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

`\theta = 23.083 degree`

Step-by-step explanation:

Given data:

yield stress
`\tau_y  = 15 Pa`

thickness t = 3 mm

`\mu = 60 cP = 60* 10^(-2) P`

G= 1.3

`\rho_(point) = G * \rho_(water)`

`=1.3 * 1000 = 1300 kg/m^3`

`\tau = \tau_y + mu (du)/(dy)`

for point flow
`(du)/(dy) = 0`

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

`\tau = (force)/(area)`

`= (weight of point . sin\theta)/(area) = (\rho_(point) . volume. sin\theta)/(area)`[/tex]

`15 = (1300 * 9.8 * A* t * sin\theta)/(A)`

solving for theta value

`sin\theta = 0.392`

`\theta =sin^(-1) 0.392`

`\theta = 23.083 degree`