How does one define the expansion ratio of an internal combustion engine? My boy is doing a science project -he chose how a 4 cycle engine works. I'd like him to show how expansion is the key to engine function/power/tq. I realized, I've never learned or known how to calc the expansion ratio of a 14.7:1 gasoline/air mix. I'm sure there isn't an "Exact" answer, but is there a method to show roughly how "big" a ~70cc charge wants to be after ignition?

 

..... You might ask the race gas folks for that kind of info ... try www.yellowbullet.com ... race fuel forum in the technical section ... most of the various race fuel vendors are in attendance i.e. : VP , Sunoco , Renegade , and others .......

 

Kindly let us know what you find.

 

Use the ideal gas law: PV = nRT.

P is pressure, V is volume. n is the number of molecules. R is a constant to make the numbers and units come out nice. T is temperature (my recollection is that it is in Kelvins).

Basically, when the fuel ignites the temperature goes way up, which causes the pressure to do the same. The piston gets pushed down in the bore and the volume expands. That relieves some of the pressure, but not much.

 

Hmm so what numbers are n and R?
What I'd like for my boy to show is; you take a typical charge in cylinder, it's about 100cc's? Ignite it and forget about the cylinder and piston -let that charge expand to what ever size it wants to at atmospheric pressure. What is that volume? That volume will be huge, I imagine, and that will show why an IC engine works so well.

 

Going back a few decades - - we used a ball park ratio of 4:1 -compression and combustion pressures. Not exactly a precise science for sure but 200 PSI compression yielded 800 PSI in combustion pressure. Advances in everything have probably moved that "ball park" ratio to 5:1 - perhaps even more -

To help your son - have him do some research on BMEP -Brake Mean Effective Pressure. BMEP is a calculated number used to determine the average (mean) pressure in a cylinder acting upon the piston to produce a given torque. Do a Google search about combustion pressures - lots to learn about this.

Also - - Check out this site. http://www.epi-eng.com/index.html Jack Kane is a very brilliant engineer and his writings are such as to be understandable by most technical folks. Your son will enjoy this place - click on the links located on the left side.

Best of luck in your project.

Jake -

 

I was a professional wrench for years several decades ago and I have never heard the term, "Expansion Ratio." So, I'm going to take a stab here and suggest that the "Expansion Ratio" is the reciprocal of the Compression Ratio. If the compression ratio is, say, 10.5:1, the expansion ratio is 1:10.5. No?

 

The ideal gas law is explained here:

http://en.wikipedia.org/wiki/Ideal_gas_law
The constant R is explained here:

http://en.wikipedia.org/wiki/Gas_constant
n is the number of molecules you're dealing with. In the above references that's generally expressed in moles, which is 6.02 x 10^23 molecules (Avogadro's number). That number of molecules gives you the "atomic weight" of an element.

This is a tough problem. I guess you could use some simplifying assumptions. The n and R don't change. You would have to figure out how much the temperature T(1) increased due to combustion. Assume initially that the volume V(1) doesn't change (it doesn't for a tiny fraction of a second) and then calculate the increase in pressure P(1). Now the pressure will go down to atmospheric pressure P(2) when the gas expands completely and the temperature will also go down to whatever the ambient temperature is T(2). That leaves only the change in volume V(2) to calculate. You would end up with two equations that must balance, one just after combustion and one after expansion:

P(1)V(1) = nRT(1)
P(2)V(2) = nRT(2)

You can rearrange the second equation to find the final volume, which would be:

The important thing in all this is to keep the units consistent. You will have to convert back and forth between °F and °K and make sure the volumes work. If you start with a volume of 100cc, then it needs to be converted to whatever is needed (like liters, quarts, gallons, etc.). The same applies to pressure as well. It could be in PSI or Pascals or something else, but needs to be consistent with the units of the other parts of the equation and the constant R.

 

If only those Stanley Steamers had caught on. Steam expands at ratio of 1,100 to one; for some reason, 1,400 sticks in my mind; it might for the diesel cycle though...now where did I put my slide rule ?