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 x1 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 (μm) 106 - 75 75 - 53 53 - 37 37 - 0

Interval No. j 1 2 3 4

Sj 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: dyidt

= {b(i, j)j=1

j=i−1

∑ .Sj.yj}− Siyi

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

Δyi = {b(i, j)j=1

j=i−1

∑ .S j.yj} − Siyi

Change of fraction in interval 1: change in mass fraction in size interval one, Δy1 = 0 − S1y1 = 0 − 0.6 × 0.3 = -0.18 Hence, new y1 = 0.3 - 0.18 = 0.12 Change of fraction in interval 2:

Δy2 = b(2,1) S1y1 S2y2− = (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: Δy3 = b(3,1) S1y1 + b(3,2) S2y2[ ] S3y3− = (0.3× 0.6 × 0.3) + (0.6 × 0.5 × 0.5)[ ]− (0.45× 0.2)

= +0.114 Hence, new y3 = 0.2 + 0.114 = 0.314 Change in fraction in interval 4: Δy4 = b(4,1) S1y1 + b(4,2) S2y2 b 4,3(+ )S3y3[ ] S4y4 −

= (0.3× 0.6× 0.3) + (0.4 × 0.5× 0.5)[ ]+ 1.0× 0.45× 0.2[ ]− (0.4 × 0)

= +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, Sj 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 (μm) 300-212 212-150 150-106 106-75 75-53 53-37 37-0

Interval no. j 1 2 3 4 5 6 7

Sj 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: Feed size distribution

Interval 1 2 3 4 5 6 7

Fraction 0.25 0.45 0.2 0.1 0 0 0

SOLUTION TO EXERCISE 12.2 Generally, from Text-Equation 12.12: dyidt

= {b(i, j)j=1

j=i−1

∑ .Sj.yj}− Siyi

For an increment in time equal to the time basis of the specific rate of breakage, Si:

Δyi = {b(i, j)j=1

j=i−1

∑ .S j.yj} − Siyi

SOLUTIONS TO CHAPTER 12: SIZE REDUCTION Page 12.3

Change of fraction in interval 1: Change in mass fraction in size interval one, Δy1 = 0 − S1y1 = 0 − 0.5 × 0.25 = -0.125 Hence, new y1 = 0.25 - 0.125 = 0.125 Change of fraction in interval 2:

Δy2 = b(2,1) S1y1 S2y2− = (0.25 × 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: Δy3 = b(3,1) S1y1 + b(3,2) S2y2[ ] S3y3− = (0.24 × 0.5 × 0.25) + (0.29× 0.45× 0.45)[ ]− (0.42× 0.2)

= +0.00473 Hence, new y3 = 0.2 + 0.00473 = 0.2047 Change in fraction in interval 4: Δy4 = b(4,1) S1y1 + b(4,2) S2y2 b 4,3(+ )S3y3[ ] S4y4 −

= (0.19 × 0.5 × 0.25) + (0.27 × 0.45 × 0.45) + 0.33 × 0.42 × 0.2( )[ ]− (0.4 × 0.1)

= +0.0661 Hence, new y4 = 0.1 + 0.0661 = 0.1661 Change in fraction in interval 5: Δy5 = b(5,1) S1y1 + b(5,2) S2y2 b 5,3( )S3y3+ + b(5,4)S4y4[ ] S5y5 −

= (0.12 × 0.5× 0.25) + (0.2× 0.45× 0.45) + 0.3× 0.42 × 0.2( )+ 0.45 × 0.4 × 0.1( )[ ]− (0.38 × 0)

= +0.0987 Hence, new y5 = 0 + 0.0987 = 0.0987 Change in fraction in interval 6: Δy6 = b(6,1) S1y1 + b(6,2) S2y2 b 6,3( )S3y3 b(6,4)S4y4+ + + b(6,5)S5y5[ ]− S6y6

= [(0.1 × 0.5× 0.25) + (0.16× 0.45× 0.45) + 0.25 × 0.42 × 0.2( )++ 0.3× 0.4 × 0.1( )+ (0.6 × 0.38× 0)] − (0.25× 0)

= +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: Δy7 = [b(7,1) S1y1 + b(7,2) S2y2 + b 7,3( )S3y3 + b(7,4)S4y4 + b(7,5)S5y5 + b(7,6)S6y6 ] −S7y7

= (0.1× 0.5 × 0.25) + (0.08× 0.45× 0.45) + 0.12× 0.42 × 0.2( )++ 0.25× 0.4 × 0.1( )+ (1.0 × 0.25× 0) − (0.2 × 0)

= +0.04878 Hence, new y7 = 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 (μm) 150-106 106-75 75-53 53-37 37-0

Interval No. (j) 1 2 3 4 5

Sj 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: dyidt

= {b(i, j)j=1

j=i−1

∑ .Sj.yj}− Siyi

For an increment in time equal to the time basis of the specific rate of breakage, Si:

Δyi = {b(i, j)j=1

j=i−1

∑ .S j.yj} − Siyi

Change of fraction in interval 1: Change in mass fraction in size interval one, Δy1 = 0 − S1y1 = 0 − (0.65 × 0.25) = -0.1625 Hence, new y1 = 0.25 - 0.1625 = 0.0875 Change of fraction in interval 2:

Δy2 = b(2,1) S1y1 S2y2− = (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: Δy3 = b(3,1) S1y1 + b(3,2) S2y2[ ] 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: Δy4 = b(4,1) S1y1 + b(4,2) S2y2 b 4,3(+ )S3y3[ ] S4y4 −

= (0.2 × 0.65 × 0.25) + (0.3 × 0.55 × 0.4) + 0.6 × 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

SOLUTIONS TO CHAPTER 12: SIZE REDUCTION Page 12.7

Change in fraction in interval 5: Δy5 = b(5,1) S1y1 + b(5,2) S2y2 b 5,3( )S3y3+ + b(5,4)S4y4[ ] S5y5 −

= (0.2 × 0.65 × 0.25) + (0.25× 0.55 × 0.4) + 0.4 × 0.4× 0.2( ) + 1.0× 0.35× 0.1( )[ ]− (0.3× 0)

= +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