For two 0.3-m-long rotating concentric cylinders, the velocity distribution is given by u (r) = 0.4/r-1000r m/s. If the diameters of the cylinders are 2cm and 4cm, respectively, calculate the fluid viscosity when the inner cylinder rotates at 500 rpm at 1W.

viscosity = 0.051725 N/m-s

Explanation:

Given data

diameter = 0.3 m

u (r) = 0.4/r - 1000r m/s

diameter d1 = 2 cm = 0.02 m

diameter d2 = 4 cm = 0.04 m

rotation = 500 rpm

power = 1 w

to find out

fluid viscosity

solution

we have given

u (r) = 0.4/r - 1000r m/s

du/dr = - 0.4/r² + 1000

here r = 0.3 / 2 = 0.01 m

du / dr = - 0.4 / (0.01) ² + 1000

du/dr = 5000

and

so velocity is

force = viscosity (2π*r*L) du/dr

0.026 / 0.02 = viscosity (2π * 0.02*0.04) 5000

viscosity = 0.051725 N/m-s