Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
https://overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding

User Menu

Custom Search

Author Topic: BACK TO BASICS.............  (Read 2323 times)

dsmer1991

  • Newbie
  • *
  • Posts: 20
BACK TO BASICS.............
« on: December 31, 2008, 07:34:54 AM »
going off of my car, a 2008 chevy Aveo with a 1.6 liter motor it requires a 1.6 liter displacement of air to complete one revolution with a fuel source to create (cumbustion,explosion,ingition)whatever. using gasoline as a good source of combustion it gets about a 14 to 1 air fuel ratio to burn with somewhat effeciency. there are many fuel sources that can get a better ratio of air to fuel. hydrogen being one of them.i even believe that a 14:1 ratio seems a little rich myself but it is a documented number and is easy to work with to get my point.

-at idle my car is running at about 650 RPM's..........that means for everyone reading that my car burns 1040 liters of compressed air with a ratio of 14 to 1 a minute..........i did the math and i am missing something. Possibly size of combustion chamber,shape and location and size of injectors and what not.

i hope i am missing something stupid because from the looks of my math the internal combustion engine itself is a OU vehicle and we need to give props where it is do.

the best answer i can come up with is it takes the 1.6 liters and compreses them to the determined size of combustion chamber and THEN addes the 14:1 ratio of fuel. because if go off of straight atmosperic pressure of air then my engine is outperforming all calculations. anyone with further insight, please post. cause using my math which once again is documented hydrogen burns with a 120 to 1 ratio, somewhat effeciently which either means i need a whole lot per revolution or a whole little.

basically what i am asking is the (mileage of a pure hydrogen powered car) fuel consumption and what not. i know someone hooked up a tank and let it run what size tank and how long did it run.

TinselKoala

  • Hero Member
  • *****
  • Posts: 13958
Re: BACK TO BASICS.............
« Reply #1 on: December 31, 2008, 07:03:15 PM »
I can't identify exactly where your error or misconception is, because you don't show your calculations. But I'll bet it is because you are not taking into account that the 14:1 ratio is concerning the ratio of air to fuel by mass, not volume.
So your 1040 liters of air is about a cubic meter, which has a mass of about 1.3 kilograms. 1.3 kilos divided by 14 is about 93 grams.
Plus, I think a 4-cycle engine has to go around twice for every displacement volume, so cut that in half, so about 45 grams of gasoline per minute at 650 rpm. That's about 2 1/2 tablespoons.
Still seems like a lot to me.

But is it really? While running at speed on the highway, maybe you are burning 8 liters per hour or so. 50 grams of gasoline is about 65 cc, so in ten minutes idling that's 650 cc, or about 4 liters per hour at idle using my numbers from the above calculation.
So we are in the ball park anyway, and the Chevy isn't Aveo-runity after all.
 ;)

fritznien

  • Sr. Member
  • ****
  • Posts: 294
Re: BACK TO BASICS.............
« Reply #2 on: January 01, 2009, 12:28:34 AM »
what you missed, first nothing OU about an otto engine.
next you forgot about manifold vacum. the throtle plate controls the air flow. at idle it will be closed and your volume of air  is much restricted