11-30-2012, 02:45 AM
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#284 (permalink)
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Master EcoModder
Join Date: Nov 2012
Location: San Diego, California
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I see what you are saying.
Quote:
Originally Posted by oil pan 4
Unfortunatly the little kubotas only manage 14% running at 1/3 load.
1/3 of a gallon of diesel in, 2kw out over 1hr.
Typically from governed speed at no load to 100% load you see fuel consumption double.
When you compair no load to 1/3 load there is usually about a 10% difference in fuel consumption. Using that rule of thumb at full load they should be about 23% to 24% efficient. This is real world output and is good considering the engine is supplying power to turn the radiator fan, gives the coolant pump its power to turn, the engine effectivly providing all its own cooling, the generator is moving air to cool its self, the battery charging alternator is turning at 2x to 3x crank speed moving air plus only generates power with about 50% efficiency at best.
When an engine is efficiency tested that engine is tested at all speeds usually only has to pump its own oil, some times its own coolant and they post the peak number. Thats it, no radiator fans, alternators and such to rob shaft power or set speed to limit efficiency.
I have one of these and work on them all the time. I know big improvements could be made with out reinventing the combustion process.
Simple things like dropping down to 50Hz at least for light loads. Installing an over sized cooling system that usually doesn't require use of a fan and add a thermal switch controled electric fan, we see good gains on belt driven fan deletes with on road vehicles. It would apply here too.
Delete the battery charging alternator and replace it with a 110 volt plug in battery charger, it does the same thing as the battery charging alt with no moving parts and 90% or better efficiency.
Replace the exhaust manifold with a header. If the engine will be ran under medium to heavy load maybe install a more efficient 3 phase generator, run mild water injection, add a turbo and intercooler.
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You are speaking of Net efficiency while I was speaking of Gross efficiency. Subtract your losses from your Gross value and your net power output will be as you say.
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