Off-Road Equipment Management TSM 262: Spring 2016 LECTURE 4: Machine Performance II and Tractor Performance I Off-Road Equipment Engineering Dept of Agricultural and Biological Engineering achansen@illinois.edu

Homework and Lab

Class Objectives Students should be able to: Explain the meaning of and determine the theoretical and effective field capacities of machines and their field efficiencies Identify factors that affect the field efficiency of machines Compare the performance of different machines with changes to field operating patterns and other variables Analyze the relationships between engine speed, torque, power, and fuel consumption for diesel engines used in off-road applications

Material Efficiency The Material Efficiency is equal to ratio: Yield/(Yield + Material Losses) Note that the yield is the actual or measured yield (e.g. grain actually gathered from the field) Concave

Factors Influencing Machine Performance Field patterns Field shape Field size Maneuverability Limitations on field speed Yield (harvesting) Soil and crop conditions System limitations

Machine maneuverability Field efficiency depends on minimizing number of turns (turning time)

Limitations on Field Speed Overloading of machine (not enough power) Operator Inability to steer along straight line Comfort level relative to skill and conditions Potential damage to machine Material handling limits

System performance Field operation dependent on a set of machines Sum of machines working together is called a system Harvesting system components Combine Unloading machine Trucks for transport Others? Efficiency of the system is combination of component efficiencies Use of cycle diagram to study system performance

Cycle diagram All machine cycles must add up to the total system time Idle times are permitted only after the completion of the intermediate specific operation Cycle times and diagrams need to be computed only for a representative cycle of each machine type used in the system for a steady state solution Transport times need to be determined for average travel distance and speeds to determine an average system performance The system capacity is limited to that of the cycle with zero idle time

Development of a cycle diagram Sketch the cycles to show the proper machine interrelationships Mark the productive and support times along the cycles Example: Planter system One planter + One Tractor + One Wagon + One Unloader Activity of each machine represented by closed loop making up complete time cycle Idle time Support time Travel to and from planter takes 5 min each way

Development of a cycle diagram Sum the times required for each cycle The longest system time determines the cycle time Example: Planter system Which machine is limiting?

Cycle diagram: Forage Harvesting Example – Forage harvesting with two harvesters H1 and H2, four trucks T1-T4, one unloader U Truck unloader, 6+1+6+1+6+1+6+1=28 min Harvester, 10+10 = 20 min Transport, 10+5+1+6+5 = 27 min The longest system time determines the cycle time (28 min) Add idle time to the other cycles to bring their system time up to the cycle time (28 min). The harvester has 28-20 = 8 min of idle time (4 min between each truck). The trucks will have 28-27 = 1 min idle time in their cycles. Source: Hunt p20-22

Tractor Power and Performance

Tractor Power and Performance Tractor is very important source of power in agriculture Review a few terms Force A Force of 1 Newton must be applied to give a mass of 1 kg an acceleration of 1 m/s2 according to Newton’s second law: F [N] = m [kg] * a [m/s2]

Tractor Power and Performance Work Work is performed when a force is applied through a distance: W [Nm] = F [N] * D [m] Also measure of energy-units of Joules [J] Power is the rate of doing work or the amount of work that can be performed per unit time P = W/t [Nm/s]= Force * Distance/time Unit [Nm/s] is called a Watt Distance/time is speed

Rotary Power and Torque In case of a PTO shaft, Work can be found by multiplying a force through a distance. Total work per revolution of shaft: The power is equal to this value divided by the time per revolution.

Rotary Power and Torque cont.  

Rotary Power and Torque cont. Rotational speed of engine normally in units of rev/min and power is in kiloWatts [kW] Therefore

Example problem How much work and power are required to lift an implement 60 cm in 3 seconds? The mass of the implement is 800 kg. Force Force = m*g = 800 * 9.81 N = 7848 N Work Work = Force * distance Work = 7848 * 0.6 Nm = 4709 Nm Power Power = Work/time = 4709/3 W = 1570 W

Example Problem  

Class Problem A Challenger MT865 tractor produces a maximum power of 390 kW at 1750 r/min. Calculate the equivalent torque.

Forms of Power Tractor can deliver power in four ways Linear (Drawbar) pulling an implement Rotary (PTO) Hydraulic (pump) Electrical (alternator) Focus on first two in TSM 262

Engine Performance Torque versus speed for diesel engine TRS Peak Torque Point represents speed and torque coinciding closely with manufacturer’s rated power specification. TP TRS Torque High Idle NHI Speed NRS

Describing Engine Performance Torque Reserve expresses the percentage increase in Torque at Peak Torque relative to Torque at Rated Speed TP Torque Reserve or Torque Rise TRS Tr = 100(TP-TRS) TRS Torque Speed

Describing Engine Performance Peak Torque Rated point of operation TP TRS Engine Torque (Nm) High Idle NP NRS NHI Engine Speed (rev/min)

Example Performance of John Deere 8320 tested at the Nebraska Tractor Test Laboratory

Example: Torque and Power John Deere 8320

Fuel Consumption Fuel consumption (FC) measured in L/h, kg/h (gal/h) Specific Fuel Consumption (SFC) Measure of engine efficiency Represents the quantity of fuel used per unit time to generate a unit of power SFC = FC [kg/h]/power [kW]

How does fuel consumption vary? at high idle sufficient to maintain rpm Fuel Consumption Engine Speed Torque Speed

How does SFC vary with speed? Fuel Consumption Engine Speed SFC (kg/kWh) Engine Speed (rpm) Torque Speed

FC and SFC Example John Deere 8320