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Using appropriate formulae, determine the minimum channel length required to settle a grit particle with a density of 2500𝑘𝑘g/𝑚𝑚3 and a diameter of 0.15𝑚𝑚𝑚𝑚. The channel depth is 2𝑚𝑚 and the horizontal flow velocity is 0.4𝑚𝑚/𝑠𝑠. Make necessary a...

Assumptions:

  1. The settling of grit particles can be approximated by Stokes’ Law.
  2. The flow in the channel is laminar.
  3. The particle size distribution is homogeneous.
  4. There is no turbulence or eddy currents in the channel.

The settling velocity (Vs) of a particle can be calculated using Stokes’ Law:

Vs = (29) * ((𝜌p - 𝜌f) / 𝜇) * g * r^2

where, 𝜌p = density of the particle 𝜌f = density of the fluid 𝜇 = dynamic viscosity of the fluid g = acceleration due to gravity r = radius of the particle

Since, d = 0.15 mm, radius ® = d/2 = 0.075 mm

Given, Density of particle (𝜌p) = 2500 kg/m^3 Channel depth (h) = 2 mm Horizontal flow velocity (v) = 0.4 m/s

Assuming that water is used as the fluid medium, its density and dynamic viscosity at room temperature are 𝜌f=1000 kg/m^3 and 𝜇=0.001 Pa*s respectively.

The critical velocity for sedimentation can be calculated using: vcrit=(Vs* h)/L

where L is length of the channel required to settle the particle.

Substituting values into Stokes’ Law equation, we get:

Vs = (29) * ((2500 - 1000) / 0.001) * 9.81 * (0.075*10^-3)^2 ≈ 1.49 ×10^-5 m/s

Using this value and substituting given values into the critical velocity formula, we get:

vcrit=(1.49 ×10^-5*2)/L=0.4 L=(1.49 ×10^-5*2)/0.4 L=7.45×10^-6 m

Therefore, the minimum channel length required to settle a grit particle with a diameter of 0.15 mm and density of 2500 kg/m^3 is approximately 7.45 mm (or 0.00745 m).

Figure:

    _______________________________________
   |                                         |
   |                                         |
h->|_________________________________________|
      <-------- L = 7.45 mm -------->

Note: The figure is not to scale

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