ok, so here's a question for the mathematically and mechanically minded...

I was wondering about the amount of fuel my stock injectors would be maxed out at. given they are rated at 275cc/min (and these guys (http://injector-rehab.com/kbse/flowrates.htm) have tested them to 318cc/min @ 3 Bar), I thought it would be a simple equation relating to the bore, stroke, compression ratio and AFR to work out how much fuel was actually going through the injector at WOT @ xxxx RPM.

it doesn't appear to be that simple though.

my calculations follow:

bore = 94mm (my pistons are +1mm overbore)
stroke = 85.8mm
swept volume = pi * 47^2 * 85.8 = ~595.4cc (* 6 cylinders = ~3573cc or 3.6L :D )

pretty simple so far...

I need to find the volume of the combustion chamber (let's call that x) above the piston at top dead centre, so:

CR = 8.8:1
which means
1/8.8 = x/595.4 + x
8.8x = 595.4 + x
7.8x = 595.4
x = 595.4/7.8 = 76.3cc

still with me...?

so the combustion chamber is 76.3cc, and the swept volume is 595.4cc, so at bottom dead centre the volume of the whole chamber is 671.7cc.

now this is where I lose the plot.

an AFR at WOT is typically around 12.0:1 (12 parts air for 1 part fuel). that means if the whole combustion chamber is filled up with the air/fuel mix, there'd be 671.7cc/13 = 51.7cc of fuel going in on each intake stroke.

in a four stroke engine, the intake stroke is every other revolution, so at 3000 RPM, you have 1500 intake strokes per min.

1500 * 51.7cc = ~77.5 litres per minute... :nuts:

if you use just the volume above the piston at TDC (76.3cc) you get 5.9cc per stroke which works out to ~8.8 litres per minute (still WAY more than the ~300cc/min you'll get out of the stock injectors).

who can let me know where I've stuffed up?

I know it's not simple and you have to take into account fuel maps vs RPM and throttle position. but I'm out by a HUGE factor and can't figure out where.

cheers,
.wook

so the combustion chamber is 76.3cc, and the swept volume is 595.4cc, so at bottom dead centre the volume of the whole chamber is 671.7cc.




If the swept volume is 595.4 cc and the CR is 8.8:1 then the combustion chamber volume will be around 67.66 cc, I think :confused: .

During the intake stroke the pressure inside the cylinder is lower than the atmospheric pressure which means that you have less air inside the cylinder than it's volume is. Temperature of the air also affects it's volume.

You need to know the time (milliseconds) each injector is open during each intake stroke.
If the injector is rated at 275 (318) cc/min, that means it's maximum flow rate, that does not mean that at WOT it will flow that amount of petrol.

If you travel 100 km in 1 hour and use 10 litres of petrol, that mean your engine will use 166.66 cc of petrol each minute, or each cylinder will use 27.77 cc of petrol per minute.
Of course you will have WOT just occasionally during above example.

Let your ECU do the maths.

People often overlook some of the fundamentals of the combustion process and fundamentally, we should not start with volumes when trying to solve this problem.

The correct place to start is a mass balance of the fuel, air and products. A given mass of fuel requires a given mass of air and produces a given mass of product. Using very basic first principles, the actual volumes of each can then be determined via the ideal gas equation. Of course that calculation increment is for that state, and the inside of an i/c engine is not necessarily a steady state.

I suppose it's a bit easier to take other people's rules of thumb and then putting it on a dyno and tuning it. I doubt there are too many performance shops who would consult the laws of thermodynamics before modifying the engine, it's more of an art.

If the swept volume is 595.4 cc and the CR is 8.8:1 then the combustion chamber volume will be around 67.66 cc, I think :confused: .


Woops, my bad here. Wookiee your calculation for clearance volume was spot on.