The cross-sectional radius of a pipeline decreases graudally as `r=r_0e^(-alphax)`, where `alpha=0.50ms^-1`, x is the distance from the pipeline inlet. Find the ratio of Reynolds numbers of two cross-sections separated by `Deltax=3.2m`.

Explanation

We know that, Reynold's number `(R_e)` is defined as, `R_e=rhovl//eta`, where v is the velocity l is the characteristic length and `eta` the coefficient of velocity. In the case of circular cross section the characteristic length is the diameter of cross-section d, and v is taken as average velocity of flow of liquid.
Now, `R_(e_1)` (Reynold's number at `x_1` from the pipe end) `=(rhod_1v_1)/(meta)` where `v_1` is the velocity at distance `x_1`
and similarly, `R_(e_2)=(rhod_2v_2)/(eta)` so `(R_(e_1))/(R_(e_2))=(d_1v_1)/(d_2v_2)`
From equation of continuity, `A_1v_1=A_2v_2`
or, `pir_1^2v_1=pir_2^2v_2` or `d_1v_1r_1=d_2v_2r_2`
`(d_1v_1)/(d_2v_2)=(r_2)/(r_1)=(r_0e^(-alphax_2))/(r_0e^(-alphax_1))=e^(-alphaDeltax)` (as `x_2-x_1=Deltax`)
Thus `(R_(e_2))/(R_(e_1))=e^(alphaDeltax)=5`