Illustrative Example 6.1: Water Content Calculation of Coal Slurry

Illustrative Example 6.1: Calculate the water content of a coal slurry after it has been dewatered using a 0.6 mm mesh screen with dimensions of 3.66 m in length and 0.914 m in width, inclined at an angle of 15°. The screen operates in a vibration mode with an amplitude of 2 cm and a frequency of 300 rpm, with the vibration inclined at 45° to the screen surface. The saturation water content of the coal is 7.35%.

Illustrative Example 6.1: Water Content Calculation of Coal Slurry After Dewatering

This example involves calculating the water content of a coal slurry after dewatering on a vibrating screen. The parameters of the screen and coal slurry are given as follows:

• Screen Dimensions:
  • Length ($L$): 3.66 m
  • Width ($W$): 0.914 m
  • Angle of Declination ($\theta$): $15^\circ$
• Vibration Parameters:
  • Amplitude ($A$): 2 cm = 0.02 m
  • Frequency ($f$): 300 rpm (rotations per minute)
  • Inclination of vibration angle ($\phi$): $45^\circ$
• Saturation Water Content of Coal ($S_{\text{sat}}$): 7.35% = 0.0735 kg water/kg slurry

Step-by-Step Solution

1. Convert Frequency to Angular Velocity ($\omega$):
   The angular velocity is calculated using the formula:

   $$\omega = \frac{2 \pi f}{60}$$

   Substituting $f = 300 \, \text{rpm}$:

   $$\omega = \frac{2 \pi (300)}{60} = 31.42 \, \text{rad/s}$$

2. Determine the Effective Acceleration in the y-Direction ($G_y$):
   The y-component of acceleration due to gravity is:

   $$G_y = 9.81 \cos \theta$$

   Substituting $\theta = 15^\circ$:

   $$G_y = 9.81 \cos(15^\circ) = 9.47 \, \text{m/s}^2$$

3. Calculate the Vibration Force Contribution ($A_y$):
   The y-component of the vibration acceleration is:

   $$A_y = \omega^2 A \sin \phi$$

   Substituting $\omega = 31.42 \, \text{rad/s}$, $A = 0.02 \, \text{m}$, and $\phi = 45^\circ$:

   $$A_y = (31.42)^2 (0.02) \sin(45^\circ) = 0.679 \, \text{m/s}^2$$

4. Determine the Parameter $b$:
   Using the formula for $b$:

   $$b = 1.44 - 2.52 (0.262)$$

   Substituting the value:

   $$b = 0.780$$

5. Calculate the Reduction Factor ($\dot{X}$):
   The reduction factor is given as:

   $$\dot{X} = 1.87 \exp(-b \, A_y)$$

   Substituting $b = 0.780$ and $A_y = 0.679$:

   $$\dot{X} = 1.87 \exp(-0.780 \times 0.679) = 1.101$$

6. Compute the x-Component of Amplitude ($A_x$):
   The x-component of the vibration amplitude is:

   $$A_x = A \cos \phi$$

   Substituting $A = 0.02 \, \text{m}$ and $\phi = 45^\circ$:

   $$A_x = 0.02 \cos(45^\circ) = 0.014 \, \text{m}$$

7. Calculate the Slurry Velocity ($u$):
   The slurry velocity is given as:

   $$u = \dot{X} \omega A_x$$

   Substituting $\dot{X} = 1.101$, $\omega = 31.42 \, \text{rad/s}$, and $A_x = 0.014 \, \text{m}$:

   $$u = 1.101 \times 31.42 \times 0.014 = 0.489 \, \text{m/s}$$

8. Determine the Residence Time ($t$):
   The residence time is calculated as:

   $$t = \frac{L}{u}$$

   Substituting $L = 3.66 \, \text{m}$ and $u = 0.489 \, \text{m/s}$:

   $$t = \frac{3.66}{0.489} = 7.485 \, \text{seconds}$$

9. Calculate the Water Content ($S$):
   Using the formula for water content:

   $$S = S_{\text{sat}} + p \, t^{-q}$$

   Where:

   $$p = 0.234, \quad q = 0.33 + 0.081 (0.6 - 1) = 0.298$$

   Substituting $S_{\text{sat}} = 0.0735$, $p = 0.234$, $t = 7.485$, and $q = 0.298$:

   $$S = 0.0735 + 0.234 \times (7.485)^{-0.298} = 0.202 \, \text{kg water/kg slurry}$$

Final Answer:

The water content of the coal slurry after dewatering is 0.202 kg water/kg slurry.