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 x2 according to this law. b) Table 12.1.1 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 12.1.2 after 30 seconds in the mill at this speed. Table 12.1.1: Specific rates of Breakage and Breakage Distribution Function for the Hammer Mill
Interval ( m)
106 - 75
75 - 53
53 - 37
37 - 0
Interval No. j
1
2
3
4
S j
0.6
0.5
0.45
0.4
b(1,j)
0
0
0
0
b(2,j)
0.4
0
0
0
b(3,j)
0.3
0.6
0
0
b(4,j)
0.3
0.4
1.0
0
Table 12.1.2: Feed size distribution
Interval
1
2
3
4
Fraction
0.3
0.5
0.2
0
SOLUTION TO EXERCISE 12.1
(b) Generally, from Text-Equation 12.12: dy i dt
j= i −1
=
∑{b(i, j). S j. y j} − Siy i
j =1
For an increment in time equal to the time basis of the specific rate of breakage, Si:
SOLUTIONS TO CHAPTER 12: SIZE REDUCTION
Page 12.1
j= i− 1
Δ yi =
∑ {b(i, j). S j. y j} − Si yi
j = 1
Change of fraction in interval 1: change in mass fraction in size interval one,
Δ y1 = 0 − S1y1 = 0 − 0.6 × 0.3 = -0.18 Hence, new y1 = 0.3 - 0.18 = 0.12 Change of fraction in interval 2:
Δ y2 = b(2,1 b( 2,1)) S1y1 − S2y 2 = (0.4 × 0. 6 × 0.3) − (0.5 × 0.5) = -0.178 Hence new y2 = 0.5 - 0.178 = 0.322 Change in fraction in interval 3: Δ y3 = [ b(3,1) b(3,1) S1y1 + b(3,2) b(3, 2) S 2 y2 ]− S3y3 0. 6 × 0.3) + (0.6 × 0.5 0. 5 × 0.5)] − (0.45 × 0.2) = [(0.3 × 0.6 = +0.114 Hence, new y3 = 0.2 + 0.114 = 0.314 Change in fraction in interval 4: b(4, 2 ) S 2y 2 + b(4,3 4, 3)S3y3 ] − S4y 4 Δ y4 = [ b(4,1) S1y1 + b(4,2 0. 6 × 0.3) + (0.4 × 0.5 0. 5 × 0.5) ] + [1. 0 × 0.45 × 0.2 0. 2] − (0.4 × 0) = [(0.3 × 0.6 = +0.244 Hence, new y4 = 0.0 + 0.244 = 0.244 Checking: Sum of predicted product interval mass fractions = y1+y2+y3+y4 = 1.000 Hence, product size distribution:
Interval No. (j)
1
2
3
4
Fraction
0.12
0.322
0.314
0.244
SOLUTIONS TO CHAPTER 12: SIZE REDUCTION
Page 12.2
EXERCISE 12.2
Table 12.2.1 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 Table12.2.2. Table 12.2.1 Results of ball mill tests on coal
Interval ( m) 300-212 300-212
212-150 212-150
150-106 150-106
106-75 106-75
75-53 75-53
53-37 53-37
37-0 37-0
Interval no. j
1
2
3
4
5
6
7
S j
0.5
0.45
0.42
0.4
0.38
0.25
0.2
b(1,j)
0
0
0
0
0
0
0
b(2,j)
0.25
0
0
0
0
0
0
b(3,j)
0.24
0.29
0
0
0
0
0
b(4,j)
0.19
0.27
0.33
0
0
0
0
b(5,j)
0.12
0.2
0.3
0.45
0
0
0
b(6,j)
0.1
0.16
0.25
0.3
0.6
0
0
b(7,j)
0.1
0.08
0.12
0.25
0.4
1.0
0
Table 12.2.2: Table 12.2.2: Feed size distribution
Interval
1
2
3
4
5
6
7
Fraction
0.25
0.45
0.2
0.1 0. 1
0
0
0
SOLUTION TO EXERCISE 12.2
Generally, from Text-Equation 12.12: dy i dt
j= i −1
=
∑{b(i, j). S j. y j} − Siy i
j =1
For an increment in time equal to the time basis of the specific rate of breakage, Si: j= i− 1
Δ yi =
∑ {b(i, j). S j. y j} − Si yi
j = 1
SOLUTIONS TO CHAPTER 12: SIZE REDUCTION
Page 12.3
Change of fraction in interval 1: Change in mass fraction in size interval one,
Δ y1 = 0 − S1y1 = 0 − 0.5 × 0.25 = -0.125 Hence, new y1 = 0.25 - 0.125 = 0.125 Change of fraction in interval 2:
Δ y2 = b(2,1 b( 2,1)) S1y1 − S2y 2 = (0.25 × 0.5 0. 5 × 0.25) − (0.45 × 0.45) = -0.1713 Hence, new y2 = 0.45 - 0.1713 = 0.2787 Change in fraction in interval 3: Δ y3 = [ b(3,1) b(3,1) S1y1 + b(3,2) b(3, 2) S 2 y2 ]− S3y3 0. 5 × 0.25) + (0.29 × 0.45 × 0.45) ] − (0.42 × 0.2) = [(0.24 × 0.5 = +0.00473 Hence, new y 3 = 0.2 + 0.00473 = 0.2047 Change in fraction in interval 4: b(4, 2 ) S 2y 2 + b(4,3 4, 3)S3y3 ] − S4y 4 Δ y4 = [ b(4,1) S1y1 + b(4,2
= [(0.19 × 0.5 × 0.25) + (0.27 × 0.45 × 0.45) + (0.33 × 0.42 × 0.2 0. 2)] − (0.4 × 0.1) = +0.0661 Hence, new y4 = 0.1 + 0.0661 = 0.1661 Change in fraction in interval 5: b( 5,1) S1y1 + b(5 b (5,, 2 ) S 2y 2 + b(5, 3 )S3 y3 + b(5 b( 5, 4 )S4 y 4 ] − S5y 5 Δ y5 = [ b(5 0. 5 × 0.25) + (0.2 × 0.45 × 0.45) + (0.3 × 0.42 × 0.2 0. 2) + (0.45 × 0.4 0. 4 × 0.1)] − (0.38 × 0) = [(0.12 × 0.5 = +0.0987 Hence, new y 5 = 0 + 0.0987 = 0.0987 Change in fraction in interval 6: b(6,1) S1y1 + b(6,2 b(6, 2 ) S 2y 2 + b(6,3 6, 3)S3y3 + b(6,4)S b(6, 4)S 4 y4 + b(6,5)S b(6, 5)S 5y5 ] − S6y6 Δ y6 = [ b(6,1) 0. 5 × 0.25) + (0.16 × 0.45 × 0.45) + (0.25 × 0.42 × 0.2 0. 2) + = [(0.1 × 0.5 0. 4 × 0.1 0.1) + (0.6 × 0.38 × 0)] − (0.25 × 0 ) +(0. 3 × 0.4 = +0.0779 Hence, new y6 = 0 + 0.0779 = 0.0779
SOLUTIONS TO CHAPTER 12: SIZE REDUCTION
Page 12.4
Change in fraction in interval 7: b(7,1) S1y1 + b(7,2) b(7, 2) S2 y2 + b (7,3)S3 y3 + Δ y7 = [ b(7,1) b(7, b (7,4)S 4)S 4y 4 + b(7 b( 7, 5)S 5y5 + b(7,6)S b(7, 6)S 6y6 ] − S7y 7 0. 5 × 0.25) + (0.08 × 0.45 × 0.45) + (0.12 × 0.42 × 0.2 ) + = (0.1 × 0.5 0. 4 × 0.1) + (1.0 × 0.25 × 0) − (0.2 × 0) +(0.25 × 0.4 = +0.04878 Hence, new y 7 = 0 + 0.04878 = 0.04878 Checking: Sum of predicted product interval mass fractions = y1+y2+y3+y4+y5+y6+y7= 0.9999 Hence product size distribution:
Interval
1
2
3
4
5
6
7
Fraction
0.125
0.2787
0.2047
0.1661
0.0987
0.0779 0.04878
EXERCISE 12.3
Table 12.3.1 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 12.3.2 after 5 revolutions of the mill. Table 12.3.1: Results of ball mill tests
Interval ( m)
150-106
106-75
75-53
53-37
37-0
Interval No. (j)
1
2
3
4
5
S j
0.65
0.55
0.4
0.35
0.3
b(1,j)
0
0
0
0
0
b(2,j)
0.35
0
0
0
0
b(3,j)
0.25
0.45
0
0
0
b(4,j)
0.2
0.3
0.6
0
0
b(5,j)
0.2
0.25
0.4
1.0
0
SOLUTIONS TO CHAPTER 12: SIZE REDUCTION
Page 12.5
Table 12.3.2: Feed size distribution
Interval
1
2
3
4
5
Fraction
0.25
0.4
0.2
0.1
0.05
SOLUTION TO EXERCISE 12.3
Generally, from Equation 12.12: dy i dt
j= i −1
=
∑{b(i, j). S j. y j} − Siy i
j =1
For an increment in time equal to the time basis of the specific rate of breakage, Si: j= i− 1
Δ yi =
∑ {b(i, j). S j. y j} − Si yi
j = 1
Change of fraction in interval 1: Change in mass fraction in size interval one,
Δ y1 = 0 − S1y1 = 0 − (0.65 × 0.25) = -0.1625 Hence, new y 1 = 0.25 - 0.1625 = 0.0875 Change of fraction in interval 2:
Δ y2 = b(2,1 b( 2,1)) S1y1 − S2y 2 = (0.35 × 0.65 × 0.25) − (0.55 × 0. 4) = -0.1631 Hence new y2 = 0.4 - 0.1631 = 0.2369 Change in fraction in interval 3: Δ y3 = [ b(3,1) b(3,1) S1y1 + b(3,2) b(3, 2) S 2 y2 ]− S3y3
= [(0.25 × 0.65 × 0.25) + (0.45 × 0.55 × 0.4)] − (0.4 × 0.2) = +0.05963 Hence, new y3 = 0.2 + 0.05963 = 0.2596 Change in fraction in interval 4: b(4, 2 ) S 2y 2 + b(4,3 4, 3)S3y3 ] − S4y 4 Δ y4 = [ b(4,1) S1y1 + b(4,2
= [(0.2 × 0.65 × 0.25) + (0.3 × 0.55 × 0.4) + (0.6 0. 6 × 0.4 0. 4 × 0.2 )]− (0.35 × 0.1) = +0.1115
SOLUTIONS TO CHAPTER 12: SIZE REDUCTION
Page 12.6
Hence, new y4 = 0.1 + 0.1115 = 0.2115 Change in fraction in interval 5: b( 5,1) S1y1 + b(5 b (5,, 2 ) S 2y 2 + b(5, 3 )S3 y3 + b(5 b( 5, 4 )S4 y 4 ] − S5y 5 Δ y5 = [ b(5 0. 4 × 0.4 0. 4 × 0.2 0. 2) + (1. 0 × 0.35 × 0.1)]− (0.3 × 0) = [(0.2 × 0.65 × 0.25) + (0.25 × 0.55 × 0.4) + (0.4 = +0.1545 Hence, new y5 = 0.05 + 0.1545 = 0.2045 Checking: Sum of predicted product interval mass fractions = y1+y2+y3+y4+y5 = 1.0 Hence product size distribution:
Interval
1
2
3
4
5
Fraction
0.0875
0.2369
0.2596
0.2115
0.2045
SOLUTIONS TO CHAPTER 12: SIZE REDUCTION
Page 12.7
