How much lower would a sea level compression result of 150 psi be at 5500 ft, e.g. Denver?
02-14-2008, 07:45 AM
#31
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Re: Atmospheric pressure at 5500 is << 1 Bar (or 1.1 bar where 1 Bar = 14.5 psi)
Not
<a href="http://en.wikipedia.org/wiki/Barometric_pressure">wikipedia page on barometric pressure</a>
<a href="http://en.wikipedia.org/wiki/Barometric_pressure">wikipedia page on barometric pressure</a>
02-14-2008, 07:50 AM
#32
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After reading your question, and the answers...
...I am separating the variables from the constants and come up with this:
Volume = constant.
Temperature = constant. (The temperature of the warmed-up metal, when the test is done properly, is >> the temp gain from compressing the gas in the cylinder.)
From the ideal gas law, there are only three variables to account for, and the above settles two of them. So, now all we have is pressure.
The mass of molecules taken into the cylinder will be less at altitude, due to the lower density of the atmosphere there. Thus, the final pressure should be lower in Denver than in San Diego, if taken on the same car, even if you use the same gauge in both places.
I don't know that 180psi in Denver makes much sense, on an AAN.
Volume = constant.
Temperature = constant. (The temperature of the warmed-up metal, when the test is done properly, is >> the temp gain from compressing the gas in the cylinder.)
From the ideal gas law, there are only three variables to account for, and the above settles two of them. So, now all we have is pressure.
The mass of molecules taken into the cylinder will be less at altitude, due to the lower density of the atmosphere there. Thus, the final pressure should be lower in Denver than in San Diego, if taken on the same car, even if you use the same gauge in both places.
I don't know that 180psi in Denver makes much sense, on an AAN.
02-14-2008, 07:51 AM
#33
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Nope, still no difference at altitude...math inside
Ideal Gas Law:
pV = nRT
so, p = V*n*R*T
where:
p = pressure
V = Volume
n = amount of gas in moles (remarkable constant up to 100km of altitude)
R = Gas Constant
T = Temperature
Conclusion:
If the volume of air your engine is compressing is the same and the temperature is the same, altitude has no effect on pressure.
pV = nRT
so, p = V*n*R*T
where:
p = pressure
V = Volume
n = amount of gas in moles (remarkable constant up to 100km of altitude)
R = Gas Constant
T = Temperature
Conclusion:
If the volume of air your engine is compressing is the same and the temperature is the same, altitude has no effect on pressure.
02-14-2008, 08:05 AM
#36
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The heat of compression should be approximately the same.
If you hold volume constant, and change the pressure by 15%, the T-delta will also be about 15%, from the ideal gas law - P & T are on opposite sides of the equals sign, so are proportional.
BUT, that's *delta* T, which is already small in comparison to the operating temp of the test.
So it may be a 15% change in delta T, which means the temp change difference between San Diego and Denver on the same car would be really small.
Use the ideal gas law first to find the temp difference from a 9.3:1 compression at sea level then at altitude. Then plug that into the ideal gas law for the same compression at altitude, and you'll get the final effect of the difference.
Hey, I was the only P-Chem student that thought this **** was fun. FWIW. The LibrariAAN calling me Mr. Wizard? Well, SWEET!
BUT, that's *delta* T, which is already small in comparison to the operating temp of the test.
So it may be a 15% change in delta T, which means the temp change difference between San Diego and Denver on the same car would be really small.
Use the ideal gas law first to find the temp difference from a 9.3:1 compression at sea level then at altitude. Then plug that into the ideal gas law for the same compression at altitude, and you'll get the final effect of the difference.
Hey, I was the only P-Chem student that thought this **** was fun. FWIW. The LibrariAAN calling me Mr. Wizard? Well, SWEET!
02-14-2008, 08:08 AM
#37
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121.67 psi is the answer
GP = (CRE^1.2 × AP) - AP
Where:
GP = gauge pressure in PSI
CRE = effective compression ratio
AP = atmospheric pressure 14.7psi @ SeaLevel
Now let's correct for elevation.
ECRE = CRE - [(altitude/1000)x 0.2]
CGP = (ECRE^1.2 × AP) - AP
Where:
CGP = Corrected gauge pressure in PSI
ECRE = Elevation Corrected Effective Compression Ratio
AP = atmospheric pressure 14.7psi @ SeaLevel
Answer to UrS4Boy's original question
GP = (CRE ^ 1.2 x AP) - AP
150.26 psi = (7.5^1.2 x 14.7) -14.7
ECRE = CRE - [(altitude/1000)x 0.2]
6.4 = 7.5-((5500/1000) * .2)
CGP = (ECRE ^ 1.2 x AP) - AP
121.67 psi = (6.4^1.2 x 14.7) -14.7
So the difference between sea level and 5500 is 150.26 - 121.67 = 28.58 psi
(thanks to 90Carat (Tom) for the equations)
Where:
GP = gauge pressure in PSI
CRE = effective compression ratio
AP = atmospheric pressure 14.7psi @ SeaLevel
Now let's correct for elevation.
ECRE = CRE - [(altitude/1000)x 0.2]
CGP = (ECRE^1.2 × AP) - AP
Where:
CGP = Corrected gauge pressure in PSI
ECRE = Elevation Corrected Effective Compression Ratio
AP = atmospheric pressure 14.7psi @ SeaLevel
Answer to UrS4Boy's original question
GP = (CRE ^ 1.2 x AP) - AP
150.26 psi = (7.5^1.2 x 14.7) -14.7
ECRE = CRE - [(altitude/1000)x 0.2]
6.4 = 7.5-((5500/1000) * .2)
CGP = (ECRE ^ 1.2 x AP) - AP
121.67 psi = (6.4^1.2 x 14.7) -14.7
So the difference between sea level and 5500 is 150.26 - 121.67 = 28.58 psi
(thanks to 90Carat (Tom) for the equations)
02-14-2008, 08:18 AM
#40
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Not sure why you don't think you have the answer...
150psi at sealevel would be approx 122.4 psi..
Depending on how hot the air is may vary the test slightly so 135-138 sounds fine....
Depending on how hot the air is may vary the test slightly so 135-138 sounds fine....