Quote:
Larry do you feel you have been successful in humidifying the slug of air in the 50 milliseconds it spends between filter and turbo inlet? Can you support that?
just my opinion that with these coarse sprays, and so little residence time, that nearly no evaporation actually occurs, especially considering the low water/alc temps used. Maybe you can change my mind.
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Interesting question --- My short answer is enough to make a useful difference.
The objective is NOT to humidify the air, the objective is to increase engine (and in this case turbocharger performance) due to the sum of multiple effects. One of which happens to be evaporative cooling.
After considering you question for a while here's my view.
First very very small amounts of evaporation are all that are necessary to make a major change in the inlet conditions of the turbocharger. The specific heat of the intake air is .25 Cal/gm deg C (give or take a bit) while the latent heat of evaporation of water is 540 Cal / gm.
The maximum intake air flow on the stock WRX turbocharger is approximately 360 CFM or about 25.2 lbs / min ( 0.42) lb/sec or about 191 grams /second.
If you are spraying at a rate of 4 gallons / hour that is approximately 4.2 grams/second of water.
If you lower the intake air by 11 degrees Farenheight or about 6 deg Centigrade you increase the air density by 1% which will give you an increased mass flow through the turbocharger of approximately the same and give a net power increase (all else being about equal) of 1%. On a 250 hp engine then, each 6 deg C drop in intake temp is worth about 2.5 hp.
Given the ratio between the specific heat of air and the latent heat of water you have to evaporte only a very small fraction of the water to bring down the intake air temp significantly. Lots of people have recorded intake air temp drops on the order of 30 - 50 deg F (17 - 28 deg C) with initial air temps near 105 deg F (40 deg C).
So just to persue your question lets look at what we know. The air flows at a rate of 360 CFM max in an intake tube of 2.5 inch ID --- this gives a peak air speed at redline of about 176 ft/second (120 mph). The spray nozzle is about 2 ft away upstream so flight time for the dropplets are on the order of 11.4 milliseconds.
Now how many drops are in that air column and what is their surface area?
One cubic centimeter of water in the form of 50 micron dropplets (if my numbers are right) consists of about 424 million dropplets with a total surface area of about 333.3 cm^2. We are spraying at a rate of 4 gal/hr or 4.2 cm^3 / second so we are putting about 1,780 million dropplets into the airstream per second, they have an effective lifetime of 11.4 milliseconds in the air column, so the air column at any moment contains about 0.0477 cm^3 of spray or 20.44 million dropplets with a total surface area of 15.9 cm^2. Plus we also have the evaporation that occurs off the wetted interior surface of the tube, which would amount to 1216 cm^2 of surface.
How much water evaporates every 11.4 milliseconds, from 1232 cm^2 of water surface when exposed to a 120 mph wind ??
That intake air column has an air volume of only 0.068 cubic ft, which converts to about 2.166 grams of air, it obviously only takes a miniscule amount of evaporation during that 11.4 millisecond flight time to make a huge change in the temperature of a slug of air that only weighs 2.166 grams.
To lower that 2.166 grams of air 6 deg C, you only require .54166 calories of energy per deg C. To get that amount of cooling by evaporation of water you only need to evaporate 0.001 gm of water in 11.4 milliseconds/ deg C, or 0.08799 grams per second/ deg C which equals .5279 grams of water evaporated per second for a 6 deg C intake air temp drop.
{ all this ignoring the higher volatility of the alcohol component and its lower latent heat of evaporation but --- lets keep it simple for discucussion sake}
The interesting thing about this exercise is that it is obvious that the wetted surface of the intake tract dominates the evaporative cooling process by a large margin over the cooling that takes place at the dropplet surfaces.
Your swap cooler comment is right on target, and I have been considering similar ideas. It might be useful to intentionally maximize the wetted intake surface both to gain this cooling and to capture large mist dropplets.
I have considered putting pieces of fine mesh brass or stainless steel screen wire placed at extremely oblique angles to the air flow. (intention to give a very large open surface area but little or no direct path through the screen) Any dropplets that couldn't make the "jog" due to their large size, as they approached a screen wire would be broken up and wet the screen wire surface. Smaller dropplets would simply pass through the screen unmolested.
In a perfect design the screen would be treated so it wetted well with water alcohol mixes and had high capillary attraction to pull liquid water up off the floor of the intake and into the air stream. This is basically what is done inside heatpipes.
Larry