Thermodynamics and evaporation rates.

[reef_research]

If given the following factors:

Voulme of total system
Surface area of the tank (Water)
Temperature of the tank water and the atmosphere above
Humidity of the atmosphere
salinity
intensity of lighting (not sure how to measure for usefulness here)

Is it possible to determine the amount of fresh water that evaporates from the system in a given amount of time?


also, given these same factors, would it be possible to predict the rate of cooling/heating between the tank and the outside environment?

Thank you.

[musty baby]

I suppose it would be, but wouldn't it be just as easy to simply measure the rate of evaporation by measuring how much top-off is needed? Also, variation in the temp of the tank, temp and humidity of the surrounding air, salinity, and any variation in lighting (change of bulbs, changing times for timers, etc) make the calculations too tedious for real use?

What are you trying to calculate this for?

[reef_research]

Thank you for your prompt response.

Well, if this were a normal setup, I would, but the kind of thing I'm planning on reqiures careful balances between marine and feshwater conditions and i cannot afford to have the water level change unpredictably.

What about the heat dynamic part of the question?
(Not just for marine either; IE if tank A has a temp of __, and the air has a temp of__ how long will it take to...)


Thank you.

[reef_research]

Originally posted by reef_research
If given the following factors:

Voulme of total system
Surface area of the tank (Water)
Temperature of the tank water and the atmosphere above
Humidity of the atmosphere
salinity
intensity of lighting (not sure how to measure for usefulness here)

Is it possible to determine the amount of fresh water that evaporates from the system in a given amount of time?

Thank you.

More factors that may be needed:

How many lights
distance from the surface
photoperiod


Using basic physics, can anyone tell me a formula for figuring out how much wil evaporate?

Also, if:

A body of water (tank) is in an atmosphere (room) and each have their own temps (w/ out heaters or cooling devices in effect), How long will it take for a hot room to heat the tank, and vise versa?

Thank you,

[TimTen]

You need to remove variable's. To even think of calculating evaporation you need to have a constant tank temp and ambient temp. I only took one intro to thermo class, several years ago, but what your asking for is crazy way to many X's and Y's.

[reef_research]

So it would not be possibel at all?

what about the heat transferal question?

Thank you.

[marsh]

http://en.wikipedia.org/wiki/Cooling_tower

The above link may be of some help to you.

[Galilean]

Assuming no fans on your lights and an open top you can reasonably model it as evaporation from a flat plane with natural convection. You just need to know the temperature of the water and the surface area. All the other factors won’t be significant (though a fan blowing over the water would be important). If you have a hood, then you need to determine the humidity of the air under the hood. If you have a fan you will need to know the air velocity and mass flow rate as well. A standard heat-transfer engineering text will have the equations you would need. Different equations would apply depending on the exact geometry of your situation. If you are not an engineer, there is probably no point in trying to do this.
No equations exist that will model the transient case of changing temperature. You would need to write a computational fluid dynamics program for that.

But that’s all irrelevant because you can just measure the evaporation rate on the side of the tank!

[Galilean]

Rereading the thread I think my response was inappropriate.

What exactly are you trying to accomplish here? Whatever your system design you can just allow enough margin in the feedback control mechanism for variable evaporation rates and adjust it as needed. Only a rough estimate of the evaporation rate is required because you want lots of margin. You do not want a precise determination which would be unstable. Depending on how sophisticated you want this to be it may be worthwhile to investigate the Chesapeake Bay Mesocosm. They varied from fresh to full salt in something like 18 gradations if I remember correctly. Levels were controlled with gates activated by electronic sensors through a computerized feedback control algorithm. Controlling the salinity will be much harder than any evaporation problems you might face whether it’s a once through or a recirculation type system.

[Randy Holmes-Farley]

Personally, I don't think that you can usefully do this without a lot of engineering tables and such with data from very similar systems. The amount evaporating will depend importantly on the flow of air over the tank, how fast the humidity actually rises in the vicinity of the tank etc.

[reef_research]

Originally posted by Galilean
Rereading the thread I think my response was inappropriate.

What exactly are you trying to accomplish here? Whatever your system design you can just allow enough margin in the feedback control mechanism for variable evaporation rates and adjust it as needed. Only a rough estimate of the evaporation rate is required because you want lots of margin. You do not want a precise determination which would be unstable. Depending on how sophisticated you want this to be it may be worthwhile to investigate the Chesapeake Bay Mesocosm. They varied from fresh to full salt in something like 18 gradations if I remember correctly. Levels were controlled with gates activated by electronic sensors through a computerized feedback control algorithm. Controlling the salinity will be much harder than any evaporation problems you might face whether it’s a once through or a recirculation type system.

Chesapeake Bay Mesocosm? Could you provide a link?

Thank you all for your tremendous responses, they were very helpful.

[AZDesertRat]

Don't forget surface agitation, this adds to evaporation significantly.

[mrbill70]

What Galilean said. I think I would model it as a radiation/convection problem to include the lights. You are supplying heat to the tank through the electromagnetic radiation given off by the lights. This will cause the temperature of the water to rise (assuming starting at equilibrium conditions) which will bring about natural convection. If you were to leave the lights on and keep the temp of the room constant, it would eventually reach a new steady state where the heat input to the lights would be balanced by the heat lost to the air. It would be pretty easy to figure out if you knew the required parameters. For the radiation part you would need to know the temp of the bulbs, surface area of the bulbs, and the emissivity. Not all the radiation given off will be absorbed by the tank. Some will go to the tank, some the air, some the light fixture, some the walls in the room, etc. For an ideal, simplifed situation the emissivity is 1 which eliminates it as a factor, but this certainly isnt ideal. Once you know the heat flux applied to your tank, you need to figure out how much heat is lost to the surroundings. To figure this part out you will need to figure out the heat transfer coefficent. For that you will need things like specific heats, thermal conductivities, and viscosity. There are some others but they can pretty much be found from some tables. Also you will need more temperature measurements. However, this will only apply to the water/air interface. The other sides of the tank will have a thermal resistance provided by the glass.

After you have the Q (heat flux) entering and the heat transfer coefficents for the heat leaving through various means you can set them equal to each other. Once the temp of the water goes up enough, the heat leaving the system will match the heat entering. This is kind of like filling a sump. Once it fills to the height of the baffle, the rate at which water leaves is equal to the rate at which the water enters. Anyway, you equate all the Qs and can solve for the temperature of the water. This is also assuming the radiation heat transfer from the tank to the walls is negligable (do you live where it is really cold? are your walls well insulated?).

Also, keep in mind that heat transfer problems are largely solved by experimental data. For a good portion of situations (geometries, flow conditions, etc.) there isnt a theoretical equation that can be solved to get the answer. It's done using correlations based on experimental data for given conditions. There are bunches and bunches of these correlations, some more accurate than others, for each different flow regime. Also, in heat transfer calculations, "good" accuracy is considered within 15-20%. All that hassle for a somewhat close estimate. Like Galilean said, any heat transfer textbook will have all that you need and much more. It really isnt that difficult. There isnt any complicated math involved, and many of the small factors (perhaps radiation heat loss to the walls of the house) can be left out.

All this mumbo jumbo relates to finding out what the water temp of the tank will be, as far as evaporation rate, it would be much more difficult.

Oh gosh, stability, automatic control systems, matlab, laplace, state space...memories.

[topQuark]

It is totally possible to calculate the the answer to your question...

But to how many decimal places do you want the answer? Are you interested in knowing a rough estimate (e.g., gallons per day) or do you want to know down to the level of mililiters? The more variables that you include, the more accurate your answer will be.

Can you program? I would recommend writing a computer program to carry out the calculations... start very simple and then add on to it in steps making it progressively more complex... For example, start with a simple rectangular tank of volume V and temp T_{tank} sitting in an infinite "room" with temp T_{room} and humidity H_{room}. Then start adding other variables...

Based on your own intuition try adding the variables that you feel are the most important first...

[Galilean]

reefresearch,

I haven’t been able to find an online description. The estuary setup I mentioned is described in the book Dynamic Aquaria. As AZDesertRat points out, surface agitation will also be a strong influence on evaporation. Calculating the temperature will be difficult if you are not trained in thermodynamics. Even if you do precisely determine the system behavior for a specific set of parameters you will have to actively control the temperature anyway. There is no need to predict it precisely. Just make the room temperature 5-10 degrees cooler than you want the water and add heaters if the lights don’t make up the difference. As mrbill70 mentions, I would write any simulation in MatLab/SIMULINK or maybe Visual BASIC. These are very easy to use say, compared to FORTRAN or C+. Keep in mind that calculations of fluid and thermodynamic behavior are never more than fancy methods of guessing. Don’t rely on them for more than two significant figures. Instead, design adjustable systems.

[reef_research]

Originally posted by Galilean
Instead, design adjustable systems.

This appears to be the better option as of now. I qish that I could say I am a fluent programmer or an engineer, but I think that, for the time being, it would be best to design an adjustable system.

Thank you all very much for your insightful and helpful responses.

Reef_research.

[crazyguy34834]

I agree with Randy Holmes-Farley, I am taking fluid systems and thermodynamics in college and while it is possible to calculate, you need a lot of information and it has to be accurate. If you have one variable in your equations that is off by a small measurment, that error will multiply through your calculations and can throw your answer way off. Even if you get a great theoretical answer it will not be the same as the actual result depending on how accurate you want to be.

The method of adding fresh water without constantly measuring the salinity is just a temporary method. The longest time I have heard of this being used was for one week and there is no way I would do that if you have anything easy to kill in the tank.

If I were you I would set up an aoutomatic way to measure the salinity and combine it with a controler that adds fresh water when needed. Add a water level controler with a saltwater source and you should have perfect salinity as I think this is what you are trying to do. you could even hook up pumps and timers and make the tank do weekly automatic water changes while you are at it.

Calculating the rate of temperature change of your tank is the easy part but it is only valid while the variables in your calculations are constant during the time the temperature is changing.

additional things to consider when calculating change of temperature rate:

-thermal energy conduction coefficient of the material your tank is made of

-thickness of the material

-density of the water in your tank

-haw much radiation is reaching the water per unit area and time(from the sun and some kinds of lights, research solar heating)

-surface area exposed to each radiation intensity and % water volume exposed (if your tank is in the dark then skip all the radiation stuff)

-flow (mixing) of the water (if the water is not moving you will get different temperatures in different parts of your tank, very true in shaded areas)

-if the water is moving there is a energy loss to the pipes and whatever the water comes in contact with (look up how to calculate temperature "head loss" of fluids through pipes)

-is it worth doing these calculations, do you really need to know what you are trying to calculate or are you just looking for a headache? (I recommend lots of drinking and the headache will follow, or a little drinking to help get through the calculations)


I assume you want to find out how long your tank can go without power. Fist consider the oxygen level in your water (no oxygen = eveything dead), just add a sensor to your tanks power source that turns on a battery and converter hooked up to whatever your tank constantly needs and turns off when the power comes back on. don't forget to hook up an airstone to that battery

If you are still wanting to calculate this then I recommend buying the books:
- Fundamentals of Fluid Mechanics, fifth edition by Munson ,Young,Okiishi
- Fundamentals of Engineering Thermodynamics 5th edition by Moran,Shapiro

for the solar/radiation heating you will have to do some research (if your tank in in the dark then don't bother with this, the energy will be zero)

good luck with your calculations and if you still want to do this just for the fun of it then commit yourself to a mental hospital

[dots]

Originally posted by reef_research


Is it possible to determine the amount of fresh water that evaporates from the system in a given amount of time?


also, given these same factors, would it be possible to predict the rate of cooling/heating between the tank and the outside environment?

Though this question has really been answered I would like to add.........

Can you determine the theorhetical rate........sure, but it would take a lot more than one "basic physics" equation, rather than a series of a dozen or so that would take you a while to figure out. The more accurate and easier way would note temperature, humidity and daily evap rates and graph them in Excel. With some Excel magic, one could find the correlation and predict you daily evap rate based on those main variables......it would be close.....close enough. But even then, its too much work.

Your second question is a little more realistic and more straight forward. As mentioned review a Thermodynamics text and treat the air and tank and the heat sinks and source........the rate of Q could be found for fun.

But this is the exact reason we have adjustable heaters/chillers and float valves.

[herostar]

I say just get an automated top-off system with a large resevior and don't worry about it again... (I know this doesn't actually answer your question...)

[BeanAnimal]

Originally posted by topQuark
It is totally possible to calculate the the answer to your question...

But to how many decimal places do you want the answer? Are you interested in knowing a rough estimate (e.g., gallons per day) or do you want to know down to the level of mililiters? The more variables that you include, the more accurate your answer will be.

Can you program? I would recommend writing a computer program to carry out the calculations... start very simple and then add on to it in steps making it progressively more complex... For example, start with a simple rectangular tank of volume V and temp T_{tank} sitting in an infinite "room" with temp T_{room} and humidity H_{room}. Then start adding other variables...

Based on your own intuition try adding the variables that you feel are the most important first...

I have to disagree here in a sense. Yes it is certainly possible to calculate... but there are simply too many variables to monitored that affect the calculations. A slight change in air movement over a body this small can drastically change the evaporation rate. The air water interface tends to become saturated and dense very easily, even if the room air itself is rather dry. Slightly increasing the airflow flow can almost double the evaporation rate. The heat input and export from the system changes as room conditions change and this will certainly be something to consider.

I think the point here is that any math you THINK you are doing is going to be no better than any guess you are going to make. It is a waste of time. Without a LOT more data and realtime input of the changing conditions the calculations are at best a shot in the dark and the same as an educated guess and both are less accurate than simply monitoring the tank under normal operating conditions and coming up with a mean evaporation for each environmental mode.

That "simple" computer program would have to be VERY complex and built with a feedback loop (your observations) to even come close to what your brain can deduct in minutes. Silly.

[fload]

Float switch? Kidding but let us know what your working on that needs to be so precise?

[hahnmeister]

Bean, Randy, and the bunch are right... the precision needed to be accurate just isnt there... too many variables. Convection is complicated enough, but the radiation, evaopration, etc... its just insane to try to figure it out... not to mention, things change over time (even in a single day) that will make any calculation worthless.

[Qcks]

First, i have to agree with pretty much every thing else that has been stated thus far: It's simply not practical.

However, your question reminded me of an idea i had a little while ago.

At the time, I had just read some reef chemistry article (which if i remember correctly was written by Randy). Specifically the article
was about ambient oxygen level in the tank and how different factors affect it. What made the article interesting (well.... it was interesting on it's own but, it made it pertinent to this thread..) is that within it a study was quoted. In that study a nebulizer was used to boost the oxygen content of water (although i don't believe nebulizer was the exact word used to describe it.).

Now, i have access to nebulizers (I have CF and asthma). I thought it wouldn't be all that difficult to modify one so as to increase the oxygen level of my tank. Luckily i thought out the pros and cons of the situation: the fog that covered the top of the tank would prevent adquate light penitration and, given gaseous state of the mixture as well as the variable amount of evaporation, it would almost certainly lead to a change in salinity.

As a result i decided that it probably wouldn't be fruitful, but if you are interested in slowly changing your salinity back and forth over time, it'd probably be worth looking into.

ah... real quick, for those of you who have never heard of what i'm talking about:

Nebulizer

The mechanism to alter salinity with a nebulizer would be fairly easy. The nebulizer would have to be able to alternate between running off of the tank water, and running off of a RO/DI (or distilled reserve) input. Running off of the tank water would slowly (painfully so) increase the salinity of the water. Running the nebulizer off of the freshwater reserve would decrease salinity (once again... very slowly).

Random and crazy thought, i know.... i still think i might try some thing like this to "stabilize" the tank at night, since it would help combat some of the pH increase associated with night time respiration. Course it's a bit of a beast to setup and since the night time pH shift isn't really a problem for me, i've avoided.

Anyway...

[mfp1016]

actually, this would have to be solved with the use of both heat transfer and mass transfer, both very difficult subjects

as stated by other posts, it would not be worth your time to do so, you could factor in so many different things that unless you had the equipment to measure the parameters required to solve the problem it wouldn't even work because without the use of equipment that can detect the different variables to a very high degree its not statistically accurate; in other words, you would need all of your measurement equipment to be accurate with at least 3 significant figures so that by the time you use the measured variables you have at least 1 significant figure left. The best you could get it down to is one digit, something like 4 mL/day. thats probably about the most accurate you could achieve

i'm a chemical engineer so i do these types of calculations for a living