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

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j