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Author Topic:   Compression Ratio, PSI, and BARs
jimh posted 01-22-2008 12:05 AM ET (US)   Profile for jimh   Send Email to jimh  
My local Evinrude certified master mechanic checked the cylinder pressure on my trusty old V6 Evinrude 225-HP and gave me the readings. The values were in pounds-per-square-inch or PSI, and the numbers were in the 90 to 95 range.

Atmospheric pressure, nominally 1-BAR, is about 14.5-PSI. So if I divide my cylinder pressure readings by 14.5, I ought to get the compression ratio:

90 / 14.5 = 6.2

This implies my engine has a compression ratio of 6.2:1.

I have heard some guys toss around PSI readings in the 150-range. That implies quite a bit higher compression:

150 / 14.5 = 10.3

or a compression ratio of 10.3:1.

Is my assumption regarding the relationship between PSI and compression ratio correct?


Cf: http://en.wikipedia.org/wiki/Pressure

an86carrera posted 01-22-2008 06:38 AM ET (US)     Profile for an86carrera  Send Email to an86carrera     
I always thought that compression ratio was a volumetric calculation such as, 500 cc volume at bottom of stroke versus 50cc at top of stroke = 10:1.
tmann45 posted 01-22-2008 06:55 AM ET (US)     Profile for tmann45  Send Email to tmann45     
Simplistically, yes.

P1*V1 = P2*V2

V1 = P2*V2/P1

So, if P2 is equal to 1, as in X:1 ratio, then V1, the initial volume is equal to P2, the compression pressure, divided by atmospheric pressure.

Peter posted 01-22-2008 07:44 AM ET (US)     Profile for Peter  Send Email to Peter     
In a 2-stroke, I don't think you are getting any meaningful compression until both ports are closed off by the piston. So is the maximum volume of the combustion chamber at the beginning of the up stroke the right basis for determining compression ratio on a 2-stroke?

In a 4-stroke, on the other hand, I believe compression happens at the beginning of the up compression stroke so the maximum volume of the combustion chamber would seemingly be the right basis for determining compression ratio.

Yamaha publishes compression ratio specifications for its outboards. Their V6 2-strokes have a compression ratio right around 6.2:1. So the 6.2:1 figure calculated for Jim's Evinrude 225 seems about right (or 6.1:1 if you use 14.7 PSI as atmospheric pressure at at sea level). Their 4-strokes tend to have a compression ratio mostly in the 9:1 to 10:1 range so you would expect to see a compression gauge reading in the 132 to 147 PSI range.

Tohsgib posted 01-22-2008 02:11 PM ET (US)     Profile for Tohsgib  Send Email to Tohsgib     
Actually it really depends on the gauge used and the engine it is used on. My 225 was just over 100 when completely rebuilt(with my gauge). My buds was between 70-90 and was getting tired(with my gauge) but the tiredness was due to the low PSI. 90-95 is good on that engine. My 40hp Evinrude 2S is over 140 on each pot. My jetski is over 180. That is why I said it depends on the engine.
Jerry Townsend posted 01-22-2008 02:12 PM ET (US)     Profile for Jerry Townsend  Send Email to Jerry Townsend     
Jim - you are basically right, but remember that the barometric pressure changes with altitude and is about 14.7 psig at sea level. The ratio of the maximum pressure (measured with a pressure gage) divided by the atmospheric pressure is the ACTUAL compression ratio. The "actual" compression ratio is also indicative of the condition of the valves and rings - that is leaking valves and/or rings will decrease the maximum pressure and hence the compression ratio.

A 4 stroke gasoline engine will have a design compression ratio of around 10 +/- -- and at sea level would give a maximum cylinder pressure of around 150 +/-. Having a cylinder pressure of 115 - 150 would be indicative of a good tight engine. I don't have experience in the 2 stroke engines.

And yes, the design of the engine cylinder gives the maximum/optimum compression ratio. This "optimum" compression ratio is the sum of the displacement volume (cylinder area times the stroke throw) plus the "clearance" volume (cylinder volume at top-dead-center) divided by the clearance volume.

tmann45 - your PV = constant formulation is applicable for a process where the temperature is constant - but that is not the case in an internal combustion engine. Indeed, diesel engine operation is based on the temeprature rise in the ccompression process for ignition of the mixture - via a high compression ratio. ---- Jerry/Idaho

jimh posted 01-22-2008 09:03 PM ET (US)     Profile for jimh  Send Email to jimh     
Interesting comments all around. First, I see that "standard" pressure and normal atmospheric pressure are different, so I agree, we ought to use 14.7-PSI as the base pressure.

Regarding how different instruments will have different readings, yes, this is a given. It does not affect how we use the data to compute the compression ratio. We just assume our instrument is accurate, and we compute the compression ratio.

Now as for the change in temperature, that is a valid point. The temperature of the gases in the cylinder rises during compression, and that might have to be considered. But I'd say that is a fine point.

Regarding the volume of the cylinder and how the intake and exhaust ports affect the compression, that is a very interesting observation. I don't know what protocol is normally used when computing compression ratios. That is, do you use the whole cylinder volumes, or do you actually measure the pressure change. Perhaps someone can comment.

jimh posted 01-22-2008 09:04 PM ET (US)     Profile for jimh  Send Email to jimh     
Oh--here is a good discussion of compression ratio.

http://en.wikipedia.org/wiki/Compression_ratio

The Wikipedia article says

Pressure at dead top center = 1-ATM X (CR)^1.4 where CR=compression ratio

Since we measure the pressure at dead top center, we want to solve for CR. Lets call pressure at dead top center PSI and use ATM=14.7:


CR = (PSI/14.7)^0.71

Thus for my engine, measured at say 90-PSI, my compression ratio works out to

CR = (90/14.7)^0.71

CR = 3.6

Hmm, that does seem rather low. But the article does say:

"Factors including late intake valve closure...can produce a misleadingly low figure from this test."

I think that describes the situation in the cylinder of a two-cycle motor--the "valve" closing is rather late.

Bella con23 posted 01-22-2008 09:40 PM ET (US)     Profile for Bella con23  Send Email to Bella con23     
I would think that the different bore and stroke of 4 stroke vs. 2 stroke block would also play into the equation.
A 225 Verado has a 3.23" bore and a 3.23" stroke.
The 225 Optimax has a 3.63" bore and a 3.00" stroke.

The Optimax is at a further disadvantage due to "the "valve" closing is rather late."
Joe

Jerry Townsend posted 01-23-2008 12:40 AM ET (US)     Profile for Jerry Townsend  Send Email to Jerry Townsend     
Jim - The compression ratio is defined as the piston displacement plus the "clearance" volume divided by the clearance volume. Generally, we don't know the clearance volume.

As I mentioned above, the process is not one of constant temperature - but is an adiabatic process (no heat transfer) and for a perfect gas, the exponent is the 1.4 that is given in the Wikipedia article. BUT, we are dealing with a fuel/air mixture and the process is not strictly adiabatic, nor strictly reversible - therefore, the exponent will be somewhere between 1.0 and 1.4.

Note that by just dividing the maximum pressure by the atmospheric pressure assumes an exponent of 1.0 - which we know is not correct - and will give a high compression ratio. But, using 1.4 is incorrect as well. The exponent is one big unknown and I suspect that the manufacturers will use something around 1.3

The actual equation is: P)1 x V)1^EXP = P)2 x V)2^EXP
where the states 1 and 2 represent, for example, the bottom of stroke or top-dead-center.

Remember - the atmospheric pressure of 14.7 at sea level is absolute pressure - but the pressure obtained from a pressure gage is gage pressure. Absolute pressure (psia) is given as the gage pressure (psig) plus the atmospheric pressure.

Also - the atmospheric pressure at sea level is the 14.7, however, many are at a higher elevation. Atmospheric pressure varies as the altitude which can be calculated as:

P)atm = 14.7 - AltitudeInFeet(0.0749/144)

so that the atmospheric pressure at 4000 feet is about 12.6 psia.

Your low calculated compression ratio is due to several things - the exponent is smaller than the 1.4 and there is some leakage around the valves/reeds and rings and late timing can also contribute. I don't think your engine is shot. ------------- Jerry/Idaho


jimh posted 01-23-2008 12:58 AM ET (US)     Profile for jimh  Send Email to jimh     
Jerry--Thanks for the excellent comments. And, no, I don't think there is any problem with the engine. It runs very well, and , according to David Zammitt, whose has been a Master Mechanic for OMC for decades, the compression readings taken on the engine were typical for that model.
jimh posted 01-23-2008 08:21 AM ET (US)     Profile for jimh  Send Email to jimh     
Following Jerry's suggestions, I recalculate the compression ratio (CR) using

90 PSI (measured) = 1-ATM X (CR)^1.3

where 1-ATM = 14.6 PSI. Solving for CR


CR = (90/14.6)^0.77

Thus for my engine, measured at say 90-PSI, my compression ratio works out to

CR = (90/14.7)^0.71

CR = 4.1

Peter posted 01-23-2008 08:45 AM ET (US)     Profile for Peter  Send Email to Peter     
Because you had a pressure reading taken on the cylinders, why isn't your compression ratio simply 7.1?

((PSIG TDC + PSIA) / (0 + PSIA)) = CR

(90 + 14.7) / (0 + 14.7) = CR

104.7 / 14.7 = 7.1

(I forgot to add 14.7 to 90 before when coming up with 6.1).

Jerry Townsend posted 01-23-2008 12:19 PM ET (US)     Profile for Jerry Townsend  Send Email to Jerry Townsend     
Jim - remember, add the atmospheric pressure to the gage pressure to get the absolute pressure - and all units must be consistent. So your 90 psig becomes 104.7 psia (at sea level).

Peter - the compression ratio is defined as the ratio of the volume at the beginning of the stroke (piston area X stroke + "clearance" volume (cylinder volume with the piston at top-dead-center) ) divided by the "clearance volume" as:

Compression Ratio = (PA * S + CV)/CV

where PA = Piston Area
S = Stroke
CV = Clearance volume

all in consistent units.

But, in general, we don't know the clearance volume. Therefore, we use a thermodynamic equation as discussed above to give us the compression ratio. Unfortunately, all is not peaches and cream there either as the ACTUAL process is not well defined (gas is not a perfect gas and the process is not adiabatic). ---- Jerry/Idaho

Peter posted 01-23-2008 12:28 PM ET (US)     Profile for Peter  Send Email to Peter     
Jerry -- I understand the theoretical compression ratio derivation to calculate an expected compression ratio but we have some actual data that suggests to me that the actual compression ratio of the motor is something on the order of 7.1:1.

Is Jim's compression ratio not 7.1 based on the compression gauge results?

L H G posted 01-28-2008 05:11 PM ET (US)     Profile for L H G    
I recently had a compression check done on my 1988 Mercury V-6 2.0 liter 150. All cylinders registered between 130-PSI and 132-PSI.

How could there be such a discrepancy between Jim's 90-PSI figures and mine, since both are carbureted 2-cycle motors, and both loop charged?

Peter posted 01-28-2008 05:57 PM ET (US)     Profile for Peter  Send Email to Peter     
The reason for the difference is not all motors are made the same.

The compression of all the cylinders of my 2.6L Johnson 150 V6 60 degree looper were 105 PSI +/- 5 when checked last Summer. This is within specification for that motor.

The 1.8L 90 degree V4 Johnson 140 looper had compression figures in the 125 +/- 5 PSI range when checked last Summer, which were in spec too.

It seems intuitively right to me that a small displacement 150 HP would have higher compression than a larger displacement 150 HP motor.

fourdfish posted 01-28-2008 06:06 PM ET (US)     Profile for fourdfish  Send Email to fourdfish     
The older Mercury engines were all tighter and had higher compression. This made them harder to start. Actually 90 to 95-PSI is not real bad. It is more important that all the cylinder readings be within 10-PSI of each other. So Jim's readings are in that range. My old 175hp Johnson had 110-115 PSI when I sold it and the buyer was happy about it.
jimh posted 01-28-2008 08:34 PM ET (US)     Profile for jimh  Send Email to jimh     
On my 16-year-old Evinrude V6 225-HP, I plan to have the compression checked again next season, after I get the motor back into commission. My feeling it that since the original check, the motor is running much better. I think it will be interesting to see if there has been a change in the compression PSI readings. I think the motor might have been full of carbon when I first got it. I have run a lot of gasoline through it, with occasional fuel additives, and I am interested to see if the compression PSI readings have changed. Cylinders and piston rings free from carbon tend to produce higher compression.

It is hard to compare readings among engines when they are measured by different gauges. Among gauges is it fairly common that there is significant variation. None of these gauges are exactly traceable to the National Bureau of Standards. If you want to compare readings, take them all with the same gauge.

As for comparisons between a 3-liter V6 Evinrude and a 2-liter V6 Mercury, I don't have the foggiest idea what inference to make of compression readings.

If I have understood all the advice, particularly that offered by Jerry, I should use the PSI readings taken to compute the compression ratio like this:

I measure about 90-PSI cylinder pressure, relative to atmospheric. Atmospheric pressure is about 14.6-PSI. So my actual pressure is 90 + 14.6 or 104.6-PSI. Thus my compression ratio, CR,is

CR = (104.6/14.6)^0.77

(Here I am using the exponent 0.77 as 1/1.3 as suggested above)

Thus for my engine, measured at say 90-PSI, my compression ratio works out to

CR = 4.55 to 1

This does seem a bit on the low side, but on the other hand, the engine turns up to 6.000-RPM without coming apart at the seams.

jimh posted 01-28-2008 08:41 PM ET (US)     Profile for jimh  Send Email to jimh     
Larry writes:

"How could there be such a discrepancy between Jim's 90-PSI figures and mine..."

Characterizing the difference in the readings as a discrepancy tends to imply there is a problem.

A discrepancy is something in the state of being discrepant. "Discrepant" means to be at variance. That sounds rather neutral, but I think there is a general interpretation that a variance is from the norm.

discrepant

Function:
adjective

Etymology: Middle English discrepaunt, from Latin discrepant-, discrepans, present participle of discrepare to sound discordantly, from dis- + crepare to rattle, creak — more at raven

Date: 15th century

: being at variance : disagreeing <widely discrepant conclusions>

In this case I do not think it is particularly discrepant that there should be a difference in the readings between these two engines.

Jerry Townsend posted 01-28-2008 09:37 PM ET (US)     Profile for Jerry Townsend  Send Email to Jerry Townsend     
Larry - the compression ratio check is ONLY a relative indication for that particular engine. The ratio is indicative of the engine condition and operation. That is, the leakage around valves and rings can lower the compression ratio as can the timing.

And as others point out - all gauges are not made equal. In all probability, the two pressure gauges used in the two tests are of different manufacture and will give different readings. And taken by different people.

One can get an indication of leakage via the rings by running a second compression test after squirting some oil into the cylinder.

I do not have engine design experience - but I suspect that the published compression ratio is the ratio of the volume at the bottom of the stroke divided by the volume at top-dead-center. The volume at top-dead-center is known as the "clearance volume" while the volume at the bottom of the stroke is the piston displacement volume plus the clearance volume. As such, this calculation ignores the actual thermodynamic process, or in effect assumes the compression process of one of constant temper3ature. In this case - the exponent would be 1 instead of 1.3 - and the reciprocal is 1 instead of 0.77.

If anyone has engine design experience - your thoughts are welcomed and encouraged. I will also get some clarification on this subject. ---- Jerry/Idaho

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