The effects of air temperatures
Discussion
I did some looking into this before, but it rapidly got a bit "dimensional"
Generally I was looking to see the average (or even just a good yard stick) for the effect of increases in temperature on the volume of oxygen in air.
For a single gas, it isn't too hard to work it out, but air is a mix, and there is also vapour pressure which means a lot of things all changing at the same time.
Does anyone have any rules of thumb, even if it is just to get a basic over-view of how much "better" cold air is?
The scope of this is to see how effective cold air intakes are, whether they make enough difference based on how much the air heats up, how much it heats up considering there will be a decent turnover of air at high revs, and how much difference it makes to power.
This also would have a bearing on dyno testing with the bonnet up vs bonnet down, whether that makes any large differences to intake temps (ie more space to 'grab' air, vs less airflow compared to driving) and finally whether there is much of a 02 density change as a result of this.
Generally I was looking to see the average (or even just a good yard stick) for the effect of increases in temperature on the volume of oxygen in air.
For a single gas, it isn't too hard to work it out, but air is a mix, and there is also vapour pressure which means a lot of things all changing at the same time.
Does anyone have any rules of thumb, even if it is just to get a basic over-view of how much "better" cold air is?
The scope of this is to see how effective cold air intakes are, whether they make enough difference based on how much the air heats up, how much it heats up considering there will be a decent turnover of air at high revs, and how much difference it makes to power.
This also would have a bearing on dyno testing with the bonnet up vs bonnet down, whether that makes any large differences to intake temps (ie more space to 'grab' air, vs less airflow compared to driving) and finally whether there is much of a 02 density change as a result of this.
There are two main factors that contribute to how much power an engine can generate, in relation to intake air temperature. The first is the amount of oxygen in a given volume of air. Colder air is denser, so you will get more oxygen molecules into your engine if you inhale the same volume of air. From what I understand, the number is somewhere around the 5-10% power per 10 degrees Celsius.
The second one is how much ignition advance your engine will be able to tolerate, before it starts pinging. Air taken into the engine will get compressed. That will warm the air up even more. Once your piston is almost at the top, you will need to ignite the mixture, so you can achieve maximal pressure when the piston is 12-15 degrees after TDC. If you start with a hotter mixture, you will get a hotter half burnt gas mixture during ignition, resulting in pinging if you go over a threshold, unless you retard ignition, so the pressure will be lower and the temperature for pinging will not be reached. In effect, this means that a car with a good cold air intake, will possibly take better to ignition advance at wide open throttle. Almost always on modern engines, you reach a point where more power can be had with more ignition advance, but pinging will prevent you from doing so. By getting colder air in, you can get closer to the optimal ignition timing for full power. This is something that needs to be explicitly tuned in to most ECUs, but some have been set up to take advantage of this automatically. This is the main reason why turbo intercoolers are so successful. If it was just about volume, you could up the boost a bit and get the same amount of oxygen in.
The second one is how much ignition advance your engine will be able to tolerate, before it starts pinging. Air taken into the engine will get compressed. That will warm the air up even more. Once your piston is almost at the top, you will need to ignite the mixture, so you can achieve maximal pressure when the piston is 12-15 degrees after TDC. If you start with a hotter mixture, you will get a hotter half burnt gas mixture during ignition, resulting in pinging if you go over a threshold, unless you retard ignition, so the pressure will be lower and the temperature for pinging will not be reached. In effect, this means that a car with a good cold air intake, will possibly take better to ignition advance at wide open throttle. Almost always on modern engines, you reach a point where more power can be had with more ignition advance, but pinging will prevent you from doing so. By getting colder air in, you can get closer to the optimal ignition timing for full power. This is something that needs to be explicitly tuned in to most ECUs, but some have been set up to take advantage of this automatically. This is the main reason why turbo intercoolers are so successful. If it was just about volume, you could up the boost a bit and get the same amount of oxygen in.
I think I figured out 3*C of air is around 1% more power.
Of course the dyno will correct to STP anyway.
Cold air is only one aspect, dont forget most OEM intakes will not allow 100% of atmospheric pressure to be seen at the intake valve. There is often a vacuum which can gain considerable power when eliminated too.
This is partly why individual throttle bodies are generally the best solution for NA.
Bonnet Up/Down can also be dependant on correction factors and where the guy puts the intake probe if it exists.
Of course the dyno will correct to STP anyway.
Cold air is only one aspect, dont forget most OEM intakes will not allow 100% of atmospheric pressure to be seen at the intake valve. There is often a vacuum which can gain considerable power when eliminated too.
This is partly why individual throttle bodies are generally the best solution for NA.
Bonnet Up/Down can also be dependant on correction factors and where the guy puts the intake probe if it exists.
ringram said:
Cold air is only one aspect, dont forget most OEM intakes (made before about the year 1995) will not allow 100% of atmospheric pressure to be seen at the intake valve.
The vast majority of intakes designed since the mid 90's are very effective in terms of pressure recovery. A lot people stick pressure guages in intakes and proclaim them to be "rubbish", even though they are just measuring the static component of pressure and not the dynamic one. Total pressure loss in a modern intake system will generally be below 5kPa, and often on high performance cars be only 1 or 2 kPa (due to effective pressure recovery in the plenum)ringram said:
This is partly why individual throttle bodies are generally the best solution for NA.
Again generally speaking, individual port throttling is used to maximise transient response (due to the small volume between throttle plate and intake value) and allow the use of more extreme cam profiles and maintain idle / low speed driveability (because the port throttle helps prevent charge dilution at low speed due to the large EVC/IVO overlap)With sufficient plenum volume and suitable throttle plate sizing you can make pretty much the same power with a single throttle as with ITB's
Lovely, there's some figures turning up.
Maths time- Using DIN 70020 for the BHP correction for a dyno, lets see what happens..
CF =*equivalent to= P/P0 = (P/P0).(T0/T)^0.5
^that's what I've read, and it seems impossible, unless it means you can correct for pressure, OR correct for pressure and temp.
P0 = 1.01325 bar
T0 = 293K
in other words, any dyNo power correction using the DIN format shows you what the power would be like at 1013 mbar and 19.85'C
So lets try a calc at 1013mbar, and 9.85'C
(1013mbar/1013mbar) X (293K/283K)^0.5 = 1.0175
Obviously if pressure remains the same P/P0 = 1.
That means maths failure if the correction is 1.017 (ie 1.7% per 10 deg) as that would mean a car/engine tested at ~10'C would get a boost of 1.7% when it would be compared against something tested at 20'C.
link to source=
www.scielo.br/pdf/jbsms/v25n3/a10v25n3.pdf
I assume a . hanging in midair means multiply and superscript 0.5 is square root...
When I did maths, we'd write an X or the SQRT Tick...
Maths time- Using DIN 70020 for the BHP correction for a dyno, lets see what happens..
CF =*equivalent to= P/P0 = (P/P0).(T0/T)^0.5
^that's what I've read, and it seems impossible, unless it means you can correct for pressure, OR correct for pressure and temp.
P0 = 1.01325 bar
T0 = 293K
in other words, any dyNo power correction using the DIN format shows you what the power would be like at 1013 mbar and 19.85'C
So lets try a calc at 1013mbar, and 9.85'C
(1013mbar/1013mbar) X (293K/283K)^0.5 = 1.0175
Obviously if pressure remains the same P/P0 = 1.
That means maths failure if the correction is 1.017 (ie 1.7% per 10 deg) as that would mean a car/engine tested at ~10'C would get a boost of 1.7% when it would be compared against something tested at 20'C.
link to source=
www.scielo.br/pdf/jbsms/v25n3/a10v25n3.pdf
I assume a . hanging in midair means multiply and superscript 0.5 is square root...
When I did maths, we'd write an X or the SQRT Tick...
TheEnd said:
Lovely, there's some figures turning up.
Maths time- Using DIN 70020 for the BHP correction for a dyno, lets see what happens..
CF =*equivalent to= P/P0 = (P/P0).(T0/T)^0.5
^that's what I've read, and it seems impossible, unless it means you can correct for pressure, OR correct for pressure and temp.
P0 = 1.01325 bar
T0 = 293K
in other words, any dyNo power correction using the DIN format shows you what the power would be like at 1013 mbar and 19.85'C
So lets try a calc at 1013mbar, and 9.85'C
(1013mbar/1013mbar) X (293K/283K)^0.5 = 1.0175
Obviously if pressure remains the same P/P0 = 1.
That means maths failure if the correction is 1.017 (ie 1.7% per 10 deg) as that would mean a car/engine tested at ~10'C would get a boost of 1.7% when it would be compared against something tested at 20'C.
link to source=
www.scielo.br/pdf/jbsms/v25n3/a10v25n3.pdf
I assume a . hanging in midair means multiply and superscript 0.5 is square root...
When I did maths, we'd write an X or the SQRT Tick...
You haven't read that right: P and p are different, with big P being power and little p being pressure. And the x in a box isn't multiply. That's why they have nomenclature at the start!Maths time- Using DIN 70020 for the BHP correction for a dyno, lets see what happens..
CF =*equivalent to= P/P0 = (P/P0).(T0/T)^0.5
^that's what I've read, and it seems impossible, unless it means you can correct for pressure, OR correct for pressure and temp.
P0 = 1.01325 bar
T0 = 293K
in other words, any dyNo power correction using the DIN format shows you what the power would be like at 1013 mbar and 19.85'C
So lets try a calc at 1013mbar, and 9.85'C
(1013mbar/1013mbar) X (293K/283K)^0.5 = 1.0175
Obviously if pressure remains the same P/P0 = 1.
That means maths failure if the correction is 1.017 (ie 1.7% per 10 deg) as that would mean a car/engine tested at ~10'C would get a boost of 1.7% when it would be compared against something tested at 20'C.
link to source=
www.scielo.br/pdf/jbsms/v25n3/a10v25n3.pdf
I assume a . hanging in midair means multiply and superscript 0.5 is square root...
When I did maths, we'd write an X or the SQRT Tick...
On my iPhone is making It trick to get to what it should be, I'll have another look in the morning.
Nick1point9 said:
You haven't read that right: P and p are different, with big P being power and little p being pressure. And the x in a box isn't multiply. That's why they have nomenclature at the start!
On my iPhone is making It trick to get to what it should be, I'll have another look in the morning.
Really? Surely P is the measured pressure on the day of the run? Nothing to do with power - if you put power figures into that equation you'd get bonkers answers!On my iPhone is making It trick to get to what it should be, I'll have another look in the morning.
He is right, it was this bit-
CF =*equivalent to= P/P0 = (P/P0).(T0/T)^0.5
The middle bit in bold is Power = P, the ones after should be lowercase p = pressure.
That explains why an equation looked like it was equal to itself with other factors added on!
Corrected-
CF =*equivalent to= P/P0 = (p/p0).(T0/T)^0.5
CF =*equivalent to= P/P0 = (P/P0).(T0/T)^0.5
The middle bit in bold is Power = P, the ones after should be lowercase p = pressure.
That explains why an equation looked like it was equal to itself with other factors added on!
Corrected-
CF =*equivalent to= P/P0 = (p/p0).(T0/T)^0.5
TheEnd said:
He is right, it was this bit-
CF =*equivalent to= P/P0 = (P/P0).(T0/T)^0.5
The middle bit in bold is Power = P, the ones after should be lowercase p = pressure.
That explains why an equation looked like it was equal to itself with other factors added on!
Corrected-
CF =*equivalent to= P/P0 = (p/p0).(T0/T)^0.5
Ah, OK, makes sense.CF =*equivalent to= P/P0 = (P/P0).(T0/T)^0.5
The middle bit in bold is Power = P, the ones after should be lowercase p = pressure.
That explains why an equation looked like it was equal to itself with other factors added on!
Corrected-
CF =*equivalent to= P/P0 = (p/p0).(T0/T)^0.5
You can use Eric's turbo calculator (set with no boost) to see the effects on HP for a varying inlet temperature:
3.9 V8 @ 20'c inlet = 222BHP:
http://www.not2fast.com/turbo/glossary/turbo_calc....
3.9 V8 @30'c inlet = 214BHP:
http://www.not2fast.com/turbo/glossary/turbo_calc....
Seems like for every 1'c increase in temperature you loose about 1 bhp at 5500RPM - according to that calculator.
3.9 V8 @ 20'c inlet = 222BHP:
http://www.not2fast.com/turbo/glossary/turbo_calc....
3.9 V8 @30'c inlet = 214BHP:
http://www.not2fast.com/turbo/glossary/turbo_calc....
Seems like for every 1'c increase in temperature you loose about 1 bhp at 5500RPM - according to that calculator.
tickious said:
All over my head.
Can anyone tell me in layman's terms, if I did a power run on a cold day with the bonnet open, will it give a higher bhp reading than on a warm day?
That would depend on the vehicle, engine, tuning, fuel, difference in temperature, difference in dyno, difference in dyno operator, tyres, tyre pressures, type of floor mats in the car, type of shoes....Can anyone tell me in layman's terms, if I did a power run on a cold day with the bonnet open, will it give a higher bhp reading than on a warm day?
tickious said:
All over my head.
Can anyone tell me in layman's terms, if I did a power run on a cold day with the bonnet open, will it give a higher bhp reading than on a warm day?
The engine would potentially produce more power, but whether that was reflected in the reading would depend what reading you are talking about. Power figures often include compensation for environmental conditions.Can anyone tell me in layman's terms, if I did a power run on a cold day with the bonnet open, will it give a higher bhp reading than on a warm day?
GreenV8S said:
tickious said:
All over my head.
Can anyone tell me in layman's terms, if I did a power run on a cold day with the bonnet open, will it give a higher bhp reading than on a warm day?
The engine would potentially produce more power, but whether that was reflected in the reading would depend what reading you are talking about. Power figures often include compensation for environmental conditions.Can anyone tell me in layman's terms, if I did a power run on a cold day with the bonnet open, will it give a higher bhp reading than on a warm day?
I had it done in summer, hoping for 200bhp just because I was a nice round number.
But it wasn't running on 99Ron and the k&n was dirty. Now both sorted so I think I'll have another cheeky little run to see if it can find another 5 horses from somewhere.
Dyno figures should always be corrected for atmospheric conditions, or the weather on the day would change the power reading, which makes comparisons with other runs a bit difficult.
If you really want a magic number (like 200 hp), run your car early in the morning on a cold winter's day, and get the dyno operator to print you a graph with no atmospheric correction
.
If you really want a magic number (like 200 hp), run your car early in the morning on a cold winter's day, and get the dyno operator to print you a graph with no atmospheric correction
.Dyno numbers are just that, the peak number at a particular point in the rev range. The area under the graph makes a fast car, thats why a turbocharged car will always be faster than a supercharged, the area under the graph is greater in a turbo car. Rather than numbers, you should focus on how it drives and the delivery of the power more
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