2))/(V_(1))=(A_(1))/(A_(2))=(r_(1)^(2))/(r_(2)^(2))=((0.1)^(2))/((0.04)^(2))=(25)/(4)` and, according to Bernoulli.s equation for a horizontal tube, `P_(1)+(1)/(2)rhoV_(1)^(2)=P_(2)+(1)/(2)rhoV_(2)^(2)`. `V_(2)^(2)-V_(1)^(2)=2((P_(1)-P_(2)))/(rho)=2xx((10N//m^(2)))/((1.25xx10^(3)kg//m^(3)))` `=16xx10^(-3)m^(2)//s^(2)` but `V_(2)=(25)/(4)V_(1)=6.25V_(1)` `therefore [(6.25)^(2)-1^(2)]V_(1)^(2)=16xx10^(-3)m^(2)//s^(2)orV_(1)~~0.0205m//s`. the rate of volume flow = `A_(1)V_(1)=pi(0.1)^(2)xx(0.02)` `=6.28xx10^(-4)m^(3)//s` And the rate of mass flow is `(dm)/(dt)=rhoAV`. `=(1.25xx10^(3)kg//m^(3))xx(6.28xx10^(-4)m^(3)//s)=0.785kg//s`