Question

7 A very large thin plate is centered in a gap of width 0.06 m with different oils of unknown viscosities above and below; one viscosity is twice the other. When the plate is pulled at a velocity of 0:3 m=s, the resulting force on one square meter of plate due to the viscous shear on both sides is 29 N. Assuming viscous flow and neglecting all end effects, calculate the viscosities of the oils.

Answer 1

`\mu_1\approx 0.967`

`\mu_2\approx 1.933`

Step-by-step explanation:

The viscosities above and below the plate are given by

`\mu_2=2\mu_1` where
`\mu_1` and
`\mu_2` are viscosities of fluid below and above plate respectively

Force on plate due to top layer of the fluid

`\tau_1=\mu_1\frac {\triangle u}{\triangle y}` where
`\triangle u` and
`\triangle y` are the velocity of plate and gap between the plate and upper surface respectively.

`\tau_1=\mu_1* \frac {0.3}{0.03}=10\mu_1`

Force on plate due to bottom layer of the fluid is given by

`\tau_2=\mu_2\frac {\triangle u}{\triangle y}` where
`\triangle u` and
`\triangle y` are the velocity of plate and gap between the plate and upper surface respectively

`\tau_2=\mu_2* \frac {0.3}{0.03}=10\mu_2`

Total force per unit area is the sum of two shear forces

`\frac {F}{A}=\tau_1 +\tau_2` hence

`\tau_1+\tau_2=29`

`10\mu_1+10\mu_2=29`

`10(\mu_1+\mu_2)=29`

`\mu_1+\mu_2=2.9` but since
`\mu_2=2\mu_1` hence

`\mu_1+2\mu_1=2.9`

`3\mu_1=2.9`

`\mu_1=\frac {2.9}{3}=0.966667\approx 0.967`

`\mu_2=2\mu_1=2*0.966667=1.933333\approx 1.933`