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.
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...
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
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