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So what happened
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<blockquote data-quote="ollie989898" data-source="post: 6490245" data-attributes="member: 54866"><p>Frankzy's graph is interesting. I presume ad blue use is metered in at low engine RPM because combustion temperature is lower or something- seems to be gently falling as engine speed increases?</p><p></p><p>The fuel consumption is interesting, too. At 1000rpm the engine is burning 25kg/hour, but only putting out 130 kW.</p><p></p><p>I always thought that your best absolute fuel consumption would be where peak torque occurs- the speed where fuel is giving you the maximum bang for your buck in terms of Nm vs grams of fuel consumed- normally some way above idle and some way below peak engine power. If this is the case, then it appear to be around 1500rpm in the case of the 939 in this example, where the engine, according to PTO power, must be producing around 250 kW and using around double the fuel (50kg/hour). But, compare this to the point where the lowest <em>specific</em> fuel consumption occurs which seems to be a sniff under 1300 rpm, interesting and bang goes my old-school theory.<img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite26" alt=":unsure:" title="Unsure :unsure:" loading="lazy" data-shortname=":unsure:" /></p><p></p><p>Fendt drivers will be able to confirm at what revs a 900 series tractor will 'settle' at once the transmission is running at top speed under load (I presume this is somewhere around the 1300 and below mark) and, how steady the revs will be when putting 24 tonnes all told.</p><p></p><p>Time for some old-fashioned calculatoring. <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite76" alt=":wacky:" title="Wacky :wacky:" loading="lazy" data-shortname=":wacky:" /></p><p></p><p>The 939 in this example is burning 210 grammes of fuel per kWh and running at a power output of around 190 kW. (Interestingly this should mean fuel use at that engine speed and power output of around 40kg/hour).</p><p></p><p>90 miles is around 145 KM. Assuming the journey was at 65kph the entire time, so the journey took about 2 hours 14 minutes or 2.23 hours.</p><p></p><p>Need to work out total energy used in this journey- relatively simple, just times that 190 kW engine power (assuming rock-steady engine speed at all times) by duration of the journey.</p><p></p><p> 2.23 hours x 190kW is 423.7 kWh.</p><p></p><p>Take the total kWh of energy used, and times that by the absolutely best specific fuel consumption on that graph (210 grams/kWh) gives 88.977 kg of fuel used.</p><p></p><p>Using the standard EC figures for off-road diesel, we can see best-case, the off road diesel will have a density of 845 kg/m3 or 0.845L/kg. Take the weight of fuel consumed (88.977kg) and times it by that density to get total fuel consumption in litres: 88.977 x 0.845= 76.185565 Litres.</p><p></p><p>Need to convert that to (UK) gallons, divide total litres by 4.54609= 16.538512216 gallons.</p><p></p><p>Divide miles (90) by gallons to give best-case fuel consumption: 90/ 16.53812216 = <u>5.4418437901 miles per gallon</u>, representing a best case scenario and with a transmission/drivetrain that gives zero loss from engine to road and with no change in engine speed whatsoever to get the vehicle moving initially (or any use of the brakes at any time).</p></blockquote><p></p>
[QUOTE="ollie989898, post: 6490245, member: 54866"] Frankzy's graph is interesting. I presume ad blue use is metered in at low engine RPM because combustion temperature is lower or something- seems to be gently falling as engine speed increases? The fuel consumption is interesting, too. At 1000rpm the engine is burning 25kg/hour, but only putting out 130 kW. I always thought that your best absolute fuel consumption would be where peak torque occurs- the speed where fuel is giving you the maximum bang for your buck in terms of Nm vs grams of fuel consumed- normally some way above idle and some way below peak engine power. If this is the case, then it appear to be around 1500rpm in the case of the 939 in this example, where the engine, according to PTO power, must be producing around 250 kW and using around double the fuel (50kg/hour). But, compare this to the point where the lowest [I]specific[/I] fuel consumption occurs which seems to be a sniff under 1300 rpm, interesting and bang goes my old-school theory.:unsure: Fendt drivers will be able to confirm at what revs a 900 series tractor will 'settle' at once the transmission is running at top speed under load (I presume this is somewhere around the 1300 and below mark) and, how steady the revs will be when putting 24 tonnes all told. Time for some old-fashioned calculatoring. :wacky: The 939 in this example is burning 210 grammes of fuel per kWh and running at a power output of around 190 kW. (Interestingly this should mean fuel use at that engine speed and power output of around 40kg/hour). 90 miles is around 145 KM. Assuming the journey was at 65kph the entire time, so the journey took about 2 hours 14 minutes or 2.23 hours. Need to work out total energy used in this journey- relatively simple, just times that 190 kW engine power (assuming rock-steady engine speed at all times) by duration of the journey. 2.23 hours x 190kW is 423.7 kWh. Take the total kWh of energy used, and times that by the absolutely best specific fuel consumption on that graph (210 grams/kWh) gives 88.977 kg of fuel used. Using the standard EC figures for off-road diesel, we can see best-case, the off road diesel will have a density of 845 kg/m3 or 0.845L/kg. Take the weight of fuel consumed (88.977kg) and times it by that density to get total fuel consumption in litres: 88.977 x 0.845= 76.185565 Litres. Need to convert that to (UK) gallons, divide total litres by 4.54609= 16.538512216 gallons. Divide miles (90) by gallons to give best-case fuel consumption: 90/ 16.53812216 = [U]5.4418437901 miles per gallon[/U], representing a best case scenario and with a transmission/drivetrain that gives zero loss from engine to road and with no change in engine speed whatsoever to get the vehicle moving initially (or any use of the brakes at any time). [/QUOTE]
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