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