Screening
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Screening
The data below were obtained on the operation of a 6-mesh (square) hummer screen at the tipple of a coal mine. The screening was done to separate a very fine refuse from a fine coal stream so that it could be reprocessed. Calculate (a) the recovery and rejection of each size fraction and (b) the screen effectiveness .Feed is 131tons/hr and a Size +1/4 in ¼ x6 mesh 6 x 14 14 x 28 28 x48 48 x 0
Sample Weight 3825 grams 1006 750 303 219 807
Overflow from screen Size sample Wt. +1’4 in 2905 grams ¼ x 6 mesh 767 14 x 28 117 28 x 48 68 48 x 0 278
Underflow from screen (9.8 Tons/hr) Dry Solids Size % Sample 1/4x 6 11.3 6x8 7.8 8 x14 6.9 14 x 48 8.6 28 x 48 3.3 48 x 0 62.1
Solution:
Very fine particles (coal) will be reprocess to fine coal, therefore Oversize= P 0.05 moisture 131 (Feed in tons/hr) 0.95 dry Fdry= 124.45 P= 114.65
9.8 (tons/hr dry solid)
the tipple of a coal mine. could be reprocessed. eness .Feed is 131tons/hr and approximately 5% moisture.
Screening
What is the power requirement to crush 100 tons per hour of limestone if 80% of the feed passes a 2-inch screen and 80% of the product passes a 1/8-inch screen? Use Bond’s law. Wi =12.74 kW-hr/ton. Solution: Dp1=
2 inch x25.4 mm/1 inch = 50.8
Dp2= 3.175mm mm
P/m = 0.3162Wi [(1/√Dp2) – (1/√Dp1)] P= ###
kW-hr/ton.
Given the screen analysis of a product, determine the particle size based on (a) weight mean diameter and (b) surface mean diameter. Size of apperture, in mm through 6 on 4 on 2 on 0.75 on 0.55 on 0.25 on 0.125 through 0.125
% product based on no. of particles 100 6 18 23 8 17 3 5
Solution: Dp ave. -6 + 4 (6+4)/2 = 5 -4 +2 (4+2)/2 = 3 -2 + 0.75 (2+0.75)/2 = 1.38 -0.75+0.5 (0.75+0.5)/2 = 0.63 -0.5+0.25 (0.5+0.25)/2 = 0.38 -0.25+0.125(0.25+0.125)/2= 0.19 -0.125+0 0.13 0.13 ∑
Ni NiDpave NiDp²ave 0.26 1.3 6.5 0.18 0.54 1.62 0.23 0.32 0.43 0.08 0.05 0.03 0.17 0.06 0.02 0.03 0.01 0 0.05 0.01 0 1 2.28 8.61
e size based
(a) Weight Mean Diameter Dpm= ∑((Ni(Dpave)⁴)/(Ni(Dpave)³)) ### = mm
(b) Surface mean diameter Dps = ∑((Ni(Dpave)³)/(Ni(Dpave)²)) # =## mm
NiDp³ave NiDp⁴ave 32.5 162.5 4.86 14.58 0.6 0.82 0.02 0.01 0.01 0 0 0 0 0 37.99 177.92
The data below were obtained on the operation of a 6-mesh (square) hummer screen at the tipple of a coal mine. The screening was done to separate a very fine refuse from a fine coal stream so that it could be reprocessed. Calculate (a) the recovery and rejection of each size fraction and (b) the screen effectiveness .Feed is 131tons/hr and a Size +1/4 in ¼ x6 mesh 6 x 14 14 x 28 28 x48 48 x 0
Sample Weight 3825 grams 1006 750 303 219 807
Overflow from screen Size sample Wt. +1’4 in 2905 grams ¼ x 6 mesh 767 14 x 28 117 28 x 48 68 48 x 0 278
Underflow from screen (9.8 Tons/hr) Dry Solids Size % Sample 1/4x 6 11.3 6x8 7.8 8 x14 6.9 14 x 48 8.6 28 x 48 3.3 48 x 0 62.1
Solution:
Very fine particles (coal) will be reprocess to fine coal, therefore Oversize= P 0.05 moisture 131 (Feed in tons/hr) 0.95 dry Fdry= 124.45 P= 114.65
9.8 (tons/hr dry solid)
the tipple of a coal mine. could be reprocessed. eness .Feed is 131tons/hr and approximately 5% moisture.
Screening
What is the power requirement to crush 100 tons per hour of limestone if 80% of the feed passes a 2-inch screen and 80% of the product passes a 1/8-inch screen? Use Bond’s law. Wi =12.74 kW-hr/ton. Solution: Dp1=
2 inch x25.4 mm/1 inch = 50.8
Dp2= 3.175mm mm
P/m = 0.3162Wi [(1/√Dp2) – (1/√Dp1)] P= ###
kW-hr/ton.
Given the screen analysis of a product, determine the particle size based on (a) weight mean diameter and (b) surface mean diameter. Size of apperture, in mm through 6 on 4 on 2 on 0.75 on 0.55 on 0.25 on 0.125 through 0.125
% product based on no. of particles 100 6 18 23 8 17 3 5
Solution: Dp ave. -6 + 4 (6+4)/2 = 5 -4 +2 (4+2)/2 = 3 -2 + 0.75 (2+0.75)/2 = 1.38 -0.75+0.5 (0.75+0.5)/2 = 0.63 -0.5+0.25 (0.5+0.25)/2 = 0.38 -0.25+0.125(0.25+0.125)/2= 0.19 -0.125+0 0.13 0.13 ∑
Ni NiDpave NiDp²ave 0.26 1.3 6.5 0.18 0.54 1.62 0.23 0.32 0.43 0.08 0.05 0.03 0.17 0.06 0.02 0.03 0.01 0 0.05 0.01 0 1 2.28 8.61
e size based
(a) Weight Mean Diameter Dpm= ∑((Ni(Dpave)⁴)/(Ni(Dpave)³)) ### = mm
(b) Surface mean diameter Dps = ∑((Ni(Dpave)³)/(Ni(Dpave)²)) # =## mm
NiDp³ave NiDp⁴ave 32.5 162.5 4.86 14.58 0.6 0.82 0.02 0.01 0.01 0 0 0 0 0 37.99 177.92
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