High altitude boost question
06-09-2016, 12:46 AM
#41
I've raced and dynoed between Bandimere @5860' and Famoso @800' elv. You will be down about 10~12% in power at same boost psi but the power range is much narrower. Spool is about 5~600 + RPM later, and turbo chokes a few hundred RPM sooner. If you have plenty of turbo left and air temps are not a problem you should be able to get away with 3~4 more psi of boost at 6000' vs sealevel. As far as KPA vs kpa a really efficient twin turbo engine will be pretty close in pk power numbers but a single turbo working hard will be down due to heat, and efficiency losses. The turbo will be spinning much faster to generate the same KPA at 6K' vs sealevel.
06-21-2016, 04:25 PM
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06-21-2016, 04:37 PM
#45
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I just went basically 600/600 SBE with a 7868 turbonetics on e85...get some boost ive since put another 2# on it without any issue besides adding a bit more fuel to it up top. Now that RTT is back ill get her really dialed in.
06-25-2016, 04:31 PM
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Let's bring some clarity to some of the confusion.
At altitude (less atmospheric pressure) there is less oxygen per given volume. There is also less of everything else that makes up air, most notably nitrogen. The percentage of oxygen to nitrogen does not change. The air is simply under less pressure and therefor less dense. So, the statement that the percentage doesn't change is correct from any perspective. But, the statement that there is less oxygen is only true when comparing unit per volume against the same volume at a different absolute pressure (and/or temperature for that matter). In which case there would be more oxygen at lower altitudes, however more nitrogen too. Bad feeling that may have confused things more than cleared them.
About the statement that there's less pressure to push the air in the engine at altitude and more at sea level. Yes... But, that's not the complete story either.
You must take into account pressure AND density to determine mass flow. Either on their own can be an indication used to get an idea of what/how things may change. However, one without the other is nothing more than that... an almost in the ball park idea.
Physics can be daunting as a whole. But, in this case it's quite simple and the equations are easy to perform to get a much better idea of what's going on and what to expect.
First, calculate air density. We're only looking at the effects of altitude change and ignoring other atmospheric variables anyway. So... good enough in my book. It's easier to calculate if we're not concerned with humidity anyways.
Then, calculate the effect of pressure through an orifice in a choked condition. Again, not completely on the mark, but close enough to compare one situation to another with a reasonable amount of accuracy... for me anyway. Anyone want to expand on this, please do.
Since where I am on the planet commonly uses Pounds to measure weight, Cubic Feet to measure air volume, PSI for pressure, and °F to measure temperature... that's what I use. I also use °R for absolute temperature instead of °K since it's easier to convert °F to °R by simply adding/subtracting 459.67 to the °F figure. If this makes no sense to you, don't sweat it. It's just a technicality. Results are the same. (Actually better IMO. I seem to get slight variances in the final solution when I convert into °K)
Air Density in pounds per cubic foot:
All you need to calculate this (for dry air, no humidity) is the conditions/measurements in the manifold, air temperature and absolute pressure.
The product of
Absolute Pressure x 144
Divided by the product of
53.35 x absolute temperature (°F plus 459.67)
Equals lb/Ft³, how much one cubic foot of air weighs at the specified conditions
So, at sea level
14.7psi x 144 = 2116.8
53.35 x 519.67 (60°F + 459.67) = 27724.3945
2116.8 / 27724.3945 = 0.076351532221921
So, let's call it 0.076 lb/Ft³ at 60°F at sea level
Using the same temp and the 12.2psi noted by the OP gives us 0.063 lb/Ft³
Adding 10psi boost to Denver's 12.2 gives us 22.2 psia and assuming 100% efficient intercooling for simplicity, it increases density to 0.115 lb/Ft³
But, what if we take into account heat added by compression and the compressor without intercooling? Just for conversations sake, lets say the compressor is 70% efficient.
Calculating temperature gain from compression (adiabatic) and compressor efficiency:
Pressure Ratio to the power of 0.286 x absolute (°F + 459.67) = adiabatic temperature after compression. ("adiabatic" simply means no thermal energy is added or taken away to/from the air as it's compressed.)
So, 60°F air compressed from 12.2psia to 22.2psia (not counting the compressor (in)efficiency) would result in:
Pressure ratio... 22.2 (final pressure) / 12.2 (atmospheric pressure) = 1.82PR
1.82 to the power of 0.286 = 1.1868
1.1868 x 519.67 (60°F + 459.67) = 616.744356°R minus 459.67 to get back to °F = 157°F
Now, account for compressor (in)efficiency
Subtract the starting temp from the final temp (in absolute) and divide it by compressor efficiency...
616.744356°R - 519.67°R = 97.074356°R divided by 0.70 = 138.68°R
Add the product back to the starting (atmospheric) absolute temp to get the final temp in °R. Subtract 459.67 to convert back to °F
519.67 + 138.68 = 658.35°R
658.35 - 459.67 = 198.68°F IAT
Now tie that into the density calculation for manifold air density in Denver...
22.2psia x 144 = 3196.8
53.35 x 658.35°R (198.68°F manifold IAT + 459.67) = 35122.9725
3196.8 / 35122.9725= 0.09101735338602107210601266735041
0.091 lb/Ft³ at 60°F atmospheric temp, 10psi boost, and 70% compressor efficiency in Denver.
Man... I'm getting tired of writing! Let's see if I can breeze through the pressure/airflow part kinda quickly.
Here's the idea, It's not about calculating actual numbers into a specific engine. It's about identifying a given mass flow rate through a given orifice and adjusting numbers in the above calculations until one scenario is relatively similar to another. Every engine is gonna have different heads, different intake, different cam, different exhaust restrictions... Way too much to take into consideration. But, if we know what one set of conditions will flow through a given restriction, we can fiddle with the variables until the results of one closely match another.
This one's a little tougher to calculate. We're actually getting into "rocket science" now, but if you follow the instructions you should be good to go. Soon to be a community of LS enthusiasts with a minor in rocket science. LOL By the way, I learned this stuff out of interest for engine related stuff and am not formally educated in the subject. Just an FYI, take it for what it's worth. You did, after all, read it on the internet. Hahaha
Actually... I'm gonna finish this post up tomorrow or Monday. This stuff takes a toll...
Until then
At altitude (less atmospheric pressure) there is less oxygen per given volume. There is also less of everything else that makes up air, most notably nitrogen. The percentage of oxygen to nitrogen does not change. The air is simply under less pressure and therefor less dense. So, the statement that the percentage doesn't change is correct from any perspective. But, the statement that there is less oxygen is only true when comparing unit per volume against the same volume at a different absolute pressure (and/or temperature for that matter). In which case there would be more oxygen at lower altitudes, however more nitrogen too. Bad feeling that may have confused things more than cleared them.
About the statement that there's less pressure to push the air in the engine at altitude and more at sea level. Yes... But, that's not the complete story either.
You must take into account pressure AND density to determine mass flow. Either on their own can be an indication used to get an idea of what/how things may change. However, one without the other is nothing more than that... an almost in the ball park idea.
Physics can be daunting as a whole. But, in this case it's quite simple and the equations are easy to perform to get a much better idea of what's going on and what to expect.
First, calculate air density. We're only looking at the effects of altitude change and ignoring other atmospheric variables anyway. So... good enough in my book. It's easier to calculate if we're not concerned with humidity anyways.
Then, calculate the effect of pressure through an orifice in a choked condition. Again, not completely on the mark, but close enough to compare one situation to another with a reasonable amount of accuracy... for me anyway. Anyone want to expand on this, please do.
Since where I am on the planet commonly uses Pounds to measure weight, Cubic Feet to measure air volume, PSI for pressure, and °F to measure temperature... that's what I use. I also use °R for absolute temperature instead of °K since it's easier to convert °F to °R by simply adding/subtracting 459.67 to the °F figure. If this makes no sense to you, don't sweat it. It's just a technicality. Results are the same. (Actually better IMO. I seem to get slight variances in the final solution when I convert into °K)
Air Density in pounds per cubic foot:
All you need to calculate this (for dry air, no humidity) is the conditions/measurements in the manifold, air temperature and absolute pressure.
The product of
Absolute Pressure x 144
Divided by the product of
53.35 x absolute temperature (°F plus 459.67)
Equals lb/Ft³, how much one cubic foot of air weighs at the specified conditions
So, at sea level
14.7psi x 144 = 2116.8
53.35 x 519.67 (60°F + 459.67) = 27724.3945
2116.8 / 27724.3945 = 0.076351532221921
So, let's call it 0.076 lb/Ft³ at 60°F at sea level
Using the same temp and the 12.2psi noted by the OP gives us 0.063 lb/Ft³
Adding 10psi boost to Denver's 12.2 gives us 22.2 psia and assuming 100% efficient intercooling for simplicity, it increases density to 0.115 lb/Ft³
But, what if we take into account heat added by compression and the compressor without intercooling? Just for conversations sake, lets say the compressor is 70% efficient.
Calculating temperature gain from compression (adiabatic) and compressor efficiency:
Pressure Ratio to the power of 0.286 x absolute (°F + 459.67) = adiabatic temperature after compression. ("adiabatic" simply means no thermal energy is added or taken away to/from the air as it's compressed.)
So, 60°F air compressed from 12.2psia to 22.2psia (not counting the compressor (in)efficiency) would result in:
Pressure ratio... 22.2 (final pressure) / 12.2 (atmospheric pressure) = 1.82PR
1.82 to the power of 0.286 = 1.1868
1.1868 x 519.67 (60°F + 459.67) = 616.744356°R minus 459.67 to get back to °F = 157°F
Now, account for compressor (in)efficiency
Subtract the starting temp from the final temp (in absolute) and divide it by compressor efficiency...
616.744356°R - 519.67°R = 97.074356°R divided by 0.70 = 138.68°R
Add the product back to the starting (atmospheric) absolute temp to get the final temp in °R. Subtract 459.67 to convert back to °F
519.67 + 138.68 = 658.35°R
658.35 - 459.67 = 198.68°F IAT
Now tie that into the density calculation for manifold air density in Denver...
22.2psia x 144 = 3196.8
53.35 x 658.35°R (198.68°F manifold IAT + 459.67) = 35122.9725
3196.8 / 35122.9725= 0.09101735338602107210601266735041
0.091 lb/Ft³ at 60°F atmospheric temp, 10psi boost, and 70% compressor efficiency in Denver.
Man... I'm getting tired of writing! Let's see if I can breeze through the pressure/airflow part kinda quickly.
Here's the idea, It's not about calculating actual numbers into a specific engine. It's about identifying a given mass flow rate through a given orifice and adjusting numbers in the above calculations until one scenario is relatively similar to another. Every engine is gonna have different heads, different intake, different cam, different exhaust restrictions... Way too much to take into consideration. But, if we know what one set of conditions will flow through a given restriction, we can fiddle with the variables until the results of one closely match another.
This one's a little tougher to calculate. We're actually getting into "rocket science" now, but if you follow the instructions you should be good to go. Soon to be a community of LS enthusiasts with a minor in rocket science. LOL By the way, I learned this stuff out of interest for engine related stuff and am not formally educated in the subject. Just an FYI, take it for what it's worth. You did, after all, read it on the internet. Hahaha
Actually... I'm gonna finish this post up tomorrow or Monday. This stuff takes a toll...
Until then
Last edited by SethU; 06-25-2016 at 06:00 PM.
06-25-2016, 04:55 PM
#47
On The Tree
Thread Starter
Bravo.....
Let's bring some clarity to some of the confusion.
At altitude (less atmospheric pressure) there is less oxygen per given volume. There is also less of everything else that makes up air, most notably nitrogen. The percentage of oxygen to nitrogen does not change. The air is simply under less pressure and therefor less dense. So, the statement that the percentage doesn't change is correct from any perspective. But, the statement that there is less oxygen is only true when comparing unit per volume against the same volume at a different absolute pressure. In which case there would be more oxygen at lower altitudes, however more nitrogen too. Bad feeling that may have confused things more than cleared them.
About the statement that there's less pressure to push the air in the engine at altitude and more at sea level. Yes... But, that's not the complete story either.
You must take into account pressure AND density to determine mass flow. Either on their own can be an indication used to get an idea of what/how things may change. However, one without the other is nothing more than that... an almost in the ball park idea.
Physics can be daunting as a whole. But, in this case it's quite simple and the equations are easy to perform to get a much better idea of what's going on and what to expect.
First, calculate air density. We're only looking at the effects of altitude change and ignoring other atmospheric variables anyway. So... good enough in my book. It's easier to calculate if we're not concerned with humidity anyways.
Then, calculate the effect of pressure through an orifice in a choked condition. Again, not completely on the mark, but close enough to compare one situation to another with a reasonable amount of accuracy... for me anyway. Anyone want to expand on this, please do.
Since where I am on the planet commonly uses Pounds to measure weight, Cubic Feet to measure air volume, PSI for pressure, and °F to measure temperature... that's what I use. I also use °R for absolute temperature instead of °K since it's easier to convert °F to °R by simply adding/subtracting 459.67 to the °F figure. If this makes no sense to you, don't sweat it. It's just a technicality. Results are the same. (Actually better IMO. I seem to get slight variances in the final solution when I convert into °K)
Air Density in pounds per cubic foot:
All you need to calculate this (for dry air, no humidity) is the conditions/measurements in the manifold, air temperature and absolute pressure.
The product of
Absolute Pressure x 144
Divided by the product of
53.35 x absolute temperature (°F plus 459.67)
Equals lb/Ft³, how much one cubic foot of air weighs at the specified conditions
So at sea level
14.7psi x 144 = 2116.8
53.35 x 519.67 (60°F + 459.67) = 27724.3945
2116.8 / 27724.3945 = 0.076351532221921
So, let's call it 0.076 lb/Ft³ at 60°F at sea level
Using the same temp and the 12.2psi noted by the OP gives us 0.063 lb/Ft³
Adding 10psi boost to Denver's 12.2 gives us 22.2 psia and 100% efficient intercooling increases density to 0.115 lb/Ft³
But, what if we take into account heat added by compression and the compressor without intercooling? Just for conversations sake, lets say the compressor is 70% efficient.
Calculating temperature gain from compression (adiabatic) and compressor efficiency:
Pressure Ratio to the power of 0.286 x absolute (°F + 459.67) = adiabatic temperature after compression. ("adiabatic" simply means no thermal energy is added or taken away to the air as it's compressed.)
So, 60°F air compressed from 12.2psia to 22.2psia (not counting the compressor (in)efficiency) would result in:
Pressure ratio... 22.2 (final pressure) / 12.2 (atmospheric pressure) = 1.82PR
1.82 to the power of 0.286 = 1.1868
1.1868 x 519.67 (60°F = 459.67) = 616.744356 minus 459.67 to get back to °F = 157°F
Now, account for compressor (in)efficiency
Subtract the starting temp from the final temp (in absolute) and divide it by compressor efficiency...
616.744356 - 519.67 = 97.074356 °R divided by 0.70 = 138.68
Add the product back to the starting (atmospheric) absolute temp to get the final temp in °R. Subtract 459.67 to convert back to °F
519.67 + 138.68 = 658.35°R
658.35 - 459.67 = 198.68°F IAT
Now tie that into the density calculation for manifold air density in Denver...
22.2psia x 144 = 3196.8
53.35 x 658.35 (198.68°F manifold IAT + 459.67) = 35122.9725
3196.8 / 35122.9725= 0.09101735338602107210601266735041
0.091 lb/Ft³ at 60°F and 10psi (70% eff) boost in Denver.
Man... I'm getting tired of writing! Let's see if I can breeze through the pressure/airflow part kinda quickly.
Here's the idea, It's not about calculating actual numbers into a specific engine. It's about identifying a given mass flow rate through a given orifice and adjusting numbers in the above calculations until one scenario is relatively similar to another. Every engine is gonna have different heads, different intake, different cam, different exhaust restrictions... Way too much to take into consideration. But, if we know what one set of conditions will flow through a given restriction, we can fiddle with the variables until the results of one closely match another.
This one's a little tougher to calculate. We're actually getting into "rocket science" now, but if you follow the instructions you should be good to go. Soon to be a community of LS enthusiasts with a minor in rocket science. LOL By the way, I learned this stuff out of interest for engine related stuff and am not formally educated in the subject. Just an FYI, take it for what it's worth. You did, after all, read it on the internet. Hahaha
Actually... I'm gonna finish this post up tomorrow or Monday. This stuff takes a toll...
Until then
At altitude (less atmospheric pressure) there is less oxygen per given volume. There is also less of everything else that makes up air, most notably nitrogen. The percentage of oxygen to nitrogen does not change. The air is simply under less pressure and therefor less dense. So, the statement that the percentage doesn't change is correct from any perspective. But, the statement that there is less oxygen is only true when comparing unit per volume against the same volume at a different absolute pressure. In which case there would be more oxygen at lower altitudes, however more nitrogen too. Bad feeling that may have confused things more than cleared them.
About the statement that there's less pressure to push the air in the engine at altitude and more at sea level. Yes... But, that's not the complete story either.
You must take into account pressure AND density to determine mass flow. Either on their own can be an indication used to get an idea of what/how things may change. However, one without the other is nothing more than that... an almost in the ball park idea.
Physics can be daunting as a whole. But, in this case it's quite simple and the equations are easy to perform to get a much better idea of what's going on and what to expect.
First, calculate air density. We're only looking at the effects of altitude change and ignoring other atmospheric variables anyway. So... good enough in my book. It's easier to calculate if we're not concerned with humidity anyways.
Then, calculate the effect of pressure through an orifice in a choked condition. Again, not completely on the mark, but close enough to compare one situation to another with a reasonable amount of accuracy... for me anyway. Anyone want to expand on this, please do.
Since where I am on the planet commonly uses Pounds to measure weight, Cubic Feet to measure air volume, PSI for pressure, and °F to measure temperature... that's what I use. I also use °R for absolute temperature instead of °K since it's easier to convert °F to °R by simply adding/subtracting 459.67 to the °F figure. If this makes no sense to you, don't sweat it. It's just a technicality. Results are the same. (Actually better IMO. I seem to get slight variances in the final solution when I convert into °K)
Air Density in pounds per cubic foot:
All you need to calculate this (for dry air, no humidity) is the conditions/measurements in the manifold, air temperature and absolute pressure.
The product of
Absolute Pressure x 144
Divided by the product of
53.35 x absolute temperature (°F plus 459.67)
Equals lb/Ft³, how much one cubic foot of air weighs at the specified conditions
So at sea level
14.7psi x 144 = 2116.8
53.35 x 519.67 (60°F + 459.67) = 27724.3945
2116.8 / 27724.3945 = 0.076351532221921
So, let's call it 0.076 lb/Ft³ at 60°F at sea level
Using the same temp and the 12.2psi noted by the OP gives us 0.063 lb/Ft³
Adding 10psi boost to Denver's 12.2 gives us 22.2 psia and 100% efficient intercooling increases density to 0.115 lb/Ft³
But, what if we take into account heat added by compression and the compressor without intercooling? Just for conversations sake, lets say the compressor is 70% efficient.
Calculating temperature gain from compression (adiabatic) and compressor efficiency:
Pressure Ratio to the power of 0.286 x absolute (°F + 459.67) = adiabatic temperature after compression. ("adiabatic" simply means no thermal energy is added or taken away to the air as it's compressed.)
So, 60°F air compressed from 12.2psia to 22.2psia (not counting the compressor (in)efficiency) would result in:
Pressure ratio... 22.2 (final pressure) / 12.2 (atmospheric pressure) = 1.82PR
1.82 to the power of 0.286 = 1.1868
1.1868 x 519.67 (60°F = 459.67) = 616.744356 minus 459.67 to get back to °F = 157°F
Now, account for compressor (in)efficiency
Subtract the starting temp from the final temp (in absolute) and divide it by compressor efficiency...
616.744356 - 519.67 = 97.074356 °R divided by 0.70 = 138.68
Add the product back to the starting (atmospheric) absolute temp to get the final temp in °R. Subtract 459.67 to convert back to °F
519.67 + 138.68 = 658.35°R
658.35 - 459.67 = 198.68°F IAT
Now tie that into the density calculation for manifold air density in Denver...
22.2psia x 144 = 3196.8
53.35 x 658.35 (198.68°F manifold IAT + 459.67) = 35122.9725
3196.8 / 35122.9725= 0.09101735338602107210601266735041
0.091 lb/Ft³ at 60°F and 10psi (70% eff) boost in Denver.
Man... I'm getting tired of writing! Let's see if I can breeze through the pressure/airflow part kinda quickly.
Here's the idea, It's not about calculating actual numbers into a specific engine. It's about identifying a given mass flow rate through a given orifice and adjusting numbers in the above calculations until one scenario is relatively similar to another. Every engine is gonna have different heads, different intake, different cam, different exhaust restrictions... Way too much to take into consideration. But, if we know what one set of conditions will flow through a given restriction, we can fiddle with the variables until the results of one closely match another.
This one's a little tougher to calculate. We're actually getting into "rocket science" now, but if you follow the instructions you should be good to go. Soon to be a community of LS enthusiasts with a minor in rocket science. LOL By the way, I learned this stuff out of interest for engine related stuff and am not formally educated in the subject. Just an FYI, take it for what it's worth. You did, after all, read it on the internet. Hahaha
Actually... I'm gonna finish this post up tomorrow or Monday. This stuff takes a toll...
Until then
06-25-2016, 06:46 PM
#48
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Thanks. 
Made a few corrections to typos and just thought I'd put up where we're at in the calculations thus far, specifically the two scenarios you mentioned.
Denver 60°F, 12.5 psi, 70% comp eff... 24.7psia... 2.03PR... 226.6°F IAT... 0.097 lb/Ft³
Sea Lvl 60°F, 10.0 psi, 70% comp eff... 24.7psia... 1.68PR... 178.7°F IAT... 0.104 lb/Ft³
The results are a pretty clear example of what KingtalOn mentioned in post 3.
Same manifold pressure, different density. If an engine processes 1000 cfm (of manifold conditions), it would consume 7lb/min more at sea level. Which, using the common rough figure of 10HP/lb, is a 70HP premium! Or 35HP at 500cfm...
Which brings up the question, how much boost in Denver to be the same?
That's where calculating mass airflow will come in handy. Another day or two. Need to reverse engineer my spreadsheet. It's been a couple years since this was all fresh in my mind. Looks like Greek now. Hahaha
A last bit of food for thought... the difference between the pressure ratios may land you in a more or less efficient area of the compressor map. Could be good, could be bad. Just a note. Lot's of variables.
Made a few corrections to typos and just thought I'd put up where we're at in the calculations thus far, specifically the two scenarios you mentioned.
Denver 60°F, 12.5 psi, 70% comp eff... 24.7psia... 2.03PR... 226.6°F IAT... 0.097 lb/Ft³
Sea Lvl 60°F, 10.0 psi, 70% comp eff... 24.7psia... 1.68PR... 178.7°F IAT... 0.104 lb/Ft³
The results are a pretty clear example of what KingtalOn mentioned in post 3.
Same manifold pressure, different density. If an engine processes 1000 cfm (of manifold conditions), it would consume 7lb/min more at sea level. Which, using the common rough figure of 10HP/lb, is a 70HP premium! Or 35HP at 500cfm...
Which brings up the question, how much boost in Denver to be the same?
That's where calculating mass airflow will come in handy. Another day or two. Need to reverse engineer my spreadsheet. It's been a couple years since this was all fresh in my mind. Looks like Greek now. Hahaha
A last bit of food for thought... the difference between the pressure ratios may land you in a more or less efficient area of the compressor map. Could be good, could be bad. Just a note. Lot's of variables.
Last edited by SethU; 06-25-2016 at 08:22 PM.
06-25-2016, 11:38 PM
#49
On The Tree
Thread Starter
I agree, alot of math to it, but for laymens sake id bet a shiney nickle its between 2 and 4 pounds for equal uncorrected hp. Everyone wants to say theres less oxygen and thats just not the case when your talking compressors. The air here at 14.7 is the same at sea level 14.7
The difference is my cubic foot of air molecules(volume) isnt as dense as most, but the percentages are the same. 1/4 and 3/12 is the same percentage and a turbo will make 1/4 into 3/12 bolume wise. Of course theres a penalty for having to compress that added volume.
The end result being i have to spin my compressor harder to reach the same boost levels and at said same boost levels i make less hp. In the instance of stock bottom turbo ls1 engines a safe tune up and roughly 600hp is generally regarded as upper end of reliable hp limit. Some would say 550ish. The point i was trying to convey was that my engine at 13 psi to make 600hp uncorrected sees the same internal stresses as a 600 hp 10 pound engine at sea level. My 12.5# gauge pressure is the same kpa as your 10# gauge pressure.
The difference is my cubic foot of air molecules(volume) isnt as dense as most, but the percentages are the same. 1/4 and 3/12 is the same percentage and a turbo will make 1/4 into 3/12 bolume wise. Of course theres a penalty for having to compress that added volume.
The end result being i have to spin my compressor harder to reach the same boost levels and at said same boost levels i make less hp. In the instance of stock bottom turbo ls1 engines a safe tune up and roughly 600hp is generally regarded as upper end of reliable hp limit. Some would say 550ish. The point i was trying to convey was that my engine at 13 psi to make 600hp uncorrected sees the same internal stresses as a 600 hp 10 pound engine at sea level. My 12.5# gauge pressure is the same kpa as your 10# gauge pressure.
06-26-2016, 12:40 AM
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I agree, alot of math to it, but for laymens sake id bet a shiney nickle its between 2 and 4 pounds for equal uncorrected hp. Everyone wants to say theres less oxygen and thats just not the case when your talking compressors. The air here at 14.7 is the same at sea level 14.7
The difference is my cubic foot of air molecules(volume) isnt as dense as most, but the percentages are the same. 1/4 and 3/12 is the same percentage and a turbo will make 1/4 into 3/12 bolume wise. Of course theres a penalty for having to compress that added volume.
The end result being i have to spin my compressor harder to reach the same boost levels and at said same boost levels i make less hp. In the instance of stock bottom turbo ls1 engines a safe tune up and roughly 600hp is generally regarded as upper end of reliable hp limit. Some would say 550ish. The point i was trying to convey was that my engine at 13 psi to make 600hp uncorrected sees the same internal stresses as a 600 hp 10 pound engine at sea level. My 12.5# gauge pressure is the same kpa as your 10# gauge pressure.
The difference is my cubic foot of air molecules(volume) isnt as dense as most, but the percentages are the same. 1/4 and 3/12 is the same percentage and a turbo will make 1/4 into 3/12 bolume wise. Of course theres a penalty for having to compress that added volume.
The end result being i have to spin my compressor harder to reach the same boost levels and at said same boost levels i make less hp. In the instance of stock bottom turbo ls1 engines a safe tune up and roughly 600hp is generally regarded as upper end of reliable hp limit. Some would say 550ish. The point i was trying to convey was that my engine at 13 psi to make 600hp uncorrected sees the same internal stresses as a 600 hp 10 pound engine at sea level. My 12.5# gauge pressure is the same kpa as your 10# gauge pressure.
Everyone wants to say theres less oxygen and thats just not the case when your talking compressors.
The air here at 14.7 is the same at sea level 14.7
id bet a shiney nickle its between 2 and 4 pounds for equal uncorrected hp.
It is kind of an intriguing thought. I'll be curious to see how the calculations pan out with different values.
Last edited by SethU; 06-26-2016 at 12:48 AM.
06-26-2016, 12:50 AM
#51
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I believe that e85 all but negates most charge temp issues. The problem with proving that theory is the fact that you would have to be able to measure iat post fuel charge.
06-26-2016, 01:00 AM
#52
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Detonation control? I can see that.
Charge cooling? I don't think it would make any difference unless it was carb or throttle body injected... and even then the cooling effect may cool either case the same, if not the more dense mixture slightly more due to additional fuel the less dense charge wont have at the same AFR. Purely speculation though.
Something else?
06-26-2016, 01:24 AM
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OH... Are you saying E85 at altitude vs. Gasoline at sea level? Hmmm. Maybe. Dunno. If ya like the theory, go with it. Won't get any argument from me. LOL
06-26-2016, 01:31 AM
#56
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Ive been doing a bit of experimentation, just need to get back on a dyno for power conformation. Then i guess time will tell. Ready to start building the new forged shortblock, then see how far i can push this thing before it pretzels the rods.
06-27-2016, 12:09 PM
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06-27-2016, 12:46 PM
#60
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Lovington? Not quite sure ive ever heard of it. We were in loveland then moved to greeley. We originally moved up from lake jackson which is about an hour south of houston, down on the coast.