Ok, so I'm looking at the MAF tables in the $32 hack(again), and I'm wonderin' why the GM engineers limited the ECM to recognizing a maximum of 255 grams/sec. of airflow. The answer of course is that the ECM uses 8-bit addressing, and the maximum stored value under that scheme(without paging) is 255.
But then I got to thinkin,... how much air would need you need to measure 255 grams. So I'm at work, in a 20' x 40' x 15' room, and I'm thinkin: could all the air in here weigh 255 grams?
So, could someone explain to me how much volume(in cubic feet) 255 grams represents? And for the sake of simplicity, we'll assume the ambient enviromental pressure is 1 bar(14.7 psi). And knowing how much air is present, can we calculate the maximum power that is available if we mixed the air with gas at a stoich mixture(14.7 parts of air: 1 part fuel)?
Gawd, I gotta find more productive things to do at work.
But then I got to thinkin,... how much air would need you need to measure 255 grams. So I'm at work, in a 20' x 40' x 15' room, and I'm thinkin: could all the air in here weigh 255 grams?
So, could someone explain to me how much volume(in cubic feet) 255 grams represents? And for the sake of simplicity, we'll assume the ambient enviromental pressure is 1 bar(14.7 psi). And knowing how much air is present, can we calculate the maximum power that is available if we mixed the air with gas at a stoich mixture(14.7 parts of air: 1 part fuel)?
Gawd, I gotta find more productive things to do at work.
Member
OK, it's just a matter of applying the ideal gas equation:
P*V = m*R*T
where
P = Pressure
V = Volume
m = mass
R = ideal gas constant for air (.287 kJ/kg*K)
T = Temperature
So, the volume of your room is 20*40*15 = 12000 ft^3 or 339.802 m^3
Therefore assuming a pressure of 100 kPa (standard atmospheric pressure) and a temperature of 298 K (25*C or 77*F), the equation looks like this:
100 kPa * 339.802 m^3 = m * .287 kJ/kj*K * 298 K
Solving for m you get m = 397.308 kg.
Now if you want to turn the equation around, if you assume a mass of 255 g, the equivalent volume is .218 m^3 or 7.702 ft^3.
So, the maximum flow rate of air the MAF can register is 7.702 ft^3/sec.
Which considering that a 350 cu in engine spinning at 5k rpm can take in up to 8.439 ft^3/sec of air at atmospheric pressure, that really shows the limitations of the MAF system I guess.
Also, for comparisons sake, 255 g is just over half a pound.
------------------
Black 88 GTA L98
261 RWHP, 345 RWTQ
ZZ4 bottom end, Edelbrock 6085 heads, LT4 HOT cam, GMPP 1.6 RR's, ported stock TPI, SLP 1 3/4" headers, no cat, Dynomax cat-back, Stock PROM
E.T.F.A Member #11
[This message has been edited by Scott 88 GTA (edited January 31, 2001).]
P*V = m*R*T
where
P = Pressure
V = Volume
m = mass
R = ideal gas constant for air (.287 kJ/kg*K)
T = Temperature
So, the volume of your room is 20*40*15 = 12000 ft^3 or 339.802 m^3
Therefore assuming a pressure of 100 kPa (standard atmospheric pressure) and a temperature of 298 K (25*C or 77*F), the equation looks like this:
100 kPa * 339.802 m^3 = m * .287 kJ/kj*K * 298 K
Solving for m you get m = 397.308 kg.
Now if you want to turn the equation around, if you assume a mass of 255 g, the equivalent volume is .218 m^3 or 7.702 ft^3.
So, the maximum flow rate of air the MAF can register is 7.702 ft^3/sec.
Which considering that a 350 cu in engine spinning at 5k rpm can take in up to 8.439 ft^3/sec of air at atmospheric pressure, that really shows the limitations of the MAF system I guess.
Also, for comparisons sake, 255 g is just over half a pound.
------------------
Black 88 GTA L98
261 RWHP, 345 RWTQ
ZZ4 bottom end, Edelbrock 6085 heads, LT4 HOT cam, GMPP 1.6 RR's, ported stock TPI, SLP 1 3/4" headers, no cat, Dynomax cat-back, Stock PROM
E.T.F.A Member #11
[This message has been edited by Scott 88 GTA (edited January 31, 2001).]
Wow! Awesome reply. Cut and Pasted your entire response into a Word file on my desktop. Thanks!
TGO Supporter
88IROCs, can you please send me an e-mail? I would like to discuss some PROM burning information privately.
ArnoldBraker, a "hack" is a document that has "reversed engineered" the machine code of the eprom. The "32 hack" is for the earlier TPI Maf systems in the mid-80s up to 87 or 88 (88s actually used 32B), the "6E Hack" is for the 89 Maf cars and the "8D Hack" is for the SD TPI from 90-92.
The hack will show you all the "tables, switches and constants" within the eprom along with the specific machine code that uses it. Very useful if you are "bit" twiddling the eprom, or using a "eprom editor" that does not have pre-defined tables like GMEPRO or TunerCat. Or if you are using TunerCat and it does not have a particular switch or table defined in the "TDF", you can define the table, switch or constant using TDF Editor.
If you have more questions or interested in PROM burning, read the article in the link below and take a visit to the PROM Board.
https://www.thirdgen.org/newdesign/tech/promintro.shtml
ArnoldBraker, a "hack" is a document that has "reversed engineered" the machine code of the eprom. The "32 hack" is for the earlier TPI Maf systems in the mid-80s up to 87 or 88 (88s actually used 32B), the "6E Hack" is for the 89 Maf cars and the "8D Hack" is for the SD TPI from 90-92.
The hack will show you all the "tables, switches and constants" within the eprom along with the specific machine code that uses it. Very useful if you are "bit" twiddling the eprom, or using a "eprom editor" that does not have pre-defined tables like GMEPRO or TunerCat. Or if you are using TunerCat and it does not have a particular switch or table defined in the "TDF", you can define the table, switch or constant using TDF Editor.
If you have more questions or interested in PROM burning, read the article in the link below and take a visit to the PROM Board.
https://www.thirdgen.org/newdesign/tech/promintro.shtml
Member
Any time 88IROCs.