A water flow transports spherical particles (diameter = 2mm; density = 3g/cm3) in suspension mode. If additional particles of density 2g/cm3 are added into the flow, then the diameter of the particles that can be transported without a change in terminal fall velocity, using Stokes law, is _____mm. (Round off to two decimal places) (Use density of water = 1g/cm3)
Correct Answer: 2.8
Solution and Explanation
To determine the diameter of particles that can still be transported without changing terminal fall velocity, we apply Stokes' Law for the terminal velocity (v) given by:
v = (2/9) * (r2(ρp - ρw)/η) * g
where ρp is the density of the particle, ρw is the density of the water, η is the dynamic viscosity of the fluid, g is the acceleration due to gravity (assumed constant), and r is the radius of the particle.
We need the new particle diameter for the same terminal velocity with ρp = 2 g/cm3.
Since v remains constant and g, η, and ρw do not change, equate velocities:
((2/9) * (r12(ρ1 - ρw))) = ((2/9) * (r22(ρ2 - ρw)))
where ρ1, ρ2 are densities, and r1, r2 the respective radii.
Given:
- ρ1 = 3 g/cm3
- ρ2 = 2 g/cm3
- d1 = 2 mm => r1 = 1 mm (since diameter = 2*radius)
By inserting known values:
(12 * (3 - 1)) = (r22 * (2 - 1))
Solving: 12 * 2 = r22 * 1
Thus, r22 = 2 → r2 = √2 mm.
So, the diameter d2 is 2√2 mm.
Approximating: d2 = 2.83 mm.
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