This is a question that i have been wondering how to calculate for a while and finally wanted to post this to get some more expert people. I have just all ways wondered how much pressure are motors build up inside when we start pushing lots of air into it.

Just remember these calculations are in a perfect world, we all no this world isn't perfect


Ok

We all know that are motors in the best of shape will do about 120-125pis compression, well what happens to this pre-ignition compression when you add X number of boost into the intake before the compression starts.

1ATM (Atmosphere) = 14.696 psi

Like a NA motor draws in 14.696 psi of air before it is compressed to the 120-125

Now when you introduce 24.696 psi into the motor (14.696 + 10psi) before you compress, with my calculations you get 204.24 psi

Now this is how i came to this final number but it probably is wrong, but thats why this thread is here.

120psi / 14.696 = 8.27

8.27 X 24.696 = 204.24 psi of total compression at TDC

So 204.24 psi with 10psi of boost.

So 286.94 psi with 20psi of boost

So is this how you would calculate the pressure at TDC

 

a bit over my head for now, but i'd like to know as well ....

 

nope.

there are a couple kinds of pressure. guage pressure(Pg) and absolute pressure(Pabs). absolute pressure Pabs = Pg + Patm where Patm = atmospheric pressure.

the 120 psi is guage pressure. look at a pressure guage when its not in use, it doesnt read 14.969 psi does it. it neglects atmospheric pressure. the 14.969 would be atmospheric pressure at sealevel. you have to use absolute pressure.

Patm = 14.969 psi
120 psi from compression (guage pressure ) is equal to 134.696 psi in absolute pressure.

now your compression ratio will be 134.696/14.969 = 9.165

now it you have 14.969 psi + 10 psi of boost. initial pressure is = 24.969 psi.

now multiplying 24.969psi*9.165 you will have 228.85psi in absolute pressure.

228.85 psi- 14.969 psi = 213.88 psi in guage pressure.

this is done assuming many things, one of which is that im not to drunk right now to type this.

 

There's a very basic flaw in this whole thing. The compression numbers are taken at cranking rpm. However, because of that, the air in each intake charge is rather less than the total volume it could be; some is going back out the intake port, because the intake ports are designed to make power at higher RPM. Put a different way, the volumetric efficiency at cranking RPM is going to be a lot lower than that at, say, 4k RPM, obviously. So, even without boost, the chamber pressure is going to vary a lot. Unless you're making full boost at idle, the 120psi figure isn't that useful.

 

That is true, one other thing i forgot to think about is this

When i check compression with my digital gauge and let the motor start on one rotor it would never go back to zero after each compression cycle. The rotor with the compression test being preformed had no plug wires on it so it wasn't firing off any fuel.

In other words there was about 14-20psi of pressure (I guess from the exhaust gas)

Other people that have forgotten to turn the fuel off when checking compression probably understands this.

 

And another more glaring flaw.

Compression doesn't cause pre-ignition or detonation, it is the resultant heat of compression (or other heat source in engine- too hot plugs, poor quench area).

What difference does that make?

Well, a turbo/supercharger is much more efficient at compressing the air (operating within its efficiency range) than the piston/rotor and that means less heat from its compression and less chance of detonation.

You can also intercool the compressed charge leaving the turbo/supercharger to further reduce the heat from compression which you cannot do with the piston/rotor.

There are so many other factors in engine design that affect the compression vs boost ratios as they pertain to correct fuel burn that you would really have to test the EXACT engine set up in question with all the various boost levels and compression for accurate results.

Who knew 50 or even 20 years ago that gasoline engines could safely boost 15-20psi in production trim as they do today? Well, they couldn't then.

 

wow you guys are good at making things more complicated then they need to be!

ok first off, this was a crude calculation in a "perfect world", so we'll make some assumptions.

assumption #1. air is ideal gas and follows the ideal gas law. PV = mRT. P= pressure, V = volume, m = mass, R = gas constant for air, T = temp

assumption #2. during the compression process, the temperature does not change. this is crude, does not really happen but simplifies calculations significantly. but we will calculate it this way for simplicity.

assumption #3 combustion does not start untill after TDC. we all know you fire spark plugs BTDC in the real world. if you fire BTDC then combustion will contribute to a pressure increase in the chamber even before TDC.

assumption #4 its a perfect world there is no loss of pressure like seal leakage. so rpm doesnt have much effect.

assumption #5 before compression starts the chamber equalizes pressure with intake. iex: when the intake port closes the chamber pressure is 24.696 psi. 14.696psi + 10psi of boost.

assumption #6 no mass of air leaves the system during the compression stroke. (perfect seals)

now for calculations:

there are 2 states of concern. air before its compressed(S1) and air after its compressed(S2). here are the ideal gas laws for each state:

S1: Pmin*Vmax = m*R*T S2: Pmax*Vmin = m*R*T

m is constant since no mass leaves the system during the process.
r is constant (gas constant for air)
T is constant because we are assuming so to make life easy.
that makes every thing on the right side of those two equations constant.

so: Pmin*Vmax = m*R*T = Pmax*Vmin
and: Pmin*Vmax = Pmax*Vmin

divide both sides by Vmin and you get: Pmin*Vmax/Vmin = Pmax
divide both sides by Pmin and you get: Vmax/Vmin = Pmax/Pmin

putting those 2 equations together you get: Pmin*Pmax/Pmin = Pmax , or in other words multiply the initial pressure by the ratio of compressed pressure to initial to get the final pressure. note that the pressure ratio is equal to the ratio of Vmax to Vmin. so as long as initial volume and final volume does not change then neither does either ratio.

from my earlier post thats where 134.696/14.969 = 9.165 comes from this is the compression ratio. dont confuse this with the theoretical compression ratio of the rotors, they are a little different. you can no multiply that ratio by any initial pressure to find the final pressure, as long as the Vmin and Vmax remain the same.

note that Volumetric efficiency has nothing to do with anything, and I think that it is actually poorly named. VE impies the ratio between the actual volume of air ingested and the theoretical maximum amount that can be injested. it should imply that higher VE's mean a large mass of air is ingested.

if you take 2 cubic meters of air injested into a 1 cubic meter engine then you have VE = 200% but you still have 1 cubic meter of air. that air is now at a higher pressure and has more mass. mass doesnt matter in this analysis. pressure does, but that only effects the final pressure.

like i said this was a very, very crude analysis. analyzing pressure inside the combustion chamber is very complex subject. this was mearly a simple model to understand what is happening to pressure.

 

inertial supercharging?

 

Another huge variable, though sort of stated in other words in previous posts, is the difference between static and dynamic compression.

Inertial Supercharging = Ram Charging = Volumetric efficiency exceeding %100.

 

inertial supercharging should be taken care of with assumption #5. what exactly do you mean by static and dynamic compression? im not familar with those terms. maybe static is the theoretical compression ratio which is volume at bdc divided by the volume at tdc? and dynamic is volume when the intake port closes divided by volume at tdc?