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 achansen@illinois.edu

Homework, Lab and Technical Sessions

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

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

Baling Two types of balers in popular use Rectangular (small and large) Round Small square baler – New Holland http://www.youtube.com/watch?v=TLW2okLSdUs Large square baler – New Holland http://www.youtube.com/watch?v=15ExSQkxMnI Small square baler – John Deere http://www.youtube.com/watch?v=THgZsrkUUO8&NR=1

Bale Sizes Rectangular Round Small Large Wt: 340 – 1000 kg 14” x 18” x 36” (355 x 457 x 914 mm): 23-36 kg 16” x 18” x 36” (406 x 457 x 914 mm): 32-41 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

Square Baler Stuffer: http://www.youtube.com/watch?v=5HRFwjkCXuM&feature=relmfu Knotter: http://www.youtube.com/watch?v=XDAmQXdTwXs&NR=1

Round Baler Forming the core of the bale - http://www.youtube.com/watch?v=b8uRXhomhJA&NR=1 Round Baler - Variable Geometry/Chamber Round Baler - Fixed Geometry

Round Baler Finished bale formation - http://www.youtube.com/watch?v=Xb7oAHNuhws&NR=1 Crop cutter - http://www.youtube.com/watch?v=miJwDWR_1HQ&feature=relmfu Complete baling process - http://www.youtube.com/watch?v=zBcUME026w0&feature=related

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

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

Chopping Two types Precision cut Non-precision cut http://www.youtube.com/watch?v=W5D-Z_d-u-g http://www.youtube.com/watch?v=zMnwzbxO-FQ

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

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

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

Harvesting Productivity Miscanthus Approx. 30-35 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

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

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

Disk Mower vs Drum Mower Drum-Type Mower Disk Mower

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

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

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 = (1.5 + 2.1·4.2)·2.4 = 25 kW For blunt blades This is a 100*(25-19)/19 = 32% increase!