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 10 N//m^(2)

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

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 10 N//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 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)))/(2)=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)` or `V_(1)~~0.0205 m//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)=rhoA.V.=(1.25xx10^(3) kg//m^(3))xx(6.28xx10^(-4)m^(3)//s)=0.785 kg//s`