mew905
New Member
There's got to be a way to do this without an oxygen sensor.
Wikipedia says the optimal ratio of fuel to air is 14.7:1, or 14.7 gallons of air to 1 gallon of gasoline.
When I do the calculations, engine CFM is SUPPOSED to be (Displacement (CI) x RPM x VE)/3456 and that will get you a CFM rating. Well assuming a 65% VE on a reed-valved engine tested at 80MPG @ 40MPH (giving you a run time of 2 hours per gallon), that means that my engine should be sucking in that full 14.7 gallons of air over 2 hours. Safe to assume, right? Well, that formula says my engine should be sucking in about 4.5CFM at 6000 RPM (4 CI). Not much, right? Well remember 4.5 CFM is cubic feet per minute. So over two hours that gives us a number of 540 cubic feet over the course of 2 hours. Guess how many gallons that is? about 4040... That's right, over the course of using 1 gallon of gasoline, every calculation I can find says my engine is sucking in 4 THOUSAND gallons of air. That's 4000:1, waaaayyyyyyyy off from 14.7:1, the motor shouldn't even come close to starting.
So the question is... where am I going wrong? It should be possible to check your air-fuel ratio using this method (or at least get close). So unless our engines have a volumetric intake efficiency of 28%, this has to be way, way way wayy off. Unless my engine is just really poorly tuned... Average 2-stroke volumetric efficiency is about 55% according to a chart I saw.
I realize this calculation would change because I'm assuming I have a perfect ratio, which I know I dont, but it has to be close enough to even fire, whereas my calculations say I'd be way too far off.
Another way I would imagine is to use the compression ratios to calculate how much air the engines are getting (13.8 PSI atmospheric to 100 PSI combustion = 7:1 ratio), assuming reed valves and no losses through the exhaust...
Cmon guys, there's gotta be a way to do this.
EDIT: apparently that CFM calculation is for 4-strokes, and very wrong. Donaldson Engine Intake System Catalog says 2-strokes is (CID x RPM / 1728) x VE.
So that gives me 9 CFM... still way, way way way too much. I should be seeing 0.0159 CFM... or a VE of about 0.11% according to that calculation. Dafuq??
EDIT 2: Apparently it is actual WEIGHT for 14.7:1, so 14.7 grams of air vs 1 gram of gasoline. so 2.83KG of gasoline to 41.6KG of air, or 1120 CF of air over 2 hours. assuming I have the perfect ratio (which I don't), gives me 9.3 CFM. Seems I've got it! Horay!
So now, instead of a much more expensive AFR meter (AKA an oxygen sensor) which can cost upward of $300 for a handheld one (but give you real-time accurate results), you can calculate it based on your milage and intake air flow, which an air flow meter only costs about $15 for a cheap one.
Wikipedia says the optimal ratio of fuel to air is 14.7:1, or 14.7 gallons of air to 1 gallon of gasoline.
When I do the calculations, engine CFM is SUPPOSED to be (Displacement (CI) x RPM x VE)/3456 and that will get you a CFM rating. Well assuming a 65% VE on a reed-valved engine tested at 80MPG @ 40MPH (giving you a run time of 2 hours per gallon), that means that my engine should be sucking in that full 14.7 gallons of air over 2 hours. Safe to assume, right? Well, that formula says my engine should be sucking in about 4.5CFM at 6000 RPM (4 CI). Not much, right? Well remember 4.5 CFM is cubic feet per minute. So over two hours that gives us a number of 540 cubic feet over the course of 2 hours. Guess how many gallons that is? about 4040... That's right, over the course of using 1 gallon of gasoline, every calculation I can find says my engine is sucking in 4 THOUSAND gallons of air. That's 4000:1, waaaayyyyyyyy off from 14.7:1, the motor shouldn't even come close to starting.
So the question is... where am I going wrong? It should be possible to check your air-fuel ratio using this method (or at least get close). So unless our engines have a volumetric intake efficiency of 28%, this has to be way, way way wayy off. Unless my engine is just really poorly tuned... Average 2-stroke volumetric efficiency is about 55% according to a chart I saw.
I realize this calculation would change because I'm assuming I have a perfect ratio, which I know I dont, but it has to be close enough to even fire, whereas my calculations say I'd be way too far off.
Another way I would imagine is to use the compression ratios to calculate how much air the engines are getting (13.8 PSI atmospheric to 100 PSI combustion = 7:1 ratio), assuming reed valves and no losses through the exhaust...
Cmon guys, there's gotta be a way to do this.
EDIT: apparently that CFM calculation is for 4-strokes, and very wrong. Donaldson Engine Intake System Catalog says 2-strokes is (CID x RPM / 1728) x VE.
So that gives me 9 CFM... still way, way way way too much. I should be seeing 0.0159 CFM... or a VE of about 0.11% according to that calculation. Dafuq??
EDIT 2: Apparently it is actual WEIGHT for 14.7:1, so 14.7 grams of air vs 1 gram of gasoline. so 2.83KG of gasoline to 41.6KG of air, or 1120 CF of air over 2 hours. assuming I have the perfect ratio (which I don't), gives me 9.3 CFM. Seems I've got it! Horay!
So now, instead of a much more expensive AFR meter (AKA an oxygen sensor) which can cost upward of $300 for a handheld one (but give you real-time accurate results), you can calculate it based on your milage and intake air flow, which an air flow meter only costs about $15 for a cheap one.
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