Particulate-technology-solution

CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] SCREENING 1. It is desired to separate a mixture of sugar crystals into two fractions, a coarse fraction retained on an 8-mesh screen, and a fine fraction into passing through it. What is the screen effectiveness? Screen analysis of feed, coarse and fine fractions show Mass fraction of +8 particles in feed =0.46 Mass fraction of +8 particles in coarse material = 0.88 Mass fraction of +8 particles in fine material = 0.32

Feed xF

8 mesh

Oversize (Coarse) xR

Undersize (Fine material) xP

REQUIRED: Screen effectiveness, E SOLUTION: To calculate the screen effectiveness, the ff formula will be used where mass fractions (xp,xF,xR) are based on the desired material �1 − 𝑥𝑥𝑝𝑝 �(𝑥𝑥𝐹𝐹 − 𝑥𝑥𝑅𝑅 ) 𝑥𝑥𝑃𝑃 (𝑥𝑥𝐹𝐹 − 𝑥𝑥𝑅𝑅 ) �1 − � 𝐸𝐸 = (1 − 𝑥𝑥𝐹𝐹 )(𝑥𝑥𝑃𝑃 − 𝑥𝑥𝑅𝑅 ) 𝑥𝑥𝐹𝐹 (𝑥𝑥𝑃𝑃 − 𝑥𝑥𝑅𝑅 ) In screening process, the material which is usually desired is the fine material so xp,xF,xR should all be based on the fine material which are -8 particles (through 8 mesh). Since the given mass fractions are in terms of +8 particles ( on 8 mesh), then 𝑥𝑥𝐹𝐹 = 1 − 0.46 = 0.54 𝑥𝑥𝑃𝑃 = 1 − 0.32 = 0.68 𝑥𝑥𝑅𝑅 = 1 − 0.88 = 0.12 Substituting, 𝐸𝐸 = 0.4517 𝑜𝑜𝑜𝑜 45.17% SIZE REDUCTION 1. In crushing a certain ore, the feed is such that 80% is less than 50.8 mm in size and the product size is such that 80% is less than 6.35 mm. The power required is 89.5 kW. Based on the Bond equation the power required using the same feed so that 80% is less than 3.18 mm is? (ans: 146.7 kW) Ore x1 =50.8 mm

Crusher Power required = 89.5 kW

Product x2 = 6.35 mm

REQUIRED: P if same feed is used and x2= 3.18 mm (using Bond’s equation) SOLUTION: 1|Page

CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] BOND’S LAW 𝑃𝑃 𝑇𝑇 𝑃𝑃 𝑇𝑇

= 1.46 𝐸𝐸𝐸𝐸 �

1

�𝑋𝑋2

= 0.3162 𝐸𝐸𝐸𝐸 �

−

1

�𝑋𝑋2

1

�𝑋𝑋1

−

� where P in hP; T in tons/min ; Ei in kW.h/ton; X1 & X2 in ft 1

�𝑋𝑋1

� where P in kW; T in tons/hr; Ei in kW.h/ton; X1 &X2 in mm

To solve for unknown, we will need the work index, Ei, which remains constant for the same material and equipment. Thus, we can get this from condition 1 Condition 1: P=89.5 kW, x1=50.8 mm, x2= 6.35 mm 𝑃𝑃 1 1 = 0.3162 𝐸𝐸𝐸𝐸 � − � 𝑇𝑇 �𝑋𝑋2 �𝑋𝑋1

1 1 89.5 = 0.3162𝐸𝐸𝐸𝐸 � − � 𝑇𝑇 √6.35 √50.8 𝐸𝐸𝐸𝐸𝐸𝐸 = 1102.3094

Since the same feed is used for condition 2, then T also remains constant Condition 2: x1=50.8 mm, x2= 3.18 mm 𝑃𝑃 = 0.3162(1102.3094) � 𝑃𝑃 = 146.5545 𝑘𝑘𝑘𝑘

1

√3.18

−

1

� √50.8

2. A material is crushed in a Blake Jaw Crusher and the average size of particles reduced from 50 mm to 10 mm with the consumption of energy at the rate of 13 kW/kg.s. What is the energy consumption needed to crush the same material of an average size 75 mm to an average size 25mm assuming (a) Kick’s Law applies (b) Rittinger’s Law applies ? (ans 8.88 kW/kg.s) FEED Crusher X1= 50mm P/T=13 kW/kg.s

PRODUCT X2=10 mm

REQUIRED: P if X1=75mm X2=25mm using (a) Kick’s law (b) Rittinger’s Law SOLUTION: KICK’S LAW

RITTINGER’S LAW

𝑋𝑋1 𝑃𝑃 = 𝑘𝑘 log � � 𝑋𝑋2 𝑇𝑇 1 1 𝑃𝑃 = 𝑘𝑘 � − � 𝑋𝑋2 𝑋𝑋1 𝑇𝑇

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CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] a. Using Kick’s Law Since no unit is specified for Kick’s eqn, then any unit can be used as long as they are consistent The value of k is constant for the same material and equipment. Condition 1: P/T=13 kW/kg.s ; X1=50 mm ; X2= 10 mm 𝑋𝑋1 𝑃𝑃 = 𝑘𝑘 log � � 𝑋𝑋2 𝑇𝑇 50 13 = 𝑘𝑘 log � � 10 𝑘𝑘𝑘𝑘 𝑘𝑘 = 18.5988 𝑘𝑘𝑘𝑘. 𝑠𝑠 Condition 2: 𝑃𝑃 75 = 18.5988 log � � 𝑇𝑇 25 𝑃𝑃 𝑘𝑘𝑘𝑘 = 8.8739 𝑇𝑇 𝑘𝑘𝑘𝑘. 𝑠𝑠

b. Using Rittinger’s Law Condition 1: P/T=13 kW/kg.s ; X1=50 mm ; X2= 10 mm 1 1 𝑃𝑃 = 𝑘𝑘 � − � 𝑋𝑋2 𝑋𝑋1 𝑇𝑇 1 1 13 = 𝑘𝑘 � − � 10 50 𝑘𝑘𝑘𝑘 𝑘𝑘 = 162.5 𝑘𝑘𝑘𝑘. 𝑠𝑠 Condition 2: 1 1 𝑃𝑃 = 162.5 � − � 25 75 𝑇𝑇 𝑘𝑘𝑘𝑘 𝑃𝑃 = 4.3333 𝑘𝑘𝑘𝑘. 𝑠𝑠 𝑇𝑇 HANDLING OF SOLIDS

1. One hundred tons per hour of anthracite coal are to be moved horizontally a distance of 120 ft. Select a conveyor of each of the three classes listed, and calculate the power required to operate the system. Choose the smallest conveyor that will do the job. Assume a bulk density of 60 lb/ft3. a. Screw conveyor b. Flight conveyor c. Belt conveyor GIVEN: Anthracite Coal T= 100 tons/hr

ΔZ =120 ft

ρbulk = 60 lb/ft3

REQUIRED: i. HP if (a) Screw (b) Flight (c) Belt

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CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] ii. Choose the Smallest conveyor that can do the job SOLUTION: a. Screw Conveyor

𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 100

𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ℎ𝑟𝑟

2000 𝑙𝑙𝑙𝑙 1ℎ𝑟𝑟 �� � 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 60 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

�

For coal, coefficient= 2.5

𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 =

𝐻𝐻𝐻𝐻 =

(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) �𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶,

= 3333.3333

𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚

33,000

𝑙𝑙𝑙𝑙 � (𝑙𝑙𝑙𝑙𝑛𝑛𝑛𝑛𝑛𝑛ℎ, 𝑓𝑓𝑓𝑓) 𝑚𝑚𝑚𝑚𝑚𝑚

2.5(3333.3333)(120) = 30.3030 ℎ𝑃𝑃 33000

To solve for diameter of the screw conveyor the table below is used. Since from the table, capacity is in cu ft/hr, then we need to convert 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 3333.3333

𝑓𝑓𝑓𝑓 3 60𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓 3 𝑙𝑙𝑙𝑙 � = 3333.3333 � �� ℎ𝑟𝑟 ℎ𝑟𝑟 𝑚𝑚𝑚𝑚𝑚𝑚 60 𝑙𝑙𝑙𝑙

Under coal, we determine the screw diameter which will be able to handle 3333.3333, thus Diameter of Screw, in. 3 4 5 6 7 8 9 10 12 14 16 18 20

Light Nonabrasive Material e.g. Grain Capacity, cu.ft/hr

Max. rpm

74 171 304 500 820 1180 1600 2050 3300 4000 7000 9000 12000

250 220 210 200 190 180 175 160 150 140 130 120 115

Heavy Nonabrasive Material e.g. Coal Capacity, Max. rpm cu.ft/hr 37 125 86 110 150 105 255 100 410 95 590 90 780 85 1030 80 1660 75 2000 70 3400 65 4500 60 5800 55

Heavy Abrasive Material e.g. Ash Capacity, Max. rpm cu.ft/hr … … 46 90 85 85 135 80 200 75 300 75 400 70 516 65 820 60 1200 55 1630 50 2100 45 2860 46

We choose screw diameter= 16 in since its maximum capacity is 3400 which is above 3333.3333 ft3/hr.

b. Flight conveyor

𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 16 𝑖𝑖𝑖𝑖 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 =

𝑎𝑎𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 + 10𝐿𝐿 1000

From the table below, we determine constants a, b Since the coal is being moved horizontally, then angle of inclination = 0 degree 4|Page

CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] Inclination w/ Horizontal 0° 5° 10° 15° 20° 25° 30° 35° 40° Anthracite 0.343 0.42 0.50 0.586 0.66 0.73 0.79 0.85 0.90 a Bituminous 0.60 0.69 0.76 0.83 0.88 0.95 1.02 1.08 1.13 Ashes 0.54 0.62 0.72 0.80 0.85 0.90 0.97 1.03 1.06 Flights & chain supported on blocks 0.03 0.03 0.03 0.029 0.028 0.027 0.026 0.025 0.023 w/c slide directly on the b track Flights supported by 3 0.004 0.004 0.004 0.004 0.004 0.004 0.003 0.003 0.003 ½ -in rollers a = 0.343 Since it was not stated whether flights have blocks or rollers, we will compare their HP; b = 0.03 or 0.004

45° 0.945 1.15 1.10 0.020 0.003

For S, since it was not stated, we will assume it to be S=100 fpm which is the common speed for flight conveyors For W, assume dimensions of 8 x 18 so width of flight= 18 in 𝑊𝑊 =

1 𝑙𝑙𝑙𝑙 (18 𝑖𝑖𝑖𝑖 )(𝟐𝟐) = 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑/𝒇𝒇𝒇𝒇 𝑖𝑖𝑖𝑖 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ. 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

Note: W was multiplied by 2 since it is the weight for both runs Flight Conveyors 4 x 10 to 6 x 18 8 x 18 to 10 x 24 Belt Conveyors

0.5 lb/in. of width per running foot 1.0 lb/in. of width per running foot 1.0 lb/in. of width per running foot

HP using flights & chain supported on blocks w/c slide directly on the track 𝑎𝑎𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 + 10𝐿𝐿 1000 0.343(100)(120) + 0.03(36)(120)(100) + 10(120) = 18.276 𝐻𝐻𝐻𝐻 = 1000

𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 =

HP using flights supported by 3 ½- in rollers 𝐻𝐻𝐻𝐻 =

0.343(100)(120) + 0.004(36)(120)(100) + 10(120) = 7.044 1000

c. Belt Conveyor

𝐹𝐹(𝐿𝐿 + 𝐿𝐿𝑜𝑜 )(𝑇𝑇 + 0.03𝑊𝑊𝑊𝑊) + 𝑇𝑇∆𝑍𝑍 990 For F, since the type of bearing was not stated, we will compare both; F= 0.05 or 0.03 For Lo, we will also compare both so, Lo = 100 or 150 For W, using the table below 𝐻𝐻𝐻𝐻 =

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CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] Flight Conveyors 4 x 10 to 6 x 18 8 x 18 to 10 x 24 Belt Conveyors

0.5 lb/in. of width per running foot 1.0 lb/in. of width per running foot 1.0 lb/in. of width per running foot

W=( 1lb/in width. ft )(in belt width) (2) Note : for W it was multiplied by 2 since the belt conveyor has 2 runs (back and forth) First run (top)

Second run (bottom)

To solve for W, we need to determine first the belt width. We can use the following tables Maximum Lump Size and Speeds for Conveyor Belts Belt width, in

14 16 18 20 14 30 36 42 48 54 60

Maximum lump size,in Uniform With Size 90% fines 2 3 2 ½ 4 3 5 3 ½ 6 4 ½ 8 6 11 8 15 10 18 12 21 14 24 16 28

Belt width, in.

Cu. yd/hr at 100fpm

14 16 18 20 24 30

23.6 31.1 39.6 49.3 72.4 116.7

Cross sectional area of load, ft2 0.11 0.14 0.18 0.22 0.33 0.53 0.78 1.09 1.46 1.90 2.40

Normal Speed, fpm 200 200 250 300 300 350 400 400 400 450 450

Maximum Belt Speeds, fpm Free-flowing material 400 500 500 600 600 700 800 800 800 800 800

Average material 300 300 400 400 500 500 600 600 600 600 600

Abrasive material 250 250 300 300 350 350 400 400 400 400 400

Maximum capacity with materials of Various Bulk Densities, tons/hr at 100 fpm 3 3 25 lb/ft 50 lb/ft 75 lb/ft3 100 lb/ft3 150 lb/ft3 8 16 24 32 48 10 21 31 42 63 13 27 40 54 81 16 33 49 66 99 24 49 73 98 147 39 79 118 158 237 6|Page

CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] 36 42 48 54 60

173.3 242.2 324.4 422.2 533.3

57 82 110 142 180

115 165 220 285 360

172 247 330 427 540

230 330 440 570 720

345 495 660 855 1080

Since the belt width that we will choose should be able to handle the given capacity, we need to determine first the maximum capacity at maximum speed of the different belt widths, thus by ratio and proportion max 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 @ max 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 @100𝑓𝑓𝑓𝑓𝑓𝑓 = 100 𝑓𝑓𝑓𝑓𝑓𝑓 max 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

But, data for capacity @ 100 fpm with bulk density= 60 lb/ft3 is not yet available, thus by interpolation we get Maximum capacity with materials of Various Bulk Densities, tons/hr at 100 fpm 50 lb/ft3 60 lb/ft3 75 lb/ft3 16 24 19.2 21 31 25 27 40 32.2 33 49 39.4 49 73 58.6 79 118 94.6 115 172 137.8 165 247 197.8 220 330 264 285 427 341.8 360 540 432 Anthracite coal is an average material . Thus, using the formula

Belt width, in. 14 16 18

max 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 @ max 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (90% 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓) @100𝑓𝑓𝑓𝑓𝑓𝑓 = max 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 100 𝑓𝑓𝑓𝑓𝑓𝑓 Maximum Speed Average material 300 300 400

Maximum capacity at 100 fpm 19.2 25 32.2

Assume 90% fines Capacity @ 100 fpm 19.2*0.9 =17.28 22.5 28.98

Maximum capacity at maximum Speed 51.84 67.5 115.92

Since the max capacity of 18 in belt is 115.92 which is greater than the feed 100 tons/hr, then we can use this. Belt width =18 in W=( 1lb/in width. ft )(18 in) (2) = 36 lb/ running ft 7|Page

CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] For S, by ratio and proportion 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 400 𝑆𝑆 = 100 115.92 𝑺𝑺 = 𝟑𝟑𝟑𝟑𝟑𝟑. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝒇𝒇𝒇𝒇𝒇𝒇

ΔZ=0 (since horizontal) For Plain Bearings

𝐹𝐹(𝐿𝐿 + 𝐿𝐿𝑜𝑜 )(𝑇𝑇 + 0.03𝑊𝑊𝑊𝑊) + 𝑇𝑇∆𝑍𝑍 990 0.05(120 + 100)(100 + 0.03(36)(345.0656) + 0 𝐻𝐻𝐻𝐻 = 990 𝐻𝐻𝐻𝐻 = 5.2519

𝐻𝐻𝐻𝐻 =

For Anti-Friction Bearings

0.03(120 + 150)(100 + 0.03(36)(345.0656) + 0 990 𝐻𝐻𝐻𝐻 = 3.8673 𝐻𝐻𝐻𝐻 =

Conclusion

Among the three conveyors, the smallest one is the screw conveyor having a diameter of 16 in. However, based on HP, the belt conveyor has the least power requirement. Thus, in terms of operating cost (for power), it is best to choose the belt conveyor. 2. A screw conveyor is to be installed to convey 800 bushels of wheat per hour over a distance of 80 ft. Determine the size (diameter), speed (rpm) and the horsepower requirements for the installation. (1 bushel= 1.2444ft3) GIVEN: 800 bushels of wheat/hour

L=80 ft

REQUIREMENT: Dscrew, S, HP SOLUTION: a. Diameter of Screw To use the table below, convert capacity to ft3/hr first 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 800

𝑓𝑓𝑓𝑓 3 𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑒𝑒𝑒𝑒𝑒𝑒 1.2444𝑓𝑓𝑓𝑓 3 � � = 995.52 1𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑒𝑒𝑒𝑒 ℎ𝑟𝑟 ℎ𝑟𝑟

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CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] Diameter of Screw, in. 3 4 5 6 7 8 9 10 12 14 16 18 20

Light Nonabrasive Material e.g. Grain Capacity, cu.ft/hr

Max. rpm

74 171 304 500 820 1180 1600 2050 3300 4000 7000 9000 12000

250 220 210 200 190 180 175 160 150 140 130 120 115

Since 1180>995.52, we choose 8 in

Heavy Nonabrasive Material e.g. Coal Capacity, Max. rpm cu.ft/hr 37 125 86 110 150 105 255 100 410 95 590 90 780 85 1030 80 1660 75 2000 70 3400 65 4500 60 5800 55

Heavy Abrasive Material e.g. Ash Capacity, Max. rpm cu.ft/hr … … 46 90 85 85 135 80 200 75 300 75 400 70 516 65 820 60 1200 55 1630 50 2100 45 2860 46

𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 8 𝑖𝑖𝑖𝑖

b. Speed in rpm Using the eqn by ratio and proportion,

𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 180 𝑆𝑆 = 995.52 1180 𝑆𝑆 = 151.8590 𝑟𝑟𝑟𝑟𝑟𝑟

c. Horsepower

𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 =

(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) �𝐶𝐶𝐶𝐶𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝, 33,000

𝑙𝑙𝑙𝑙 � (𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ℎ, 𝑓𝑓𝑓𝑓) 𝑚𝑚𝑚𝑚𝑚𝑚

For grain, coefficient =1.3 Since density is not given, from Perry’s HB ρave= 48 lb/ft3. Assume ρave≈ρbulk = 48 lb/ft3 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 800

1ℎ𝑟𝑟 𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑒𝑒𝑒𝑒𝑒𝑒 1.2444𝑓𝑓𝑓𝑓 3 48𝑙𝑙𝑙𝑙 � = 796.416 � �� 3 �� 𝑓𝑓𝑡𝑡 1𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑒𝑒𝑒𝑒 60𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 ℎ𝑟𝑟 𝐻𝐻𝐻𝐻 =

1.3(796.416)(80) = 2.5099 33000

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CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] 6. A belt conveyor is required to deliver gravel at a rate of 175 tons/hr. The conveyor is to be 180ft between centers if pulleys with a rise of 18 ft and discharge over the end. Choose the smallest conveyor that will do the job and calculate the power required to operate the system. Compare using plain bearings and anti-friction bearings. GIVEN: BELT CONVEYOR L=180 ft Δz = 18 ft T= 175 tons/hr

180 ft

18 ft

REQUIRED: a. smallest belt conveyor that can do the job b. HP for plain bearings and anti-friction bearings SOLUTION: a. Width of belt conveyor In order to determine the belt width, we will first need to determine the maximum capacity (at maximum speed) at different belt widths. We can do this by using the formula from ratio and proportion which is max 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 @ max 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 @100𝑓𝑓𝑓𝑓𝑓𝑓 = max 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 100 𝑓𝑓𝑓𝑓𝑓𝑓 Maximum Lump Size and Speeds for Conveyor Belts Belt width, in

14 16 18 20 14 30 36 42 48 54 60

Maximum lump size,in Uniform With Size 90% fines 2 3 2 ½ 4 3 5 3 ½ 6 4 ½ 8 6 11 8 15 10 18 12 21 14 24 16 28

Cross sectional area of load, ft2 0.11 0.14 0.18 0.22 0.33 0.53 0.78 1.09 1.46 1.90 2.40

Normal Speed, fpm 200 200 250 300 300 350 400 400 400 450 450

Maximum Belt Speeds, fpm Free-flowing material 400 500 500 600 600 700 800 800 800 800 800

Average material 300 300 400 400 500 500 600 600 600 600 600

Abrasive material 250 250 300 300 350 350 400 400 400 400 400

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CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] Belt width, in.

Cu. yd/hr at 100fpm

14 16 18 20 24 30 36 42 48 54 60

23.6 31.1 39.6 49.3 72.4 116.7 173.3 242.2 324.4 422.2 533.3

Maximum capacity with materials of Various Bulk Densities, tons/hr at 100 fpm 25 lb/ft3 50 lb/ft3 75 lb/ft3 100 lb/ft3 150 lb/ft3 8 16 24 32 48 10 21 31 42 63 13 27 40 54 81 16 33 49 66 99 24 49 73 98 147 39 79 118 158 237 57 115 172 230 345 82 165 247 330 495 110 220 330 440 660 142 285 427 570 855 180 360 540 720 1080

For gravel (abrasive material), we take the bulk density ρb= 105 lb/ ft3. Since the capacity at 100 fpm for this density is not directly available from the table, we interpolate. Belt Width 14 16 18 20 24 30 36 42 48 54 60

Maximum capacity, tons/hr at 100 fpm 100 lb/ft3 105 lb/ft3 150 lb/ft3 32 33.6 48 42 44.1 63 54 56.7 81 66 69.3 99 98 102.9 147 158 165.9 237 230 241.5 345 330 346.5 495 440 462 660 570 598.5 855 720 756 1080

We can now solve the maximum capacity @ maximum speed. Assume that the feed contains 90% fines, thus

14 16 18 20

max 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑦𝑦 @ max 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝟎𝟎. 𝟗𝟗𝟗𝟗)(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 @100𝑓𝑓𝑓𝑓𝑓𝑓) = 100 𝑓𝑓𝑓𝑓𝑓𝑓 max 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Max Capacity at Max Belt width, in Max speed, fpm Capacity @ 100 fpm Speed, tons/hr 250 33.6 75.6 250 44.1 99.225 300 56.7 153.09 300 69.3 187.11

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CHE 511A: SEPARATION PROCESSES AND INTRO TO PARTICULATE TECHNOLOGY [PARTICULATE TECHNOLOGY] Since the maximum capacity at 20 in belt width, 187.11 tons/hr, is higher than the actual capacity, 175 tons/hr, it will be able to deliver the gravel. Thus 𝑆𝑆𝑚𝑚𝑎𝑎𝑙𝑙𝑙𝑙𝑒𝑒𝑠𝑠𝑡𝑡 𝑏𝑏𝑒𝑒𝑙𝑙𝑡𝑡 𝑤𝑤𝑖𝑖𝑑𝑑𝑡𝑡ℎ 𝑡𝑡ℎ𝑎𝑎𝑡𝑡 𝑐𝑐𝑎𝑎𝑛𝑛 𝑑𝑑𝑜𝑜 𝑡𝑡ℎ𝑒𝑒 𝑗𝑗𝑜𝑜𝑏𝑏 = 20 𝑖𝑖𝑛𝑛

b. HP for plain bearings and anti-friction bearings 𝐹𝐹(𝐿𝐿 + 𝐿𝐿𝑜𝑜 )(𝑇𝑇 + 0.03𝑊𝑊𝑊𝑊) + 𝑇𝑇∆𝑍𝑍 𝐻𝐻𝐻𝐻 = 990 For plain bearings F=0.05, Lo=100 For anti-friction bearings F=0.03; Lo=150 𝑊𝑊 = Flight Conveyors 4 x 10 to 6 x 18 8 x 18 to 10 x 24 Belt Conveyors

1.0𝑙𝑙𝑏𝑏 𝑙𝑙𝑏𝑏 (20 𝑖𝑖𝑛𝑛)(2𝑟𝑟𝑢𝑢𝑛𝑛𝑠𝑠) = 40 𝑖𝑖𝑛𝑛 𝑜𝑜𝑓𝑓 𝑤𝑤𝑖𝑖𝑑𝑑𝑡𝑡ℎ. 𝑟𝑟𝑢𝑢𝑛𝑛𝑛𝑛𝑖𝑖𝑛𝑛𝑔𝑔 𝑓𝑓𝑜𝑜𝑜𝑜𝑡𝑡 𝑟𝑟𝑢𝑢𝑛𝑛𝑛𝑛𝑖𝑖𝑛𝑛𝑔𝑔 𝑓𝑓𝑜𝑜𝑜𝑜𝑡𝑡 0.5 lb/in. of width per running foot 1.0 lb/in. of width per running foot 1.0 lb/in. of width per running foot

For S, se use the formula from ratio and proportion 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 300 S = 175 187.11 𝑆𝑆 = 280.5836 fpm for Plain Bearings 0.05(180 + 100)�175 + 0.03(40)(280.5836)� + 175(18) = 10.4180 990 for Anti-Friction Bearings 0.03(180 + 150)�175 + 0.03(40)(280.5836)� + 175(18) = 8.2988 𝐻𝐻𝑃𝑃 = 990 𝐻𝐻𝑃𝑃 =

Thus, in terms of power requirement, it is better to use anti-friction bearings

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