1. Off-Road Equipment Management TSM 262: Spring 2016
LECTURE 16: Hay and Forage Harvesting II
Off-Road Equipment Engineering
Dept of Agricultural and Biological Engineering

2. Homework, Lab and Technical Sessions

3. Hay & Forage Harvesting: Objectives
Students should be able to:
Identify and explain the processes involved in hay and forage harvesting
Understand what type of equipment can be used for this type of harvesting
Calculate power requirements for PTO-driven implements with particular reference to mowers

4. Functional Processes Chop FORAGE Windrow Wilt Transport Store CUT
Condition
HAY
Swath
Wilt
Rake
Dry
Bale
Tran.
Store

5. Baling Two types of balers in popular use
Rectangular (small and large)
Round
Small square baler – New Holland
Large square baler – New Holland
Small square baler – John Deere

6. Bale Sizes Rectangular Round Small Large Wt: 340 – 1000 kg
14” x 18” x 36” (355 x 457 x 914 mm): kg
16” x 18” x 36” (406 x 457 x 914 mm): kg
Large
3’ x 3’ x 3’ (0.9 x 0.9 x 0.9 m): 385 kg
3’ x 4’ x 6’ (0.9 x 1.2 x 1.8 m): 510 kg
4’ x 4’ x 8’ (1.2 x 1.2 x 2.4 m): 907 kg
Round
Wt: 340 – 1000 kg
Up to 1.7 m wide
Up to 2 m diameter

7. Square Baler
Stuffer:
Knotter:

8. Round Baler
Forming the core of the bale -
Round Baler - Variable Geometry/Chamber
Round Baler - Fixed Geometry

9. Round Baler
Finished bale formation -
Crop cutter -
Complete baling process -

10. Adjustments Sensitivity to feed rate (forward speed) Bale density
Ensure even feeding into bale chamber
Bale density
Adjust sides of rectangular bale chamber
Bale tying
Type of twine – refer to ASAE Standard S315.3: classification according to material, knot strength and minimum tensile strength
Pickup height

11. Functional Processes Chop FORAGE Windrow Wilt Transport Store CUT
Condition
HAY
Swath
Wilt
Rake
Dry
Bale
Tran.
Store

12. Chopping Two types Precision cut Non-precision cut

13. Forage Harvester Power Requirement
Pfh = Forage harvester power requirement
Pc=chopping power
Pf=power absorbed from friction
Paccel=power to accelerate forage
Pair=power to move air
Ph=power consumed by header

14. Forage Harvester Power Requirement
Power to
accelerate
forage
Power to
air
Chopping
Power
Power
Consumed
By header
Power
Absorbed
From friction

15. Power Requirements EP496.3 section 4.1.2
PTO power required by implement
Ppto=a + bw + cF
Ppto = PTO power required by implement (kW)
w = implement working width (m)
F = material feed rate (Mg/h)
a, b, c = machine specific parameters from ASAE D497.7 Table 2

16. Harvesting Productivity
Miscanthus
Approx kW required to operate tractor and implement separate from cutting and conditioning biomass
Higher travel speeds generate a higher proportion of useful work relative to baseline power input
Tractor should operate as higher travel speed as possible subject to maintaining quality of cut crop

17. Power Requirements ASAE Data D497.7 (Table 2) Compare mowers
Cutterbar mower, 1.2 kW/m
Disk mower, 5.0 kW/m
Flail mower, 10.0 kW/m
Reasons for differences?
impact cutting
air pumping done by rotor
Compare mower-conditioners

18. Power Requirement for Rotary Mower
Much higher than for sickle bar
forage not only cut but also accelerated by knives during impact
NIAE suggest following equation:
Pmt = (PLs + Esc·vf) wc
where
Pmt = total PTO power to mower (kW)
PLs = specific power losses due to air, stubble
and gear train friction (kW/m of width, 1.5< PLs<4)
Esc = specific cutting energy (kJ/m2), 1.5< Esc <2.1
wc = width of mower (m)
Disk Mower
Drum-Type Mower
Sharp blade
Worn blade

19. Disk Mower vs Drum Mower
Drum-Type Mower
Disk Mower

20. Example Problem
Consider a disk-type rotary mower that has six disks, each cutting a 0.4 m width. The mower is traveling at 15 km/h. Stating other assumptions:
Estimate PTO power requirement if the blades are sharp
Estimate PTO power requirement for same mower after blades become worn

21. Solution Given: Assumptions Pmt = (PLs + Esc·vf) wc
Mower with 6 disks at 0.4m per disk
Total width, wc = 6 x 0.4 = 2.4 m
Travel speed, vf = 15 km/h = 15/3.6 = 4.2 m/s
Assumptions
Power losses for disk mower, PLs = 1.5 kW/m
Specific cutting energy (sharp), Esc = 1.5 kJ/m2
Specific cutting energy (blunt), Esc = 2.1 kJ/m2
Pmt = (PLs + Esc·vf) wc

22. Solution (a) Pmt = (PLs + Esc·vf) wc Pmt = (1.5 + 1.5·4.2)·2.4 = 19 kW
For sharp blades
(b) Pmt = (PLs + Esc·vf) wc
Pmt = ( ·4.2)·2.4 = 25 kW
For blunt blades
This is a 100*(25-19)/19 = 32% increase!