10.1 SECTION 10 MATERIALS HANDLING Choosing Conveyors and Elevators for Specific Materials Transported 10.1 Determining Equipment Design Parameters for Overhead Conveyors 10.4 Bulk Material Elevator and Conveyor Selection 10.9 Screw Conveyor Power Input and Capacity 10.13 Design and Layout of Pneumatic Conveying Systems 10.15 CHOOSING CONVEYORS AND ELEVATORS FOR SPECIFIC MATERIALS TRANSPORTED Determine the maximum allowable product weight between supports that can be handled by a belt conveyor at any one time when it conveys 100,000 lb /h (4540 kg/h) of pulverized aluminum oxide in an abrasive state at a belt speed of 50 ft/ min (15.2 m/min) with a center-to-center distance of 32 ft (9.75 m) between belt supports. Compare this capacity with that at belt speeds of 150, 250, and 350 fpm (45.7, 76.2, and 106.7 m /min). Choose the type of conveyor and elevator to handle this material under the conditions given. Calculation Procedure: 1. Find the maximum allowable product weight at the given belt speed Use the relation, P ϭ KC /60 S, where P ϭ maximum product weight on the belt at any one time, lb (kg) between belt supports; K ϭ load, lb /h (kg / h); C ϭ center- to-center distance between belt supports, ft (m); S ϭ belt speed, ft/min (m / min). Substituting, we have, P ϭ (100,000)(32)/50 (60) ϭ 1066.7 lb (484.3 kg). 2. Determine the maximum allowable product weight at other belt speeds Typical conveyor belt speeds vary from a low of 150 ft/min (45.7 m /min) to a high 800 ft / min (243.8 m / min), depending on belt width, type of material con- veyed, belt construction, etc. For this belt, using the data given earlier, P ϭ (100,000)(32) /150(60) ϭ 355.6 lb (161.4 kg) when the speed is 150 ft/min (45.7 m /min). Likewise for the two higher speeds, respectively, P ϭ (100,000)(32) / 250(60) ϭ 213.3 lb (96.9 kg); P ϭ (100,000)(32)/(350(60) ϭ 152.4 lb (69.2 kg). 3. Verify the type of conveyor and elevator to use With such a wide variety of conveyors and elevators to choose from, it is wise to verify the choice before a final decision is made. Table 1, presented by Harold V. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 10.2 TABLE 1 Preferred Types of Conveyors and Elevators for Bulk and Packaged Materials Material Physical condition Av wt / volume lb/ft 3 kg / m 3 Reaction on conveyor Preferred conveyors* Preferred elevators* Comment Acid phosphate Alum Aluminum oxide Ammonium nitrate Ammonium nitrate Arsenic salts Ashes: dry wet Bone meal Borax Bran Brewers grains, hot Carbon black (pellets) Cement, dry Clays Coal: anthracite steam sizes bituminous, lump bituminous, slack Chalk Coffee beans Copra, ground Cork, ground Corn, shelled Cottonseed Cullet Flaxseed Flue dirt Fly ash, clean Glass batch Glue Graphite (flour) Gravel Gypsum Heavy ores Hog fuel Lead salts Lime, pebble Damp Granular Pulv. Pulv. Damp Pulv. Granular Sticky Pulv. Pulv. Granular Granular Granular Pulv. Pulv. Lumpy Granular Lumpy Granular Pulv. Granular Pulv. Pulv. Granular Granular Granular Granular Pulv. Pulv. Granular Granular Pulv. Granular Pulv. Lumpy Stringy Pulv. Granular 90 60–65 60 62 65 ϩ 100 35–40 45–50 55–60 50–70 16–20 55 40 90–118 35–60 50–54 50–60 50–60 50–60 70–75 40–45 40 5–15 45 35–40 80–100 45 100 35–45 80 ϩ 45 40 95–135 60 100 ϩ 15–30 60–150 55–80 1,440 960–1,040 960 990 1,040 ϩ 1,600 560–640 720–800 880–960 800–1,120 260–320 880 640 1,440–1,890 560–960 800–860 800–960 800–960 800–960 1,120–1,200 640–720 640 80–240 720 560–640 1,280–1,600 720 1,600 560–720 1,280 ϩ 720 640 1,520–2,160 960 1600 ϩ 240–480 960–2,400 880–1,280 Adheres Abrasive Abrasive Hygroscopic Adheres Heavy Abrasive Abrasive Abrasive Corrosive Adheres May be abra- sive Abrasive shell Sometimes sticky Abrasive Shell abrasive Abrasive Mild abrasive Abrasive Lubricant Abrasive May jam a, e a, b, c, e a, e b, c, e c, e c, e d, f f a, b, c, d, e a, b, c, d, e a, b, c, d, e c, e a, e a, c, d, e a, b, c, e a, b, c, e a, b, c, d, e a, b, e a, b, c, d, e a, b, c, d, e a, c, e a, b, c, e a, b, c, d, e a, c, e a, b, c, d, e a, b, e a, b, c, d, e b, d, e, f a, b, c, d, e a, b, e a, c, e a, b, c, d, e a, e, f a, b, c, e a, b, f a, b, d, e a, b, c, e a, b, c, e b g, b g g, b g, b g, b b b g, b, c g, b g, b g, b g, b g, b g, b g, b g, b, c b g, b, c g, b, c g, b g, b g, b g, b, c g, b g, b g, b, c g, b g, b, c g, b g, b, c g, b, c g, b g, b g, b g Sticky Explosive Sticky Poisonous Dusty Corrosive Sometimes sticky Fragile Packs Sluggish Sluggish Fragile Sticky Sluggish Corrosive Free-flowing Free-flowing Keep cool May be tough Poisonous Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MATERIALS HANDLING 10.3 Limestone dust Malt Manufactured products Merchandise: Packaged Garments Metallic dusts Mica, pulverized Molybdenum conc’ts Petroleum coke Pumice Quartz (ground) Rubber scrap Salt: coarse cake Sand: dry damp Sawdust Sewage sludge Silica flour Soap flakes Soda ash: light heavy Soybean flour Starch Sugar: raw refined Sulfur Talc Tobacco stems Wheat Wood chips Zinc oxide Zinc sulfate Pulv. Dry Boxed Boxed Hanging Pulv. Pulv. Pulv. Lumpy Pulv. Pulv Stringy Granular Pulv. Granular Granular Granular Pulv. Pulv. Granular Pulv. Pulv. Pulv. Pulv. Granular Granular Pulv. Pulv. Stringy Granular Granular Pulv. Pulv. 85–95 45 1–200 15 5 50–100 20–30 110 42 45 110 50 50 75–95 90–110 90–110 15–20 60 80 10–20 25–35 55–65 30 30–40 55–65 50–55 55 50–60 25 48 18–20 20–35 70 1,360–1,520 720 16–3,200 240 80 800–1,600 320–480 1,760 670 720 1,760 800 800 1,200–1,520 1,440–1,760 1,440–1,760 240–320 960 1,280 160–320 400–560 880–1,040 480 480–640 880–1,040 800–880 880 800–960 400 770 290–320 320–560 1,120 Sluggish Abrasive May be sticky Abrasive Free-flowing Abrasive Mild abrasive Mild abrasive Very abrasive Sluggish Hygroscopic Flows freely Abrasive Sticky Sticky if wet Sluggish Fragile Flows freely Flows freely Sticky Sticky Corrosive if wet Mild abrasive Sluggish Free-flowing May arch May pack May pack a, b, e a, b, c, d, e a, i, j a, b, i, j i, j a, b, c, d, e a, b, c, d, e a, b, d a, b, c, e a, b, c, d, e a, b, c, d a, b, e a, b, c, e a, b, c, d, e a, e, f a, e, f a, b, c, d, e a, b, e, f a, d, e a, c, e a, b, c, d, e a, b, c, d, e a, b, c, e a, b, c, e a, b, c, e a, b, c, e a, b, c, e a, b, c, d, e a, b, d, e a, c, d, e a, c, d, e a, b, c, d, e a, b, c, d, e g, b g, b g, b, c b g, b g, b, c g g, b g, b g, b g, b g, b g, b, c g g g g, c g, c g, c g, c g g g, b g, b g g, c g, c g g Sometimes difficult Dusty Sticky Polisher Difficult Corrosive if wet Abrasive Abrasive Sticky if hot Caustic Caustic Explosive dust Explosive dust Handle gently Explosion risk Adheres to metal Keep clean Corrosive if wet Avoid discoloration *Explanation of letter symbols: a—belt, b—flight, c—continuous flow, d—pneumatic, e—screw, f—drag chain, g—belt and bucket, h—chain and bucket, i—overhead straight power, j—overhead power and free. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MATERIALS HANDLING 10.4 PLANT AND FACILITIES ENGINEERING Hawkins, Manager, Product Standards and Services, Columbus McKinnon Corpo- ration, lists preferred types of conveyors and elevators for a variety of materials in both bulk and packaged forms. Entering this table at aluminum oxide in pulverized form shows that a belt or screw conveyor is preferred, while a flight elevator is recommended for vertical lifts of this material. While Table 1 gives general recommendations, the engineer should remember that a careful economic study is required to keep the capital investment to the minimum consistent with safe and dependable conveying of the material. Along with capital cost, the operating and maintenance costs must also be evaluated before a final choice of the conveyor and elevator is made. Related Calculations. Choose the belt length to accommodate the maximum expected product capacity. Belt speed should be compatible with the process equip- ment served and with the other materials-handling equipment associated with the conveyor belt. Belt conveyors are suitable for bulk materials of many types. However, char- acteristics of the material conveyed must be considered before a final choice of the belting material is made. Thus, as outlined by K. W. Tunnell Company: (a) Material stickiness may prevent materials handled from discharging completely from the conveyor belt, or may interfere with the belt drive components: motors, chains, etc. (b) Ambient temperatures exceeding 150 ЊF (83ЊC) could cause deterioration or dam- age to the belt materials. (c) Chemical reactions between the conveyed product and the belt material can cause damage. Thus, oils, chemicals, fats, and acids can dam- age belts. (d) Excessively large lump size may require an oversize belt system to handle the conveyed product safely. One way around these problems is use of metal-belt conveyors. These are similar in design to conventional rubber and composite-material conveyor belts except that their surface is made of woven or solid metal. Popular materials include carbon steel, galvanized steel, stainless steel, and other metals and alloys. With today’s emphasis on environmental and safety aspects of engineering de- cisions, it is wise for the design engineer to refer to the appropriate codes and specifications governing the particular type of equipment being considered. Thus, in the materials handling field, ANSI B 20.1 ‘‘Conveyors, Cableways, and Related Equipment’’ should be consulted before any final design choices are made. Likewise, state and city codes should be checked before a firm equipment se- lection decision. In certain instances the local code may be more restrictive than the national code. OSHA—Occupational Safety and Health Administration— regulations are important where human safety is involved. Since these regulations vary so widely with material handled, type of equipment used, and location, no generalizations about them can be made other than to recommend strongly that the regulations be studied and followed. DETERMINING EQUIPMENT DESIGN PARAMETERS FOR OVERHEAD CONVEYORS Select suitable equipment for the overhead conveyor shown in Fig. 1. Determine the total chain pull and horsepower required if the conveyor is 700 ft (213 m) long, the coefficient of friction is 0.03, the total chin pull is 60 lb/ft (89.4 kg/m) com- prised of the components detailed below. Use the design method presented the K. W. Tunnell Company. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MATERIALS HANDLING MATERIALS HANDLING 10.5 SI values 150' 45.7 m 3' 0.91 m 130' 39.6 m 5' 1.5 m 3' 0.91 m 8' 2.4 m 10' 3.0 m 4' 1.2 m 30' 9.1 m 280' 85.3 m 20' 6.1 m 40' 12.2 m 60' 18.3 m FIGURE 1 (a) Plan of conveyor layout. (b) Elevation of conveyor layout. Calculation Procedure: 1. From the process flow charts, determine all the operations to be serviced by the conveyor The process flow charts will be provided by the manufacturing enginner or the process engineer, depending on the type of installation the conveyor is serving. To assist in the conveyor layout and design a listing of each process served by the conveyor should be prepared. 2. Determine the path of the conveyor on a scaled plant layout Draw a plan and elevation of the conveyor, Fig. 1, on a scaled layout of the plant. Show all obstructions the conveyor will encounter, such as columns, walls, ma- Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MATERIALS HANDLING 10.6 PLANT AND FACILITIES ENGINEERING FIGURE 2 Clearance design for overhead conveyor turns and rises. chinery, and work aisles. Indicate the loading and unloading zones, probable drive location, and passage through walls. 3. Develop a vertical elevation to determine incline and decline dimensions Show the inclines and declines, and their dimensions, Fig. 1b. A three-dimensional view of the installation can be prepared at this point to help people better visualize the final installation and the various routes of the conveyor. 4. Determine the material movement rate, unit load size, spacing, and carrier design for the conveyor Information for these variables can be obtained from the flow chart and the per- sonnel in charge of the process being served by the conveyor. It is important that the conveyor be designed for the maximum anticipated load and material size. 5. Modify turn radii to provide adequate clearances Prepare drawings showing needed load spacing on turns, Fig. 2. Without adequate clearnaces, the conveyor may not provide the desired transportation capability needed to serve properly the process for which the conveyor is being designed. 6. Design the load spacing for clearances on inclines and declines As inclines and declines get steeper, Fig. 2 load spacing has to be increased to provide a constant clearance or separation between loads. Table 2 gives selected clearances on inclined track for overhead conveyors for a given separation at various incline angles. 7. Redraw the conveyor path and vertical elevation views using newly deter- mined radii and incline information Show the new radii and incline information as determined by the redesign of the system layout, Figs. 1 and 2. 8. Compute the chain pull in the conveyor The chain pull is the total weight of the chain, trolleys, Fig. 3, and other compo- nents, plus the weight of the carriers and load. Thus, for the given system, the tenative chain pull can be found from C p ϭ L ϫ P L ϫ ƒ, where C p ϭ tentative Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MATERIALS HANDLING MATERIALS HANDLING 10.7 TABLE 2 Selected Load Clearance on Inclined Track for Overhead Conveyors Load spacing, in Incline angle, deg 10 20 30 40 50 60 Horizontal centers, in 12 16 18 24 11 7 ⁄ 8 15 3 ⁄ 4 17 3 ⁄ 4 23 5 ⁄ 8 11 1 ⁄ 4 15 1 ⁄ 8 17 22 5 ⁄ 8 10 3 ⁄ 8 13 7 ⁄ 8 15 5 ⁄ 8 20 7 ⁄ 8 9 1 ⁄ 4 12 1 ⁄ 4 13 7 ⁄ 8 18 3 ⁄ 8 7 3 ⁄ 4 10 3 ⁄ 8 11 3 ⁄ 8 15 1 ⁄ 2 6 8 9 12 cm Horizontal centers, cm 30.5 40.6 45.7 60.9 29.9 40.0 45.1 60.0 28.6 38.7 43.2 57.5 26.4 35.2 39.7 53.0 23.5 31.1 35.2 46.7 19.7 26.4 28.9 39.4 15.2 20.3 22.9 30.5 chain pull, lb (kg); L ϭ conveyor length, ft (m); P L ϭ chain load, lb/ ft (kg/m); ƒ ϭ coefficient of friction ϭ 0.03 for this installation. The given chain load of 60 lb/ft (89.4 kg/m) is comprised of 10.0 lb/ ft (14.9 kg/m) for the chain and trolleys, 12.5 lb/ft (18.6 kg/m) for the carriers, and 37.5 lb/ft (55.9 kg/m) for the line load. Substituting, C p ϭ 700(60.0)(0.03) ϭ 1260 lb (572 kg). For this initial calculation, inclines and declines are assumed to be level sections if the number of declines balances out the number of inclines. However, for each additional incline, the total line load rise has to be added to determine the total chain pull. If, for example, a vertical incline in this installation raises the line load 8 ft ((2.4 m), then the additional chain pull ϭ 37.5 lb line load ϫ 8ftϭ 300 lb (136.2 kg). The total chain pull then becomes 1260 ϩ 300 ϭ 1560 lb (708.2 kg). 9. Select the tenative conveyor size based on the trolley load and chain pull Use the manufacturer’s data to choose the tenative conveyor size. In making your choice, try to comform to standard conveyor sizes and layouts because this will reduce the capital cost of the installation. Further, the installation will probably be made faster because there will be less customizing required. 10. Select vertical curve radii Again, work with the standard radii available from the manufacturer, if possible. This will reduce installation costs and time. 11. Determine the conveyor power requirements and drive locations Make point-to-point calculations of the chain pull around the complete path of the conveyor, Fig. 1. Use the following equations to compute point-to-point chain pull: (a) Pull for each horizontal run, lb (kg), P H ϭ XWL, where X ϭ 0.02 for standard ball-bearing trolleys; W ϭ total moving weight, lb/ft (kg/m), empty or loaded as the case may be; L ϭ length of straight run, ft (m). (b) Pull for each traction wheel or roller turn, lb (kg), P T ϭ YP, where Y ϭ 0.02 for traction wheel or roller turn; P ϭ pull at turn, lb (kg). (c) Pull for each vertical curve, lb (kg), P V ϭ XWS ϩ ZP ϩ HW(1 ϩ Z), where X ϭ 0.02 for standard ball- bearing trolleys; W ϭ total moving weight, lb / ft (kg/ m); S ϭ horizontal span of Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MATERIALS HANDLING 10.8 PLANT AND FACILITIES ENGINEERING FIGURE 3 Power- and free-trolley overhead conveyors. vertical curve, ft (m); H ϭ total change of level of conveyor, ft (m) (plus, when conveyor is traveling up the curve; minus when conveyor is traveling down the curve); Z ϭ 0.03 for 30Њ incline; 0.045 for 45Њ incline; 0.06 for 60Њ incline; 0.09 for 90 Њ incline; P ϭ pull at start of curve, lb (kg). Drive horsepower (kW) can be calculated from: Drive hp ϭ (drive capacity, lb)(maximum speed, ft / min)/0.6(33,000). Thus, if the drive capacity required is 6000 lb, the maximum speed is 50 ft/min, the horsepower required ϭ (6000)(50) /0.6(33,000) ϭ 15.2 hp (11.3 kW). 12. Design the conveyor supports and superstructures Refer to the manufacturer’s data for suitable supports and superstructures. It is best, if possible, to use standard supports and superstructures. This will save money and time for the firm owning the plant being fitted with the conveyor. 13. Design guards required by laws and codes Federal, state, and applicable codes require guards of various types under high trolley runs, particularly over aisles and work areas. Guard panels are normally Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MATERIALS HANDLING MATERIALS HANDLING 10.9 made from woven or welded wire mesh with structural angles and channels to suit the size and weight of the material being handled. Related Calculations. The general procedure presented here is valid for over- head conveyors handling a variety of materials: manufactured goods, parts for as- sembly, raw materials, etc., in plants in many different industries. Since conveyor layout, sizing, and safety design are a specialized skill, the engineer should consult carefully with the conveyor manufacturer. The manufacturer’s wide experience will be most helpful to the engineer in achieving an economical and safe design for the installation being considered. The steps, illustrations, and table in this procedure are the work of the K. W. Tunnell Company. SI values were added by the handbook editor. BULK MATERIAL ELEVATOR AND CONVEYOR SELECTION Choose a bucket elevator to handle 150 tons/h (136.1 t /h) of abrasive material weighing 50 lb/ft 3 (800.5 kg/m 3 ) through a vertical distance of 75 ft (22.9 m) at a speed of 100 ft/min (30.5 m/min). What hp input is required to drive the elevator? The bucket elevator discharges onto a horizontal conveyor which must transport the material 1400 ft (426.7 m). Choose the type of conveyor to use, and determine the required power input needed to drive it. Calculation Procedure: 1. Select the type of elevator to use Table 3 summarizes the various characteristics of bucket elevators used to transport bulk materials vertically. This table shows that a continuous bucket elevator would be a good choice, because it is a recommended type for abrasive materials. The second choice would be a pivoted bucket elevator. However, the continuous bucket type is popular and will be chosen for this application. 2. Compute the elevator height To allow for satisfactory loading of the bulk material, the elevator length is usually increased by about 5 ft (1.5 m) more than the vertical lift. Hence, the elevator height ϭ 75 ϩ 5 ϭ 80 ft (24.4 m). Related Calculations. The procedure given here is valid for conveyors using rubber belts reinforced with cotton duck, open-mesh fabric, cords, or steel wires. It is also valid for stitched-canvas belts, balata belts, and flat-steel belts. The re- quired horsepower input includes any power absorbed by idler pulleys. Table 6 shows the minimum recommended belt widths for lumpy materials of various sizes. Maximum recommended belt speeds for various materials are shown in Table 5. 3. Compute the required power input to the elevator Use the relation hp ϭ 2CH /1000, where C ϭ elevator capacity, tons/h; H ϭ ele- vator height, ft. Thus, for this elevator, hp ϭ 2(150)(80) / 1000 ϭ 24.0 hp (17.9 kW). The power input relation given above is valid for continuous-bucket, centrifugal- discharge, perfect-discharge, and super-capacity elevators. A 25-hp (18.7-kW) mo- tor would probably be chosen for this elevator. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MATERIALS HANDLING 10.10 PLANT AND FACILITIES ENGINEERING TABLE 3 Bucket Elevators 4. Select the type of conveyor to use Since the elevator discharges onto the conveyor, the capacity of the conveyor should be the same, per unit time, as the elevator. Table 4 lists the characteristics of various types of conveyors. Study of the tabulation shows that a belt conveyor would prob- ably be best for this application, based on the speed, capacity, and type of material it can handle. Hence, it will be chosen for this installation. 5. Compute the required power input to the conveyor The power input to a conveyor is composed of two portions: the power required to move the empty belt conveyor and the power required to move the load horizontally. Determine from Fig. 4 the power required to move the empty belt conveyor, after choosing the required belt width. Determine the belt width from Table 5. Thus, for this conveyor, Table 5 shows that a belt width of 42 in (106.7 cm) is required to transport up to 150 tons/ h (136.1 t/h) at a belt speed of 100 ft / min (30.5 m/min). [Note that the next larger capacity, 162 tons /h (146.9 t/h), is used when the exact capacity required is not tabulated.] Find the horsepower required to drive the empty belt by entering Fig. 4 at the belt distance between centers, 1400 ft (426.7 m), and projecting vertically upward to the belt width, 42 in (106.7 cm). At the left, read the required power input as 7.2 hp (5.4 kW). Compute the power required to move the load horizontally from hp ϭ (C/ 100)(0.4 ϩ 0.00345L), where L ϭ distance between conveyor centers, ft; other symbols as before. For this conveyor, hp ϭ (150 /100)(0.4 ϩ 0.00325 ϫ 1400) ϭ 6.83 hp (5.1 kW). Hence, the total horsepower to drive this horizontal conveyor is 7.2 ϩ 6.83 ϭ 14.03 hp (10.5 kW). The total horsepower input to this conveyor installation is the sum of the elevator and conveyor belt horsepowers, or 14.03 ϩ 24.0 ϭ 38.03 hp (28.4 kW). Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MATERIALS HANDLING [...]... (mmH2O) per 10 0 ft (30.5 m) of each duct by entering with the air quantity and diameter of that duct Enter the frictional resistance thus found in column 12 , Table 11 Compute actual friction in each duct by multiplying the friction per 10 0 ft (30.5 m) of duct, column 12 , Table 11 , by the total duct length, column 11 Ϭ 10 0 Thus for duct run A, actual friction ϭ 5.4 (10 / 10 0) ϭ 0.54 in (13 .7 mm) H2O... Conveyor Characteristics MATERIALS HANDLING 10 .11 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website MATERIALS HANDLING 10 .12 PLANT AND FACILITIES ENGINEERING TABLE 5 Capacities of Troughed Rest [tons / h (t / h) with Belt Speed of 10 0 ft... Terms of Use as given at the website TABLE 11 (Continued ) MATERIALS HANDLING 10 .18 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website MATERIALS HANDLING MATERIALS HANDLING 10 .19 TABLE 12 Recommended Exhaust Air Quantities TABLE 13 Duct... 11 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website TABLE 11 Exhaust System Design Calculations MATERIALS HANDLING 10 .17 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006... of each elbow in the duct runs in column 10 , Table 11 For convenience, assume that the equivalent length of an elbow is 12 times the duct diameter in ft Thus, an elbow in a 6-in (15 2.4-mm) diameter duct has an equivalent resistance of (6-in diameter / [ (12 in / ft) (12 )]) ϭ 6 ft (1. 83 m) of straight duct When making this calculation, assume that all elbows have a radius equal to twice the diameter of. .. 90Њ elbows Note that branch ducts are usually arranged to enter the main duct at an angle of 45Њ or less These assumptions are valid for all typical industrial exhaust systems and pneumatic conveying systems Find the total equivalent length of each duct by taking the sum of columns 9 and 10 , Table 11 , horizontally, for each duct run Enter the result in column 11 , Table 11 7 Determine the actual friction... Conveyors of this type can be sloped upward to angles of 35Њ with the horizontal However, the capacity of the conveyor decreases as the angle of inclination is increased Thus the reduction in capacity at a 10 Њ inclination is 10 percent over the horizontal capacity; at 35Њ the reduction is 78 percent The capacities of screw and spiral conveyors are generally stated in ft3 / h (m3 / h) of various classes of. .. H2O Add the filter resistance to the main and branch duct resistance as shown in Table 11 Find the sum of each column in the table, as shown This is the total resistance in each branch, inH2O, Table 11 10 Balance the exhaust system Inspection of the lower part of Table 11 shows that the computed branch resistances are unequal This condition is usually encountered during system design To balance the... this is slightly less than 4.5 in (11 4.3 mm) H2O, a fan developing 4.5 in (11 4.3 mm) H2O static pressure will be chosen A 10 percent safety factor is usually applied to these values, giving a capacity of 3600 ft3 / min (10 1.9 m3 / min) and a static pressure of 5.0 in (12 7 mm) H2O for this system 12 Select the duct material and thickness Galvanized sheet steel is popular for industrial exhaust systems,... balance can be obtained by decreasing the size of ducts E and F to 4.75 in (12 0.7 mm) and 4.375 in (11 1 .1 mm), respectively Duct B would be increased to 6.5 in (16 5 .1 mm) in diameter 11 Choose the exhaust fan capacity and static pressure Find the required exhaust fan capacity in ft3 / min from the sum of the airflows in the ducts, A through H, column 3, Table 11 , or 3300 ft3 / min (93.5 m3 / min) Choose . in 12 16 18 24 11 7 ⁄ 8 15 3 ⁄ 4 17 3 ⁄ 4 23 5 ⁄ 8 11 1 ⁄ 4 15 1 ⁄ 8 17 22 5 ⁄ 8 10 3 ⁄ 8 13 7 ⁄ 8 15 5 ⁄ 8 20 7 ⁄ 8 9 1 ⁄ 4 12 1 ⁄ 4 13 7 ⁄ 8 18 3 ⁄ 8 7 3 ⁄ 4 10 3 ⁄ 8 11 3 ⁄ 8 15 1 ⁄ 2 6 8 9 12 cm Horizontal centers,. sulfate Pulv. 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