1. Off-Road Equipment Management TSM 262: Spring 2016
LECTURE 17: Materials Handling I
Off-Road Equipment Engineering
Dept of Agricultural and Biological Engineering

2. Homework and Lab

3. Materials Handling: Class Objectives
Students should be able to:
Explain what materials handling entails and why it is important
Identify different types of material conveyor systems for agricultural applications
Identify different parts of a screw/auger conveyor
Analyze the performance of auger conveyors
Evaluate the safety of auger conveyors

4. What is Materials Handling?
Movement and handling of materials and products in a systematic manner
Can be horizontal, vertical or a combination of the two

5. Why is it Important?
Materials handling costs account for as much as 25% of total production cost for certain crops
Costs can be lowered with efficient handling systems properly integrated into the overall production system
Examples of materials handling?

6. Introduction: Conveyors
Materials
Liquid, granular, powder, fibrous, or combination
Methods of conveying
Mechanical, inertial, pneumatic, gravity
Types of conveyors
Auger (screw), belt and mass conveyors (mechanical force)
Pneumatic conveyors (aerodynamic drag)
Bucket elevators (inertial and gravity forces)
Forage blowers (inertial and aerodynamic forces)

7. Examples of Conveyors
Belt Conveyor
3D Bucket Conveyor

8. Types of Conveyors used in Agric.

9. Advantages & Disadvantages

10. Examples of applications
Combines
Header
Convey clean grain from cleaning shoe to elevator
Convey grain from combine to grain wagon
Portable units at silos and grain elevators
Other examples?

11. Examples of Conveyors in Combine

12. Auger Conveying Most common method of handling grain and feed on farms
ASABE Standards
ASAE S374 MAR1975 (R2006)
Terminology and Specification Definitions for Agricultural Auger Conveying Equipment
ASAE S361.3 APR1990 (R2005)
Safety for Portable Agricultural Auger Conveying Equipment

13. Safety for Portable Auger Conveyors
ASAE S361.3 standard (on Compass)
Purpose
Electrical specifications
Conductors, connectors and grounding
Electrical controls and wiring
Electric motors
General specifications
Shields and guards for drive components
Shields and guards for functional components
Lateral stability
Tube restraint
System for lifting and varying the angle of the auger
Overhead power lines
Safety signs

14. Auger Conveyor

15. Auger Conveyors Shaft carries helicoid flightings on outer surface
Standard auger pitch
Ranges from 0.9dsf to 1.5dsf (dsf= outside screw diameter or flighting diameter)
Assume “standard pitch” = dsf
Std pitch up to 20º auger inclination angle
½ std pitch for > 20º
Double, triple, variable pitch also available
pitch
dsf

16. Auger Conveyors Theoretical Capacity:
Qt = .25 p (dsf2 - dss2) lp n (m3/s)
Qt = volumetric capacity (m3/s)
dsf = flighting diameter (m)
dss = shaft diameter (m)
lp = pitch (m)
n = rotational speed (rev/s)

17. Auger Conveyor Equations
Volumetric Efficiency (% Fill):
hv = Qa/Qt
Qa = actual capacity (m3/s)
Qt = theoretical capacity (m3/s)
Power for small-diameter, fully loaded augers
Where P=total power to operate at full load (kW)
ρb=material bulk density (kg/m3)
L=auger length (m)
f=power factor from table

18. Factors affecting Auger Performance
Capacity affected by:
Auger diameter
Auger speed
Angle of elevation
Length of intake exposure
Type of grain
Moisture content of grain
Length of auger affects power requirement but not the capacity
Qt = .25 π(dsf2 - dss2) lp n (m3/s)

19. Power Requirements
Almost all power is used to overcome friction and energy losses between flighting and tube
Small proportion of power (≈1%) used to lift weight of grain for inclined auger

20. Auger Conveyor Performance
Incline,
deg [rad]
% Fill at given rpm
Power Factor
100
300
500
700
0 [0] 85 84 69 54
1500
20 [0.35] 82 70 40
1085
40 [0.70] 74 59 43 33
845
57.3 [1] 65 49 36 28
630
70 [1.22] 47 29 22
612

21. Auger Conveyor Performance

22. Auger Conveyor Performance

23. Class Problem
A horizontal 10-m standard pitch auger with a 10-cm diameter tube and a 2.5-cm shaft turns at 400 rpm. Calculate the following:
Estimated capacity for shelled corn in kg/h (bulk density of shelled corn = 719 kg/m3)
Estimated power required in kW

24. Solution Given Find Qa and P Length, L=
Flighting diameter (assume same as tube diameter), dsf =
Pitch same as diameter, lp =
Shaft diameter, dss =
Shaft rotational speed =
Bulk density of corn, ρb =
Find Qa and P
Qt = .25 π(dsf2 - dss2) lp n (m3/s)
hv = Qa/Qt

25. Solution
Qt = .25 p (dsf2 - dss2) lp n (m3/s) Qt = And hv = Qa/Qt Therefore Qa = Qt· hv Determine % fill from average values for 300 and 500 rpm = ____ Qa = Qt· hv = Qa = Estimated capacity of auger = Power factor from table = P=60·Qa·ρb·L/f =