Question

Calculate the force F (in N) required for ethyl alcohol to flow at 5.13 cm s between two plates separated 0.893 mm with a cross sectional plate area of 2.42 cm2. Useful information: nalcohol 1.2 x 10 Pa s.

Answer 1

The Force required (F) is obtained by

$F= \eta \frac{Av}{D}$

$\eta$ = coefficient of viscosity = 1.2 x 10^-3 Pa S

A = area of cross section = 2.42 cm² = 2.42 x 10^-4 m²

v = velocity = 5.13 cm/s = 0.0513 m/s

D = Distance between the plates = 0.893 mm = 0.893 x 10^-3 m

F = ( 1.2 x 10^-3 x 2.42 x 10^-4 x 0.0513)/ 0.893 x 10^-3 N = 1.67 x 10^-8 N------------------(Answer)