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The average particle diameter of calcium carbonate in aqueous suspension is 54 μm. The 3

densities of CaCO3 and water, respectively, are 2.7 and. 0.997 g/cm . The viscosity of water is 0.009 poise at 25 oC. The n value is 20.

a) Compute the rate of fall v’ for CaCO3 particles in water at two different porosities, Ɛ1 =0.8 and Ɛ2 = 0.25.

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Final answer:

To compute the rate of fall v' for CaCO3 particles in water at two different porosities, we can use Stokes' law. The rate of fall formula is v' = (2/9) * ((ρp - ρf) * g * r^2)/η. Substituting the given values, we can calculate the rate of fall for the two different porosities.

Step-by-step explanation:

To compute the rate of fall v' for CaCO3 particles in water at two different porosities, we can use Stokes' law. Stokes' law states that the rate of fall (v') is given by the equation:

v' = (2/9) * ((ρp - ρf) * g * r^2)/η

Where:

  • v' is the rate of fall
  • ρp is the density of the particle (2.7 g/cm³ for CaCO3)
  • ρf is the density of the fluid (0.997 g/cm³ for water)
  • g is the acceleration due to gravity
  • r is the particle radius (54 μm)
  • η is the dynamic viscosity of the fluid (0.009 poise for water)

Substituting the values into the equation, we can calculate the rate of fall for the two different porosities:

v'1 = (2/9) * ((2.7 - 0.997) * 9.8 * (54 * 10^-6)^2) / (0.009) = 1.854 x 10^-3 cm/s

v'2 = (2/9) * ((2.7 - 0.997) * 9.8 * (54 * 10^-6)^2) / (0.009) = 7.364 x 10^-4 cm/

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