Tutorial 3 Solution MARCH 2023 Mean Particle Question 1 A material consisting originally of 46 mm particles is crushed to an average size of 8 mm and requires 25 ki/ke for this size reduction. Determine the energy required to crush the material from 46 mm to 4.5 mm assuming: (a) Rittinger’s law, (b) Kick’s law and (c) Bond’s law. Solution a) Applying Rittinger’s law as expressed by the equation Cy = 242.1 ‘Therefore, applying Cx applying the to the desired 4.5mm and determining its energy requirements: 11 = 221(5-Z) E = 48.5 k/kg b) Applying Kick’s Law as expressed by the equationTherefore, applying Cx applying the to the desired 4.5mm and determining its energy requirements: E =-(-143)In ($) B= 3.2 ky /kg ©) Applying Bond's law as expressed by the equation 14 £=20(—-F) 25 266( 4) v8 v46) Cy = 606 ‘Therefore, applying Ce applying the to the desired 4.5mm and determining its energy requirements: 14 25 = 2+606(7=-—2) E = 393kI/kgQuestion 2 Values of breakage distribution function b(i, j) and specific rates of breakage Sj for a particular material in a ball mill are shown in Table 1 below. Size interval (um) 212-150 150-106 106-75. 75-53 1 Interval no. 2 3 4 07 06 05 035 0 0 0 0 032 0 0 0 03 o4 0 0 ous 02 os 0 0.2 02 0.25 06 0.12 02 025 04 To test the validity of these values, a sample of the material with the size distribution indicated in Table 2 below is to be ground in a ball mil, Interval no. () 1 2 3 4 5 6 Fraction 02 oa 03 0.06 a) Use the information in these tables to predict the size distribution of the product after one minute in the mill. (Note: Sj values in Table 1 are based on 1-minute grinding time.) Solution Ietct a _'$ Gem Y bensml-s Change of fraction in interval 1 ays Ge = bADSnI-Sin oy 2 2 SE = [007 +02] -07+02 =0-0.7402 =-014 Hence, new y; = 0.2 — 0.14 = 0,06Change of fraction in interval 2 dy ‘ Fe = WeDsy.1-Sey2 dy, _ P= [0.32+0.7 +02]-06-04 0.1952 Hence, new yz = 0.4 ~ 0.1952 = 0.2048 Change of fraction in interval 3 2 Wa.ns.y1 + B.2Say_1-S95 (0.3 + 0.7 + 0.2) + 04+06+0.4)]- 05 +03 = 0.012 Hence, new y3 = 0.3 ~ 0.012 = 0.288 Change of fraction in interval 4 ae = WGDSin + b4.2)Sa¥2 + bE4,3)Sa¥a]-Sa¥s ayy [0.14 +07 +02) + (0.206 +04) + (05 +05 +0.3)]— 0.35 +0.06 = 40.1216 Hence, new yy = 0.06 + 0.1216 = 0.1816 Change of fraction in interval 5 a FE = GAS ys + bS.2)Si¥2 + DEA)SaYs + bOGAVS,ysI—Seysays at = [(0.12 + 0.7 + 0.2) + (0.2 *0.6+ 0.4) + (0.25 + 0.5 + 0.3) + (0.6 + 0.35 + 0.06)] 0.3 0.04 = +0.1029 Hence, new ys = 0.04 + 0.1029 = 0.1429 Change of fraction in interval 6 dys at [b(6,1)Siyi + b(6,2)Sz¥2 + B(6,3)Sa¥s + b(GA)Sa¥4 + B(6,5)Ss¥5)-Se¥s, dys ae [(0.12 + 0.7 * 0.2) + (0.2 + 0.6 + 0.4) + (0.25 + 0.5 + 0.3) + (0.4 + 0.35 * 0.06) + (1.0 +03 *0.04)] -0+0 = +0.1227 Hence, new ys = 0+ 0.1227 = 0.1227 Checking: ‘Sum of predicted product interval mass fractions MtVe + Vs +Vs + yst¥e = 1.0 0.06 + 0.2048 + 0.288 + 0.1816 + 0.1429 +0.1227 = 1.0Question 3 Power of 3 kW is supplied to @ machine crushing material at the rate of 0.3 kg/s from 12.5 mm toa product having the following sizes: 80 per cent 3.175 mm, 10 per cent 2.5 mm and 10 per cent 2.25 mm. if the material mass is 50 kg, what power (kW) should be supplied to this machine to crush 0.3, kg/s of the same material from 7.5 mm cube to 2.0 mm cube: i) Assuming Rittinger’s law? ii) Assuming Kick’s law? iii) Which of the above results would be regarded as being more reliable and why? Solution a) We first determine the energy required from the power that was supplied Po 3 Bata 3 = 10K /m Sa ggn 1iikg The we determine the mean (average) particle size from the values given, and the mean partcicle size by mass is given by: = 04250 , 0150) “Gay * Gay, 04-8, 01-50) (5) * 225) ( 5.75566 200872 Sy = 2.863 i) Applying Rittinger’s law: = 13.6ki/kgii) Applying Kick’s law: hence Cx = 6.78 kJ/kg and so with x; = 7.5 mm and x7=2mm E = (6.78) In(75/y) E = 9kj/kg iy Results calculated using kick’s Law are more reliable since it applies for large particle sizes and the material being crushed has large particles.