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PROBLEMS 2 9 . 1 . Trap rock is crushed in a gyratory crusher. The feed is nearly uniform 2 - in . spheres. The

PROBLEMS
29.1. Trap rock is crushed in a gyratory crusher. The feed is nearly uniform 2-in. spheres. The differential screen analysis of the product is given in column (1) of Table 29.5. The power required to crush this material is 400kW? ton. Of this 10kW is needed to operate the empty mill. By reducing the clearance between the crushing head and the cone, the differential screen analysis of the product becomes that given in column (2) in Table 29.5. From (a) Rittinger's law and (b) Kick's law, calculate the power required for the second operation. The feed rate is 110tonh.
TABLE 29.5
Data for Prob. 29.1
\table[[,Product],[Mesh,\table[[First grind],[(1)]],\table[[Second grind],[(2)]]],[46,3.1,],[68,10.3,3.3],[810,20.0,8.2],[1014,18.6,11.2],[1420,15.2,12.3],[2028,12.0,13.0],[2835,9.5,19.5],[3548,6.5,13.5],[4865,4.3,8.5],[-65,0.5,],[65100,,6.2],[100150,,4.0],[-150,,0.3]]
29.2. Using the Bond method, estimate the power necessary per ton of rock in each of the operations in Prob. 29.1.
29.3. Solve Example 29.2 for a mill charge consisting entirely of 46-mesh material. Plot the results and compare them with Fig. 29.1.
29.4. Solve Prob. 29.3 with (a)=1.1 and (b)=1.6. Plot the results. Do they vary as you expected?
29.5. Solve Prob. 29.3 with =1.3 and with S varying with Dp3 for material coarser than 10-mesh and with Dp2 for the finer sizes. Are the results as you predicted?
29.6. What rotational speed, in revolutions per minute, would you recommend for a ball mill 1200mm in diameter charged with 75-mm balls?
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