Off-Road Equipment Management TSM 262: Spring 2016 LECTURE 5: Tractor Performance II Off-Road Equipment Engineering Dept of Agricultural and Biological Engineering achansen@illinois.edu

Homework and Lab

Class Objectives Students should be able to: Analyze the relationships between engine speed, torque, power, and fuel consumption for diesel engines used in off-road applications Estimate average fuel consumption for a tractor or combine Understand the “shift up and throttle back” process for achieving better fuel economy Understand the basic working principles of an engine Analyze the pulling performance and tractive efficiency of a tractor

Brake Thermal Efficiency How is brake thermal efficiency calculated? From fuel PPTO = Power measured at PTO [kW] FC = Fuel consumption [kg/h] HV = Gross calorific value for fuel [kJ/kg] HV - diesel fuel = 45,500 kJ/kg HV - biodiesel = 38,000 kJ/kg HV – ethanol = 29,000 kJ/kg

Example Problem Engine manufacturers strive to achieve 40% or higher brake thermal efficiency. If an engine runs on diesel fuel with a gross calorific value of 45,500 kJ/kg, what SFC should the manufacturer be targeting? What happens to SFC when burning biodiesel or ethanol?

Engine Performance Map Add contours of constant SFC and constant power to load-speed graph

“Shift Up & Throttle Back” Concept

“Shift Up & Throttle Back” Concept When tractor is operating at less than full load, operator is advised to shift up to a higher gear and reduce engine speed to retain original ground speed How does this improve fuel economy? At Pb = 20 kW: BSFC > 400 g/kWh @ full speed BSFC = 290 g/kWh @ 1500 rev/min Savings in fuel = (400-290)/400 = 27.5% !!

Map Examples Ford 2.5L DI engine Ford 2.5L DI turbocharged engine

Estimation of Average Fuel Consumption Annual average fuel requirements for tractors or other diesel engine driven machinery may be used in calculating overall machinery costs for particular enterprise Fuel requirement for particular operation should be based on actual power required If available, make use of data from Nebraska tests: http://tractortestlab.unl.edu/testreports.htm

Estimation of Average Fuel Consumption Method of estimation given in ASAE Standard D497.7 Equation for predicting SFC (specific fuel consumption in L/kWh) based on Nebraska tractor test data 1980-2005 Applies to diesel tractor and combine engines Standard includes: Fraction of equivalent PTO power available Partial throttle multiplier that accounts for adjustments in engine speed Why are these two factors included?

Estimation of Average Fuel Consumption SFC (L/kWh) for Diesel: X = equivalent PTO Power required by operation ÷ maximum available from PTO nPT = partial throttle engine speed, rpm nFT = full throttle engine speed, rpm

Shift up-throttle back Vary X Vary N X = Equivalent PTO Power required by operation/Rated PTO Power N = Partial throttle engine speed/full throttle engine speed

Example Problem An AGCO Allis 9775 tractor fitted with a Cummins engine has a rated PTO power of 152 kW at 2200 r/min and its maximum high idle speed is 2430 r/min. While cultivating a field on average the tractor uses the equivalent PTO power of 80 kW with the throttle setting at 2000 r/min Calculate the SFC in L/kWh

Solution Given Solution Rated PTO power = 152 kW Rated speed = 2200 r/min High idle speed = 2450 r/min For cultivating tractor uses only 50 kW Throttle setting at 2000 r/min Solution X = 50/152 = 0.33 N=2000/2450 = 0.82 PTM = 1-(N-1)(0.45*X-0.877) PTM =1-(0.82-1)(0.45*0.33-0.877)= 0.869 SFC =(0.22+0.096/X)*PTM=(0.22+0.096/0.33)*0.869 SFC =0.444 L/kWh

Another Part of the Problem Compare this SFC with the SFC that the tractor would obtain if the throttle remained at the full setting, i.e. N = 1 SFC =(0.22+0.096/X)*PTM SFC =(0.22+0.096/0.33)*1 = 0.511 L/kWh Improvement with reduced throttle setting % reduction in SFC = 100*(0.511-0.444)/0.511 =13%

Shift up-throttle back Vary X Vary N X = Equivalent PTO Power required by operation/Rated PTO Power N = Partial throttle engine speed/full throttle engine speed

Brief Introduction to Engines Engines covered in detail in TSM 464 Four-stroke engine Intake Compression Power Exhaust

Engines Many types & sizes of IC engines Most common type of tractor engine has vertical cylinders arranged side by side or in line V engines have been used as well

Engine Structure

Engine Example: John Deere Diesel, Industrial, Power Tech Series, 6068T, 185 HP Identify components? Turbocharger Electric Starter Oil filter Alternator Fan Fuel filter

Tractor Pulling Performance How effective are tractors in using engine power to pull implements? Depends on: Configuration of tractor 2WD vs MFWD vs 4WD vs Track Firmness of ground Soft vs firm vs concrete Can be described by Tractive Efficiency

Tractive Efficiencies

Power Flow from Fuel to Soil 40 COMBUSTION 32 80 GROSS FLYWHEEL Percentages 92 25-26 NET FLYWHEEL 77-80 91 82-85 84-88 99 89-91 TRANSMISSION INPUT Tractive Efficiency tractor type conc Firm Tilled Soft 87 72 67 55 2WD 76 64 MFWD 88 77 75 70 4WD 74 Track 85-90 90-92 AXLE PTO 96 See table DRAWBAR

Example Problem A farmer has a 6 m wide offset disk harrow that he uses for primary tillage. This harrow requires a drawbar pull force of 46 kN while traveling in the field at 8 km/h on average. Assuming that a MFWD (mechanical front wheel drive) tractor will be used to pull the implement and that the soil condition is firm on average, calculate the following: Drawbar power, Equivalent PTO power for tractor to be able to pull implement at the given speed

Example Problem: Solution Given: Drawbar force=46 kN Speed=8 km/h Tractor: MFWD and soil is firm (a) Pdb=Drawbar force x travel speed Pdb=46 [kN] x 8 [km/h] x 1000/3600 kW Pdb= 102 kW (b) etractive= Pdb /PPTO PPTO= Pdb / etractive PPTO=102/0.76=134 kW Equivalent PTO power required is 134 kW to pull this implement For a given implement, how do we determine the drawbar pull?