Testing Engine Peak Cylinder Pressure

[jimh]

Many reports are often made of engine "compression", that is, a report on a reading of peak cylinder pressure measured in a particular cylinder when the engine was not actually running by using a pressure gauge inserted in place of a spark plug. These pressure readings are reported typically in units of PSI or lbs-per-square-inch.

The nominal standard pressure in the atmosphere at ground level is 14.7-PSI. A report of a cylinder having a peak pressure of, say, 120-PSI can be interpreted as a "compression" reading if the peak pressure is divided by the standard atmospheric pressure. This suggests the compression ratio will

120 / 14.7 = 8.2-to-1

This rather low compression ratio is typical for older two-stroke-power-cycle engines.

More modern engines using four-stroke-power-cycle may use higher compression. For example, the SUZUKI DF350 engine is said to have a compression ratio of 12:1. If making a peak cylinder pressure measurement on that engine, one would expect to measure

14.7 × 12 = 176.4-PSI

Many gauges for measuring pressure are also calibrated in a scale of BAR. One BAR is equal to 14.5-PSI. A reading on the gauge of the BAR value will give a very good approximation of the compression ratio.

[GoldenDaze]

Jim, a typical compression gauge measures the differential between the input and atmospheric pressure, correctly denoted PSIG (pounds per square inch gauge), though this is often incorrectly simplified to PSI. Such a gauge measures zero with atmospheric pressure on the input.

Consider a cylinder with a 2:1 compression ratio (where the piston compresses the cylinder gasses to half their original volume). With the piston at bottom, a compression gauge would read zero. At top dead center, the gauge would read 14.7 PSIG.

Consequently, 120 PSIG would equate to (120+14.7)/14.7, or approximately 9.2:1 compression ratio.

-Bob

[jimh]

Bob--Thanks for your comments. Your thinking about compression ratio calculation from pressure measurement is the same as was my initial thinking about the topic. But after searching for some opinions about it on the web, I did not find any confirmation; and what I did find seemed to suggest the method I described in my initial post. A couple youTube presentations suggested that different version, and I read many comments about them looking for a critique of an error. So I figured I must have been on the wrong track with my first thoughts on the topic.

As you mention, that a PSI measuring gauge must measure a pressure in relation to the atmospheric pressure seemed reasonable to me. And that makes the cylinder pressure reading as a differential increment from standard atmospheric pressure.

Assuming your comments are correct--and they make sense to me--then we could make a table of PSI readings and Compression Ratio (based on atmospheric standard pressure being 14.7-PSI) as follows

Cylinder  Imputed
Peak      Compression 
PSI       Ratio
 90 ===>  7.1
100 ===>  7.8
110 ===>  8.5
120 ===>  9.2
130 ===>  9.8
140 ===> 10.5
150 ===> 11.2
160 ===> 11.9
170 ===> 12.6

[jimh]

Most of the calculators I found on the net to compute compression ratio were based on the geometry of the cylinder and the difference between volume at top of piston travel and bottom of piston travel. But that method ignores any pressure loss due to poor seal of the piston rings or with the inlet and outlet valves or ports. This suggests that measuring cylinder pressure might be a better test, as it should detect poor sealing at the piston rings or inlet and outlet valves or ports.

On the other hand, when actual combustion occurs in a cylinder, the peak pressure is probably--perhaps certainly--going to be higher than just the pressure increase resulting from the compression of the air in the cylinder. Here I am assuming the volume of the exhaust gases created by combustion is going to be larger than the volume of air taken into the cylinder. Those pressures might create a leak that would not be apparent at pressures of around 100 to 150-PSI as occur in a dry cylinder test.

[Jefecinco]

Decades ago we were instructed to take both wet and dry compression tests. A wet test was taken after adding a little engine lubricating oil to a cylinder using a squirt can. It has been several years since I've seen any discussion of providing both readings and the reasons for any differential between them. Are wet compression tests no longer used?

[jimh]

In my article "Assessing Used Outboard Engines" (in the REFERENCE section) I include a procedure from an 1992 OMC service manual for taking cylinder pressure readings. The procedure specifies the readings should be taken after the engine has been run and has reached normal operating temperature. An engine that had just been running would probably have some lubricating oil on the cylinder walls. A warm engine would also probably have a tighter fit of the piston rings to the cylinder walls, improving the seal.

[Jefecinco]

For checking the compression on an engine which can't be started a wet test would more closely approximate the conditions of an engine which has run long enough to reach normal operating temperature.

[GoldenDaze]

when actual combustion occurs in a cylinder, the peak pressure is probably--perhaps certainly--going to be higher

It's certainly going to be higher, since that's the force that generates the engine's torque and power. How much higher? 1400 PSI is an upper-limit estimate based on chemistry. Read on for details if you're into it, but if you're bored, hey I warned you.

Let's start by assuming we have an engine with 0.5 liter cylinders (science and engineering is easier in metric), which would make for a 1.5 liter (93 CID) 3-cylinder or a 2.0 liter (124 CID) 4-cylinder engine, which is pretty close to the current Mercury 115 4-stroke engine. Let's further assume a 10:1 compression ratio.

Each cylinder will ingest 0.5 liters of fuel/air mixture at normal atmospheric pressure. If we assume a stoichiometric air:fuel mixture of 14.7:1 by weight (that 14.7 is coincidental and has nothing to do with atmospheric pressure at sea level), that 0.5 liter fuel/air charge will contain 1.2g of air and 0.082g of gasoline. That's our first simplifying assumption, since there will be throttling and induction losses, and those losses will increase with increasing engine RPM, so the actual mass of air and fuel in the cylinder will be somewhat less.

This thread is about boating, so we'll assume a nice starting temperature of 20 degrees C (293 degrees K). During the 10:1 compression stroke, pressure in the cylinder increases (neglecting leakage) to 900 kPa (130 PSI). If the fuel charge burns completely (a simplifying estimate, but it will probably be 90%+) then 3.8 kJ of energy will be added to the gas mixture in the cylinder. The specific heat of air is approximately 1kJ/kg*K (that is, 1 kJ of energy will heat 1 kg of air by 1 degree K). This 3.8 kJ of energy will heat the air in the cylinder by approximately 2923 degrees K, to 3216 K.

If the cylinder is at top-dead-center and the charge burns instantly ([silly expression deleted] simplifying assumption, since the burn will not happen instantly and by the time it completes the volume in the cylinder will be somewhat larger) then according to the ideal gas law (PV = nRT) this temperature change while holding volume and gas quantity constant will change the pressure by a factor of about 11, so our 900 kPa cylinder pressure rises to 9900 kPa, or around 1400 PSI.

This is an overestimate for a few reasons, but the main reason is that the fuel/air charge will not instantly convert its chemical energy to heat. The burn will take some time, during which the piston is already moving downward and the cylinder volume is increasing. Nevertheless compares pretty well to estimates from a variety of online resources, like https://www.performancetrends.com/Definitions/Cylinder-Pressure.htm.

Well that was fun, wasn't it?

[jimh]

BOB--great information. Don't worry, no millennial readers here--we can go more than eight seconds into a post to get information.