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Old 10-26-2012, 09:26 AM
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coax coax is offline
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Sorry to quasi-hijack the thread, but could someone give a quick explanation of why you have to adjust for altitude when doing a compression test? I find this stuff really interesting, but have no background in fluid dynamics.

From the reading I did, the Pressure ={nRT}/{V}


P is the absolute pressure of the gas
n is the amount of substance
T is the absolute temperature
V is the volume
R is the ideal gas constant.

So if the volume decreases by half, the pressure decreases by half. And I realize this is not an Ideal gas. SO for altitude, the "amount of substance" will be decreased, IE the amount of air in the cylinder will be less. However, we'd be comparing this pressure in the cylinder to the atmospheric pressure outside, at whatever elevation we are at.

So 14.7 psi at sea level and 12.2 at 5000 feet is atmosphere. So if we are taking the "gauge pressure" as a ratio to the pressure in the cylinder, why would we need to adjust for altitude? I could see having to adjust for altitude if our gauge was measuring the compression compared to absolute pressure, but since its measuring the rise against atmospheric pressure, isn't it already adjusted?

According to this sheet, the volume correction factor at 5k feet is 1.17. As we go up in elevation, the volume goes up, density goes down. So a given mass of air 0 feet elevation now has a volume of 1.17 times what it had at sea level. So we get in 85% of the air into a cylinder at 5k feet.

This is the correction factor that everyone quotes. So its correct that the cylinder gets that much air mass into it, but its not then being measured against sea level. Its the ratio back to atmospheric at that given altitude. Which is also about 85% here at 5k feet.

But this is a linear function. I can't see any non-linear functions in any of this so how do we have to correct?

14.7 : {nRT}/{V}

12.2 : {.85nRT}/{V} --> ON this one we get only 85% of the air into the cylinder, hence we multiply the "n" (amount of gas) by .85. [Do we have to correct by an additional .85 for the R value of absolute pressure? Is that where we get an extra .85 reduction?]

What am I missing? I guess what I am saying is I don't see why there is any correction necessary unless using an absolute pressure gauge. /quasi-hi-jack-over
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