The radius of a pipe decreases according to r=r_(o)e^(-propx) , where alpha=0.50m^(-1)andxis the distance of a cross - section from the first end (x=0) . Find the ratioof Reynolds number for two cross -sections lying at the distance of 2m from each other . (take e=2.718)

The radius of a pipe decreases according to r=r_(o)e^(-propx) , where

1.	The radius of a pipe decreases according to r=r_(o)e^(-propx) , where alpha=0.50m^(-1)andxis the distance of a cross - section from the first end (x=0) . Find the ratioof Reynolds number for two cross -sections lying at the distance of 2m from each other . (take e=2.718)
Answer» Solution :Reynold number `N_(R)=(rhovD)/(eta)` `THEREFORE` For a given liquid `N_(R)propvD` `therefore((N_(R))_(1))/((N_(R))_(2))=(v_(1))/(v_(2))XX(D_(1))/(D_(2))`….(1) ACCORDING to the equation of continuity, `A_(1)v_(1)=A_(2)v_(2)` `thereforepir_(1)^(2)v_(1)=pir_(2)^(2)v_(2)` `therefore(v_(1))/(v_(2))=(r_(2)^(2))/(r_(1)^(2))=((D_(2))/(D_(1)))^(2)`....(2) From equation (1) and (2) , `((N_(R))_(1))/((N_(R))_(2))=(r_(o)e^(-propx2))/(r_(o)^(-propx1))=e^(-alpha(x_(2)-x_(1)))` `=e^(-(0.5)(2))` `=e^(-1)` `(1)/(2.718)` `therefore((N_(R))_(1))/((N_(R))_(2))=0.368`