Solid Waste Engineering
Solid Waste Engineering
3rd Edition
ISBN: 9781305635203
Author: Worrell, William A.
Publisher: Cengage Learning,
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Chapter 4, Problem 4.11P
To determine

(a)

The characteristic size of the product.

Expert Solution
Check Mark

Answer to Problem 4.11P

The characteristic size of the product is 2.08in .

Explanation of Solution

Calculation:

Draw the table to calculate the corresponding Y value for use in the Rosin Rammler equation from the sieve analysis result and the linearized Y values.

    Sieve Size (inches)(a) Percent by weightFiner than Feed(Y)(b) X=ln(a) Linearised Yf=log(ln(11Y))
    4 80% 1.39 0.21
    2 15% 0.69 0.79
    1 5% 0 1.29
    0.5 0% 0.69 0

The slope calculated by the slope function in the excel sheet is,

n=0.165

The intercept calculated by the intercept function in the excel sheet is,

i=0.525

Write the expression to calculate the characteristic size.

x0=e(in) ...... (I)

Here, the characteristic size is x0 , the intercept is i and the slope is n .

Substitute 0.525 for i and 0.165 for n in Equation (I).

x0=e( 0.525 0.165 )=e3.182=24in

Calculate the characteristic size for product.

    Sieve Size (inches)(a) Percent by weightFiner than Product(Y)(b) X=ln(a) Linearised Yf=log(ln(11Y))
    4 95% 1.39 0.476
    2 65% 0.69 0.02
    1 25% 0 0.54
    0.5 10% 0.69 0.98

The slope calculated by the slope function in the excel sheet is,

n=0.70

The intercept calculated by intercept function in the excel sheet is,

i=0.511

Calculate the characteristic size for the product.

Substitute 0.511 for i and 0.70 for n in Equation (I).

x0=e( 0.511 0.70 )=e0.73=2.08in

Conclusion:

Thus, the characteristic size of the feed is 24in .

The characteristic size of the product is 2.08in .

To determine

(b)

If both the distributions fit the Rosin-Rammler particle size distribution function.

Expert Solution
Check Mark

Answer to Problem 4.11P

Only the product distribution fits the Rosin-Rammler distribution function.

Explanation of Solution

Only the distribution for the product fits the Rosin-Rammler distribution function as in the feed distribution the characteristic size is of very high order of 2ft which is not possible.

Thus, only the product distribution fits the Rosin-Rammler distribution function.

To determine

(c)

The diagram for the particle size distribution curve for the given cases.

Expert Solution
Check Mark

Answer to Problem 4.11P

  1. The feed becomes wetter (higher moisture content).

The particle size distribution curve is shown below.

    Solid Waste Engineering, Chapter 4, Problem 4.11P , additional homework tip  1

  • The shredder is run at a higher speed.
  • The particle size distribution curve is shown below.

        Solid Waste Engineering, Chapter 4, Problem 4.11P , additional homework tip  2

    Explanation of Solution

  • The feed becomes wetter (higher moisture content).
  •     Solid Waste Engineering, Chapter 4, Problem 4.11P , additional homework tip  3

        Figure (1)

    When the moisture content is increase the product size also increases considerably.

  • The shredder is run at a higher speed.
  •     Solid Waste Engineering, Chapter 4, Problem 4.11P , additional homework tip  4

        Figure (2)

    When the shredder speed is increased the product size decreases considerably.

    To determine

    (d)

    The effective power requirement.

    Expert Solution
    Check Mark

    Answer to Problem 4.11P

    The effective power requirement is 4.445kWh/ton .

    Explanation of Solution

    Calculation:

    Write the expression to calculate the effective power requirement.

    E=10×E1( x0 ( 1.61 ) ( 1n ) )1210E1( LF )12 ...... (II)

    Here, the bond work index is E1 , the effective power requirement is E and the diameter of particle in micrometer is LF .

    Substitute 10in for LF , 400 for E1 , 0.70 for n and 2.08in for x0 in Equation (II).

    E=[10×400 ( ( 2.08in)( 25400μm 1in)× ( 1.61) ( 1 0.70)) 1 210×400 ( ( 10in)( 25400μm 1in)) 1 2]kWh/ton=(12.3857.94)kWh/ton=4.445kWh/ton

    Conclusion:

    Thus, the effective power requirement is 4.445kWh/ton .

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