Question

A film of paint 3 mm thick is applied to a surface inclined above the horizontal at an angle \theta. The paint has rheological properties of a Bingham plastic with yield stress 15 Pa and \mu[infinity]= 60 cP. With a specific gravity of 1.3, at what angle \ theta will the paint start to flow down the surface?

Answer 1

Answer:

$\theta = 23.083 degree$

Explanation:

Given data:

yield stress $\tau_y  = 15 Pa$

thickness t = 3 mm

$\mu = 60 cP = 60\times 10^{-2} P$

G= 1.3

$\rho_{point} = G \times \rho_{water}$

                     $=1.3 \times 1000 = 1300 kg/m^3$

$\tau = \tau_y + mu \frac{du}{dy}$

for point flow $\frac{du}{dy} = 0$

$\tau = \tau_y = 15 N/m^2$

$\tau = \frac{force}{area}$

       $= \frac{weight of point . sin\theta}{area} = \frac{\rho_{point} . volume. sin\theta}{area}$[/tex]

$15 = \frac{1300 \times 9.8 \times A\times t \times sin\theta}{A}$

solving for theta value

$sin\theta = 0.392$

$\theta =sin^{-1} 0.392$

$\theta = 23.083 degree$