Is there a calculation for.... - The Planted Tank Forum
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post #1 of 24 (permalink) Old 08-27-2014, 02:07 PM Thread Starter
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Is there a calculation for....

Water displacement when changing water by using an overflow system rather than vacuuming old water out first?

Suppose I add 15 gallons of water at 74° to a 40 gallon tank at 76°; is there a formula here to determine how much water I've actually changed? If it matters, let's say that it's at 10gph.
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post #2 of 24 (permalink) Old 08-27-2014, 03:12 PM
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A. you would need that GPH number to do any calculations like that.
B. pump manufacturers state the GPH as given them by engineering calculations on
each pump so those numbers are inaccurate once a hose is attatched to the pump discharge especially if the hose is virtical, say from the floor to the tank.
C. a siphon would not work because it would just run till the end got clear of the ank water so you would need a built in overflow in the tank like sumps have in order for this to work.
But @ the 10GPH rate you stated in one hr you would change 25% of a 40g tank.

The shortest distance between any two points is a straight line...in the opposite direction...
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post #3 of 24 (permalink) Old 08-27-2014, 03:45 PM Thread Starter
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But won't the temperatures of the water going in and out impact how much is displaced? I imagine that not as much water would be changed if I fed the tanks with warmer water. Also, some new water is bound to make its way to the overflow. How do I account for that?

My shrimp rack aquariums each have 4gph drip emitters over them that are fed by a pump from a reservoir. Water exits through overflows on each tank. It's a slow water change.

My other rack also has tank overflows, and valves (instead of drip emitters) above each tank. I'm thinking of adding a flow meter to get a better idea of how much is going in.

So, this is what I have to work with:

Temperature (both in and out)
Volume (both in and out)
GPH
Time
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post #4 of 24 (permalink) Old 08-27-2014, 04:31 PM
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Have you taken into account evaporation?
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post #5 of 24 (permalink) Old 08-27-2014, 05:03 PM
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Any calculation will be ballpark at best, because it is impossible to know how efficiently you are mixing the old and new water. Temperature doesn't really tell us anything, because its effect will be minute compared to the location of the added water and tank flow.

To solve this problem you need to make an assumption about the ratio of new/old water that is overflowing. The generally used assumption here is that the ratio of the overflowing water is identical to the ratio of water in the tank. In other words, the water you add to the tank is being mixed in perfectly.

As a general rule, the greater the ratio of tank flow to added water, the closer the tank will approximate perfect mixing. For a typical drip overflow system I would assume perfect mixing.


Under that assumption, this becomes a textbook calculus problem (Search calculus mixing problem).
Quote:
Example 1. A tank has pure water flowing into it at 10 l/min. The contents of the tank are kept
thoroughly mixed, and the contents flow out at 10 l/min. Initially, the tank contains 10 kg of
salt in 100 l of water.
http://www.math.washington.edu/~conr...Examples01.pdf

You can back out a percentage of water changed by comparing salinity in this example before and after your chosen time period has elapsed.


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post #6 of 24 (permalink) Old 08-27-2014, 05:12 PM
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Difference in volume from 68F to 77F is about .15 ounces, so nothing to fret there. I think you're over thinking this. How much water are you trying to change? 100% a week? More?
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post #7 of 24 (permalink) Old 08-27-2014, 05:27 PM
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The temperature difference is insignificant for this application. Rather than trying to calculate this using hydrodynamics, why not use TDS as an indicator? Seems a whole lot less complicated to me. The acronym K.I.S.S seems applicable here.
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post #8 of 24 (permalink) Old 08-27-2014, 06:41 PM
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Quote:
Originally Posted by 691175002 View Post
Any calculation will be ballpark at best, because it is impossible to know how efficiently you are mixing the old and new water. Temperature doesn't really tell us anything, because its effect will be minute compared to the location of the added water and tank flow.

To solve this problem you need to make an assumption about the ratio of new/old water that is overflowing. The generally used assumption here is that the ratio of the overflowing water is identical to the ratio of water in the tank. In other words, the water you add to the tank is being mixed in perfectly.

As a general rule, the greater the ratio of tank flow to added water, the closer the tank will approximate perfect mixing. For a typical drip overflow system I would assume perfect mixing.


Under that assumption, this becomes a textbook calculus problem (Search calculus mixing problem).
http://www.math.washington.edu/~conr...Examples01.pdf
This is the right approach. If you are adding water at a high flow rate you probably get very rapid mixing, so the theoretical answer is nearly correct. If you add the water at a very low flow rate, you get slower mixing, but you also have more time for the mixing to occur. So, the theoretical answer is still nearly correct. It would be a very complicated calculation to get the exact answer.

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post #9 of 24 (permalink) Old 08-27-2014, 06:45 PM Thread Starter
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I didn't expect 100% accuracy. I already have a good idea of how much water I'm changing, but figured there had to be some loss of new water somewhere, from temp difference, how well the water is mixed, evaporation, or whatever else.

If I add 15 gallons to a 40 gallon tank using the overflow method, I know I'm not changing exactly 15 gallons of water.

A TDS meter might work, but I don't want to have to use one for every tank on every water change. On my shrimp rack, all I do is press an ON button, and come back later when the water is changed in all 8 tanks. It's pretty simple, and works well. I'd just like a better idea of just how much water I'm changing.

The calculus mixing problem that 691175002 referred to might be what I need (might be sufficient), but it's been a while since I've done any math like that. I'll need to study that a bit closer.
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post #10 of 24 (permalink) Old 08-27-2014, 08:05 PM
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I didn't mean to have a TDS controller. Simply measure the TDS (or anything you can measure NO3, PO4 etc.) to start then how long it takes to reduce it to where you want it. The math is easy to decipher how much water you changed that way. Then set a timer to that time frame for future water changes. Periodically check the TDS to make sure you're still in the ballpark. That would be about as close as you would ever need to come and it doesn't require a PhD in physics lol.
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post #11 of 24 (permalink) Old 08-27-2014, 11:09 PM Thread Starter
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I get what you're saying. Not a bad idea. It still might not be as accurate as I'd like, but I think I'll try it anyway.
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post #12 of 24 (permalink) Old 08-28-2014, 01:51 AM
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I think you can make a pretty accurate guess, and avoid the calculus, if you have some idea of how quickly pollutants (nitrates, whatever) are building up in your tank.

I'm thinking that using a drip overflow system, you want to eventually have the amount of pollutants leaving the tank be equal, or very close to the amount generated by your tanks occupants.

Just for the sake of argument, I'm going to make up some numbers. Say, after a week without any waterchanges, your tank has ~100ppm NO3 in 40 gallons.

Assuming the pollutant generation rate is equal, you're tank is building up 100ppm over 168 hours, about ~.59ppm/hour

Adding a gallon of water to the tank over the course of an hour, and allowing it to over flow should remove pretty close to 1/40th (2.5%) of the tank volume, and with it remove approximately 1/40th (2.5%)of the current buildup of pollutant.

Things should about balance out when you have about 24ppm or so, as 2.5% of that is about ~.6ppm. If NO3 level in the tank is higher, more will be removed with each gallon, so it should gradually approach the 24ppm balance.

This is a pretty rough estimate, and is only considering rates, and assumes constant addition/removal of water and pollutants, but you should be able to get a fairly close estimate based on previous NO3/TDS counts. It's been quite a while since I've did any of these types of problems, so there is a very good chance I've overlooked or omitted some important detail.
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post #13 of 24 (permalink) Old 08-28-2014, 02:15 AM
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Too much thought on this will lead you to a headache before the correct answer. Tanks are in constant flux. fish and plants grow or die, décor gets moved or changed. Filter flow changes with different amounts of cleaning. Fish swim in different patterns when the ambient light changes. Too many variables to even begin to think the answer is correct and steady so you might as well just go with making sure you are changing enough to fit what you are trying to do.
Otherwise you might as well set your calculations up there alongside the weather forecast and assume it will be just as accurate!!
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post #14 of 24 (permalink) Old 08-28-2014, 02:26 AM
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Quote:
Originally Posted by lochaber View Post
Adding a gallon of water to the tank over the course of an hour, and allowing it to over flow should remove pretty close to 1/40th (2.5%) of the tank volume, and with it remove approximately 1/40th (2.5%)of the current buildup of pollutant.
Unfortunately, this assumes no "new" water exits the tank. The position relative to the over flow is one variable. Another and more important is laminar versus turbulent flow. The video below starts with a laminar flow then becomes turbulent.

www.youtube.com/watch?v=nl75BGg9qdA&index=1&list=PL624A469C91F23914
The more turbulent, the more "new" water is mixed with "old" so the ratio of old to new changes based on this. A direct laminar flow pointed at the overflow will remove more "new" water. Calculating the flow of two different fluids in a mixed volume is no easy task. In fact, this is one of the reasons engineers use wind tunnels. The fact that the scape changes alters this laminar to turbulent flow. So we can never predict what patterns we will see based on simple positions of the inhabitants.

One sure way to remove a given percentage is to remove it first then add new water slowly over time. More accurate and FAR less wasted water but that was not the question.
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post #15 of 24 (permalink) Old 08-28-2014, 04:23 PM
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At one time, I was more into the thrill of problem solving and did as Zorfox says. I had a system of barrels where I pumped out a given amount the lifted a float and shut off the pump. Then I had a timer to pump a given amount of water back into the tank. I did it to let the chlorine gas off so that there was no need of water conditioning. This was before the use of chloramine and it worked very well for giving lots of new water at the correct temperature.
But then as I learned more, I found that I was doing a lot of "work" that really wasn't that important to keeping the fish happy. Any chance of me doing it again? NO WAY!
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