Presentation on material (grain) conveying system.

1. BAHIR DAR UNIVERSITY
BAHIR DAR INSTITUTE OF TECHNOLOGY
Faculty of Mechanical and industrial Engineering
Agricultural Machinery Engineering
Presentation on: Grain storage facilities
July 15/2022
1

2. Presentation outline
 Introduction
 Types of conveyor
 Belt Conveyor

3. Introduction
 Conveyors are one of the transport equipment
in material handling
 Transport equipment
Conveyors
Cranes
Industrial Trucks

4. Conveyors are used
 Where material is to be moved frequently
between specific points
 To move materials over a fixed path
 When there is a sufficient flow volume to
justify the fixed conveyer investment

5. Classification of conveyors
 Type of product being handled unit
bulk
 Location of the conveyor overhead
on-floor
in-floor
Whether or not loads can accumulate on the
conveyor
 (Pls follow this web link:
http://www.mhi.org/learning/cicmhe/resources/tax
onomy/TransEq/Conv/index.htm

6. Type of Conveyors
1. Chute Conveyors
2. Wheel Conveyors
3. Roller Conveyors
a) Gravity roller conveyor
b) Live (powered) conveyor
4. Chain Conveyors
5.Slat Conveyors
6.Belt Conveyors
a) Flat belt conveyor
b) Magnetic belt conveyor
c) Trough belt conveyor

7. 7.Bucket Conveyors
8. Vibrating Conveyors
9. Screw Conveyors
10. Pneumatic Conveyors
a) Dilute-phase pneumatic conveyors
b) Carrier-system pneumatic conveyors
11. Vertical Conveyors
a) Vertical lift conveyors
b) Reciprocating vertical conveyor

8. 12. Cart-on-track conveyors
13. Tow conveyor
14.Trolley conveyor
15.Power-and-free conveyor
18.Monorail
19.Station conveyor
a) Diverter d) Tilting device
b) Pop-up device e) Cross-belt transfer device
c) Sliding shoe device

9. Belt Conveyors
Characteristics
 Very Efficient
 Damage to the products is very low
 High Carrying capacity
 Long distance conveying
 Long service life

10. Elements of a belt conveyor
Belt
Drive
Pulley
Idler Pulley
Feeder
Discharge

12. 1) Belt -consists of one or more layers of
material. Many belts -two layers. An under
layer (carcass) for linear strength and shape
an over layer (cover). The carcass is often a
woven fabric -polyester, nylon and cotton.
The cover is often various rubber or plastic
compounds specified by use of the belt.
2) Drive
- Drive the mechanism
-Located at the end of the discharge to
prevent the sag

13. 3. Pulley
- Large enough to provide enough contact
surface with the belt to ensure a positive
drive
4. The Take Up
- Automatic or manual adjustment for
contraction or expansion of the belt due to
moisture and temperature.

14. 5. Idler Pulleys
- Material – Plain wood or light steel
- Multiples to increase carrying capacity
6. Feeder
- Free flow – Single funnel with a gate valve
- Not a free flow - Screw feeder, vibrating
feeder , Star feeder

15. 7) Discharge
- Placed at the end of the belt
- Types
a) Tripper
b) Angle Scrapper

16. Types of belt conveyors
 Flat belt conveyor

17.  Transport Medium and light weight loads
 Inclined or decline when it is required
 Provide considerable control over the
orientation and placement of the load
 No smooth accumulation, merging and
sorting on the belt
 The belt is roller or slider bed supported

18.  Magnetic belt conveyor
 For transporting ferrous materials

19.  Trough belt conveyor

20. Designing
 Belt Capacity Determination
BLTCAP = 0.8 * CSA * BLTS
BLTCAP – Conveyor belt capacity (bu/min)
SI units can be used m3, m2 & ms-1
Belt capacity vary according to the belt width and
surcharge angle
CSA – Cross sectional area (ft2)
BLTS – Belt speed (ft/min)
(1bushel = 1.2444 cubic feet)

21.  Horse Power Requirement Determination
1. Horse power to drive = BLTS * (A+BL)/100
empty conveyor
BLTS – Belt speed (ft/min)
L - Conveyor length in feet
A, B – Constants depend on belt width
2.Horse power to
convey material = tons of material * (0.48 +0.00302 L)/100
on level per hour
(Metric tonne = 1000kg ton = 2240 pounds or 1016kg)
1 t = 1.1023 t

22. 3. Horse power to = Lift * 1.015 * tons of
lift the material in ft material per h/1000
Simplified SI calculation
Power (kW) = (Capacity (t/h)* lift (m)* 3.75)/1000
**** True for <10% slope
**** If efficiency of drive mechanism is < 95% use 4
instead 3.75
1hp =745.7 W

23. SI detailed calculation

24. The basics of the Calculations of
Conveyor Belt Design Parameters
 Belt tension: The belt of the conveyor always
experience a tensile load due to the rotation of
the electric drive, weight of the conveyed
materials, and due to the idlers. The belt
tension at steady state can be calculated as:
 Tb = 1.37*f*L*g*[2*mi+ (2*mb + mm)*cos (δ)]
+ (H*g*mm)…….eqn.1.1

25.  Where,
 Tb is in Newton.
 f = Coefficient of friction
 L = Conveyor length in meters. Conveyor length
is approximately half of the total belt length.
 g = Acceleration due to gravity = 9.81 m/sec2
 mi = Load due to the idlers in kg/m.
 mb = Load due to belt in kg/m.
 mm = Load due to the conveyed materials
in kg/m.
 δ = Inclination angle of the conveyor in Degree.
 H = vertical height of the conveyor in meters.

26.  Load due to idlers (mi): This can be
calculated as below:
 mi = (mass of a set of idlers) / (idlers
spacing) ……………..eqn.1.2
 Power at drive pulley: The power required at
the drive pulley can be calculated from the belt
tension value as below:
 Pp = (Tb*v)/1000……………..eqn.1.3
 Where,
 Pp is in kW.
 Tb = steady state belt tension in N.
 v = belt speed in m/sec.

27.  Belt tension while starting the
system: Initially during the start of the
conveyor system, the tension in the belt will be
much higher than the tension in steady state.
The belt tension while starting can be
calculated as:
 Tbs =Tb*Ks………………..eqn.1.4
 Where,
 Tbs is in N.
 Tb = the steady state belt tension in N.
 Ks = the start-up factor

28.  Sizing of the motor:
 The minimum motor power can be calculated
as:
 Pm = Pp/Kd………………eqn.1.5
 Where,
 Pm is in kw.
 Pp = the power at drive pulley in kw
 Kd = Drive efficiency.

29.  Acceleration :
 The acceleration of the conveyor belt can be
calculated as:
 A= (Tbs – Tb)/ [L*(2*mi + 2*mb+mm)]………eqn.1.6
 Where,
 A is in m/sec2
 Tbs = the belt tension while starting in N.
 Tb = the belt tension in steady state in N.
 L = the length of the conveyor in meters.
 mi = Load due to the idlers in kg/m.
 mb = Load due to belt in kg/m.
 mm = Load due to the conveyed materials in kg/m.

30.  An Example of Conveyor Belt Calculations
 Input data:
 Conveyor capacity (Cc) = 1500 t/h = 416.67 kg/sec
 Belt speed (V) = 1.5 m/sec
 Conveyor height (H) = 20 m
 Conveyor length (L) = 250 m
 Mass of a set of idlers (m’i) = 20 kg
 Idler spacing (l’) = 1.2 m
 Load due to belt (mb) = 25 kg/m
 Inclination angle of the conveyor (δ) = 5 0
 Coefficient of friction (f) = 0.02

31.  Start-up factor (Ks) = 1.5
 Drive efficiency (Kd) = 0.9
 Friction factor (Cr) = 15
 Breaking strength loss factor (Cv) = 0.75
 First, we will use the eqn.1.2 for finding out
the load due to idlers:
 mi = (20/1.2) = 16.67 kg/m
 We will use the eqn.1.1 for finding out the belt
tension in steady state:

32.  Tb = 1.37*0.02*250*9.81*[16.67+ {2*25+
(416.67/1.5)}*cos (5)] + (20*9.81* (416.67/1.5)) =
77556.88 N.
 The belt tension while starting the system can be
calculated by using the eqn.1.4:
 Tbs = 1.5 * 77556.88 = 116335.32 N
 For calculating the power at drive pulley, we will use
the eqn.1.3:
 Pp = (77556.88*1.5)/ 1000 = 116.335 kw
 We will use the eqn.1.5 estimate the size of the motor:
 Pm = 116.35/0.9 = 129.261 kw

33.  We will use the eqn.1.6 to find out the
acceleration of the motor:
 A = (116335.32 - 77556.88)/
[250*{(2*16.67) + (2*25) + (416.67/1.5)}]
 = 0.429 m/sec2

34. Chain Conveyors

35.  Having a powered continuous chain
arrangement
 If the chain at bottom is called the drag
The chain at the top is called as the flight
 Operate at less than 200ft/min (Slow process)
 Noise is a problem

36. Trolley conveyor
 Overhead trolleys fastened by a chain
 Can have 1800 turn and steep elevations
 Commonly used in processing, assembly,
packaging, and storage operations

37. Scraper conveyor
A type of flight conveyor. It consists of a trough in
which a continuous driven chain with flights is
running. The flights are scraping the material
over the bottom of the casing. The material is
moving forward to the discharge point
 Used for granular
or non abrasive materials
 Power requirement is
high
 Use for many raw
products

38. Apron conveyor
 Flights in scraper conveyors are replaced
with the flat slats, steel plates or boards
 Have a moving platform or apron to convey
sacked materials and large units
 The apron conveyor is an economical design
for horizontal and inclined conveying up to
28°. Special features may be added to make
inclinations of up to 60° possible. An
extremely low-cost unit preferred for
applications of up to 45° inclination.

40. Chain conveyor design
 Generally flight speed – 75 -125ft/min
 Capacity decreases with inclination
 Horse power requirement
Horse power = 2* V *Lc* Wc* Fc + Q(L *Fm +H)
33000
V – Speed of the conveyor
Lc - Horizontal projection length of conveyor (ft)
Lc = cos ( c) * TCL
[ =cos(angle of elevation conveyor) * Total conveyor
length]

41.  Wc – Weight of flight and chain (lb/ft)
 Fc – Coefficient of friction for chain and
flight
 Q – Material to be handled (lb/min)
 L - Horizontal projected length of loaded
conveyor (ft)
L = cos (c ) TLL =cos(angle of elevation
conveyor) * Total Loaded length of conveyor
H
Total conveyor length
Conveyor length
after loading

42.  Fm – Coefficient of friction for material
 H – Height of lift (ft)
H = Sin (angle of elevation conveyor) * Total
length of conveyor

43. Bucket Elevators

44.  A special adaptation of both belt and chain conveyors
 Very efficient in moving things vertically
It consists of:
 Buckets to contain the material
 A belt or chains to carry the buckets and transmit the
pull
 Means to drive the belt
 Accessories
for loading the buckets or picking up the material
for receiving the discharged material
for maintaining the belt tension
for enclosing and protecting the elevator.

45. Components
of a Bucket
Elevator
Take -ups
Elevator Buckets
Leg Belts
Leg
Boot spout
Boot pulley
Boot
Deflector pulley
Garner Throat
motor
Head pulley
Head drive
(Reducer)
Head drive

48. Discharging mechanisms
Gravity action Centrifugal action

49. Types of buckets

50. Applications
A bucket elevator can elevate a variety of
bulk materials from light to heavy and
from fine to large lumps
Eg:
 Food products – grain, sugar, flour,
coffee, salt..
 Rock products – sand, gravel, cement,
gypsum, limestone
 Chemical processing products – fertilizer,
phosphate, agricultural lime , soda ash
 Pulp and paper products

51. Designing
 Power requirement determination
Horse Power Requirement = Q.H / 33,000
Q = material handled (lb/min)
= Belt speed * No of buckets per foot * Bucket capacity
H = Lift (Ft)

52. - Theoretical value is increased by 10 to 15 %
for,
friction
power for loading
- Also additional power is needed for,
starting under load
peak load

53. Advantages
 Low power requirement
 Long service life
 Minimum maintenance requirement
 Ability to handle wet grains as easy as dry
grains with negligible increase of power
 Relative quietness
 But Grain damage may be high in handling
seed purpose grains!!!!

54. The Screw conveyor

56.  The capacity of a screw conveyor
depends on the screw diameter, screw
pitch, speed of the screw and the loading
efficiency of the cross sectional area of
the screw.
 The capacity of a screw conveyor with a
continuous screw:
 Q = V. ρ
 Q = 60. (π/4).D2.S.n.ψ.ρ.C
 Where,
 Q = capacity of a screw conveyor

57.  V = Volumetric capacity in m3/h
 ρ = Bulk density of the material, kg/m3
 D = Nominal diameter of Screw in m
 S = Screw pitch in m
 N = RPM of screw
 Ψ = Loading efficiency of the screw
 C = Factor to take into account the inclination
of the conveyor
 Screw Pitch:
 Commonly the screw pitch is taken equal to
the diameter of the screw D. However it may
range 0.75 – 1.0 times the diameter of the
screw.

58.  RPM of Screw:
 The usual range of RPM of screw is 10 to
165. It depends on the diameter of screw
and the type of material (Max RPM of
screw conveyor is 165)
 Loading efficiency:
 The value is large for free flowing and non
abrasive,
 Ψ = 0.12 to 0.15 for abrasive material
 = 0.25 to 0.3 for mildly abrasive material
 = 0.4 to 0.45 for non abrasive free flowing
materials

59.  Inclination Factor:
 The inclination factor C is determined by
the angle of screw conveyor with the horizontal.
http://www.mechanicalengineeringblog.com/tag/screw-conveyor-design-calculation/

60. Oscillating conveyors

61.  Metallic trough carried
on inclined arms fitted
with rubber bushes to
handle the reciprocating
motion of the trough
 The oscillating motion
of the trough is achieved
via specially designed
inclined arms and an
eccentric shaft driven by a
motor through v – belts

62. PRINCIPLE
CONVEYOR TROUGH IMPARTS
VELOCITIES IN x & y DIRECTIONS

63. F1
F2
F4
F3
FORCES ON ‘ Y ‘ DIRECTION
F1 - ACCELARATION FORCE
F2 - PARTICLE WEIGHT
FORCES ON ‘ X ‘ DIRECTION
F3 - FORCE IN THE X DIRECTION
F4 - FRICTION
Figure 1 Figure 2

64. F1>F2
PARTICLE LIFTS &
MOVES AXIALLY BY A SHORT TRAJECTORY
F3>F4
PARTICLE MOVES MAXIMUM X DISTANCE DURING
FLIGHT

65. DESIGN CRITERIA

67. Belt conveyor design following
site for SI
 http://www.brighthubengineering.com
/manufacturing-technology/83551-
onsite-calculations-for-conveyor-belt-
systems/