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