Ways to increase water evaporation?

[SaltyNuts]

If one wants to cool a body of water by increasing water evaporation, what all could one do to increase evaporation?

This site:

http://www.evaptainers.com/updates/2014/8/6/the-science-of-evaporative-cooling

Lists these:

• Lowering ambient humidity
• Decreasing atmospheric pressure
• Increasing ambient temperature (though this one is obviously counterproductive)
• Increasing surface area of evaporation
• Choosing different evaporative media
• Adding air movement/wind

I don't think the first one applies as I am talking about evaporating water outside, it would make no sense for me to spend electricity somehow trying to lower the himidity.

Second one similar - I cannot do.

Third one is counterproductive as set forth.

Fourth one I can do - have long, wide, thin bins with the water passing through them rather than taller bins with less surface area.

Fifth one - what they heck different "evaporative media" are they talking about? This would be in some outside aquariums, so I need to use H20. Anything I could add to it to make it evaporate faster, without hurting fish/plants?

Sixth one - what is the best way to position a fan vis-a-vis some water to encourage some evaporation? Directly into it, 90 degree angle? Like at a 45 degree angle, helping to push the water along and causing evaporation at same time? Not at all INTO the water, but instead OVER it? Some places seem to suggest this last one, but that seems weird to me.

Also, if I am causing a bunch of evaporation to lower water temps, obviously if I do nothing this is water gone. That cost can add up over time. Is there any way to recapture that water without signficant costs? Like if above the aquarium where the water is being evaporated I have a steel plate or something. I make this just a bit colder than the hot ouside air. Might the water evaporating (or otherwise in the air) collect on the underside and drip down back into the aquarium?

Thanks!

 

[RPD]

Are you unfamiliar with evaporate coolers, aka swamp coolers?

Edit also re-reading your request hurts my brain. You obviously don't understand how evaporation works in regards to the goal. What you are trying to accomplish with this method won't work.

 

[SaltyNuts]

Are you unfamiliar with evaporate coolers, aka swamp coolers?

Well, I know a little about evaporative cooling - i.e. when water evaporates, it takes some heat with it, this heat (in an aquarium) comes out of the water remaining, thus decreasing water temperature. I know what a swamp cooler is generally - shoots out a mist of water with air blowing.

I am just trying to maximize evaporative cooling in one or more outside aquariums. I don't know how a swamp cooler would play into that exactly...

 

[SaltyNuts]

Are you unfamiliar with evaporate coolers, aka swamp coolers?

Edit also re-reading your request hurts my brain. You obviously don't understand how evaporation works in regards to the goal. What you are trying to accomplish with this method won't work.

Huh? Pray tell why...

 

[RPD]

Huh? Pray tell why...

You are trying to get the benefits of evap cooling without any of the undesirables, i.e. cooling w/out water loss.
Just get a tank chiller like everyone else before you with the same problem.

 

[SaltyNuts]

You are trying to get the benefits of evap cooling without any of the undesirables, i.e. cooling w/out water loss.
Just get a tank chiller like everyone else before you with the same problem.

Yes, I realize getting the cooling effect, but somehow capturing the lost water, might be a pipe dream. But I am not an engineer so I thought I would ask. So if I have a metal plate above the tank where I am doing the evaporation that I have ever so slightly colder than the outside air, will it not cause water to condense on the bottom and drop back into the tank?

Chillers cost a ton, and, more importantly, are no fun. I want to make some badass thing myself!!!

Thanks!

 

[RPD]

Yes, I realize getting the cooling effect, but somehow capturing the lost water, might be a pipe dream. But I am not an engineer so I thought I would ask. So if I have a metal plate above the tank where I am doing the evaporation that I have ever so slightly colder than the outside air, will it not cause water to condense on the bottom and drop back into the tank?

Chillers cost a ton, and, more importantly, are no fun. I want to make some badass thing myself!!!

Thanks!

How is this metal plate going to be colder than the outside air? Why not just cut out the middle man and put said plate into the water directly?

 

[SaltyNuts]

How is this metal plate going to be colder than the outside air? Why not just cut out the middle man and put said plate into the water directly?

Ice? Or a very, very small thermo-electric cooler? I am asking the question if whether having said plate above the water might somehow, with a minimum of electricity (because you just have to get the plate a very little bit colder than the air), "catch" the evaporating water (or just water vapor in the air generally) and cause it to drip back down into the tank. I think the answer is very likely "no", I'm just brainstorming here!

Thanks RPD!

 

[RPD]

Why not just put the ice in the tanks to reduce the temperature? By coming up with some Rube Goldberg set up you are just getting more inefficiencies to your desired outcome.

Also the closet "solution" you seem to want would be a closed circuit fluid cooler. Water cooled PC's function as such, but they don't use evaporation to cool, but fans.

 

[SaltyNuts]

Why not just put the ice in the tanks to reduce the temperature? By coming up with some Rube Goldberg set up you are just getting more inefficiencies to your desired outcome.

Also the closet "solution" you seem to want would be a closed circuit fluid cooler. Water cooled PC's function as such, but they don't use evaporation to cool, but fans.

1. Again, I am not talking about using the ice to cool the water per se. I am JUST talking about using it to try and recapture some of the evaporated water, if that is possible, which I doubt it is. In other words - let's say, just made up, incomplete numbers here, by using $5 of electricity I can sufficiently cool 20 gallons of water. But by doing so, I lose $1 worth of water to evaporation. Let's say that for $.50 of chilling an ice cube I could capture that evaporation and return it to the mixture. Spending $.50 would save me $1, so I might well want to do it. Again, I doubt this is the case, but I'm a complete novice in this area, hence my questions.

2. How is a closed circuit fluid cooler cooled? A thermoelectric (peltier I think) unit? If so I think this would be very inefficient, I have another thread talking about different chilling methods and this method seems like the least efficient. I assume also it cannot get the benefit of any evaporative cooling? If its a refrigeration type cooler, I suspect that is indeed the best way to go efficiently, especially in a humid environment like that here.

Thanks!

 

[RPD]

1. Again, I am not talking about using the ice to cool the water per se. I am JUST talking about using it to try and recapture some of the evaporated water, if that is possible, which I doubt it is. In other words - let's say, just made up, incomplete numbers here, by using $5 of electricity I can sufficiently cool 20 gallons of water. But by doing so, I lose $1 worth of water to evaporation. Let's say that for $.50 of chilling an ice cube I could capture that evaporation and return it to the mixture. Spending $.50 would save me $1, so I might well want to do it. Again, I doubt this is the case, but I'm a complete novice in this area, hence my questions.

2. How is a closed circuit fluid cooler cooled? A thermoelectric (peltier I think) unit? If so I think this would be very inefficient, I have another thread talking about different chilling methods and this method seems like the least efficient. I assume also it cannot get the benefit of any evaporative cooling? If its a refrigeration type cooler, I suspect that is indeed the best way to go efficiently, especially in a humid environment like that here.

Thanks!

Ice isn’t free. Do your own research on water cooled PC’s. You are just trying to reinvent the wheel here. You won’t be able to effectively cool any decent amount of water without mechanical cooling whether it’s from a fan, compressor or you use ice in some Goldberg machine which just doubles back to compressor energy anyways.

 

[skull]

Whats the deal with all this large outside half buried fish tanks that you need to cool anyway? Can't be just for yourself can it?

I couldn't keep fish every time I see some poor little fish in an aquarium I feel sad for them, its like a bird meant to fly thats stuck in a tiny cage.

 

[Red Squirrel]

I would just have water go through a radiator with fans blowing over it. It's not evaporative cooling but might still have the same end result. If you need it colder than room temp you can put it outside. If you want it more controlled, put it inside a fridge or freezer. (freezer can be setup to be slightly above freezing so water won't freeze).

I plan to play around with this when I do my server room hvac. I'll put the rad in the garage as I want to see if I can heat the garage with server room heat, and generally, house heat. (house air will be added to server room if it gets too cold).

 

[Scarpozzi]

Whats the deal with all this large outside half buried fish tanks that you need to cool anyway? Can't be just for yourself can it?

I couldn't keep fish every time I see some poor little fish in an aquarium I feel sad for them, its like a bird meant to fly thats stuck in a tiny cage.

You may be right, but it's far easier to find Nemo when he's in only a quart of water...


Trying to cool water....remove heat from the tank.... I've looked into a few different technologies to heat liquids. I'd probably put a few ports on the tank or use hoses to pump water through a chiller circuit...then use a heat exchanger to cool the water outside of the tank. In my case, I was looking at evacuated tube solar panels to heat a circuit of liquid and transfer 400+ at a lower temp to another loop.

In your case, you'll need to use a thermostat-controlled pump to control the temperature (in case it does too good of a job). The downside is that while pumps can be cheap, good digital thermostats aren't cheap.... and that's only half of the solution. You'll either need coils and a fan or some other chilled circuit to remove the heat and I'm not sure how efficient those systems are. I'd do a cost analysis on what a decent fan and coil array would cost and whether or not actual coolant should be used for the volume of water you're working with. in my proposed heat exchanger configuration, I wasn't working with water, so temperature ranges were skewed for freeze and boil range.

 

[Ken g6]

Fifth one - what they heck different "evaporative media" are they talking about?

Once upon a time, I bought a portable "evaporative cooler" for my house. When I looked inside it I saw a tank for water, with a towel sewn together at the ends and looped over two rollers, one in the tank and one at the top of the machine. The machine rolled the towel through the water tank and then blew air over it with a fan. I think the towel is the kind of "media" they mean.

I sent the thing back to the store immediately, unused. The towel had too much potential for undesirable bacterial growth. And if you're thinking about using that system it would be even worse with a fish tank, and probably less effective cooling the water than the air.

 

[Paperdoc]

Your idea of a cold metal plate to re-condense the water and recover it cannot work. The reason such a plate can cause water vapour to condense on it is that the water must release its Heat of Vaporization to the metal for the change of state. That means the metal must absorb that heat, thus getting warmer. Immediately that means the metal is now WARMER than the water vapour reaching it, and hence this process will stop completely. The only way to make this work is to have some other process that removes heat from the metal plate at least as fast as the rate of heat removal by the plate from the water vapour. In other words, you need a refrigeration system. Or, you MIGHT be able to remove some heat slowly by blowing a big fan onto the BACK side (upper dry side) of the plate. You'd still be using a fan and electricity, but compared to just blowing the fan over the water tank you MIGHT reduce the amount of water "lost". HOWEVER, the amount of electricity you would consume to do this for any significant temperature reduction would probably be a LOT more than actually using a small refrigeration system to chill a loop of tank water flowing through a cooler.

A closed-loop liquid cooling system such as those used to cool a computer CPU cannot work, either. That type of system depends on the fact that the source of heat (the CPU) is at a VERY much higher tmperature that the source of cooling (room air), so heat flows easily from hot to cold items. In your case, OP, the problem heat source is a large open-top tank of water at current air temperature, and the source of cooling in the radiator section is the SAME air at the same temperature. NO net heat flow!

The real answer for your main question is that the maximum rate of heat removal from the tank by evaporation at the water surface is achieved by increasing the rate of air flow over the surface. The more rapidly you can remove the water vapour already evaporated over the water surface, the more rapidly more water will evaporate to replace it. So you need LOTS of air movement across the surface and away from the tank.

Yes, the more successful you are at this, the more water will be required to replace it. And that certainly is a problem in keeping fish tanks. Any water source you would be likely to use contains dissolved trace amounts of materials like inorganic salts (electrolytes). When the water evaporates, almost all of those remain in the liquid water in the tank, and only pure water vapour with NO dissolved additives escapes into the air. In reality, you are operating a very large still that removes pure water and leaves behind a solution that is becoming incresingly enriched with the materials originally dissolved in the source water. That is definitely harmful to the fish as it changes their water quality. THAT is why this type of cooling system is NOT going to be a good solution. You do need a way to change the TEMPERATURE of the water without changing the composition of it. That is why your best method is a way to use refrigeration to cool the water without removing the water.

 

[SaltyNuts]

Ice isn’t free. Do your own research on water cooled PC’s. You are just trying to reinvent the wheel here. You won’t be able to effectively cool any decent amount of water without mechanical cooling whether it’s from a fan, compressor or you use ice in some Goldberg machine which just doubles back to compressor energy anyways.

Thank you RPD!

 

[SaltyNuts]

Whats the deal with all this large outside half buried fish tanks that you need to cool anyway? Can't be just for yourself can it?

I couldn't keep fish every time I see some poor little fish in an aquarium I feel sad for them, its like a bird meant to fly thats stuck in a tiny cage.

I want to grow a ton of fish and plants and other things. Mainly for shits and giggle, but if it works I suppose I might sell them. Small fish are just fine in not-too-small aquariums, they love it there!

 

[SaltyNuts]

You may be right, but it's far easier to find Nemo when he's in only a quart of water...


Trying to cool water....remove heat from the tank.... I've looked into a few different technologies to heat liquids. I'd probably put a few ports on the tank or use hoses to pump water through a chiller circuit...then use a heat exchanger to cool the water outside of the tank. In my case, I was looking at evacuated tube solar panels to heat a circuit of liquid and transfer 400+ at a lower temp to another loop.

In your case, you'll need to use a thermostat-controlled pump to control the temperature (in case it does too good of a job). The downside is that while pumps can be cheap, good digital thermostats aren't cheap.... and that's only half of the solution. You'll either need coils and a fan or some other chilled circuit to remove the heat and I'm not sure how efficient those systems are. I'd do a cost analysis on what a decent fan and coil array would cost and whether or not actual coolant should be used for the volume of water you're working with. in my proposed heat exchanger configuration, I wasn't working with water, so temperature ranges were skewed for freeze and boil range.

Haha Scarpozzi, you are faaaar above my head!

How would the "chiller circuit" you mention be designed?

Same for the "heat exchanger".

Thanks!

 

[SaltyNuts]

Once upon a time, I bought a portable "evaporative cooler" for my house. When I looked inside it I saw a tank for water, with a towel sewn together at the ends and looped over two rollers, one in the tank and one at the top of the machine. The machine rolled the towel through the water tank and then blew air over it with a fan. I think the towel is the kind of "media" they mean.

I sent the thing back to the store immediately, unused. The towel had too much potential for undesirable bacterial growth. And if you're thinking about using that system it would be even worse with a fish tank, and probably less effective cooling the water than the air.

Interesting, thanks Ken!

 

[SaltyNuts]

Your idea of a cold metal plate to re-condense the water and recover it cannot work. The reason such a plate can cause water vapour to condense on it is that the water must release its Heat of Vaporization to the metal for the change of state. That means the metal must absorb that heat, thus getting warmer. Immediately that means the metal is now WARMER than the water vapour reaching it, and hence this process will stop completely. The only way to make this work is to have some other process that removes heat from the metal plate at least as fast as the rate of heat removal by the plate from the water vapour. In other words, you need a refrigeration system. Or, you MIGHT be able to remove some heat slowly by blowing a big fan onto the BACK side (upper dry side) of the plate. You'd still be using a fan and electricity, but compared to just blowing the fan over the water tank you MIGHT reduce the amount of water "lost". HOWEVER, the amount of electricity you would consume to do this for any significant temperature reduction would probably be a LOT more than actually using a small refrigeration system to chill a loop of tank water flowing through a cooler.

A closed-loop liquid cooling system such as those used to cool a computer CPU cannot work, either. That type of system depends on the fact that the source of heat (the CPU) is at a VERY much higher tmperature that the source of cooling (room air), so heat flows easily from hot to cold items. In your case, OP, the problem heat source is a large open-top tank of water at current air temperature, and the source of cooling in the radiator section is the SAME air at the same temperature. NO net heat flow!

The real answer for your main question is that the maximum rate of heat removal from the tank by evaporation at the water surface is achieved by increasing the rate of air flow over the surface. The more rapidly you can remove the water vapour already evaporated over the water surface, the more rapidly more water will evaporate to replace it. So you need LOTS of air movement across the surface and away from the tank.

Yes, the more successful you are at this, the more water will be required to replace it. And that certainly is a problem in keeping fish tanks. Any water source you would be likely to use contains dissolved trace amounts of materials like inorganic salts (electrolytes). When the water evaporates, almost all of those remain in the liquid water in the tank, and only pure water vapour with NO dissolved additives escapes into the air. In reality, you are operating a very large still that removes pure water and leaves behind a solution that is becoming incresingly enriched with the materials originally dissolved in the source water. That is definitely harmful to the fish as it changes their water quality. THAT is why this type of cooling system is NOT going to be a good solution. You do need a way to change the TEMPERATURE of the water without changing the composition of it. That is why your best method is a way to use refrigeration to cool the water without removing the water.

Thank you so much Paperdoc, very helpful! A few questions:

"The real answer for your main question is that the maximum rate of heat removal from the tank by evaporation at the water surface is achieved by increasing the rate of air flow over the surface. The more rapidly you can remove the water vapour already evaporated over the water surface, the more rapidly more water will evaporate to replace it. So you need LOTS of air movement across the surface and away from the tank."

So then, the fan or fans should be set as low as possible to the water in the tank (without causing water to splash out) and blow completely horizontally to maximize evaporation? Not further above the water, and not pointing any degrees down into the water?

"Yes, the more successful you are at this, the more water will be required to replace it. And that certainly is a problem in keeping fish tanks. Any water source you would be likely to use contains dissolved trace amounts of materials like inorganic salts (electrolytes). When the water evaporates, almost all of those remain in the liquid water in the tank, and only pure water vapour with NO dissolved additives escapes into the air. In reality, you are operating a very large still that removes pure water and leaves behind a solution that is becoming incresingly enriched with the materials originally dissolved in the source water. That is definitely harmful to the fish as it changes their water quality. THAT is why this type of cooling system is NOT going to be a good solution. You do need a way to change the TEMPERATURE of the water without changing the composition of it. That is why your best method is a way to use refrigeration to cool the water without removing the water."

So, what I am imagining is many clear plastic bins connected with tubing. Some of those bins would hold fish, but many would be plants. If I had sufficient numbers of plants to consume the excess nutrients left by the water evaporating wouldn't that fix the problem? Of course, it would require a good bit of trial and error to truly get it fixed.

The more I think about it buying a very energy efficient freezer and running the water from the tanks through there is probably going to be the cheapest by far. I think the analysis to compare evaporative approach versus refrigaration (freezer) approach would be to compare in the former the cost/per one degree drop, plus the cost in lost water you have to replace, versus in the latter just cost per one degree drop.

A further question on the evaporative approach - let's say I have a full enclosed very tall box. Water comes in at one side at the top, and trickles down to the bottom. At the bottom, the pooled water moves out of the box via some piping. Down the side of the box are some fans that blow air on the water from outside. Let's assume it is enclosed "perfectly" so there is no evaporation lost, some how, some way, it all stays in the system. Does this mean that there will be no temperature drop, no matter how fast the fans blow air?

Thanks!!

 

[sdifox]

... op please don't think, you are hurting my brain.

 

[Sgt. York]

OP reminds me of the tiny house guy.

 

[SaltyNuts]

... op please don't think, you are hurting my brain.

To be sure, some peoples' capacity for critical thinking and analysis is severely restricted...

 

[SaltyNuts]

OP reminds me of the tiny house guy.

??? Oh, and right, there is only one tiny house guy, lol... it was like a nutty national craze, dude...