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SIZE REDUCTION Large particles of solids are cut or broken into small piecesin food-processing industry - eg. grind wheat to flourin ore-processing & cement industries - eg copper ores, nickel, cobaltores & iron ores are ground before chemical processing reasons:1. occurs in sizes that are too large to be used2. so separation can be carried out3. increases the reactivity4. reduces bulk of materials for easier handling and for waste disposal

SIZE REDUCTION methods:1. Compression or crushing- coarse reduction of hard solids to coarse sizes2. Impact/hammer- gives coarse, medium or fine sizes3. Attrition or rubbing- yields very fine products4. Cutting- give definite sizes, sometimes a definite shape

SIZE REDUCTION An ideal crusher would:1. have a large capacity2. require a small power input per unit of product3. yield a product of the single size distribution desired cost of power is a major expense in size reduction equipment,so the factors that control this cost are important

EFFICIENCYCrushing efficiency, C C surface energy created by crushingenergy absorbed by the solidwhere C eS (Awb Awa )WneS surface energy per unit areaWn energy absorbed by a unit massAwa,Awb areas per unit mass of feed and productMechanical efficiency, m m energy absorbed by the solidtotal energy input W e (A A ) m n S wb waWn CWwhere W energy input eS (Awb Awa ) m C

POWER REQUIRMENTPower requirement by the size reduction machine is . 6m eS 11 P Wm C m p D b sb a Dsa where P power required Ý feed ratemṁ Dsa,Dsb volume-surface mean dia. of feed & product, respectively C crushing efficiency m mechanical efficiency p density of particleeS surface energy per unit area a, b sphericity of feed and product, respectively 1N xiDs 6 6andAw s Aw p N x s p i 1 i D pi i 1 D pi

SIZE REDUCTIONHighly energy intensive- 5% of all electricity generated usedMost inefficient unit operation in terms of energy99% goes to heat and noise1% goes to creating new interfacial areaFiner sizes much more costly in term of energyEquations to estimate energy due to :Rittinger (1867)Kick (1885)Bond (1952)Kick’s law – better for larger particlesRittinger’s law – better for fine grinding

POWER REQUIRED IN SIZE REDUCTIONRittinger’s law :work required in crushing is proportional to the new surface createdwhere r P K 1 1. mDsb DP power requiredKr Rittinger’s coefficient sa 11Kr b a c m6es p Ý feed ratemṁ Dsa,Dsb volume-surface meandia. of feed & product, respectively C crushing efficiency m ratio of energy absorbed to energy input p density of particleeS surface energy per unit area a, b sphericity of feed and product, respectively

Example 1A certain crusher accepts a feed rock having a volume-surface meandiameter of 2 cm and discharges a product of volume-surface meandiameter of 0.5 cm. The power required to crush 10 ton/h is 8 HP. Whatshould be the power consumption if the capacity is increased to 12 ton/hand the volume-surface mean diameter is reduced to 0.4 cm? UseRittinger’s law. r sa P K 1 1. mDsb D

Example 2A crusher was used to crush a material with a feed size of -5.08 cm 3.81 cmand the power required was 3.73 KW/ton. The screen analysis of the productwas as follows:What would be the power required to crush 1 ton/h of the samematerial from a feed of - 4.44 cm 3.81 cm to a product of average product size 0.051cm? Use Rittinger’s law.Size of aperture (cm)% product0.630.380.2030.0760.0510.0250.0132618238178 r sa P K 1 1. mDsb D

POWER REQUIRED IN SIZE REDUCTIONKick’s Law :Energy required to reduce a material in size was directly proportional tothe size-reduction ratio P K ln Dsa. kmDsbwhereP power required Kk is the Kick’s coefficientṁÝ feed ratem Dsa,Dsb volume-surface mean dia. of feed & product, respectively

CRUSHING EFFICIENCYBond’s Law :work required using a large-size feed is proportional to the square root of thesurface/volume ratio of the productP K.m b whereP power requiredKb constant pa 1 1DpbDÝ feed ratemṁIf 80% of the feed passes a mesh size of Dpamm and 80% of the product passes a meshsize of Dpb mm, P 0.3162W11 .i mDpa Dpbwhere Wi work index Dpa,Dpb dia. of feed & product, respectively (mm)

WORK INDEXGross energy (kW/h) required per ton of feed needed to reduce a very large feed ( Dpi ) to such a size that 80% of the product passes a 100 μm screen.Include friction in the crusher & powerMaterialSpecific gravityWork Index, WiBauxite2.208.78Cement clinker3.1513.45Cement raw 5.13Granite2.6615.13Gravel2.6616.06Gypsum rock2.696.73Iron ore (hematite)3.5312.84Limestone2.6612.74Phosphate 14.30Trap rock2.8719.32

Example 31. What is the power required to crush 100 ton/h of limestone if 80% of the feedpass a 2-in screen and 80% of the product a 1/8 in screen?Solution:The work index for limestone is 12.74. i P 0.3162W.m1 1DpbDm 100 ton/hṁ D pa 2 * 25.4 50.8mmD pb 0.125 * 25.4 3.175mmThe power required is :P 100 * 0.3162 *12.74( 169.6kW pa 11 )3.17550.8

Example 32. 80% of feed (ore) is less than 5.08 cm in size and the product size is suchthat 80% is less than 0.685 cm. The power required is 89.5 kW. What willbe the power required using the same feed so that 80% is less than 0.3175cm? i P 0.3162W.m pa 1 1DpbD

Solution Example 3-Q2 P1 89.5 kWWi 19.32Dpa1 50.8 mmDpb1 6.35 mmmflowrate ?Dpb2 3.18 mmP2 ?Size ReductionSlide15

Example 4Granite rock is crushed with a uniform feed of 2 in-spheres. The screenanalysis is given in the table below. The power required to crush this material is500 kW; of this 20kW is needed to operate the empty mill. The feed rate is at150 ton/hr. Calculate the power required for the second operation using:a) Rittinger’s Lawb) Kick’s lawChange 50to 150Meshx1(%)x2 29.635/480.53.048/65-1.865/100-0.9100/150-0.3

Example 5Trap 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 1 below The power required to crush this material is 400 kW. Of this10 kW is needed to operate the empty mill. By reducing the clearancebetween the crushing head and the cone, the differential screen analysis ofthe product becomes that given in column (2) in Table1 below. From (a)Rittinger’s law and (b) Kick’s law, calculate the power required for the secondoperation. The feed rate is 110 ton/h.

6565Product 4/2020/2828/3535/4848/6565/100100/150150Product 1.166389Size 3560.2520.1780.1260.003 8Sxi2/Dpiav2.241335Slide18

Rittinger’s Law 1P1 K r m ṁ Dsb Dsa Kick’s LawDsaP K k ln ṁmDsbDsbave 1/1.166389 (produk 1) 0.857347 mmDsaave 2 inci 50.8 mmP 390 kW/tonmflow rate 110 ton/hrKr 3.039Size ReductionSlide19

Dsbave 1/2.241335 (produk 2) 0.446163mm Dsaave 2 inci 50.8 mmmflow rate 110 ton/hrKr 3.039P 751 kWTotal Power required 751 kW 10 kW 761 kWSize ReductionSlide20

SIZE REDUCTION EQUIPMENTSselection of equipments:1) input size2) product size3) hardness4) brittleness5) plasticity6) flammabilitymajor types:crushers, ultrafine grinders, grinders & cutting machines

CRUSHERS Slow-speed machine for coarse reduction of large quantities of solids break large pieces of solid material into small lumpsPrimary crusher - accepts anything from mine & breaks into 150 - 250 mmSecondary crusher - reduces lumps into 6 mmmain types:1) Jaw crushers2) Gyratory crushers3) smooth-roll crushers4) toothed-roll crushersGyratory crusher

GRINDERS for intermediate duty (from crushers to grinders for further reduction) reduce crushed feed to powder product from intermediate grinder might pass a 40-mesh screen product from fine grinder would pass a 200-mesh screen (74 m screen)commercial grinders:1) Hammer mills & impactor2) rolling-compression machines3) Attrition mills4) Tumbling millsSpin millhammer

SIZE REDUCTION EQUIPMENTSCrushers (coarse and fine)- Jaw crushers- Gyratory crushers- Crushing rollsGrinders (intermediate and fine)- Hammer mills; impactors- Rolling-compression mills- Attrition mills- Tumbling millsUltrafine grinders- hammer mills with internal classification- Fluid-energy mills- Agitated millsCutting machines- Knife cutters; dicers; slitters

CRUSHERBlake jawcrusher

GRINDERSImpactorRoller mill

CUTTERRotary knife cutter

SIZE REDUCTION EQUIPMENTSBall Mills:Ball mill in Mining IndustryIn Cement Industry