Assignment UOP 1 (Answer Sheet)

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ASSIGNMENT 1: UNIT OPERATIONS
Particle Size
Average particle
Screen Opening
Mass Fraction
Dpi (mm)
Retained, xi
Μ… π’‘π’Š mm
𝑫
4
4.699
0.0000
-
6
3.327
0.0251
4.013
8
2.369
0.1250
2.845
2.007
Mesh
10
1.651
0.3207
14
1.168
0.2570
20
0.833
0.1590
diameter
1.409
< 1.5mm
π’™π’Š
Μ… π’‘π’Š
𝑫
π’™π’Š
Μ… π’‘π’Š πŸ‘
𝑫
0
0
0.0063
0.0004
0.0439
0.0054
0.1598
0.0397
0.1824
0.0919
0.1588
0.1585
0.0757
0.1497
0.0418
0.1650
0.0287
0.2261
0.0306
0.4812
0.0326
1.0284
0.0325
2.0496
0.0348
4.3974
0.2027
148.0663
1.0305
156.8595
1.001
28
0.589
0.0538
0.711
35
0.417
0.0210
0.503
48
0.295
0.0102
0.356
65
0.208
0.0077
0.252
100
0.147
0.0058
0.178
150
0.104
0.0041
0.126
200
0.074
0.0031
0.089
Pan
-
0.0075
0.037
Sample calculations of average particle diameter
Μ… π’‘π’Š (π’Žπ’†π’”π’‰ πŸ’) =
𝑫
πŸ”
πŸ’.πŸ”πŸ—πŸ—+πŸ‘.πŸ‘πŸπŸ•
𝟐
= πŸ’. πŸŽπŸπŸ‘
1. The size of 35 kg crushed rocks has been analyzed using sieving method. The results of
the sieving process are tabulated in the Table 1. The density of the particle is 1550 kg/m3
and the shape factor are a =0.8 and Φ = 0.69
Calculate:
(a) The specific area of particles ( Aw) (mm2/g) for the material between 4-mesh and pan.
π΄π‘Š =
6
π‘₯𝑖
∑
Μ…Μ…Μ…Μ…
𝛷𝑆 πœŒπ‘
𝐷𝑝𝑖
π΄π‘Š =
6
(1.0305) = πŸ“πŸ•πŸ–πŸ. 𝟐𝟏 π‘šπ‘š2 /𝑔
0.69(0.001550)
(b) Number of particles in mixture (Nw) (particles/g) for the material between 4-mesh and
pan
π‘π‘Š =
1
π‘₯𝑖
∑
Μ…Μ…Μ…Μ…Μ…Μ…3
π‘ŽπœŒπ‘
𝐷
𝑝𝑖
π‘π‘Š =
1
(156.8595) = πŸπŸπŸ”πŸ’πŸ—πŸ—. πŸ” π‘šπ‘š2 /𝑔
0.8(0.001550)
(c) Volume surface mean diameter (Ds) (mm)
̅𝑠 =
𝐷
1
1
=
= 0.97π‘šπ‘š
π‘₯𝑖
1.0305
∑ Μ…Μ…Μ…Μ…
𝐷𝑝𝑖
(d) If 70% of the crushed glass must have size less than 1.5mm, please state whether or not
the particles need to be crushed again?
The particle need to be crushed again because, only 52.92% of the
average particle size is less than 1.5mm
Μ… π’‘π’Š > 1.5mm = (xi =0.0251+0.1250+0.3207=
Refer to the table οƒ  particle with 𝑫
Μ… π’‘π’Š < 1.5mm = (1-0.4708=
0.4708x100= 47.08%) whereas particle with 𝑫
0.5292x100=52.92%).Thus it need to be crushed again to achieve 70%.
Size Reduction
2. State four 4 types of equipments employed in particles size reduction. Elaborate the
principles of those equipments.
Crusher
-
Do the heavy work of breaking large pieces of solid material into small pieces
-
A slow speed machines for coarse reduction of large quantities of solids
-
Primary crusher can break the solids into 150-250mm size
-
Secondary crusher could reduced the particle from primary crusher to 6mm in size
-
Crusher reduces the solid size by compression
Grinder
ο‚— Reduce crushed feed to powder
ο‚— Product from a crusher is often fed to a grinder for further reduction.
ο‚— The product from an intermediate grinder might pass a 40-mesh screen.
ο‚— Most products from a fine grinder would pass a 200- mesh screen with 74 μm opening.
ο‚— Reduce the solid size by impact and attrition, sometimes combine with compression
Ultrafine grinder
ο‚— Reduce solids to fine particles
ο‚— Accepts feed particles not larger than 6 mm.
ο‚— The product size is typically 1 to 50 μm.
ο‚— Reduce the solid size by attrition
Cutting machine
ο‚— Give particles of definite size and shape 2-10mm in length
ο‚— Reduce the size by cutting, dicing and slitting
3. 110 tons/h of trap rock is crushed in a gyratory crusher. The feed is nearly uniform 2-in
spheres and produced the product of 0.08 in.size. Using the Bond method, estimate the
work necessary for the crushing process.
π‘šΜ‡ = 110 π‘‘π‘œπ‘›/β„Žπ‘Ÿ
π·π‘ƒπ‘Ž = 2 𝑖𝑛 π‘₯ 25.4 = 50.8π‘šπ‘š
𝐷𝑃𝑏 = 0.08 π‘₯ 25.4 = 2.032 π‘šπ‘š
π‘Šπ‘– (π‘‘π‘Ÿπ‘Žπ‘ π‘Ÿπ‘œπ‘π‘˜) = 19.32
Thus by using Bonds Law
𝑃
𝐾𝑏
=
π‘šΜ‡
√𝐷𝑃
𝐾𝑏 = 0.3162 π‘ŠπΌ
𝑃
1
1
= 0.3162 π‘ŠπΌ (
−
)
π‘šΜ‡
√𝐷𝑃𝑏 √π·π‘ƒπ‘Ž
𝑃
1
1
= 0.3162 (19.32) (
−
)
110
√2.032 √50.8
𝑷 = πŸ‘πŸ•πŸ•. 𝟏𝟐 π’Œπ‘Ύ
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