Respuesta :
Answer:
Force per unit plate area is 0.1344 [tex]N/m^{2}[/tex]
Solution:
As per the question:
The spacing between each wall and the plate, d = 10 mm = 0.01 m
Absolute viscosity of the liquid, [tex]\mu =1.92\times 10^{- 3} Pa-s[/tex]
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''
[tex]F = \tau A[/tex]
where
[tex]\tau[/tex] = shear stress
A = Cross - sectional Area
Therefore,
Net Drag Force, F = [tex](\tau ' +\tau '')A[/tex]
[tex]\frac{F}{A} = \tau ' +\tau ''[/tex]
Also
F = [tex]\frac{\mu v}{d}[/tex]
where
[tex]\mu[/tex] = dynamic coefficient of viscosity
Pressure, P = [tex]\frac{F}{A}[/tex]
Therefore,
[tex]\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}[/tex]
[tex]\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}[/tex]
