# Water Change % math challenge (Math/Excel Wiz pleae help!)



## fatkinglet (May 8, 2010)

Hi, 

The scenario is 1/3 (33.333...%) water change weekly.

eg:

1st water change:
100% - 33.33% = 66.67% (old water) + 33.33% (new water) = 100% (OLD + NEW water)

2nd water change:
100% (OLD + NEW water) - 1/3 (OLD + NEW water) + 1/3 (new water) 

3rd water change:
loop .... etc


I would like to find out how many times of water change needed to remove all the old water ? How many % of old water remains after 100th water change?!

I cannot come up this math formula, preferable to have it shown on Excel, thanks!!!!


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## infolific (Apr 10, 2016)

The general formula would be:

gallons * (1 - percent_change)^number_of_changes

For example after 4 changes of 50% on a 100 gallon tank you're left with 6.25 gallons of old water. The formula for this would be:

100 * (1-0.5)^4

In answer to your question, after 12 changes you'd be left with less than 1% of the old water. I think you effectively never get to 100% new water.


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## mitko1994 (Dec 12, 2012)

Your "new" water in the previous water change becomes "Old" by the time you change the water again. Plus it's not like you are actually always taking out water that you didn't add the previous water change, so you are almost certainly guaranteed to take out what was considered "new" water in the previous water change, this time around. Also there's no guarantee the proportion of the new and old water you will take out. In WC2, you say, 1/3 (Old+new) but that assumes you are taking out 1/3 of the new water from WC1 and 1/3 of the Old in WC2. So you think that your Old water in WC2 would be 2/3-1/3*2/3 = 6/9 -2/9 = 4/9 of what you started with, but that's not guaranteed. In fact it's almost certain that in you tank right now, there is water left from the first time you filled it no matter how many times you changed your water. It's just that the Old water is really awful water that gets diluted with clean water and its bad effects are minified to a level that is harmless.


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## default (May 28, 2011)

You're trolling right?.. Tell me you're trolling. 
Perhaps it's for school? Tell me it's for school.

If you're serious about running a regime to rid 'old' water from your aquarium, it's not going to work. Remember once new water enters and mixes with the existing volume, what will happen is the dilution of whatever was in your 'old' water; e.g: If you did a 50% (5g) water change on a 10g aquarium that had a nitrate reading of 2 ppm, and you added new water with 0 ppm nitrate - you would hopefully have 1 ppm of nitrate diluted in the entire body of water now once full.


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## infolific (Apr 10, 2016)

mitko1994 said:


> So you think that your Old water in WC2 would be 2/3-1/3*2/3 = 6/9 -2/9 = 4/9 of what you started with, but that's not guaranteed. In fact it's almost certain that in you tank right now, there is water left from the first time you filled it no matter how many times you changed you water.


True. The math assumes a perfect mixing of old vs. new. The substrate probably has water that doesn't mix well, for example. The formula also shows that you never get rid of 100% of the old water, but the percentage of it gets smaller and smaller with each change.

Also, the implication of the original question was about the stuff in the water and the impact to the dilution of that "stuff" after multiple partial water changes. Again, the assumption is that the "stuff" is evenly distributed throughout the water when the change happens.



default said:


> You're trolling right?.. Tell me you're trolling.
> Perhaps it's for school? Tell me it's for school.


I think understanding the math can be useful. Sure, for most of us that do manual water changes every week or two there might not be a practical application. But consider the case of someone doing a 10% daily change using an automated system. In such a case someone might very well want to know what new vs. old water remains in the tank. Or to put it another way, how much of the water is older than a week.

Similar math is also important for understanding how the water change in the EI method puts a cap on the PPM of fertilizers in your tank even though you're constantly overdosing.


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## c31979839 (Nov 26, 2014)

infolific said:


> True. The math assumes a perfect mixing of old vs. new. The substrate probably has water that doesn't mix well, for example. The formula also shows that you never get rid of 100% of the old water, but the percentage of it gets smaller and smaller with each change.
> 
> Also, the implication of the original question was about the stuff in the water and the impact to the dilution of that "stuff" after multiple partial water changes. Again, the assumption is that the "stuff" is evenly distributed throughout the water when the change happens.
> 
> ...


Some pretty impressive insight right here!

Did you study maths or engineering in school? Or is this knowledge obtained just from reading up on aquarium articles?

To the original questions, you'll never really get out all of the old water as you cycle through partial water changes. But you have to ask yourself, what is a satisfactory percentage of diluted 'old water' you want to have in your tank. Can you live with 1-5% of 'old water' or would you rather try for <1% after numerous water changes. And if your aiming for a low number as possible, why? Changing too much too quickly can upset the equilibrium in the tank.

As long as your water chemistry doesn't drift into undesirable levels, you shouldn't worry too much about the theoretical dilution of 'old water' into new.


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## default (May 28, 2011)

infolific said:


> I think understanding the math can be useful. Sure, for most of us that do manual water changes every week or two there might not be a practical application. But consider the case of someone doing a 10% daily change using an automated system. In such a case someone might very well want to know what new vs. old water remains in the tank. Or to put it another way, how much of the water is older than a week.
> 
> Similar math is also important for understanding how the water change in the EI method puts a cap on the PPM of fertilizers in your tank even though you're constantly overdosing.


I agree, understanding the math is useful.

However, there's almost no practical advantage in utilizing this calculation in reality, IMO it's a 'wrong' question to be solving, over simplified really.
The issue is there is no way of removing all the 'old' water, you would need to do 100% WC to accomplish that, so why stress about a calculation (if not for other purposes, like school or a sudden 'what-if' moment) that doesn't have use to it?

From systems I've seen that work with auto WC systems, you would normally have smaller yet more frequent changes. Some systems are completely trickle fed, so a constant supply of new water is added, not a traditional WC. EI also doesn't work well with 'old' water calculations, instead knowing the values of (NPK+trace); base ppm, desired ppm, and remaining ppm. Then calculating how much water dilutes the remaining ppm after X time, how much is needed to raise it back to desired, and continue.


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## mitko1994 (Dec 12, 2012)

default said:


> You're trolling right?.. Tell me you're trolling.
> Perhaps it's for school? Tell me it's for school.


That wasn't at me, was it?


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## default (May 28, 2011)

mitko1994 said:


> That wasn't at me, was it?


Lol no, it was a joke towards the original question.


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## mitko1994 (Dec 12, 2012)

That's what I thought, but wanted to make sure .


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## Fisheye (Jan 13, 2015)

42

(amazing calculations and insight from smart people though)


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