Sorry but your numbers complete lost me.
What does this mean :
Quote:
0 C, 0/0 g (0/0 ml) max of water can be vaporised per kg air
50 C, 10.8/8.9 g (10.8/8.9 ml) max of water can be vaporised per kg air
100 C, 27/25.8 g (27/25.8 ml) max of water can be vaporised per kg air
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10.8 divided by 8.9 grams of what ?? 10.8 what divided by 8.9 ml of what ?? --- I am assuming the grams and ml values are the amount of water, but what is the 10.8 refer to ????
Sorry, maybe your using a british notation format that is obvious to you, but to me its not at all clear what your units are, that go with the numbers.
Even air that is below freezing can hold some water vapor, so the zero grams at 0 deg C is not correct as I read this.
At 100 deg C. and atmospheric pressure you can have unlimited amounts of water in the air.
at 50 deg C, ( 122 deg F) air can hold 85.4 grams of water / kg of dry air.
If you start with 50 % RH then you should be able to evaporate up to about 43 grams of water, if you have enough heat available to hold your air temp at 50 deg C.
I believe your trying to over think this. The typical temperature drop with WI and common alcohol mixes is about 30 deg F ( 17 deg C ) depending on outside temp and humidity. With water only the temp drop is frequently around 20 deg F ( 10 deg C) at low RH initial conditions.
The amount of water vapor air can hold varies both with temperature and with pressure, so depending on your injection point, current weather conditions, and the boost level at that injection site -- all your numbers change.
The basic rules of thumb for injection rates are a whole lot easier to work with IMHO, unless you want to solve a bunch of multi variable equations for each possible boost, initial RH and temperature condition your likely to run into.
If I'm missing the whole point here, please take another shot at it !
Larry