Question

A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled with a liquid of absolute viscosity 1.92 x 10^-3 Pa.s. Determine the force per unit plate area required to move the plate upward at a speed of 35 mm/s. Assume linear variation of velocity between the plate and the walls.

Answer 1

Answer:

Force per unit plate area is 0.1344 $N/m^{2}$

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, $\mu =1.92\times 10^{- 3} Pa-s$

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

$F = \tau A$

where

$\tau$ = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = $(\tau ' +\tau '')A$

$\frac{F}{A} = \tau ' +\tau ''$

Also

F = $\frac{\mu v}{d}$

where

$\mu$ = dynamic coefficient of viscosity

Pressure, P = $\frac{F}{A}$

Therefore,

$\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}$

$\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}$