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» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to continuity equation <br/>`(V_(2))/(V_(1))=(A_(1))/(A_(2))=(r_(1)^(2))/(r_(2)^(2))=((0.1)^(2))/((0.04)^(2))=(25)/(4)` <br/>and, according to Bernoulli.s equation for a horizontal tube, `P_(1)+(1)/(2)rhoV_(1)^(2)=P_(2)+(1)/(2)rhoV_(2)^(2)` <br/>`V_(2)^(2)-V_(1)^(2)=2((P_(1)-P_(2)))/(2)=<a href="https://interviewquestions.tuteehub.com/tag/2xx-1840186" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX">2XX</a>(10N//m^(2))/((1.25xx10^(3)kg//m^(3)))=16xx10^(-3)m^(2)//s^(2)` <br/>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` <br/> the <a href="https://interviewquestions.tuteehub.com/tag/rate-1177476" style="font-weight:bold;" target="_blank" title="Click to know more about RATE">RATE</a> of volume flow=`A_(1)V_(1)=pi(0.1)^(2)xx(0.02)=6.28xx10^(-4)m^(3)//s` <br/>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`</body></html>


Discussion

No Comment Found

Related InterviewSolutions