Humidity from large Fish tanks ,heat savings- 600 gallon tank example

  • We are currently upgrading MFK. thanks! -neo
Drstrangelove,

I just came up with the idea in effort to retain water. I've been messaging and posting back and forth about filtration and imagine that if I had enough of it, I would want to hold on to the water longer if all I had to do is trap it. It wouldn't be dirty so it wouldn't need changing. I would still do water changes because I wouldn't expect to have absolutely clean water forever. I just wouldn't need to do as many water changes as I currently do or ideally top off as much as I do due to evaporation.

Air exchange: That's why I posted before trying it. I was curious to find out if someone had worked out how to bubble a tank like this or even how they operate an upside down tank maybe? I figure an upside down tank would have similar problems unless there were also an open area. I got believe enough bubblers or even introducing air before water re-enters the aquarium would provide oxygen. The excess air would go out through the overflow or at least where the openning is. It wouldn't be air tight throughout the aquarium and I don't want to permanently fix the glass anyways. I'd be removing the pieces to clean the tank so there would be space for air to pass around the edge of each piece when they are in place. Just not fish.

skjl47,

I hadn't thought about the inverse square rule. Baed on what you described, the larger tank should hold temperature even now as it has more volume and less surface area. That's why the heater is being replaced. Doesn't seem to be holding up anymore. On that same note, I think I saw someone post somewhere about insulating an aquarium. I'm not completely sold but maybe between the bottom and the back side, it would help. The inside of the cabinet and hood would seem to be opportunity as well. I haven't given it time to think how to place it or what materials would be best for insulating something humid or wet like that.

pk705,

I don't think I have enough knowledge of chemistry or my own water to sit down to that. I had trouble following the example I googled. One velocity described was 10 to the negative 11th power meters squared a second. I don't think that's too terribly fast. I wouldn guess that the math would work out to suggest I would need a pretty good sized area or maybe just a bubbler deep enough to let air diffuse over the correct amount of time before escaping.

Likewise, if air is trapped above the water, wouldn't that air also need to be exchanged at the same interval? It may contain more oxygen but the level of oxygen in the water still has to be maintained. If it doesn't exchange fast enough after depleting the amount trapped above the water, the level in the water will drop until the fish die. Right? So, if there's no air trapped above the water but air exchanges through the water more than trapped air above the water, it would just be a matter of finding the amount of air to put into the aquarium and the amount of time it needs to be present in the water. Right? Maybe someone has figured this out on an upside down aquarium or a near air-tight aquarium?
 
The oxygen (as well as all other gasses) in the air will always be in an equilibrium with the oxygen dissolved in the water. That is, a certain (very small (around 8mg/L concentration of oxygen in freswhater at 25C, if well oxygenated)) percent of oxygen is going to diffuse into water. So, if the above air decreases in oxygen concentration by 1 percentage point (from 20% of all air to 19% of all air for ease of calculation), the decrease in dissolved oxygen that will follow (with good oxygenation/surface movement/etc.) will be around 5% (19/20 = 0.95), which shouldn't be a major problem.

Now, assuming you have the tiniest of holes in your lid, this will result in a concentration gradient in oxygen concentration between the air in your tank and the air above your tank. We know that 1 mole (6.02x10^23 molecules) of an ideal gas (which air at room temperature is close enough to) has a volume of 22.4 liters. We also know that the air inside our tank is 19% oxygen by volume; meaning oxygen has a concentration of (0.19x1mol)/22.4L=0.00848mol/L. And we know that the air outside our tank has a concentration of (0.20x1mol)/22.4L=0.00892 mol/L. A difference of 4.464x10^-4 mol/L. This is the concentration gradient that is propelling oxygen into our tank.

Now, we can do the same for water. Going by http://www.tis-gdv.de/tis_e/misc/klima.htm, we can assume that our tank, which in this case is kept at 25C has air the temperature of 25C above it. And at 25C, we see that air holds 23 grams of water for every cubic meter (or 1000 liters) at 100% relative humidity (we assume it's very humid under our lids with all that evaporation). We assume our room has a temperature of for example 20C and a relative humidity of for example 40%. The table says the room holds 5.2 grams of water for every 1000 liters of air. We need to figure out the concentration gradient. We can calculate this from the weight concentration. Water has a molar mass (the mass of 1 mole of molecules) of 18g/mol. In other words, our in tank air has a water concentration of (23g/18g/mol)x0.001L=0.001mol/L. Our outside concentration of water in the air is (5.2g/18g/mol)x0.001L=2.888x10^-4. The difference is 9.888x10^-4mol/L.

Now, going by Fick's 1st law, we can calculate how much actual water and oxygen can pass through a PVC tube of for example a diameter of 25mm (roughly 1"). 1st we need to determine the area. This is easily done with the formula for the area of a circle: A=pi x r^2. In our case, A=pi x 1.25cm^2 = 4.909cm^2. And for the purposes of this example, we will say that the tube is 20cm long. This will result in very little net movement of either substance; however, the length and diameter can be changed as needed. What we want to know is how much of each substance is going to diffuse per a unit of area of our exchanger. This is called diffusion flux. J(flux)=D(constant)x(difference in concentration)/(length of barrier). For oxygen diffusing into air, D equals 0.176 cm^2/s. So, J=0.176 cm^2/s x((4.464x10^-7mol/cm^3)/(20cm)=3.92832x10^-9mol/cm^2 x s. D for water vapor into air is around 2.5x10^-5m^2/s=0.250cm^2/s. so, J=0.250cm^2/s x ((9.888x10^-7mol/cm^3)/20cm)=1.236x10^-8mol/cm^2 x s.

So, if we multiply this by 4.9cm^2, we get a diffusion rate of 1.9x10^-8 mol/s of oxygen into the tank and 6.1x10^-8 mol/s of water out of the tank. Or in useful terms, 0.011 moles of oxygen into the tank and 0.037 moles of water out of the tank every week. Which is obviously not going to be enough to sustain our fish, but this is an example case. What does this amount of oxygen or water effectively mean?

0.037 moles of water equal 0.66 grams of water, which is not a lot, but the example conditions of this mental exercise are a bit stupid.
1 mole of glucose yields −2805 kJ of energy if burned. A body can realistically harness around 40% of that. And 1 mole of glucose needs 6 moles of oxygen to get oxidized completely. This covers around 80% of all the oxygen take up by an organism. So by this logic, 7.2 moles of oxygen would result in a body being able to gain around 1122kJ or 267 kcal. We have 0.011 moles of oxygen, enough for 3 kcal every week.

Assuming our water needs to be well oxygenated and that our fish exert around 7000kcal every week (half a human and they're cold blooded (so a very very very very well stocked tank), an optimal passive diffusion system would let out 1540 grams of water - 1.5 liters or roughly 1/2 a gallon. Since a lot of our energy goes into heat, this would realistically mean a mass of around 100lbs in fish (don't quote me on that, now we're in speculation territory).

My chemistry is a bit rusty, so if anybody spots any mistakes, they are more than welcome to correct them. I think the general logic behind it; however, is sound.

Some thoughts. :).

Edit: the shorter the area where diffusion is the only motor force, the lower the area needed for it to occur. In realistic terms, usually a number of small holes are enough. Posted above is just an example of how you can go about calculating the rate if you are very paranoid. Also note that unless the sole point of air exchange are twisted pvc pipes that more or less put an end to all temperature gradients, temperature is also going to help move your air. Taking that into account; however, features a number of a lot more complex equations and is difficult to demonstrate on a made up case. In practice, if you are paranoid about suffocating your fish, lids that are quite a bit larger than the holes and a small spacing between the tank and the lid is likely going to give you enough oxygen on diffusion alone.
 
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