View Full Version : Q - Compression Ratio to lb/in2 conversion?

Does anyone know a way to convert compression readings in lb/in2 to a compression ratio? I'm sure there must be a simple conversion table, or rule-of-thumb to use, but I've never seen it.

Thanks in advance,

Rob.

static - don't you just divide by atmospheric (14.7psi)?

or am i missing something obvious.

aeronca65t

05-12-2006, 11:22 AM

[ QUOTE ]

static - don't you just divide by atmospheric (14.7psi)?

or am i missing something obvious.

[/ QUOTE ]

That would be the logical assumption, but cam timing comes into play (overlap, duration and so forth).

The "trapped time" can differ with different cams.

Also, combustion chamber volume affects this.

For example, my race Spridget (with about 10:1 compression ratio) reads 205 psi on a compression tester.

[ QUOTE ]

static - don't you just divide by atmospheric (14.7psi)?

or am i missing something obvious.

[/ QUOTE ]

At 195lb/in2 that wiuld put my TR at 13:1 - seems a bit high. Would be getting pre-ignition at that ratio.

I suspect something closer to 10:1.

Dave Russell

05-12-2006, 12:51 PM

Hi Rob,

It's not possible to accurately convert CR to PSI. Nial is correct. To carry

it a bit further, the following explanation may help.

Question: “If an engine has compression readings around 165 PSI, and

that is taken at face value; that means the compression ratio is

165/14.7 ~ 11.2:1. This is much higher than the published numbers.

That air must get really warm when you ‘squish’ it.”

Answer: That calculation is correct for "isothermal" compression which

assumes that no heat is gained or lost during the compression cycle, and

if the actual compression ratio is the same as the rated ratio.

To complicate the calculation, at low RPM, such as at cranking speeds, the

actual engine compression ratio is never the design calculated ratio

because in reality the piston cannot begin to compress anything until

the intake valve gets closed. If the intake valve were to be completely

closed at BDC -- which by design it never is -- the actual and rated

compression ratios would be the same. At higher RPMs and thus higher

intake gas velocities, the late closing of the intake valve is offset by the

gas inertia in the intake tract and the engine can approach or exceed its

design compression ratio.

The formula for "adiabatic" compression takes heat loss or gain

into consideration. For example, as the air is compressed it is heated

by the compression and the final pressure is higher than would be

predicted by isothermal compression. The variables are rather large and

the difference between the two calculations can vary by a factor of

almost two to one in the final pressure. The pressure rise due to

heating is an "exponential" function (i.e., more heat = more pressure =

more heat = more pressure). The "exponent" for air will vary from one

with complete heat loss to about 1.4 with no heat loss.

For a hypothetical engine with 9:1 compression ratio, assume that the

actual mechanical compression ratio is 6:1 due to late intake valve

closing. We can calculate that the pressure for isothermal compression

would be 14.7 x 6 = 88 PSI. With an adiabatic exponent of 1.14 this

would give a compression pressure of 165 PSI. (88 raised to the 1.14 power)

Whether heat is gained or lost during compression depends on several

things such as how quickly the compression takes place (starter and

engine cranking speed), temperature of the engine during the test (is

the air in the cylinder hotter or cooler than the air compressed in

it?), and a few other variables. The end result is usually that the

measured compression pressure is higher than would be calculated by

isothermal compression.

The compression pressures given at "face value" are empirically

determined from measurements and experience. There are just too many

variables to make accurate calculations.

D

Wow - Thanks Dave for the very detailed explination. My college thermodynamics is starting to come back to me. I see your point - without knowing the heat transfer rate of the compression chamber you cannot know where between isothermal the and adiababatic compression you are. All I know is that entropy has to be increasing. At least mine is!

Rob.

Rusticus

05-12-2006, 02:40 PM

Good thread, gents!

How would these numbers compare to measuring the ratio of cylinder/combustion chamber volume at TDC and BDC? Would that ratio be closer to the nominal compression ratio?

Perhaps the easiest way to "solve" for this is if folks who know both their compression ratio and pressure could post them, together with the engine type.

We have 1 entry so far - aeronca65t's Spriget - 10:1 = 205lb

This is the "brute force" approach I know, but might prove quite informative.

Rob.

Dave Russell

05-12-2006, 05:03 PM

[ QUOTE ]

Good thread, gents!

How would these numbers compare to measuring the ratio of cylinder/combustion chamber volume at TDC and BDC? Would that ratio be closer to the nominal compression ratio?

[/ QUOTE ]

The standard calculation for "design" compression ratio is combustion chamber volume at TDC plus cylinder volume at BDC, divided by combustion chamber volume.

The "real, effective" compression ratio at cranking & moderate rpm would be the cylinder volume at the point of intake valve closing plus the chamber volume divided by the chamber volume. Not sure what you mean by "nominal" ratio.

The cam timing is such that the intake valve doesn't actually close until the piston is already traveling back up on the compression stroke. Typical cams will have the intake valve closing somewhere between 30 & 70 degrees after BDC. This is the point where actual compression begins at low engine speeds. The cylinder is actually pumping some of the intake charge back out past the intake valve until it closes.

At higher rpm, the intake charge's inertia keeps the piston from pumping the charge back out even though the valve is still open & the piston is rising. This raises the effective compression ratio to somewhere around the design ratio. The more radical the cam timing, the more rpm is required to prevent this reversal. Larger ports & valves will also raise the rpm at which inertial cylinder filling takes place.

In other words for an engine designed for 9 to 1, the "real - effective - actual" compression ratio may be something like 5 to 1 at cranking speeds & lower rpm. It will approach 9 to 1 at higher rpm as the intake charge inertia gives more complete cylinder filling. The "real or effective" compression ratio for a given cam timing, is a constantly changing value depending on engine rpm, throttle opening, & intake inertia. As said, it also varies greatly with different cam designs.

Hope I haven't confused things more.

D

PS; Air density decreases about 3.4% for each 1,000 ft of elevation increase, the compression readings would be 10% lower at 3,000 ft than at sea level.

D

Rippthrough

05-13-2006, 12:53 PM

Well, to help the rule of thumb method, mine on 9.3:1 gives 175psi. On 10.8:1 it gives 210psi.

kennypinkerton

05-13-2006, 05:16 PM

Wow, I like this thread... Should be in ALL the car forums. /ubbthreads/images/graemlins/thumbsup.gif

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