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

Answer» <html><body><p></p>Solution :According to <a href="https://interviewquestions.tuteehub.com/tag/continuity-932023" style="font-weight:bold;" target="_blank" title="Click to know more about CONTINUITY">CONTINUITY</a> equation `(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 Bernouli.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))/(<a href="https://interviewquestions.tuteehub.com/tag/rho-623364" style="font-weight:bold;" target="_blank" title="Click to know more about RHO">RHO</a>)=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)` <br/>`therefore [(6.25)^(2)-t^(2)]V_(1)^(2)=16xx10^(-3)m^(2)//s^(2)` or `V_(1)~~0.0205m//s` <br/>So,the rate of <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> flow=`A_(1)V_(1)=<a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a>(0.1)^(2)xx(0.02)=6.28xx10^(-4)m^(3)//s` <br/>And the rate of mass flow is `(dm)/(dt)=rhoAV=(1.25xx10^(3)kg//m^(3))xx(6.28xx10^(-4)m^(3)//s)=0.785 kg//s`</body></html>


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