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Chamber Pressure During Cranking and Running

In an earlier essay, I determined what the effective compression ratio would be if you were using boost. However, the assumptions made make that somewhat a phoney exercise. Why? Well, the compression stroke would have to be so slow that the heat of compression would be removed. It would be like turning the engine over, getting to the highest pressure then making a reading when everything cooled down. What is neglected is the thermodynamics of the situation. To get a better feel, turn on your air compressor and grab the pump to tank hard line...hot isn''t it? Well, I'll now tell you why that is and by how much...if you really wanted to know.

Chamber Pressure. You need to know the mechanical compression ratio and a guestimate of the volumetric efficiency of the engine before you begin this exercise. Let's assume for the moment that the volumetric efficiency of the engine is 0.9, temperature is a nice 60o F, and that barometric pressure is 14.7 psia. Look in the previous essay for the method of determining mechanical compression ratio. Here are the two equations needed:

P1 * V1 * VE / T1 = P2 * V2 / T2

and

T2 = T1 * (P2 * P1)0.283

where:

P1 = atmospheric pressure (ie baro pressure) * VE
P2 = pressure in the combustion chamber during cranking or running
V1 = Swept displacement of the engine (bore x stroke) + V2
V2 = combustion chamber space
T2 = air temperature after compression stroke
T1 = Ambient air temperature
VE = volumetric efficience, say around 0.9 or so

The two equations are needed because there are more unknowns than can be solved using just one. In this case, I chose to solve the first equation for the ratio of P /P1 and plug that back into the temperature equation.

P2 / P1 = T2 * V1 / T1 * V2

Now, recognise that the ratio of V1 / V2 is just the mechanical compression ratio and we have eliminated one of the variables. Then plug this back into the temperature equation and get:

T2 = T1 * (T2 * V1 / T1 * V2)0.283 = T1 * (T2 *(8.5) / T1)0.283

Remember, that when dealing with thermodynamis, we need to deal in absolute pressures and temperatures. I grew up using the Rankine/ Fahrenheit system of measurement so, by damn, that's what we will use. None of the metric silliness for me! Now we have one equation and one unknown, T2, becaue we know T1, because it is the ambient temperature.

T2 = T1 * (T2 *(8.5) / T1)0.283

T2 = (460 + 60) * (T2 *(8.5) / (460 + 60))0.283

T2 = 520 * (T2)0.283 * (8.5 / 520)0.283

T2 = 520 * (T2)0.283 * 0.312174

T2 = 162.33 * (T2)0.283

T2 / (T2)0.283 = 162.33

T2(1 - 0.283) = 162.33

T20.717 = 162.33

T2 = 1210.16o R

The air temperature, in Farhenheit is:

T2 = 1210.16o R - 460o R + 60 o F = 810.16 o F
The temperature rise, above ambient is 750.16 o F

So, now we just plug T2 back into the original equation and find P2

P2 / P1 = T2 * V1 / T1 * V2

P2 = P1 * T2 * V1 / T1 * V2

P2 = 14.7 * 0.9 * 1210.16 * 8.5 / 520 = 261.79 psia

So this is the real instantaneous pressure in a combustion chamber given the conditions I specified or 8.5 CR, 60o F, barometric pressure of 14.7 psia and a volumetric efficiency of 0.9. Of course you can use any numbers you want, just plug them into the equation. Finding T2 is the hardest.

If we use the same analogy of using pressure to compute the pressure compression ratio (PCR) then:

PCR = New pressure / Old pressure 261.79 / 14.7 = 17.8: 1
This is more than twice the mechanical compression ratio!

Now would I use this? No because it has no common usage anywhere except as the starting point for an otto cycle analyses. Also, this would exist only momentarily in the cranking combstion chamber because of the heat transfered from the compressed charge to the cyliner head, piston top, valves, etc. So this is mainly an academic exercise and not highly useful.

Copyright (C) 1998 - 2004, all dates inclusive, L.E. Mayfield - All Rights Reserved
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