Double looped Bong?

[bigben2k]

Hot Damn!

I wouldn't assume that because sub-ambient is a possibility, that a bong can do better than a heatercore, but yours is fairly large, so I'm sure that it can take the heat.

Ok, so if you add a float valve for a refill, what kind of maintenance is required?

Man, I wish I had the time to try this out!

[nuclear]

BB2K
You have the problem that if you use tap water, every salt (calcium, sodium) will start to deposit or clog the block. You will also have a higher elecrtical conducitivity.
Using disitilled water would solve the problem, but then, you need another thank (bigger) that would be connected to the float.
Same system as using the 1/4 hose you told, but instead of using the pipe system of the house but instead using a bigger (and at a higher height) tank so you don't have too much water at the bottom of the bong( which will hinder the perfomance of the bong to go below ambient but will also act as a buffer).

I think Xjinn, when he develloped is bong, calculated that it could cool around 700 watt of heat.

[airspirit]

In a double loop bong setup, the calcium deposits and such will not be an issue if the radiator is submerged. You'd just have to CLR your showerhead every once in a while. Hooking up to a float valve would be a briliant way of doing this.

Having evaporation happen on your core, though, is a big ass no-no. Think of the buildup of minerals you'd suffer. Seriously consider a coiled 1/2" copper tube submerged. Since there are no elbows, it would have much less resistance, and in the end you'll have plenty of surface area. Also, it would be submerged in the coolest part of the system: the place where all the freshly cooled water is chillin'. One other problem with having the core suspended above the fan is that the core would lower airflow, requiring a more powerful and louder fan to get the desired CFM through the unit.

Seriously, suspending a core is bad news, requiring distilled water or failure will come soon. Submerging a core will hamper water flow around and through it. Submerging a coil, on the other hand, there is no real resistance in any flow direction. There was a guy on OC forums, I think, that was setting up something similar to what I am describing. Check that out.

[nicozeg]

Ben: I don’t know the precise physics explanation, but an evaporative cooling system just take out the heat because that’s the energy needed to turn water into vapor. That energy is stored in the vapor and is released when it condensates and return to liquid state. It’s just the same as in a phase change cooling loop, where cold is generated at the evaporator, and heat is released at the condenser. Only difference is that here you forget about the condensation part, and dispose off the coolant in its vapor state; and refill it with liquid one. You just give Mother Nature the job of the condensation stage.

Volenti: How much water evaporates daily that setup? I bet that several liters a day
And how did you solder the vertical pipes? I don’t see the tees.
Very efficient design! You don’t waste pump power with a big height.

OT: Ben, when I first read the forum in the morning, the radius thread was dead on the 10K reads mark.

[waterrat]

to volenti can you explain your setup, it look like a heat exchanger that should seat inside your reservoir, but, it is insulated and the fans are siting on top maybe a picture of the complete bong will help

[bikr]

Uhh.. I read most of the posts.. and I'm just trying to make sure I'm getting this right and if I am , then help others see what's going on.. Normally room temp air is passed through a radiator to cool water inside.. in a bong setup , water is able to get below room tempature when showered down.. Instead of passing air through the radiator which is room temp , this setup would send open system water below ambient through a radiator in which a closed system is functioning.. using the showering water , which is seperate from the liquid that is cooling the cpu , Am I on ? lol hope so --bikr

[Albigger]

Originally posted by bikr
Uhh.. I read most of the posts.. and I'm just trying to make sure I'm getting this right and if I am , then help others see what's going on.. Normally room temp air is passed through a radiator to cool water inside.. in a bong setup , water is able to get below room tempature when showered down.. Instead of passing air through the radiator which is room temp , this setup would send open system water below ambient through a radiator in which a closed system is functioning.. using the showering water , which is seperate from the liquid that is cooling the cpu , Am I on ? lol hope so --bikr

i believe that is what is being suggested here.


@ unloaded - i think i agree with airspirit and some others - the rad in the middle is bad news. either submerge it at the bottom and have the pump pull water through it, or use a coil of copper tube (very cheap). I don't think you will get much evaporation off of the radiator anyhow (if its in the middle position as pictured) because, at least on my heatercore, the through passages are VERY small, and I think the surface tension of the water would keep it in between there instead of lots of little drops trickling over the surface of the core, which would be ideal.

Plus, as others pointed out - the coldest water is at the bottom anyhow, and it would help avoiding the problem of mineral deposits.

If you want to try short term, please post results, but I feel it wouldn't work out as a long term solution.

[Arcturius]

Originally posted by nicozeg

But if you say that water is colder, and there are no more heat sources, that cold water should cool the air al least, how is going to be hot? The fan motor?

Ben: I don’t know the precise physics explanation, but an evaporative cooling system just take out the heat because that’s the energy needed to turn water into vapor. That energy is stored in the vapor and is released when it condensates and return to liquid state.

Actually, there are other heat sources: the heatercore cooling the main cooling loop. Also, the water is cool because part of it has evaporated, and in the process of becoming water vapor, has absorbed large amounts of energy. Now this energy-laden water vapor is suspended in the air, making it warm. The water does not cool the air, the air cools the water.

Make sense yet?

[freeloadingbum]

In a proper working bong, the water is about 3C below room temp.Through convection, the water cools the air (the air heats the water). Through evaporation, the air cools the water. This is one of the reasons why you can only get the water a few degrees below room temp.
I measured the exaust air of a bong and it was about 2C below room temp for a bong that was about 3C below room temp.

[jamicon]

Originally posted by nicozeg
Ben: I don’t know the precise physics explanation, but an evaporative cooling system just take out the heat because that’s the energy needed to turn water into vapor. That energy is stored in the vapor and is released when it condensates and return to liquid state. It’s just the same as in a phase change cooling loop, where cold is generated at the evaporator, and heat is released at the condenser. Only difference is that here you forget about the condensation part, and dispose off the coolant in its vapor state; and refill it with liquid one. You just give Mother Nature the job of the condensation stage.

Volenti: How much water evaporates daily that setup? I bet that several liters a day
And how did you solder the vertical pipes? I don’t see the tees.
Very efficient design! You don’t waste pump power with a big height.

OT: Ben, when I first read the forum in the morning, the radius thread was dead on the 10K reads mark.

Ok, here is what actually happens. First you must realize, heat cannot be destroyed, only transferred.

First the water that drips is at room temp. This water when it drips evaporates, and the air becomes humid (and hot). The reason it becomes hot is because the hottest bits of water will evaporate first (this is the same as when you hop out of the pool, and it feels cold). So the water will eventually reach an equilibrium with the air temperature, a temperature where the water wont get any colder, and the air wont be heated up as much, if the air gets cooled by the water, the water then evaporates and heats the air up again. So to cool air, you would need to run air through pipes in the water at the bottom.

Or, in short, if the water gets cooled by the air normally, and you want the water to cool the air (when it gets cold) where will the heat go? As it cannot be destroyed.

[UnloadeD]

Well I don't know if I agree with submerging the core. I know that will slow mineral build up on it, but I don't think it will cool as well. I think the most cooling will take place where the evaporation is happening (hopefully on the warmer core). If you think about a swimming pool when you stick your hand in the water, it feels cool, but when you take it out and the water starts evaporating its much cooler. Now I could be wrong and it just feels cooler, but I'm pretty sure that as the water changes from liquid to vapor it pulls heat from whereever it can get it, for the extra boost to change phase. Also I'm not concerned with cooling the bong water, number one priority is to cool the water in the core. I think I really am gonna try this cause the curiosiy is now killing me. I should be off work the whole week of Xmas, that should be a good time to try it.

peace.
unloaded

[nicozeg]

Beyond theory, I’m sure that in the real world things work the way I’m telling. An example is a swamp cooler. There you have a small ambient temp water reservoir that wets some kind of flat pad. Then there’s a fan that blows ambient temp air thru this wet pad. The result: Cool moist air. Can someone find a better explanation for this?

I think freeloadingbum proves my point, and that has to be the same for everyone with a bong that produces colder than ambient water.

Arcturius: I was explaining the case without the main cooling loop.

heat cannot be destroyed, only transferred.

That would be in a convection only world. The more precise statement is “Energy cannot be destroyed, only transformed”

In a car for example you start with the fuel, it has a lot energy stored in it, regardless of the temperature. With combustion, that energy is transformed to heat. The engine transforms heat into kinetic energy. A moving car has a lot of energy in it, and if you want to stop it you use brakes, that transform all that kinetic energy into heat again.

Vapor can be compared to a moving car: it hold energy in itself, regardless of the temperature.

Taking this back to the thread origin, as others have said, the core under the shower is not the best solution; Submerged would be better. But still this is not the best use for a HC; a pipe coil is the ideal.

So why not use the HC the way it’s designed? Blowing only air across it, but cool air from the bong? It’s not difficult to setup the bong, and test the core in the three different positions.

[freeloadingbum]

A well working rad system usually gets the water about 3C above the air temp going through it (under load). Since the bong air is only 2C cooler than normal, thats all you would gain. Submerging the rad should get the rad water about 1C (2C max) warmer than the bong water because water removes heat better than air. This brings the rad water to -2C (-1C max) as opposed to +1C (with rad on top of bong)

Personally, I would just design a bong using a large air filter (car air filter) to lessen dirt buildup, then run it to the block.

[koslov]

How about using two separate loops and connect them with one of these:



http://lytron.com/standard/he_liquid.asp

That would preserve the purity of your waterblock coolant while allowing you to use tap water for the bong loop. These have an extremely low thermal resistance, so you don't have to worry about an extra thermal gradient.

Then again, they are a bit pricy, starting at $159, but I suppose you could make your own liquid-to-liqiud heat exchanger fairly easily.

[airspirit]

The reason the water at the bottom is coldest is for one reason: energy is drawn from the coolant as it evaporates. As a portion of the droplet goes up in smoke, it takes energy with it as the water becomes a more excited state: steam. This leaves the water droplet colder than it was before, and the steam is warmer than the droplet was. No energy is lost: some is just transferred during the phase change.

As those cooled droplets accumulate in the bottom of the reservoir, the coolant mass is colder than the surrounding air. By re-warming it via a heat exchanger (removing energy from the secondary PC loop), it will now be easier to evaporate when re-sprayed through the evaporation chamber.

There is more than just temperature involved, though. At any given room temperature, there will be a desired humidity factor that the water will try to balance at. The less humid the room, the more water will evaporate, and the cooler the water that is collected will be. Similarly, the warmer the room, the cooler the water will be in comparison. Ideally, if you put a vacuum over a water mass, the water will evaporate like a mofo, leaving nothing but the minerals that was dissolved in it. You can get your best results if you run a dehumidifier in your room as well, or if you live in a dry environment, by pumping the steam outside.

By submerging the heat exchanger, you are simply putting it in the place where the primary bong coolant is at its coldest. By putting it at the coldest point, you will get the most effective heat transfer possible. The fact that it would also reduce scale buildup is just an added benefit.

BTW, one of those heat exchangers used before the pump and after the bong res would give you spectactular results. HEs are something that can be found by many manufacturers all over the place, and would not only increase overall efficiency of the combination system, but would give you one less point of failure. When it comes time to remove scale from the HE (every six months or less), just flush it with CLR and you're all set.

[airspirit]

Check out this for the ultimate in scientifically designed uber-bongs. The best part is that over 90% of the evaporate will be completely recovered. Imagine only topping off your bong every couple of weeks at the most ....

http://forums.procooling.com/vbb/sho...4304#post54304

[redleader]

The reason the water at the bottom is coldest is for one reason: energy is drawn from the coolant as it evaporates. As a portion of the droplet goes up in smoke, it takes energy with it as the water becomes a more excited state: steam. This leaves the water droplet colder than it was before, and the steam is warmer than the droplet was. No energy is lost: some is just transferred during the phase change.

Actually both the water and air end up colder (though the air may not be significantly so on such a small evaporator). If the air ended up warmer how could evaporative coolers work so well. Here in tucson the watervapor typically leaves the cooler ("bong") a good 10C below ambinet!

This change in energy is called enthalpy of vaporization. Basically you are moving the water from a more orderly state (liquid) into a more chaotic state (gas). This change requires heat, and in so doing transforms the heat into potiential energy in the coolant. Thus both the drop of water as well as the evaporating fluid lose some of their heat energy.

[Alchemy]

Well, everyone has bits and pieces of this whole deal, but here goes my quick sum-up on evaporative cooling. Hope it's helpful.

At room temperature, water will evaporate. It's a very slow process, but water vapor will leave the surface of a pool of water. The amount of water vapor coming off can be greatly improved by increasing the amount of surface area (spraying the water does this very well) and by blowing air over the surface to remove any water vapor lingering there.

The air isn't there to heat up the water to cause evaporation. It doesn't need to. The air merely aids the evaporation by giving water vapor a place to go.

Energy is conserved - in this case, enthalpy is conserved. Ambient air meets ambient water. Water evaporates. The water that evaporates has an increase in enthalpy due to its becoming a vapor. The liquid water left must decrease its enthalpy accordingly, and does so by going down in temperature. (Due to equilibrium, the water vapor will decrease in temperature as well, but you get the general idea.)

The enthalpy of vaporization is so high that it takes very little evaporation to cool a CPU.

As for the heatercore placement, the liquid in the bottom will give you the best heat transfer. Even that will be very poor, given the water at the bottom of the bong isn't moving very much. Koslov is on track with the liquid/liquid heat exchanger. If that's not an option, coiled copper tubing or the heatercore inside the bong would work about as well, I'd expect. Or hey, try a cheapo low-pressure $5 pump at the bottom just to swirl the water around the heatercore.

On another note I just finished my finals (except the ones that finished me ), one of which almost entirely on distillation and vaporization, so I'm intrigued by airspirit's idea in the other thread. Might mosey over there soon . . .

Alchemy

[redleader]

I think [my] previous explanation was a bit off. Its more accurate to think of a bong as a system in equalibrium. For every bit of heat that enters the coolant, the same amount leaves instantly.

In this case there is no change in heat at all. The added energy from the CPU simply pushes a really tiny bit of vapor out of the liquid. Because the system is in equalibrium, the input heat has no effect on temperature.

So in this case, the vapor has the same temp as the water because no change in temperature is possible without throwing off the equailbrium. In this case the vapor temp just depends on how warm the coolant is and thus how effective the bong.

This is a logical outcome. A small bong would not be able to transfer enough heat at a low temp to maintain equalibrium, so temps would rise until the vapor pressure of the coolant was sufficent to match the incoming heat. A larger bong would the evaporate coolant more effectively, thus requiring a less favorable vapor pressure (and thus cooler water/water vapor).

Edit: Added my to first sentance.

[airspirit]

If you like distillation, basically what that rig is is a massive super-powerful distiller of ... distilled water! It serves no purpose other than to cool a bucket of water held under a vacuum. I used to be a chemistry buff.

[MadDogMe]

HC's are three a penny, who cares if one 'silts' up every 3~6months?, if it gives superior cooling (as I think it would with the evaporation) then why haggle?.

I understand your reasoning Nicozeg!, not sure if it works out though, if there's no heat left to absorb, ie, the H2o is sub ambiant already then the tables turn and the water cools the air until enough heat is absorbed by the H2o to be the 'cooled' again, instead of the cooler , I'd like someone with abit more aurthority to answer it for all time...

This idea got shot down last time it was voiced did'nt it?, I seem to remember a thread asking the same question before a couple of months ago?, or simular anyway.
It seems like alot of trouble to me, why not just use the bong?, you'll get better temps, and you still have to maintane the bong whatever, if you're worried about scale on/in the block use distilled water, it's cheap enough if you buy it bulk is'nt it?, or filter it yourself with a 'carbon' filter/jug~thingie...

[Since87]

Originally posted by Alchemy

The enthalpy of vaporization is so high that it takes very little evaporation to cool a CPU.

To put some specific numbers on it...

Liquid water has a specific heat of 1 Calorie/(gram*Kelvin). The heat absorbed by 1 gram of water in transitioning from liquid to vapor is about 600 Calories.

Roughly speaking, removing 0.17% of a body of water through evaporation, drops the body of water 1C

[bigben2k]

RedLeader's post really cleared it up for me, thanks!

In short, if we had a body of water standing still, some water would naturally evaporates into the air. This should happen until the air is saturated (100% humidity???).

The bong is a device that allows this process to happen at a much higher rate of speed. It does so by increasing the surface area of the water, and by giving the air some speed/flow.

As Alchemy stated (and quantified by Since87), the energy from the CPU is dumped by the evaporation process.

Now, being in Houston, Texas, where the relative humidity is in excess of 95% through 3/4 of the year, I wouldn't expect much to happen, except having to run a dehumidifier in the house, to compensate!

Gotta move to Colorado, or Arizona!

My next question would be: what's preventing the water from getting below 2-3 degrees under ambient?


Sidenote: the "Little Giant" pump company was established sometimes in the 40's, and was known as the "Little Giant" evaporator company. Their product was a compact pump for evaporative coolers.

They still make that pump.

[airspirit]

Nothing is preventing the water from becoming colder besides relative humidity. The humidity saturation is variable depending on a) the temperature of the liquid, and b) the temperature of the ambient air.

If you ran a dehumidifier, you'd be able to drop further below ambient because it would allow more water to evaporate during the cascade from the shower head.

Keeping the atmosphere above the water in a vacuum will result in massive evaporatiion and massive water chilling to temps far below ambient.

This effect is why these work better in dry environments, and why by and large (w/o a dehumidifier) you'll get better results running it outside where you cannot appreciably change ambient humidity with your bong.

[Alchemy]

Originally posted by bigben2k
My next question would be: what's preventing the water from getting below 2-3 degrees under ambient?

Mass transfer!



You know the correlations mentioned before about convective heat transfer? Reynolds number, Nusselt number, etc?

Mass transfer problems like this carry the same sort of calculations with them.

The saturation point of the water vapor in the air is an issue, but I sincerely doubt you're ever going to get 100% humidity air coming out of the bong. The amount of water going from the bulk liquid to the bulk gas will be instead limited by the surface area of the droplets, the air temperature, the amount of air going through, the contact time, etc. And yes, the humidity of the air going in will have a large impact on the amount of vaporization, so bongs in Denver will probably work better than ones in Phoenix.

Alchemy