Thread: cylinder compression 3.5l 9.35:1

With 300KK any compression would be good, but all cylinders should be within 10% of each other.
I would think an engine in good condition with 9.35:1 should have at least 150PSI . My LC 3.9 had 150PSI+- 5 on all cylinders.

The cause of the low PSI could be anything from leaking valves, worn camshaft, to worn out rings or a combination of all of the above. Or even an artifact if you didn't hold the throttle open and crank at least 4 or 5 times.
Regards Philip A

My 3.9 9.35:1 was up between 170 and 185, but it had new rings, valves and gaskets 10kkm ago.

At sea level and standard pressure and temp, air pressure is 14.7 lbs per square inch so a compression ratio of 9.35:1 should have a total compression of 137.5 psi. (9.35 x 14.7=137.5)

So unless is is very cold and high atmospheric pressure - or a wrongly reading gauge there is not way you can get 150 - 180 psi.

Garry

Interesting Garry. My notes that I have here are quite clear, and a search around the net have managed to find similar results to mine.

I'll go and attempt to do it again over the weekend, but I followed the instructions on here in a post by Blknight.aus, with a tried and trusted gauge.

I can certainly agree with your maths, so I need to find and eliminate which variable is producing dodgy results.

Just continuing to hunt around, and came across this post from JC: V8 compression test?

Indicating that 150-170 is ok. So perhaps mine with it's mods and rebuild aren't too far over the top.

EDIT
The relationship between pressure and volume P1 x V1 =P2 x V2 is "Boyle's Law" and on this basis a 9.35:1 compression would indeed yield a pressure of 137.4 psi.

However, this assumes an "ideal" gas compressed at constant temperature.

Then Charles (of Charles' Law) came along and said that for a constant pressure, if you heat a given volume of gas, it will expand in proportion to the amount of heat you put in (as measured by the increase in temperature) V1 x T1= V2 x T2


In measuring engine compression we have changing pressures, changing volumes and changing temperatures.

The reason for the increase in pressure above and beyond a simple calculation of compression ratio x atmospheric pressure is due to an increase in temperature caused by an input of energy units into the air-fuel mix from compressing it (think.. heat generated using air compressor....)

So, combining the formulas above we get (P1 x V1)/T1 =(P2 x V2)/T2

P1 is atmospheric pressure (14.7 psi)
V1 is the normal volume of the cylinder (let's call this volume 9.35)
Ti is ambient temp. (usually expressed in deg absolute but let's approximate...)

when the air-fuel charge is compressed the resulting pressure @TDC is P2

V2 =1 ( as in 9.35:1)

T2 ... let's say the temp increases 10% ...this will vary depending on how many revolutions are used to obtain a steady reading, the admix of fuel vapour can also effect the heating issues)

So, solving the equation for P2 we get P2 = ( P1 x V1 x T2)/ (T1 xV2)

Remember : We don't need to know the exact absolute value for T2 or V1 or V2, just the relative values of V1/V2 = compression ratio =9.35:1 and T2 = 110% of T1 (we assume this)

So P2 (the reading on the gauge) we would expect to be (14.7 x 9.35 x 1.1) /(1 x1)
= 151 psi (approx)

This assumes the temp increases only 10% e.g. on a 25 deg day i.e 2.5 deg, while the engine is turning over. Lets say the temp went up 5 deg because of the "work" being done ... this equates to a 20% increase in relative temperature..

Then the result would be 165 psi... etc

Unfortunately checking compression pressures does not relate to the compression ratio.
The compression ratio is as it states, a ratio. IE The swept volume compared to the static volume at top dead centre, generally the cylinder head volume.
Any calculations should also use absolute values, temperature should be in Kelvin, Pressure Absolute not gauge - Absolute being gauge + 1 atmosphere.

I love that i asked a simple question and get such a complex and comprehensive answer.

Originally Posted by Hoges

EDIT
The relationship between pressure and volume P1 x V1 =P2 x V2 is "Boyle's Law" and on this basis a 9.35:1 compression would indeed yield a pressure of 137.4 psi.

However, this assumes an "ideal" gas compressed at constant temperature.

Then Charles (of Charles' Law) came along and said that for a constant pressure, if you heat a given volume of gas, it will expand in proportion to the amount of heat you put in (as measured by the increase in temperature) V1 x T1= V2 x T2


In measuring engine compression we have changing pressures, changing volumes and changing temperatures.

The reason for the increase in pressure above and beyond a simple calculation of compression ratio x atmospheric pressure is due to an increase in temperature caused by an input of energy units into the air-fuel mix from compressing it (think.. heat generated using air compressor....)

So, combining the formulas above we get (P1 x V1)/T1 =(P2 x V2)/T2

P1 is atmospheric pressure (14.7 psi)
V1 is the normal volume of the cylinder (let's call this volume 9.35)
Ti is ambient temp. (usually expressed in deg absolute but let's approximate...)

when the air-fuel charge is compressed the resulting pressure @TDC is P2

V2 =1 ( as in 9.35:1)

T2 ... let's say the temp increases 10% ...this will vary depending on how many revolutions are used to obtain a steady reading, the admix of fuel vapour can also effect the heating issues)

So, solving the equation for P2 we get P2 = ( P1 x V1 x T2)/ (T1 xV2)

Remember : We don't need to know the exact absolute value for T2 or V1 or V2, just the relative values of V1/V2 = compression ratio =9.35:1 and T2 = 110% of T1 (we assume this)

So P2 (the reading on the gauge) we would expect to be (14.7 x 9.35 x 1.1) /(1 x1)
= 151 psi (approx)

This assumes the temp increases only 10% e.g. on a 25 deg day i.e 2.5 deg, while the engine is turning over. Lets say the temp went up 5 deg because of the "work" being done ... this equates to a 20% increase in relative temperature..

Then the result would be 165 psi... etc