Illustrative Example 4.5: Effect of Particle Density

 Illustrative example 4.5 Effect of particle density

A hydrocyclone was found to have a corrected classification curve given by e(dp) = 1/(1 + x– ) where x = dp/d50c. When processing a quartz slurry, d50c = 16.5 m and SI = 0.72. Calculate the recovery of 10 m particles to overflow when this hydrocyclone treats a magnetite slurry under comparable conditions. Density of quartz is 2670 kg/m3 and that of magnetite is 5010 kg/m3. Recovery of water to underflow is 18.5% in both cases.

Illustrative Example 4.5: Effect of Particle Density

This problem involves calculating the recovery of $10 \, \mu\text{m}$ particles to the overflow of a hydrocyclone when the feed is changed from a quartz slurry to a magnetite slurry. The corrected classification curve, particle densities, and recovery of water to the underflow are provided.

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Given Data

• Classification function:

  $$e(d_p) = \frac{1}{1 + x^{-SI}}$$

  Where $x = \frac{d_p}{d_{50c}}$.

• For quartz slurry:
  $d_{50c} = 16.5 \, \mu\text{m}, \, SI = 0.72$.

• Densities:
  $\rho_{\text{quartz}} = 2670 \, \text{kg/m}^3$,
  $\rho_{\text{magnetite}} = 5010 \, \text{kg/m}^3$.

• Particle size:
  $d_p = 10 \, \mu\text{m}$.

• Recovery of water to underflow ($R_w$) = $18.5\%$ in both cases.

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Steps to Solve

1. Adjust $d_{50c}$ for Magnetite Slurry

The cut size ($d_{50c}$) depends on the particle density and can be adjusted using the following relationship:

$$d_{50c,\text{magnetite}} = d_{50c,\text{quartz}} \sqrt{\frac{\rho_{\text{quartz}} - \rho_{\text{fluid}}}{\rho_{\text{magnetite}} - \rho_{\text{fluid}}}}$$

Assume the fluid density is approximately that of water ($\rho_{\text{fluid}} = 1000 \, \text{kg/m}^3$).

$$d_{50c,\text{magnetite}} = 16.5 \cdot \sqrt{\frac{2670 - 1000}{5010 - 1000}}$$

1. Calculate the density difference for quartz:

   $$\rho_{\text{quartz}} - \rho_{\text{fluid}} = 2670 - 1000 = 1670 \, \text{kg/m}^3$$

2. Calculate the density difference for magnetite:

   $$\rho_{\text{magnetite}} - \rho_{\text{fluid}} = 5010 - 1000 = 4010 \, \text{kg/m}^3$$

3. Substitute into the formula:

   $$d_{50c,\text{magnetite}} = 16.5 \cdot \sqrt{\frac{1670}{4010}} = 16.5 \cdot \sqrt{0.4164} = 16.5 \cdot 0.6451 = 10.64 \, \mu\text{m}$$

Thus, $d_{50c,\text{magnetite}} = 10.64 \, \mu\text{m}$.

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2. Calculate the Classification Function for 10 $\mu\text{m}$ Particles

For $d_p = 10 \, \mu\text{m}$:

$$x = \frac{d_p}{d_{50c,\text{magnetite}}} = \frac{10}{10.64} = 0.940$$

The classification function is:

$$e(d_p) = \frac{1}{1 + x^{-SI}}$$

Substitute $x = 0.940$ and $SI = 0.72$:

$$e(d_p) = \frac{1}{1 + (0.940)^{-0.72}}$$

1. Calculate $0.940^{-0.72}$:

   $$0.940^{-0.72} = 1.064$$

2. Substitute into $e(d_p)$:

   $$e(d_p) = \frac{1}{1 + 1.064} = \frac{1}{2.064} = 0.485$$

Thus, $e(d_p) = 0.485$.

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3. Adjust for Water Recovery

The recovery of water to the underflow ($R_w = 18.5\%$) affects the partitioning of particles. The recovery of particles to the overflow ($R_o$) is related to the classification function as follows:

$$R_o = e(d_p) \cdot (1 - R_w)$$

Substitute $e(d_p) = 0.485$ and $R_w = 0.185$:

$$R_o = 0.485 \cdot (1 - 0.185) = 0.485 \cdot 0.815 = 0.395$$

Thus, the recovery of 10 $\mu\text{m}$ particles to the overflow is 39.5%.

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Final Answer

• Adjusted $d_{50c}$ for magnetite slurry: $10.64 \, \mu\text{m}$
• Recovery of 10 $\mu\text{m}$ particles to overflow: $39.5\%$