It seems this is a typical gas spray cooling with humidification issue.
The maximum amount of water evaporated will depend on the final temperature of the cooled air. To estimate the amount of water that evaporates one can:
1. Look for the water content at the adiabatic saturation of the air at the required final temperature deducting the moisture it contains at the beginning, and multiply this by the mass of air processed, or
2. Make a heat balance assuming the air cooling comes entirely from the latent heat of evaporation of water.
The final configuration of the column contactor will be arrived at upon estimating the time needed for droplets of a given size to evaporate knowing the air flow rate and linear velocity.
25362 - Thank you for you replies. Although the difficulty I'm having is that the tables (In Perry's) for saturation points in air only go up to about 150-200degrees C. Also, I should note that this is the first time I've done thermodynamics for about 15years so I'm a bit rusty on the subject.
Any further advice?
I know that Perry VI has a humidity diagram with "dry bulb" temperatures of up to 800oF.
Besides, at 425oC, water is a supercritical fluid and would mix with air in all proportions from zero to 100%.
Regarding "dew points" and water "saturation" of gases, or air, I'm sure there are past threads that dealt with this subject in detail. Any chemical engineering book and even a search through Google would help to find articles that may remind you of whatever air humidity-related items you may have forgotten.
Kindly tell us what is the precise information or calculation you are after, for us to be able to help.
To EdStainless, you are right on the heat balance. However, when using the conventional definition of saturation, as long as the outlet air temperature is sufficiently high no air "saturation" is expected. Saturation, in this case, is meant to say that the air is at its water dew point.
Let's make an exercise on air at atmospheric pressure.
When cooling the air from 800 K down to 500 K the amount of water evaporated would be about 160 g/kg bone-dry air. If the cooled air leaves at 400 K, the amount of water evaporated would be about 200 g/kg. In either case the air isn't "saturated".
The mole fractions of water in the leaving gas mix would be 0.21 and 0.25, respectively. The dew points would be 62 and 65 deg C, respectively. These are the temperatures for (the leaving) air saturation to take place.
In practice, the amount of evaporated water is a function of the contacting residence time and the contacting efficiency between water and hot air. Time is needed for water droplets to evaporate. This factor and the air flow rate, among others, would help to determine the configuration of the quenching column.
25362: You're right that a good appraoch is to calculate the "adiabatic saturation" as if this were a quench operation.
In such case, assuming you start with almost dry air at 425°C the final temperature would be 59.9°C with 19.6 mole percent water.
I do not think that the fact that at 425° water is supercritical is relevant. In your case you have plenty of water, so the bulk of the water will nor reach high temperatures.
As you point, the real issue is whether the waterfall ensures enough contact/heat transfer.
To siretb, you are absolutely right, in hot air quenching and humidifying the supercriticality of water vapor is not a factor.
However, the context of the question obliged me to mention this fact. The questioner asked what would be the maximum amount of water that air could hold while at 425 deg C, and I said that water would mix with air in all proportions and could range from zero to 100%, because by being supercritical, water acts as any "permanent" gas would.
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