Answer to Question #143109 in Math for Usman

Question

A plate 0.025 mm distant from a fixed plate, moves at 50 cm/s and requires a force of 1.471 N/m^2 to maintain this speed. Determine the fluid viscosity between the plates in the poise.

Answer 1

Distance between plates

$dy=0.025mm=2.5\times10^{-5}m$

Velocity of upper plate

$u=50cm/s=0.5m/s$

Force on upper plate

$F=1.471\ N/m^2=\tau$

Hence

The value of the share stress

$\tau=1.471\ N/m^2$

Change of velocity

$du=u-0=0.5m/s$

Change of distance

$dy=2.5\times10^{-4}m$

$\tau=\mu\cdot\dfrac{du}{dy}$

where $\mu$ is the fluid viscosity between the plates

$\mu=\dfrac{\tau\cdot dy}{du}$

$\mu=\dfrac{1.471\ N/m^2 \cdot 2.5\times10^{-5}m}{0.5m/s}=$

$=7.355\times10^{-5}N\cdot s/m^2=7.355\times10^{-4} \text{poise}$

$7.355\times10^{-4}$ poise