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 it ends are 0.1 m and 0.04 m 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 it ends are 0.1 m and 0.04 m and pressure drop across its length is 10N//m^(2).
Answer» <P> 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 Bernoull.s equation for ahorizontal tube `P_(1)+1/2rhoV_(1)^(2)=P_(2)+1/2rhoV_(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)=25/4V_(1)=6.25V_(1)` `[(6.25)^(2)-1^(2)]V_(1)^(2)=16xx10^(-3)m^(-2)//s^(2)` or `V_(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)=rho AV` `=(1.25xx10^(3)kg//m^(3))xx(6.28xx10^(-4)m^(3)//s)` `=0.785kg//s`