"Darkbrador" (darkbrador)
02/18/2019 at 14:32 • Filed to: None
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"Darkbrador" (darkbrador)
02/18/2019 at 14:32 • Filed to: None | 0
| 8 |
So t he other day I was home sick, stuck in bed with some level of fever. T hinking about this long term motorcycle project of mine (“long term” is a positive word to actually say “unfinished”) and more specifically about the idea to relocate the gas tank inside the frame, not on top of the frame, for better weight distribution and to free up some space for a larger airbox and provisions for the electricals . A nd also to justify why this project is not completed yet.
Mental calculation, how to size a fuel pump to bring the juice from the tank to the carburetor... hmmm ... so let’s say ... an ambitious 6,000 rpm , the engine displaces 500 cc every other rotation, 13:1 air to gas ratio ...
6,000 x 0.5 x 0.5 / 13 = 116
I ’d need to flow 116 liter per minute ... errr ... the gas tank is about 6 liters, meaning I’d run 3 seconds at full throttle, that can’t be true.
I really had fever, and I’m not too smart to start with, but was I missing .... ?
Tekamul
> Darkbrador
02/18/2019 at 14:49 |
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That would be at WOT assuming no vacuum. The cylinders are sucking in much less than 500cc of air.
RamblinRover Luxury-Yacht
> Darkbrador
02/18/2019 at 14:51 |
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I believe the 13:1 ratio is a ratio of gases, not actual liquid fuel. For starters.
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RamblinRover Luxury-Yacht
> RamblinRover Luxury-Yacht
02/18/2019 at 14:57 |
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To elaborate: gasoline vapor is, best I can tell, less than a half a percent as dense as liquid fuel.
BigBlock440
> Darkbrador
02/18/2019 at 15:09 |
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You’re treating it as 13 liters of air, 1 liter liquid gasoline, but that’s not how it works. It’s 13 (lbs, g, kg, etc.) of air per 1 (lb, g, kg, etc.) of gasoline. Assuming 1 atm (16 psi), air weighs 0.0012929 gram per cubic centimeter, or .64 g/every other rev., 0.04973 g of gasoline. So at 6k RPM, you’ll be burning 149.2 g of gas per minute, or .052204 gallons. So assuming you’re screaming down the highway at 6k rpm and 60 mph, you’ll burn through 3.13 gallons in an hour and get 19.15 mpg (assuming constant 13:1 AFR).
3point8isgreat
> RamblinRover Luxury-Yacht
02/18/2019 at 15:14 |
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I agree it seems like an issue of assuming equal density.
The air/fuel ratio in engines at least is mass ratio. So OP would need to figure the mass of air flow, and then use the ratio to find mass of fuel flow. And then use density with the fuel mass to get volume flow.
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RamblinRover Luxury-Yacht
> 3point8isgreat
02/18/2019 at 15:27 |
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Yeah, I was forgetting about it being mass ratio for some reason.
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3point8isgreat
> RamblinRover Luxury-Yacht
02/18/2019 at 15:42 |
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I had to actually check, since it’s also sometimes a molecule ratio. And apparently changes based on application and what that industry finds most useful.
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Darkbrador
> BigBlock440
02/18/2019 at 16:02 |
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There you go ! Thanks amigo