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

Explanation

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`