Question 29.2: A batch grinding mill is charged with material of the compos......

A batch grinding mill is charged with material of the composition
shown in Table 29.2. The grinding-rate function S_u is assumed to be 0.001 s^{- 1} for the 4/6-mesh particles. Breakage function B_u is given by Eq. (29.13) with β = 1.3. Both S_u and B_u are assumed to be independent of time. (a) How long will it take for the fraction of 4/6-mesh material to diminish by 10 percent? (b) Tabulate the individual breakage functions ΔB_{n,u} for the 28/35-mesh fraction and for all coarser fractions. (c) How will the values of x_n. vary with the time during the first 6 h of operation? Use a time interval Δt of 30 s in the calculations.

B_{n, u}=\left(\frac{\bar{D}_n}{\bar{D}_u}\right)^β   (29.13)

TABLE 29.2
Initial mass fractions and grinding-rate functions for Example 29.2
Mesh n or u \begin{aligned}& D_{p n} \text { or } D_{p u} \\ & mm\end{aligned} x_{n,0} \begin{aligned}& S_n \text { or } S_u \times 10^4 \\ & s^{-1} \\ &\end{aligned}
4/6 1 3.327 0.0251 10.0
6/8 2 2.362 0.1250 3.578
8/10 3 1.651 0.3207 1.222
10/14 4 1.168 0.2570 0.4326
14/20 5 0.833 0.1590 0.1569
20/28 6 0.589 0.0538 0.0554
28/35 7 0.417 0.0210 0.0196
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(a) For the 4/6-mesh material there is no input from coarser material and Eq. (29.11) applies. At the end of time t_T, x_1 \text { will be } 0.0251 \times 0.9=0.02259 \text {. Thus }

\frac{d x_u}{d t}=-S_u x_u    (29.11)

-S_u \int_0^{t T} d t=\int_{0.0251}^{0.02259} \frac{d\left(x_1\right)}{x_1}

or

t_T=\frac{1}{S_u} \ln \frac{0.0251}{0.02259}=\frac{1}{0.001} \ln 1.111=105.3~ s

(b) Assume S_u \text { varies with } D_p^3 \text {. Let } S_1 \text { and } S_2 be the values for 4/6- and 6/8-mesh material, respectively. Then S_1=10 \times 10^{-4} ~s ^{-1},

S_2=S_1\left(\frac{D_2}{D_1}\right)^3=10^{-3}\left(\frac{2.362}{3.327}\right)^3=3.578 \times 10^{-4}~ s ^{-1}

Values of S_3 to S_7 are calculated similarly; the results are given in Table 29.2.
The breakage function ΔB_{n,u} is found as follows. When n and u are equal, or whenever n<u, \Delta B_{n, u}=0. The total mass fraction smaller than 6/8-mesh resulting from breakage of 4/6-mesh particles, B_{2,1} from Eq. (29.13), is

B_{2,1}=\left(\frac{2.362}{3.327}\right)^{1.3}=0.6407

Then ΔB_{2,1}, the fraction of broken material retained on the 8-mesh screen, is 1 – 0.6407, or 0.3593.

The total mass fraction smaller than 8/10-mesh resulting from breakage of 4/6-mesh material,B_{3,1}, is

B_{3,1}=\left(\frac{1.651}{3.327}\right)^{1.3}=0.4021

In general, the individual breakage functions are found from the relation

\Delta B_{n, u}=B_{n-1, u }-B_{n, u }    (29.14)

Thus the mass fraction of the broken 4/6-mesh material retained on the 10-mesh screen, \Delta B_{3,1} \text {, is } 0.6407-0.4021=0.2386 . \text { Other values of } B_{n, u} \text { and } \Delta B_{n, u} are found in the same way, to give the results shown in Table 29.3. Note that when n=u, B_{n, u} is unity, by definition. When u = 1, as shown in Table 29.3, 0.6407 of the broken particles from the 4/6-mesh material is smaller than 8-mesh, 0.4021 smaller than 10-mesh, 0.2564 smaller than 14-mesh, and only 0.0672 smaller than 35-mesh.

(c) Let x_{n,t} be the mass fractions retained on the various screens at the end of t time increments \Delta t \text {. Then } x_{1,0}, x_{2,0} \text {, } etc., are the initial mass fractions given in Table 29.2.

The left-hand side of Eq. (29.12) is approximated by \Delta x_n / \Delta t, \text { where } \Delta t in this example is 30 s and \Delta x_n=x_{n, t+1}-x_{n, t} Successive values of x on the screens can then be calculated from the following form of Eq. (29.12):

\frac{d x_n}{d t}=-S_n x_n+\sum\limits_{u=1}^{n-1} x_u S_u \Delta B_{n, u}    (29.12)

\begin{aligned}x_{n, t+1} & =x_{n, t}-S_n \Delta t x_{n, t}+\Delta t \sum_{u=1}^{n-1} x_{u, t} S_u \Delta B_{n, u} \\ & =x_{n, t}\left(1-S_n \Delta t\right)+\Delta t \sum_{u=1}^{n-1} x_{u, t} S_u \Delta B_{n, u}\end{aligned}    (29.15)

For the top screen n = 1 and ΔB = 0. Hence Eq. (29.15) becomes

\begin{aligned}x_{1, t+1} & =x_{1, t}\left(1-S_1 \Delta t\right)=x_{1, t}\left[1-\left(10 \times 10^{-4}\right)(30)\right] \\ & =0.970 x_{1, t}\end{aligned}

After 30 s, then, the mass fraction on the top screen is

x_{1,1}=0.970 \times 0.0251=0.02434

After 30 s more,

x_{1,2}=0.970 \times 0.02434=0.02360

and so forth. On the 8-mesh screen (n = 2), the mass fractions are, from Eq. (29.15),

x_{2,1}=x_{2,0}\left(1-S_2~ \Delta t\right)+\Delta t ~x_{1,0} S_1 ~\Delta B_{2,1}

Substituting the values of S_1, S_2 \text {, and } x_{1,0} \text { from Table } 29.2 \text { and } \Delta B_{2,1} from Table 29.3 gives

\begin{aligned}x_{2,1} & =x_{2,0}\left[1-\left(3.578 \times 10^{-4}\right) \times 30\right]+30 \times 0.0251 \times\left(10 \times 10^{-4}\right) \times 0.359 \\ & =0.98926 x_{2,0}+0.00027\end{aligned}

Thus

x_{2,1}=(0.98926 \times 0.1250)+0.00027=0.12393

Similarly,

\begin{aligned} x_{2,2} & =(0.98926 \times 0.12393)+\left(30 \times 0.02434 \times 10 \times 10^{-4} \times 0.359\right) \\ & =0.12285\end{aligned}

The values of x_ 3 ~to~ x_7 are found in the same way. The results are given in Table 29.4†and illustrated in. Fig. 29.1. Initially x_ 1 , x_2 , ~and ~x _3 all decrease with time and the other mass fractions increase. At the end of 1 h, 99.95 percent of the 4/6-mesh material ( x_1 ) has disappeared and x_7 has more than doubled. The fraction of material finer than 35-mesh has increased from 0.0384 to 0.0931. During the first hour the changes in x_3 and x_7 are almost linear with time. At about 70 min x_4 reaches a maximum and then diminishes with time. If the grinding were continued, the still finer fractions would eventually do the same.

29.2
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TABLE 29.3
Breakage functions for Example 29.2
u B_{n, u} \text { and } \Delta B_{n, u} \text { for } n=
1 2 3 4 5 6 7
1 1.0 0.6407 0.4021 0.2564 0.1652 0.1053 0.0672
0 0.3593 0.2386 0.1457 0.0912 0.0599 0.0381
2 0 1.0 0.6277 0.4003 0.2579 0.1643 0.1049
0 0 0.3723 0.2274 0.1424 0.0936 0.0594
3 0 0 1.0 0.6376 0.4109 0.2618 0.1671
0 0 0 0.3624 0.2267 0.1491 0.0947
4 0 0 0 1.0 0.6444 0.4106 0.2621
0 0 0 0 0.3556 0.2338 0.1485
5 0 0 0 0 1.0 0.6372 0.04067
0 0 0 0 0 0.3628 0.2305
6 0 0 0 0 0 1.0 0.6383
0 0 0 0 0 0 0.3617
7 0 0 0 0 0 0 1.0
0 0 0 0 0 0 0

 

TABLE 29.4
Mass fractions, Example 29.2
Time, min x_1 x_2 x_3 x_4 x_5 x_6 x_7
0 0.251 0.1250 0.3207 0.2570 0.1590 0.0538 0.0210
0.5 0.0243 0.1239 0.3202 0.2575 0.1596 0.0542 0.0213
1 0.0236 0.1228 0.3197 0.2580 0.1602 0.0546 0.0216
2 0.0222 0.1206 0.3187 0.2590 0.1614 0.0554 0.0222
5 0.0185 0.1143 0.3153 0.2618 0.1644 0.0578 0.0240
10 0.0137 0.1042 0.3093 0.2659 0.1685 0.0619 0.0267
20 0.0074 0.0859 0.2961 0.2724 0.1788 0.0695 0.0317
30 0.0040 0.0703 0.2819 0.2772 0.1871 0.0765 0.0363
60 0.0006 0.0376 0.2379 0.2840 0.2074 0.0946 0.0485
90 0.0000 0.0197 0.1967 0.2832 0.2226 0.1097 0.0590
120 0.0000 0.0104 0.1610 0.2777 0.2341 0.1228 0.0682
180 0.0000 0.0028 0.1058 0.2585 0.2495 0.1442 0.0839
240 0.0000 0.0008 0.0687 0.2342 0.2576 0.1611 0.0971
300 0.0000 0.0002 0.0444 0.2087 0.2608 0.1748 0.1084
360 0.0000 0.0001 0.0286 0.1839 0.2605 0.1860 0.1183

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