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

Homework and Lab

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

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

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?

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)

Examples of Conveyors Belt Conveyor 3D Bucket Conveyor

Types of Conveyors used in Agric.

Advantages & Disadvantages

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?

Examples of Conveyors in Combine

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

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

Auger Conveyor

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

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)

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

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)

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

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

Auger Conveyor Performance

Auger Conveyor Performance

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

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

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 =