Required Energy For Air Compression for Cumbustion Engine

I'm not going to be able to answer, but I do want to refine the question so I'll understand the answer.

Are you asking what percent of loss an internal combustion engine is attributable to the compression cycle?

Its a bit late at night and I feel sleepy and still have to put my child to sleep, so I'm not going to make calculations. But as I read this post I can remember an interesting number I once was taught during an introductory class about thermodinamics in university, I think it was in physics 2. After all the calculations, the efficiency of an ordinary car in respect to the fuel energy after the burn, transmission, noise, heat, etc was something that could not get to 20%. Specifically your question is most likely to be answered if you look for the Carnot cycle for thermal machines. A typical 4 time engine cycle representation explains the amount of energy spent in each engine cycle in terms of a simple 2D graphic, where you can compare visually the areas representing energy. With it, you can also see clearly the differences between engines with Otto cycle, 2 times, diesel cycle, and atkinson.

Been there, done that.

Typically... 1 cubic ft / 1 gallon (close enough to 1 liter) of compressed air under as little as 100 psi will push on piston down. This is true (rounded off) for nearly all useful sized IC engines.

4cyl, you'd need 4CF of air for one revolution of full power. (roughly the entire tank of your typical portable nail gun compressor. I've done this and it is cool, but waiting for the refill of the air tanks or snail pacing along winds down the enthusiasm.

Keep in mind, that a 4 cyl engine becomes an 8 cyl air engine as it can use both strokes as compression! Lots of power to be had. ((I patented a device that will supply the air to do just what you are thinking!))

Good luck, my friend, I know where your going with this. MealerAMC (the next major US Automaker) is about to produce a vehicle that runs on a system (patented by myself) that began with these very concepts.

Stay in touch, I like your way of thinking.

It looks very interesting. Will it be on models 2010?.

Hello Guest.

Since is already patented could you tell us about it, or where to look for the info?

There is a clarification .He is asking the energy required to compress the air .This is the first phase of Carnot Cycle.It can be calulated for a specific case or can be generalised.The power developed in the Carnot Cycle (IDEAL).

= Power Developed in expansion-Power Spent in Compression.

After multiplying with various losses it is of the order of 25-30 %.

If you consider the losses power consumed may be 50-60%.

In a specific case it can be calculated accurately.

Native

Ah, I happen to have such a program...

Here are some numbers...if you want...

I'll put them in units of work (hp or kw). Work expansion includes combution and an adiabatic blow down

Work expansion, work compression, net work, percent of Work expansion. Ok? Note that Otto cycle needs to be close to 100% due to ignition issues, and that the throttle drops the effective compression ratio...

16:1 diesel @ 100% of max fuel burn, 1043, -236, 807, 23%

16:1 Diesel @ 50% of max fuel burn, 697, -236, 461, 34%

16:1 Diesel @ 7% of max fuel burn, 306, -236, 70, 77%

9:1 Otto, 1046, -147, 899, 14%

4:1 Otto, 888, -108, 780, 12%

2:1 Otto, 750, -82, 668, 11%

So, the % compression work for a Diesel goes from 23% to 95% and Otto stays around 10-15%. Efficiency? That is another issue entirely...

I know a company that got $1.5 million of your tax dollars to model an engine just to get these answers...on CR4...free. What a deal...

Hope that answers your question.

Seaplaneguy

I found some formulas are more adjusted to gasoline engines, you must assume cilinders are not full of air when compression starts.If i didn't fail (me tired too):

E=(Po*Vo/(g-1))*(1-(1/r**(g-1))), taking Po=,55atm;g=1,4;r=compression ratio.So, just for easy calculation take r=10: E=0,83*Vo (atm*L),Vo=cilinder volumen or in SI units: E=(81*Vo) joules (keep using litres number for Vo).

I think the comparison between this energy with delivered by the engine must be made by you, course may be you want to compare related to the total energy in the gasoline carburated in one cilinder in one stroke knowing 11000 cal are contained in one gram of gasoline (i don't remember exactly but think 11000 is OK)(multiply by 4,19 to get joules) and knowing too 17 grams in air is needed for 1 gram gasoline...Could supose : Et= gas.mass*11000=n*29/17, where n=0,55n',n'=P*V/R*T...

Should be all and not exact.-

Check out http://www.engsoft.co.kr/index_r1_e.htm

The prgm you want to use is ES_Stable. Just by fixing two properties, you can retrieve the properties of the other states of the Rankine cycle.