Handbook of Mechanical Engineering Calculations ar Episode 1 Part 10 pot

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.
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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
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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.
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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.
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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-
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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
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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
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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
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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.
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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).
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10.11
TABLE 4 Conveyor Characteristics
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MATERIALS HANDLING
10.12 PLANT AND FACILITIES ENGINEERING
TABLE 5 Capacities of Troughed Rest [tons / h (t / h) with Belt Speed of
100 ft / min (30.5 m / min)]
TABLE 6 Minimum Belt Width for Lumps
FIGURE 4 Horsepower (kilowatts) required to move an
empty conveyor belt at 100 ft / min (30.5 m / min).
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MATERIALS HANDLING 10.13
TABLE 7 Maximum Belt Speeds for Various Materials
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 adsorbed by idler pulleys.
Table 5 shows the minimum recommended belt widths for lumpy materials of
various sizes. Maximum recommended belt speeds for various materials are shown
in Table 6.
When a conveyor belt is equipped with a tripper, the belt must rise about 5 ft
(1.5 m) above its horizontal plane of travel.
This rise must be included in the vertical-lift power input computation. When
the tripper is driven by the belt, allow 1 hp (0.75 kW) for a 16-in (406.4-mm) belt,
3 hp (2.2 kW) for a 36-in (914.4-mm) belt, and 7 hp (5.2 kW) for a 60-in (1524-
mm) belt. Where a rotary cleaning brush is driven by the conveyor shaft, allow
about the same power input to the brush for belts of various widths.
SCREW CONVEYOR POWER INPUT
AND CAPACITY
What is the required power input for a 100-ft (30.5-m) long screw conveyor han-
dling dry coal ashes having a maximum density of 40 lb/ ft
3
(640.4 kg /m
3
) if the
conveyor capacity is 30 tons/h (27.2 t/h)?
Calculation Procedure:
1. Select the conveyor diameter and speed
Refer to a manufacturer’s engineering data or Table 8 for a listing of recommended

screw conveyor diameters and speeds for various types of materials. Dry coal ashes
are commonly rated as group 3 materials, Table 9, i.e., materials with small mixed
lumps with fines.
To determine a suitable screw diameter, assume two typical values and obtain
the recommended rpm from the sources listed above or Table 8. Thus, the maximum
rpm recommended for a 6-in (152.4-mm) screw when handling group 3 material is
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MATERIALS HANDLING
10.14 PLANT AND FACILITIES ENGINEERING
TABLE 8 Screw Conveyor Capacities and Speeds
TABLE 9 Material Factors for Screw Conveyors
90, as shown in Table 8; for a 20-in (508.0-mm) screw, 60 r / min. Assume a 6-in
(152.4-mm) screw as a trial diameter.
2. Determine the material factor for the conveyor
A material factor is used in the screw conveyor power input computation to allow
for the character of the substance handled. Table 9 lists the material factor for dry
ashes as F
ϭ 4.0. Standard references show that the average weight of dry coal
ashes is 35 to 40 lb /ft
3
(640.4 kg/m
3
).
3. Determine the conveyor size factor
A size factor that is a function of the conveyor diameter is also used in the power
input computation. Table 10 shows that for a 6-in (152.4-mm) diameter conveyor
the size factor A
ϭ 54.

4. Compute the required power input to the conveyor
Use the relation hp ϭ 10
Ϫ
6
(ALN ϩ CWLF), where hp ϭ hp input to the screw
conveyor head shaft; A
ϭ size factor from step 3; L ϭ conveyor length, ft; N ϭ
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MATERIALS HANDLING
MATERIALS HANDLING 10.15
TABLE 10 Screw Conveyor Size Factors
conveyor rpm; C ϭ quantity of material handled, ft
3
/h; W ϭ density of material,
lb/ft
3
; F ϭ material factor from step 2. For this conveyor, given the data listed
above, hp
ϭ 10
Ϫ
6
(54 ϫ 100 ϫ 60 ϩ 1500 ϫ 40 ϫ 100 ϫ 4.0) ϭ 24.3 hp (18.1
kW). With a 90 percent motor efficiency, the required motor rating would be
24.3/0.90
ϭ 27 hp (20.1 kW). A 30-hp (22.4-kW) motor would be chosen to drive
this conveyor. Since this is not an excessive power input, the 6-in (152.4-mm)
conveyor is suitable for this application.
If the calculation indicates that an excessively large power input, say 50 hp (37.3

kW) or more, is required, then the larger-diameter conveyor should be analyzed. In
general, a higher initial investment in conveyor size that reduces the power input
will be more than recovered by the savings in power costs.
Related Calculations. Use the procedure given here for screw or spiral con-
veyors and feeders handling any material that will flow. The usual screw or spiral
conveyor is suitable for conveying materials for distances up to about 200 ft (60.9
m), although special designs can be built for greater distances. Conveyors of this
type can be sloped upward to angles of 35
Њ 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 35
Њ the reduction is 78 percent.
The capacities of screw and spiral conveyors are generally stated in ft
3
/h (m
3
/
h) of various classes of materials at the maximum recommended shaft rpm. As the
size of the lumps in the material conveyed increases, the recommended shaft rpm
decreases. The capacity of a screw or spiral conveyor at a lower speed is found
from (capacity at given speed, ft
3
/h) [(lower speed, r /min)/(higher speed, r / min)].
Table 8 shows typical screw conveyor capacities at usual operating speeds.
Various types of screws are used for modern conveyors. These include short-
pitch, variable-pitch, cut flights, ribbon, and paddle screws. The procedure given
above also applies to these screws.
DESIGN AND LAYOUT OF PNEUMATIC

CONVEYING SYSTEMS
A pneumatic conveying system for handling solids in an industrial exhaust instal-
lation contains two grinding-wheel booths and one lead each for a planer, sander,
and circular saw. Determine the required duct sizes, resistance, and fan capacity for
this pneumatic conveying system.
Calculation Procedure:
1. Sketch the proposed exhaust system
Make a freehand sketch, Fig. 5 of the proposed system. Show the main and branch
ducts and the booths and hoods. Indicate all major structural interferences, such as
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MATERIALS HANDLING
10.16 PLANT AND FACILITIES ENGINEERING
FIGURE 5 Exhaust system layout.
building columns, deep girders, beams, overhead conveyors, piping, etc. Draw the
layout approximately to scale.
Mark on the sketch the length of each duct run. Avoid, if possible, vertical drops
or rises in the main exhaust duct between the hoods and the fan. Do this by locating
the main duct centerline 10 ft (3 m) or so above the finished floor.
Number each hood or booth, and give each duct run an identifying letter. Al-
though it is not absolutely necessary, it is more convenient during the design process
to have the hoods in numerical order and the duct runs in alphabetical order.
2. Determine the required air quantities and velocities
Prepare a listing, columns 1 and 2, Table 11, of the booths, hoods, and duct runs.
Enter the required air quantities and velocities for each booth or hood and duct in
Table 11, columns 3 and 4. Select the air quantities and velocities from the local
code covering industrial exhaust systems, if such a code is available. If a code does
not exist, use the ASHRAE Guide or Table 12.
Use extreme care in selecting the air quantities and velocities, because insuffi-

cient flow may cause dangerous atmospheric conditions. Harmful process wastes
in the form of dust, gas, or moisture may injure plant personnel.
3. Size the main and branch ducts
Determine the required duct area by dividing the air quantity, ft
3
/min (m
3
/min),
by the air velocity in the duct, or column 3/ column 4, Table 11. Enter the result
in column 5, Table 11.
Once the required duct area is known, find from Table 13 the nearest whole-
number duct diameter corresponding to the required area. Avoid fractional diame-
ters at this stage of the calculation, because ducts of these sizes are usually more
expensive to fabricate. Later, if necessary, two or three duct sizes may be changed
to fractional values. By selecting only whole-number diameters in the beginning,
the cost of duct fabrication may be reduced somewhat. Enter the duct whole-number
diameter in column 6, Table 11.
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MATERIALS HANDLING
10.17
TABLE 11 Exhaust System Design Calculations
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MATERIALS HANDLING
10.18
TABLE 11 (Continued )
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MATERIALS HANDLING
MATERIALS HANDLING 10.19
TABLE 12 Recommended Exhaust Air Quantities
TABLE 13 Duct Diameters and Areas
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MATERIALS HANDLING
10.20 PLANT AND FACILITIES ENGINEERING
4. Compute the actual air velocity in the duct
Use Fig. 6 to determine the actual velocity in each duct. Enter the chart at the air
quantity corresponding to that in the duct, and project vertically to the diameter
curve representing the duct size. Read the actual velocity in the duct on the velocity
scale, and enter the value in column 7 of Table 11.
The actual velocity in the duct should, in all cases, be equal to or greater than
the design velocity shown in column 4, Table 11. If the actual velocity is less than
the design velocity, decrease the duct diameter until the actual velocity is equal to
or greater than the design velocity.
5. Compute the duct velocity pressure
With the actual velocity known, compute the corresponding velocity pressure in the
duct from h
v
ϭ (v / 4005)
2
, where h
v
ϭ velocity pressure in the duct, inH
2

O; v ϭ
air velocity in the duct, ft / min. Thus, for the duct run A in which the actual air
velocity is 4300 ft/min (1310.6 m / min), h
v
ϭ (4300/4005)
2
ϭ 1.15 in (29.2 mm)
H
2
O. Compute the actual velocity pressure in each duct run, and enter the result in
column 8, Table 11.
6. Compute the equivalent length of each duct
Enter the total straight length of each duct, including any vertical drops, in column
9, Table 11. Use accurate lengths, because the system resistance is affected by the
duct length.
Next list the equivalent length 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 (152.4-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 the duct. Consider 45
Њ bends as having the same
resistance as 90
Њ elbows. Note that branch ducts are usually arranged to enter the
main duct at an angle of 45
Њ 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 in each duct
Using Fig. 6, determine the resistance, inH
2
O (mmH
2
O) per 100 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 100 ft
(30.5 m) of duct, column 12, Table 11, by the total duct length, column 11
Ϭ 100.
Thus for duct run A, actual friction
ϭ 5.4(10/100) ϭ 0.54 in (13.7 mm) H
2
O.
Compute the actual friction for the other duct runs in the same manner. Tabulate
the results in column 13, Table 11.
8. Compute the hood entrance losses
Hoods are used in industrial exhaust systems to remove vapors, dust, fumes, and
other undesirable airborne contaminants from the work area. The hood entrance
loss, which depends upon the hood configuration, is usually expressed as a certain
percentage of the velocity pressure in the branch duct connected to the hood, Fig.
7. Since the hood entrance loss usually accounts for a large portion of the branch
resistance, the entrance loss chosen should always be on the safe side.
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10.21

FIGURE 6 Duct resistance chart. (American Air Filter Co.)
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MATERIALS HANDLING
10.22 PLANT AND FACILITIES ENGINEERING
FIGURE 7 Entrance losses for various types of exhaust-system intakes.
List the hood designation number under the ‘‘System Resistance’’ heading, as
shown in Table 11. Under each hood designation number, list the velocity pressure
in the branch connected to that hood. Obtain this value from column 8, Table 11.
List under the velocity pressure, the hood entrance loss form Fig. 7 for the particular
type of hood used in that duct run. Take the product of these two values, and enter
the result under the hood number on the ‘‘entrance loss, inH
2
O’’ line. Thus, for
hood 1, entrance loss
ϭ 1.15(0.50) ϭ 0.58 in (14.7 mm) H
2
O. Follow the same
procedure for the other hoods listed.
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MATERIALS HANDLING 10.23
9. Find the resistance of each branch run
List the main and branch runs, A through F, Table 9. Trace out each main and
branch run in Fig. 5, and enter the actual friction listed in column 3 of Table 11.
Thus for booth 1, the main and branch runs consist of A, D, G, H, and I. Insert
the actual friction, in (mm) H

2
O, as shown in Table 9, or A ϭ 9.54(242.3), D ϭ
0.42(10.7), G ϭ 0.19(4.8), H ϭ 0.20(5.1), I ϭ 0.50(12.7).
Determine the filter friction loss from the manufacturer’s engineering data. It is
common practice to design industrial exhaust systems on the basis of dirty filters
or separators; i.e., the frictional resistance used in the design calculations is the
resistance of a filter or separator containing the maximum amount of dust allowable
under normal operating conditions. The frictional resistance of dirty filters can vary
from 0.5 to 6 in (12.7 to 152.4 mm) H
2
O or more. Assume that the frictional
resistance of the filter used in this industrial exhaust system is 2.0 in (50.8 mm)
H
2
O.
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, inH
2
O, 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 system, certain duct sizes must be changed to produce equal resistance in all
ducts. Or, if possible, certain ducts can be shortened. If duct shortening is not
possible, as is often the case, an exhaust fan capable of operating against the largest
resistance in a branch can be chosen. If this alternative is selected, special dampers
must be fitted to the air inlets of the booths or ducts. For economical system op-
eration, choose the balancing method that permits the exhaust fan to operate against
the minimum resistance.

In the system being considered here, a fairly accurate balance can be obtained
by decreasing the size of ducts E and F to 4.75 in (120.7 mm) and 4.375 in (111.1
mm), respectively. Duct B would be increased to 6.5 in (165.1 mm) in diameter.
11. Choose the exhaust fan capacity and static pressure
Find the required exhaust fan capacity in ft
3
/min from the sum of the airflows in
the ducts, A through H, column 3, Table 11, or 3300 ft
3
/min (93.5 m
3
/min). Choose
a static pressure equal to or greater than the total resistance in the branch duct
having the greatest resistance. Since this is slightly less than 4.5 in (114.3 mm)
H
2
O, a fan developing 4.5 in (114.3 mm) H
2
O static pressure will be chosen. A 10
percent safety factor is usually applied to these values, giving a capacity of 3600
ft
3
/min (101.9 m
3
/min) and a static pressure of 5.0 in (127 mm) H
2
O for this
system.
12. Select the duct material and thickness
Galvanized sheet steel is popular for industrial exhaust systems, except where cor-

rosive fumes and gases rule out galvanized material. Under these conditions, plastic,
tile, stainless steel, or composition ducts may be substituted for galvanized ducts.
Table 14 shows the recommended metal gage for galvanized ducts of various di-
ameters. Do not use galvanized-steel ducts for gas temperatures higher than 400
ЊF
(204
ЊC).
Hoods should be two gages heavier than the connected branch duct. Use supports
not more than 12 ft (3.7 m) apart for horizontal ducts up to 8-in (203.2-mm)
diameter. Supports can be spaced up to 20 ft (6.1 m) apart for larger ducts. Fit a
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MATERIALS HANDLING
10.24 PLANT AND FACILITIES ENGINEERING
TABLE 14 Exhaust-System Duct Gages
duct cleanout opening every 10 ft (3 m). Where changes of diameter are made in
the main duct, fit an eccentric taper with a length of at least 5 in (127 mm) for
every 1-in (25.4-mm) change in diameter. The end of the main duct is usually
extended 6 in (152.4 mm) beyond the last branch and closed with a removable cap.
For additional data on industrial exhaust system design, see the newest issue of the
ASHRAE Guide.
Related Calculations. Use this procedure for any type of industrial exhaust
system, such as those serving metalworking, woodworking, plating, welding, paint
spraying, barrel filling, foundry, crushing, tumbling, and similar operations. Consult
the local code or ASHRAE Guide for specific airflow requirements for these and
other industrial operations.
This design procedure is also valid, in general, for industrial pneumatic convey-
ing systems. For several comprehensive, worked-out designs of pneumatic convey-
ing systems, see Hudson—Conveyors, Wiley.

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