Came up in a conversation today. The person stated that evaporative cooling, using water, will make the air (oxygen) less dense. Whatchya all think about that?
Info: Discussion on oxygen density
- Thread starter Killerbee
- Start date
x2..
the only thing that does not get denser when forzen is ice.
everything else the colder it gets the denser the said material gets.
The problem is some of that density is water. Whenever dry air passes over water, some of the water will be absorbed by the air. The hotter and drier the air, the more water that can be absorbed. This happens because the temperature and the vapor pressure of the water and the air attempt to equalize. The liquid water molecules change there physical state to gas. So even though its cooler is it really packed with more oxygen?
i would say cold dry air is denser then cold wet air.
keeping the humudity level as low as possible
keeping the humudity level as low as possible
As the humidity goes up, the density of air drops. Don't seem to make sense, but it does.
Air is N2 and 02 and H20 mostly, and in that order.
N2 is 28, O2 is 32, and H20 is 18 on the atomic weight (all gases have the same # of molecules in a given volume)
As the percentage of O2 goes up, the weight of the air does. As the percentage of H2O goes up, the weight of the air drops.
Really. I actually stayed awake in HS chemistry...
Air is N2 and 02 and H20 mostly, and in that order.
N2 is 28, O2 is 32, and H20 is 18 on the atomic weight (all gases have the same # of molecules in a given volume)
As the percentage of O2 goes up, the weight of the air does. As the percentage of H2O goes up, the weight of the air drops.
Really. I actually stayed awake in HS chemistry...
In the intake tract, depending on where you inject the water, the air can become pretty hot, the hotter the air the more water can be absorbed. So the humidity goes up and up. And actually the southern members here (or any other humid region) will see less benefits of water cooling as the outside air is already saturated with water.
Some more food (or drink) for thought:
At normal room temperature at 50% humidity, 100lb of air has about 1lb of water in it.
100% would give you 2 pounds. So the potential loss of oxygen is not that great. But when you spray the water into hot air, a lot more can dissolve.
At normal room temperature at 50% humidity, 100lb of air has about 1lb of water in it.
100% would give you 2 pounds. So the potential loss of oxygen is not that great. But when you spray the water into hot air, a lot more can dissolve.
Temperature has way more effect on air density than moisture content.
So I say probably not. There likely isn't a way that evaporative cooling can make air less dense, unless the temperature drop is small enough that the increased moisture content can then decrease the air density.
That's my feeling any way.
So I say probably not. There likely isn't a way that evaporative cooling can make air less dense, unless the temperature drop is small enough that the increased moisture content can then decrease the air density.
That's my feeling any way.
Its awfully humid an hot here in Kansass. With only the smaller of my 2 M/W nozzles on, and approx 25% methanol, after a nice hot drive home one day with the M/W active, i jumped out when i pulled in my driveway and grabed the CAC pipes...well pre-CAC was hot enough to only want to touch briefly and post-CAC it was still fairly warm but post nozzle was actually pretty cool. If i had to guess id say around 70-60*F or less. If a can of Pepsi was that temp, I'd drink it!:rofl: Cools very effectively. Not as much without methanol but still much better of than nuthin!
Our racecar makes a lot more hp on a cool humid night, so I am going to say that since water is not a fuel, it must be adding more air. Fuel and air are what make hp and water is not a fuel......atleast according to me amigo going through fire fighting class.:rofl:
Well I'm a bit hungover.... And I hated chemistry in college but here goes.
PV=nRT. ideal gas law
P = pressure
V=volume
n = # of moles
R= gas constant..... 287ish for dry air, 461ish (read on interent) of water vapor
T= temperature
D = ([P dry air]/RT) + ([P water vapor]/RT)
D = density
So expanded out we have:
D= ({nRT/V dry air}/RT) + ({nRT/V water vapor}/RT)
Basically the density of the charge never decreases when water vapor is added into the intake. You can see that from the equation. So addition of water vapor will not change the density of the dry air if the temperatures are constant. It will increase the density of the entire charge. Now if the charge of water decreases the temperature then there is a huge effect on the density.
Thats the simplistic answer to the question asked. But in an intake there are other factors and i don't think I want to do any fluid mechanics today. If I even remember them.
So maybe someone can let me know if I'm correct or not...
Well what the hell. Ok so now we have this charge (water vapor and dry air mixed) in the intake track. As the intake valve is opened a portion of this charge enters the combustion chamber. For sake of my hangover we will say that this charge is the same, pressure and density for each test. (this is where I may be wrong and need a real engine builder to chime in).
Test 1 we have a charge where the water vapor equals 0. then 100% of the charge entering the combustion chamber is dry air.
Test 2 we have a charge where the water vapor equals 20% then 80% of the charge entering the combustion chamber is dry air.
Basically the water vapor did take up some the space in the combustion chamber that dry air would have. So you tehcnically did not change density you just changed the percentage of dry air entering the chamber.
BUT I feel the real trick to water/meth is that the temperature changes. So you've got to be able to lower the temp. enough to make up for this difference, whcih it can do. Thats where you get your power.
So am i right on that one guys...?
PV=nRT. ideal gas law
P = pressure
V=volume
n = # of moles
R= gas constant..... 287ish for dry air, 461ish (read on interent) of water vapor
T= temperature
D = ([P dry air]/RT) + ([P water vapor]/RT)
D = density
So expanded out we have:
D= ({nRT/V dry air}/RT) + ({nRT/V water vapor}/RT)
Basically the density of the charge never decreases when water vapor is added into the intake. You can see that from the equation. So addition of water vapor will not change the density of the dry air if the temperatures are constant. It will increase the density of the entire charge. Now if the charge of water decreases the temperature then there is a huge effect on the density.
Thats the simplistic answer to the question asked. But in an intake there are other factors and i don't think I want to do any fluid mechanics today. If I even remember them.
So maybe someone can let me know if I'm correct or not...
Well what the hell. Ok so now we have this charge (water vapor and dry air mixed) in the intake track. As the intake valve is opened a portion of this charge enters the combustion chamber. For sake of my hangover we will say that this charge is the same, pressure and density for each test. (this is where I may be wrong and need a real engine builder to chime in).
Test 1 we have a charge where the water vapor equals 0. then 100% of the charge entering the combustion chamber is dry air.
Test 2 we have a charge where the water vapor equals 20% then 80% of the charge entering the combustion chamber is dry air.
Basically the water vapor did take up some the space in the combustion chamber that dry air would have. So you tehcnically did not change density you just changed the percentage of dry air entering the chamber.
BUT I feel the real trick to water/meth is that the temperature changes. So you've got to be able to lower the temp. enough to make up for this difference, whcih it can do. Thats where you get your power.
So am i right on that one guys...?
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A diversity of views. I was wanting to get a cross section of opinions, and try to see where the confusion exists.
Based on a deep look, evaporative cooling does increase oxygen densiity. At 1 atm and 40% humidity, it does so an average of 7%
At 3 atm and 500 F (COT), it will do so an average of 15-20%, though it is hard to find data at high temp/pressure. Nothing to sneeze at. I'll explain.
If you are in Mississippi, 90% humidity, when that air is heated in the compresor to 500 F, it goes to 3% humidity. That same lb of air can hold 30 times more water, well kinda/sorta. As water is evaporated in that 3% hot air, temp comes down quickly, and that RH goes up fast. RH is very highly dependent on temp.
Anyone really wanting to understand how that works, get a text with a psychrometric chart, and work a simple evap cooling problem. It describes the relationship between moisture and air.
This may clear up some details. True, 90 degree 100% RH air has less O2 than 90 degree dry air. But that is irrelevant. We will take that 90 degree dry air, fully sturate it to 100%, which brings THAT air temp down to 40 degrees (guess). The air experiences an increase in density from the temp change, AND a smaller decrease in density from the space occupied by the added water vapor molecules. Hope that is clear. The overall impact is increased O2 density, and increased N2 density for that matter.
Methanol is a separate discussion, since it has fuel considerations.
good discussion.
Based on a deep look, evaporative cooling does increase oxygen densiity. At 1 atm and 40% humidity, it does so an average of 7%
At 3 atm and 500 F (COT), it will do so an average of 15-20%, though it is hard to find data at high temp/pressure. Nothing to sneeze at. I'll explain.
If you are in Mississippi, 90% humidity, when that air is heated in the compresor to 500 F, it goes to 3% humidity. That same lb of air can hold 30 times more water, well kinda/sorta. As water is evaporated in that 3% hot air, temp comes down quickly, and that RH goes up fast. RH is very highly dependent on temp.
Anyone really wanting to understand how that works, get a text with a psychrometric chart, and work a simple evap cooling problem. It describes the relationship between moisture and air.
This may clear up some details. True, 90 degree 100% RH air has less O2 than 90 degree dry air. But that is irrelevant. We will take that 90 degree dry air, fully sturate it to 100%, which brings THAT air temp down to 40 degrees (guess). The air experiences an increase in density from the temp change, AND a smaller decrease in density from the space occupied by the added water vapor molecules. Hope that is clear. The overall impact is increased O2 density, and increased N2 density for that matter.
Methanol is a separate discussion, since it has fuel considerations.
good discussion.
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If I read it correctly, the O2 increased from an overall increase in charge density.