Consider the flow of water through a horizontal pipe with `R= 3.0 cm" and "varv = 3 cms^(-1)`. If `eta= 1.008, cP" at "20^(@)C" and "p= 0.9994g cm^(-3)`, calculate the Reynolds number.

Explanation

Reynolds number , `N_(R) = 2 R barv_(rho) // eta`
`= (2 (3.0 xx 10^(-2) m) (3 xx 10^(-2) ms^(-1)) ( 0.9994 xx 10^(-3) xx 10^(6) kg m^(-3)))/((1.008 xx 10^(-2) "Poise") (10^(-1) kg m^(-1) s^(-1) "Poise"^(-1))) = 1784.4`
Since `N_(R)` is less than 2100 , the flow is laminar .