Table : Specific rates of Breakage and Breakage Distribution Function for the Hammer Mill

Transcription

1 SOLUTIONS TO CHAPTER 12: SIZE REDUCTION EXERCISE 12.1 a) Rittinger's energy law postulated that the energy expended in crushing is proportional to the area of new surface created. Derive an expression relating the specific energy consumption in reducing the size of particles from x 1 to x 2 according to this law. b) Table below gives values of specific rates of breakage and breakage distribution functions for the grinding of limestone in a hammer mill. If values of specific rates of breakage are based on 30 seconds in the mill at a particular speed, determine the size distribution of the product resulting from the feed described in Table after 30 seconds in the mill at this speed. Table : Specific rates of Breakage and Breakage Distribution Function for the Hammer Mill Interval (μm) Interval No. j Sj b(1,j) b(2,j) b(3,j) b(4,j) Table : Feed size distribution Interval Fraction SOLUTION TO EXERCISE 12.1 (b) Generally, from Text-Equation 12.12: dy i dt = {b(i, j).s j.y j } S i y i For an increment in time equal to the time basis of the specific rate of breakage, S i : SOLUTIONS TO CHAPTER 12: SIZE REDUCTION Page 12.1

2 Δy i = {b(i, j).s j.y j } S i y i Change of fraction in interval 1: change in mass fraction in size interval one, Δy 1 = 0 S 1 y 1 = = Hence, new y 1 = = 0.12 Change of fraction in interval 2: Δy 2 = b(2,1) S 1 y 1 S 2 y 2 = ( ) ( ) = Hence new y 2 = = Change in fraction in interval 3: Δy 3 = [ b(3,1) S 1 y 1 + b(3,2) S 2 y 2 ] S 3 y 3 = [( ) + ( ) ] ( ) = Hence, new y 3 = = Change in fraction in interval 4: Δy 4 = [ b(4,1) S 1 y 1 + b(4,2) S 2 y 2 + b4,3 ( )S 3 y 3 ] S 4 y 4 = [( ) + ( ) ]+ [ ] (0.4 0) = Hence, new y 4 = = Checking: Sum of predicted product interval mass fractions = y 1 +y 2 +y 3 +y 4 = Hence, product size distribution: Interval No. (j) Fraction SOLUTIONS TO CHAPTER 12: SIZE REDUCTION Page 12.2

3 EXERCISE 12.2 Table below gives information gathered from tests on the size reduction of coal in a ball mill. Assuming that the values of specific rates of breakage, S j are based on 25 revolutions of the mill at a particular speed, predict the product size distribution resulting from the feed material, details of which are given in Table Table Results of ball mill tests on coal Interval (μm) Interval no. j Sj b(1,j) b(2,j) b(3,j) b(4,j) b(5,j) b(6,j) b(7,j) Table : Feed size distribution Interval Fraction SOLUTION TO EXERCISE 12.2 Generally, from Text-Equation 12.12: dy i dt = {b(i, j).s j.y j } S i y i For an increment in time equal to the time basis of the specific rate of breakage, S i : Δy i = {b(i, j).s j.y j } S i y i SOLUTIONS TO CHAPTER 12: SIZE REDUCTION Page 12.3

4 Change of fraction in interval 1: Change in mass fraction in size interval one, Δy 1 = 0 S 1 y 1 = = Hence, new y 1 = = Change of fraction in interval 2: Δy 2 = b(2,1) S 1 y 1 S 2 y 2 = ( ) ( ) = Hence, new y 2 = = Change in fraction in interval 3: Δy 3 = [ b(3,1) S 1 y 1 + b(3,2) S 2 y 2 ] S 3 y 3 = [( ) + ( ) ] ( ) = Hence, new y 3 = = Change in fraction in interval 4: Δy 4 = [ b(4,1) S 1 y 1 + b(4,2) S 2 y 2 + b4,3 ( )S 3 y 3 ] S 4 y 4 = [( ) +( ) + ( ) ] ( ) = Hence, new y 4 = = Change in fraction in interval 5: Δy 5 = b(5,1) S 1 y 1 + b(5,2) S 2 y 2 + b5,3 ( )S 3 y 3 + b(5,4)s 4 y 4 [ ] S 5 y 5 = [( ) + ( ) + ( )+ ( ) ] (0.38 0) = Hence, new y 5 = = Change in fraction in interval 6: Δy 6 = [ b(6,1) S 1 y 1 + b(6,2) S 2 y 2 + b6,3 ( )S 3 y 3 + b(6,4)s 4 y 4 + b(6,5)s 5 y 5 ] S 6 y 6 = [( ) + ( ) + ( )+ + ( )+ ( )] (0.25 0) = Hence, new y 6 = = SOLUTIONS TO CHAPTER 12: SIZE REDUCTION Page 12.4

5 Change in fraction in interval 7: Δy 7 = [b(7,1) S 1 y 1 + b(7,2) S 2 y 2 + b( 7,3)S 3 y 3 + b(7,4)s 4 y 4 + b(7,5)s 5 y 5 + b(7,6)s 6 y 6 ] S 7 y 7 = ( ) + ( ) + ( )+ + ( )+ ( ) (0.2 0) = Hence, new y 7 = = Checking: Sum of predicted product interval mass fractions = y 1 +y 2 +y 3 +y 4 +y 5 +y 6 +y 7 = Hence product size distribution: Interval Fraction EXERCISE 12.3 Table gives information on the size reduction of a sand-like material in a ball mill. If the values of specific rates of breakage Sj are based on 5 revolutions of the mill, determine the size distribution of the feed materials shown in Table after 5 revolutions of the mill. Table : Results of ball mill tests Interval (μm) Interval No. (j) Sj b(1,j) b(2,j) b(3,j) b(4,j) b(5,j) SOLUTIONS TO CHAPTER 12: SIZE REDUCTION Page 12.5

6 Table : Feed size distribution Interval Fraction SOLUTION TO EXERCISE 12.3 Generally, from Equation 12.12: dy i dt = {b(i, j).s j.y j } S i y i For an increment in time equal to the time basis of the specific rate of breakage, S i : Δy i = {b(i, j).s j.y j } S i y i Change of fraction in interval 1: Change in mass fraction in size interval one, Δy 1 = 0 S 1 y 1 = 0 ( ) = Hence, new y 1 = = Change of fraction in interval 2: Δy 2 = b(2,1) S 1 y 1 S 2 y 2 = ( ) ( ) = Hence new y 2 = = Change in fraction in interval 3: Δy 3 = [ b(3,1) S 1 y 1 + b(3,2) S 2 y 2 ] S 3 y 3 = [( ) + ( ) ] ( ) = Hence, new y 3 = = Change in fraction in interval 4: Δy 4 = [ b(4,1) S 1 y 1 + b(4,2) S 2 y 2 + b4,3 ( )S 3 y 3 ] S 4 y 4 = [( ) + ( ) + ( ) ] ( ) = SOLUTIONS TO CHAPTER 12: SIZE REDUCTION Page 12.6

7 Hence, new y 4 = = Change in fraction in interval 5: Δy 5 = b(5,1) S 1 y 1 + b(5,2) S 2 y 2 + b5,3 ( )S 3 y 3 + b(5,4)s 4 y 4 [ ] S 5 y 5 = [( ) + ( ) + ( ) + ( ) ] (0.3 0) = Hence, new y 5 = = Checking: Sum of predicted product interval mass fractions = y 1 +y 2 +y 3 +y 4 +y 5 = 1.0 Hence product size distribution: Interval Fraction SOLUTIONS TO CHAPTER 12: SIZE REDUCTION Page 12.7