coltsneckbob

I work in an R&D environment and so naturally my buddies and I usually have stupid (oh, lets say irrelevant), but interesting, technical discussion about things we really know little about...but think we can figure out!

So, today the question that came up is what is the peak PSI reached in a 455 cylinder. We figured that the absolute peak was when the fuel had fully ignited so this would perhaps milliseconds after the spark.

We had a formula to calculate that PSI from engine torque, but there were problems with that. First we needed torque as measured at the crankshaft. Secondly, since in a V-8 two cylinders fire very close to each other discounting the torque properly also became tricky. We also assumed a 10:1 compression ratio.

We did our best and arrived at something like 600-700 PSI. Given that we also calculated the square inches of a piston about 12 that gave us a tremendous amount of force on the piston, rod and crankshaft - as u can see it was over 8K pound of force at at max.

Does anyone know:
-the torque at the engine/crankshaft of a 70 455?
-Or, the actual force on the piston immediately after ignition (I googled it but didn't find anything authoritative.)

oldcutlass

I found this to be the best explaination of torque:

Torque to horsepower conversion

Now we need just a little math, it's easy but you will have to pay attention. Suppose we attach that one foot wrench to the end of a crankshaft and the engine rotates one revolution against that one pound of resistance. The end of the wrench will move 6.2832 feet (Pi * a two foot diameter circle) against a one pound weight. The end result is 6.2832 foot-pounds of work done at one foot-pound of torque.
Remember Pi? That's the ratio of the circumference of a circle to its diameter. Pi is a constant equal to 3.14159 carried for as many decimal points as you wish.
OK, here we go:
1 horsepower = 550 foot-pounds/second = 33,000 foot-pounds/minute
33,000 foot-pounds / 6.2832 foot-pounds = 5252
So, if the engine rotates against the one pound resistance at 5252 rpm:
6.2832 X 5252 = 33,000 foot-pounds/minute = 1 horsepower
because the one pound of resistance was moved 33,000 feet in one minute
(1 foot-pound X 5252) / 5252 = 1
Therefore, to convert torque to horsepower:

(Torque X RPM) / 5252 = Horsepower




For pressure you can also review this:

http://www3.fs.cvut.cz/web/fileadmin...gujevac-Bl.pdf

dmcianfa

If the nominal compression ratio of an engine is given, the pre-ignition cylinder pressure can be estimated using the following relationship:


where is the cylinder pressure at bottom dead center (BDC) which is usually at 1 atm, CR is the compression ratio, and is the specific heat ratio for the working fluid, which is about 1.4 for air, and 1.3 for methane-air mixture.
For example, if an engine running on gasoline has a compression ratio of 10:1, the cylinder pressure at top dead center (TDC) is


This figure, however, will also depend on cam or valve timing. Generally, cylinder pressure for common automotive designs should at least equal 10 bar, or, roughly estimated in pounds per square inch (psi) as between 15 and 20 times the compression ratio, or in this case between 150 psi and 200 psi, depending on cam timing. Purpose-built racing engines, stationary engines etc. will return figures outside this range.

This would be pre-ignition though, so I'm still looking up in my old textbooks post-ignition. One would have to assume that the potential/kinetic energy of the combustion of the fuel equals a particular pressure translation. So, I am looking at how much kinetic energy a given amount of fuel under this pressure gives off.

BlackGold

Bob, your estimate is pretty good. I've seen graphs of cylinder pressure in seveal technical books. The combustion peak for a performance street engine seems to be in the 800 to 1000 psi range. Of course, that's at WOT and within the power band. It'll be in the low hundreds if you don't cram as much mixture into the cylinder.

One of these books mentioned that race engines (didn't get real specific) see over 1000 psi. It also mentioned that Top Fuel engines see "several thousand" psi.

As an aside, several years ago I was playing the parlour physics game and wanted to prove or disprove the claim that Top Fuel cars drift in the direction of any cylinders which are not firing. The claim seemed a little out there but feasible. Using REALLY rough calculations, I figured that, yes indeed, the zoomies provide maybe 100 - 200 pounds of force as the combustuion pressure escapes, so a dropped cylinder really could result in a car -- on the verge of losing traction -- to drift sideways. That also tells you that the amount of downforce (and propulsion down the track) provided by the zoomies is not insignificant.