1. 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

2. Homework and Lab

3. 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

4. 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

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

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

7. 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

8. 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

9. 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

10. 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

11. 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?

12. Cycle diagram: Forage Harvesting
Example – Forage harvesting with two harvesters H1 and H2, four trucks T1-T4, one unloader U
Truck unloader, =28 min
Harvester, = 20 min
Transport, = 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 = 8 min of idle time (4 min between each truck). The trucks will have = 1 min idle time in their cycles.
Source: Hunt p20-22

13. Tractor Power and Performance

14. 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]

15. 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

16. 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.

17. Rotary Power and Torque cont.

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

19. 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

20. Example Problem

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

22. 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

23. 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

24. 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

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

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

27. Example: Torque and Power
John Deere
8320

28. 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]

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

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

31. FC and SFC Example
John Deere
8320