Mathematical approach to waterchanges

[pascal]

Hello,

I'm new to the forum. I have kept discus and Pt. altum in the past, about 15 years ago. Today, I have one tank left with wildcaught Pterophyllum sp. 'Santa Isabel'. My real interest goes to filtration, water quality, and so on. So I think that from that perspective, this forum can be interesting for me.
I've made a temptation to set up a formula, estimating (calculating is too correct) waste build up as a function of population density and waterchanges.

I have the article available as a .pdf file, but I can't upload it because the format is not compatible. Can somebody give some advise?

[Second Hand Pat]

The only suggestion I have is to embed the article within a thread. You would need to copy and paste the text and upload the pictures for placement within the text.
Pat

[pascal]

The only suggestion I have is to embed the article within a thread. You would need to copy and paste the text and upload the pictures for placement within the text.
Pat

OK thanks. It will need some copy/paste work, especially for the formulations.

[Larry Bugg]

You might want to do a search here first before going to all the trouble. This has been presented on here many times and very heavily debated.

[Second Hand Pat]

You might want to do a search here first before going to all the trouble. This has been presented on here many times and very heavily debated.

Here is the "simple" approach which does not include bio-load (from the SimplyDiscus Library) http://www.simplydiscus.com/library/..._formula.shtml.

[pascal]

Thanks, the problem with most temptatioons is that they try to describe the cinetics, waste as a function of time. When we include the bioload, the equation is still easy to set up, but quite complicated to solve, even for somebody with a scientific background. The way to aboard the problem is to consider the problem at equilibrium, then it becomes much easier from a mayhematical point of vue.

[rickztahone]

Thanks, the problem with most temptatioons is that they try to describe the cinetics, waste as a function of time. When we include the bioload, the equation is still easy to set up, but quite complicated to solve, even for somebody with a scientific background. The way to aboard the problem is to consider the problem at equilibrium, then it becomes much easier from a mayhematical point of vue.

would love to see your take on the matter.

[pascal]

Mathematical approach to waterchanges

There have been many articles about waterchanges but very few about an analytical approach. Articles have been published by Randy Holmes-Farley (1) in 2005 and later by David Boruchowitz (2), (3) in 2009. These articles were very interesting and describe the build-up of waste when applying a waterchange schedule. In these articles the author applied an iteration method simulating what happens in the tank. The result was reported under the form of tables and graphs.

Until today, I never read an analytical mathematical approach to the problem, resulting in a single formula describing the concentration of nitrates that is expected as a result of fish population density (or food administration quantity) combined with a waterchange schedule.

The principle of dynamic equilibrium

The approach applied to derive this formula is based on the principle of a dynamic equilibrium. This can be illustrated with an example.
We start a tank with a perfect running filter and pure water. Every day I feed the fish, which will result in a slight build-up of nitrates. Let’s forget the steps involving nitrite and ammonium, we want to keep it simple. After a week, I apply a water change. I will almost eliminate no nitrates, since the water is still very clean. I continue to apply my water change schedule. After two months, I will eliminate much more nitrates during the same waterchange, just because the waste concentration in the water is much higher.
At a certain moment, the absolute value (expressed in gram) of nitrates I will eliminate with my periodic waterchange will be exactly the same as the absolute amount of nitrates I added during this same period by feeding the fish. From then on, there will be a dynamic equilibrium and the situation will remain constant in time.
The concentration of nitrates can’t increase, because we eliminate as much as we added during the same period. It will not decrease, because we never eliminate more than we added in the same period. Between two waterchanges, there will be a slight increase, but this situation will remain constant indefinitely.

What is the relation between mass of food and nitrate concentration?

Food has mainly 4 components : water, proteins, lipids and carbohydrates. In this article, we will only consider dry food. I mean if I feed 10g of frozen artemia, containing 90% of moisture, I will only add 1g of dry food.

As a mean, we can say that fishfood contain 45% of proteins. Some contains more (artemia more than 60%, some less like daphnia), but most of the flakes and pellets contain this ratio of proteins. Let’s take that value. This means that 1 g of dry food contains 0.45 g of proteins.

Proteïns are build up by aminoacids. These molecules contain nitrogen and will result in the build up of nitrates. They contain as a mean value 16% of nitrogen, this value can be found on the internet. So 1 g of dry food will result in maximum 1 x 0.45 x 0.16 = 0.072g of nitrogen.

When this nitrogen has passed the nitrogen cycle, resulting in nitrate (NO3-), these 0.072g of nitrogen will give rise to 0.32g of nitrate. So now I know that 1g of dry food can result in an amount of 0.32g of nitrate when the nitrogen cycle has done his job.

Building up the equation

We have explained that after a certain period, a dynamic equilibrium will be established. That means that with the applied waterchanges, the amount of waste (say nitrates) that is eliminated is equal to the amount of waste (nitrates) that was accumulated during this same period. Only then, we have a dynamic equilibrium and a constant situation. Simply said, when the amount of waste (nitrates) that I eliminate with a waterchange is equal to the amount that was added (by feeding fish) during the same period, the amount of waste will remain constant in time.

Consider we feed every day D gram of dry food, this results in an equivalent situation where we would add (D x 0,32) gram nitrates daily to the tank. Expressed in mg, this value will be 1000 times higher : (D x 320) mg.

Consider I apply a regular waterchange since a long time (I’m at equilibrium) and every p days (period “p” is expressed in days and is the time between 2 waterchanges, once a week means p = 7 days) I perform a waterchange of v%, expressed in % of the total volume. The total volume of water in the tank is V, expressed in liter.
The amount of water I change is then (v/100 x V). Example, if I have a tank of 300 l (V) and I’m changing 25% (v), the amount of water I’m changing is indeed 25/100 x 300 = 75l.

Then, at the equilibrium, the amount of nitrates I will eliminate is equal to the amount of nitrates that I added since the last waterchange. If I add daily an amount of D x 320 mg nitrates by feeding the fish, I will have cumulated in p days (since my last waterchange) an amount of p x D x 320 mg. Indeed, I know that at equilibrium, I’m eliminating with a waterchange exactly the amount of nitrates that I added in that same period.

Example, if p = 7 days, D= 0.2g, then the amount of nitrates eliminated will be p x D x 320 = 7 x 0.2 x 320 = 448 mg of nitrates.
I can then calculate what was the concentration of nitrates that was present in the tank just before the waterchange, by dividing the amount of nitrates I know (mg) by the volume of water in which this amount was present (liter).

If there is (p x D x 320) mg nitrate present in a volume of (v/100 x V) liter, the concentration of nitrates (mg/l) at that moment is the amount of nitrate (p x D x 320), divided by the volume of water (v/100 x V) in which this amount of nitrates was present.
This gives us our first formula, expressing the maximum nitrate concentration as a function of food administration and waterchange schedule (amount and frequency) :

n = (p x D x 32000) / (v x V)

Where
n : maximum concentration of nitrates present in the tank, due to administration of food, expressed in mg/l
p : period between two waterchanges, expressed in days
v : percentage of waterchange, expressed as % of the total volume
D : Amount of dry food administrated, expressed in gram
V : Volume of water in liter. Make the correction for filter, decoration and sand.

An example :
I apply a weekly waterchange of 50% on a 300l tank and feed 1g of dry food daily.

p = 7 days
D = 1 gram
v = 50%
V = 300 l

The maximum nitrate concentration will be n = 15 mg/l. After the waterchange, concentration will be 15/2 = 7,5 mg/l.

If I apply a daily waterchange of 7 %, resulting in almost 50% per week :

p = 1 day
v = 7%

I obtain the same maximum concentration n = 15 mg/l before each waterchange daily. After the waterchange, concentration will be 15 x (100-7)/100 = 14 mg/l. That means that changing daily water is less efficient than equivalent large waterchanges. Less efficient doesn’t mean worse in my opinion. It is sufficient to change frequently a little bit more.

A tank can eliminate by itself a certain degree of nitrates by plants, anaerobic zones in sand, decoration, filtration substrate. The concentration of nitrates will be lower than expected. We express this by multiplying the formula with a correlation factor k, which will have a value between 0 and 1. We can estimate this value by measuring the nitrate concentration on a tank where the waterchange schedule is applied since a longer period on one hand and calculating the theoretical value on the other hand. I did the exercise for my tank and I estimated k equal to 0.4 – 0.6.

The formula becomes :

N = (k x p x D x 32000) / (v x V)

Where k is the correction factor for the tanks ability to reduce a part of the nitrates. If you want to overestimate the nitrateconcentration, use k = 1. N becomes the estimated concentration of nitrates expressed in mg/l.

Some of us use tapwater, having a startconcentration of nitrates. Since this concentration of nitrates is present at the start and also in the water that is used for waterchanges, we can add this value as a constant to the formula. Be N0 the startconcentration of nitrates of the tapwater expressed in mg/l. The final formula becomes :

N = [(k x p x D x 32000) / (v x V)] + N0

Where :

N : the estimated amount of nitrates present in the tank, due to administration of food, expressed in mg/l
p : period between two waterchanges, expressed in days
v : percentage of waterchange, expressed as % of the total volume
D : Amount of dry food administrated, expressed in gram
V : Volume of water, expressed in l.
N0 : the beginconcentration of nitrates in the water used for waterchanges, mg/l
k : a correlationfactor with a value 0 < k < 1.

I want to highlight that this is just an estimation of the waste content at equilibrium due to addition of food on one hand (bioload of the tank, more fish means more food) and waterchange schedule on the other hand. I hope it is usefull to somebody.


(1) Randy Holmes-Farley (2005), “Waterchanges in Reef Aquaria”, http:// reefkeeping.com, November 2005.
(2) - Boruchowitz, David E. (2009), “Time for a Change: A Mathematical Investigation of Water Changes – part I,” Tropical Fish Hobbyist, November 2009
(3) - Boruchowitz, David E. (2009), “Time for a Change: A Mathematical Investigation of Water Changes – part II,” Tropical Fish Hobbyist, December 2009

[rickztahone]

Mathematical approach to waterchanges

There have been many articles about waterchanges but very few about an analytical approach. Articles have been published by Randy Holmes-Farley (1) in 2005 and later by David Boruchowitz (2), (3) in 2009. These articles were very interesting and describe the build-up of waste when applying a waterchange schedule. In these articles the author applied an iteration method simulating what happens in the tank. The result was reported under the form of tables and graphs.

Until today, I never read an analytical mathematical approach to the problem, resulting in a single formula describing the concentration of nitrates that is expected as a result of fish population density (or food administration quantity) combined with a waterchange schedule.

The principle of dynamic equilibrium

The approach applied to derive this formula is based on the principle of a dynamic equilibrium. This can be illustrated with an example.
We start a tank with a perfect running filter and pure water. Every day I feed the fish, which will result in a slight build-up of nitrates. Let’s forget the steps involving nitrite and ammonium, we want to keep it simple. After a week, I apply a water change. I will almost eliminate no nitrates, since the water is still very clean. I continue to apply my water change schedule. After two months, I will eliminate much more nitrates during the same waterchange, just because the waste concentration in the water is much higher.
At a certain moment, the absolute value (expressed in gram) of nitrates I will eliminate with my periodic waterchange will be exactly the same as the absolute amount of nitrates I added during this same period by feeding the fish. From then on, there will be a dynamic equilibrium and the situation will remain constant in time.
The concentration of nitrates can’t increase, because we eliminate as much as we added during the same period. It will not decrease, because we never eliminate more than we added in the same period. Between two waterchanges, there will be a slight increase, but this situation will remain constant indefinitely.

What is the relation between mass of food and nitrate concentration?

Food has mainly 4 components : water, proteins, lipids and carbohydrates. In this article, we will only consider dry food. I mean if I feed 10g of frozen artemia, containing 90% of moisture, I will only add 1g of dry food.

As a mean, we can say that fishfood contain 45% of proteins. Some contains more (artemia more than 60%, some less like daphnia), but most of the flakes and pellets contain this ratio of proteins. Let’s take that value. This means that 1 g of dry food contains 0.45 g of proteins.

Proteïns are build up by aminoacids. These molecules contain nitrogen and will result in the build up of nitrates. They contain as a mean value 16% of nitrogen, this value can be found on the internet. So 1 g of dry food will result in maximum 1 x 0.45 x 0.16 = 0.072g of nitrogen.

When this nitrogen has passed the nitrogen cycle, resulting in nitrate (NO3-), these 0.072g of nitrogen will give rise to 0.32g of nitrate. So now I know that 1g of dry food can result in an amount of 0.32g of nitrate when the nitrogen cycle has done his job.

Building up the equation

We have explained that after a certain period, a dynamic equilibrium will be established. That means that with the applied waterchanges, the amount of waste (say nitrates) that is eliminated is equal to the amount of waste (nitrates) that was accumulated during this same period. Only then, we have a dynamic equilibrium and a constant situation. Simply said, when the amount of waste (nitrates) that I eliminate with a waterchange is equal to the amount that was added (by feeding fish) during the same period, the amount of waste will remain constant in time.

Consider we feed every day D gram of dry food, this results in an equivalent situation where we would add (D x 0,32) gram nitrates daily to the tank. Expressed in mg, this value will be 1000 times higher : (D x 320) mg.

Consider I apply a regular waterchange since a long time (I’m at equilibrium) and every p days (period “p” is expressed in days and is the time between 2 waterchanges, once a week means p = 7 days) I perform a waterchange of v%, expressed in % of the total volume. The total volume of water in the tank is V, expressed in liter.
The amount of water I change is then (v/100 x V). Example, if I have a tank of 300 l (V) and I’m changing 25% (v), the amount of water I’m changing is indeed 25/100 x 300 = 75l.

Then, at the equilibrium, the amount of nitrates I will eliminate is equal to the amount of nitrates that I added since the last waterchange. If I add daily an amount of D x 320 mg nitrates by feeding the fish, I will have cumulated in p days (since my last waterchange) an amount of p x D x 320 mg. Indeed, I know that at equilibrium, I’m eliminating with a waterchange exactly the amount of nitrates that I added in that same period.

Example, if p = 7 days, D= 0.2g, then the amount of nitrates eliminated will be p x D x 320 = 7 x 0.2 x 320 = 448 mg of nitrates.
I can then calculate what was the concentration of nitrates that was present in the tank just before the waterchange, by dividing the amount of nitrates I know (mg) by the volume of water in which this amount was present (liter).

If there is (p x D x 320) mg nitrate present in a volume of (v/100 x V) liter, the concentration of nitrates (mg/l) at that moment is the amount of nitrate (p x D x 320), divided by the volume of water (v/100 x V) in which this amount of nitrates was present.
This gives us our first formula, expressing the maximum nitrate concentration as a function of food administration and waterchange schedule (amount and frequency) :

n = (p x D x 32000) / (v x V)

Where
n : maximum concentration of nitrates present in the tank, due to administration of food, expressed in mg/l
p : period between two waterchanges, expressed in days
v : percentage of waterchange, expressed as % of the total volume
D : Amount of dry food administrated, expressed in gram
V : Volume of water in liter. Make the correction for filter, decoration and sand.

An example :
I apply a weekly waterchange of 50% on a 300l tank and feed 1g of dry food daily.

p = 7 days
D = 1 gram
v = 50%
V = 300 l

The maximum nitrate concentration will be n = 15 mg/l. After the waterchange, concentration will be 15/2 = 7,5 mg/l.

If I apply a daily waterchange of 7 %, resulting in almost 50% per week :

p = 1 day
v = 7%

I obtain the same maximum concentration n = 15 mg/l before each waterchange daily. After the waterchange, concentration will be 15 x (100-7)/100 = 14 mg/l. That means that changing daily water is less efficient than equivalent large waterchanges. Less efficient doesn’t mean worse in my opinion. It is sufficient to change frequently a little bit more.

A tank can eliminate by itself a certain degree of nitrates by plants, anaerobic zones in sand, decoration, filtration substrate. The concentration of nitrates will be lower than expected. We express this by multiplying the formula with a correlation factor k, which will have a value between 0 and 1. We can estimate this value by measuring the nitrate concentration on a tank where the waterchange schedule is applied since a longer period on one hand and calculating the theoretical value on the other hand. I did the exercise for my tank and I estimated k equal to 0.4 – 0.6.
The formula becomes :

N = (k x p x D x 32000) / (v x V)

Where k is the correction factor for the tanks ability to reduce a part of the nitrates. If you want to overestimate the nitrateconcentration, use k = 1. N becomes the estimated concentration of nitrates expressed in mg/l.

Some of us use tapwater, having a startconcentration of nitrates. Since this concentration of nitrates is present at the start and also in the water that is used for waterchanges, we can add this value as a constant to the formula. Be N0 the startconcentration of nitrates of the tapwater expressed in mg/l. The final formula becomes :

N = [(k x p x D x 32000) / (v x V)] + N0

Where :

N : the estimated amount of nitrates present in the tank, due to administration of food, expressed in mg/l
p : period between two waterchanges, expressed in days
v : percentage of waterchange, expressed as % of the total volume
D : Amount of dry food administrated, expressed in gram
V : Volume of water, expressed in l.
N0 : the beginconcentration of nitrates in the water used for waterchanges, mg/l
k : a correlationfactor with a value 0 < k < 1.

I want to highlight that this is just an estimation of the waste content at equilibrium due to addition of food on one hand (bioload of the tank, more fish means more food) and waterchange schedule on the other hand. I hope it is usefull to somebody.


(1) Randy Holmes-Farley (2005), “Waterchanges in Reef Aquaria”, http:// reefkeeping.com, November 2005.
(2) - Boruchowitz, David E. (2009), “Time for a Change: A Mathematical Investigation of Water Changes – part I,” Tropical Fish Hobbyist, November 2009
(3) - Boruchowitz, David E. (2009), “Time for a Change: A Mathematical Investigation of Water Changes – part II,” Tropical Fish Hobbyist, December 2009

I will have to disagree with the above bold/red statement. While some of these filter media items, like ceramic rings full of bb do help with nitrate removal, others inversely help to TRAP feces within the filter. There is no accounting for this within your formula.

Now, if the tank were a more sterile environment with no external filtration, I can see how your proposed idea could possibly be more accurate if you vacuumed the feces off the tank floor while doing your water changes. Even then, a sponge filter still serves as an area to trap detritus.

For this reason alone, I believe that in a day to day, regular hobbyist tank, the variables are not/can not be fully accounted for. If waste breaks down out of site, then it has to be removed completely during every water change in order to fall in line with your proposition.

I may be off base here with grasping the totality of your proposed tabulation, but I fear that you have not taken this situation in to consideration.

[pascal]

It is only a correction factor. In some cases, tanks are able to eliminate nitrate. Think about live rocks and sand in reef tanks. Even if it is only a small part.

Even if filtration media trap feces, the amount of waste can NEVER be higher than the the amount of waste introduced (considering your fish and plants stay alive of course). So k will never be higher than 1. If bare bottom tanks are used, with almost no anaerobic zone (bioball filtration material), then consider k=1 and it's done.

Elimination of feces by vacuum cleaning is not considered in this equation. take into account that a major part of the ammonia is not released by the feces in the filter, but is coming true the gills of the fish.

This formula does not take into account cleaning of the bottom, filter or anything else. It consider the equilibrium status between addition of waste and elimination of waste.

[John_Nicholson]

I love science and I appluad you thoguts but in reality iy just does not really work this way. Lots of people have rolled in here and tried these things. At the end of the day just change water and you will be miles ahead.

-john

[rickztahone]

It is only a correction factor. In some cases, tanks are able to eliminate nitrate. Think about live rocks and sand in reef tanks. Even if it is only a small part.

Even if filtration media trap feces, the amount of waste can NEVER be higher than the the amount of waste introduced (considering your fish and plants stay alive of course). So k will never be higher than 1. If bare bottom tanks are used, with almost no anaerobic zone (bioball filtration material), then consider k=1 and it's done.

Elimination of feces by vacuum cleaning is not considered in this equation. take into account that a major part of the ammonia is not released by the feces in the filter, but is coming true the gills of the fish.

This formula does not take into account cleaning of the bottom, filter or anything else. It consider the equilibrium status between addition of waste and elimination of waste.

It most certainly CAN. If there is trapped feces in the filter and it does not get removed, then the next time you feed it will be additive to the amount you previously left in the tank, which is still trapped in the filter. I will simply respectfully disagree, but applaud you the effort.

[pascal]

In the approach here, consider there is NO filter. We are at equilibrium and what is added is equal to what is removed. Is this totally correct? Probably not, but is an approach. Yes, there might be accumulated feces that will be transformed in nitrates somewhat later and removed with later waterchanges. But there will, as a mean value, never be more waste than you put in, that's impossible. It is a rough approach and the advantage is the simplicity, though it consider as well bioload as waterchanges. But it is nothing more than an estimation due to the simplifications, I totally agree with that

I appreciate your comments, honestly, it proves your interest in the matter. I just want to be sure I'm well understood (english is not mother language, I'm from Belgium).

[Woodduck]

I love science and I appluad you thoguts but in reality iy just does not really work this way. Lots of people have rolled in here and tried these things. At the end of the day just change water and you will be miles ahead.

-john

I agree with John. I change 50-100% every day depending on what's in that tank ie. adults vs grow outs. Works for me. I do enjoy
the precise calculation aproach for someone who can't be there for daily wc.
Woodduck

[rickztahone]

In the approach here, consider there is NO filter. We are at equilibrium and what is added is equal to what is removed. Is this totally correct? Probably not, but is an approach. Yes, there might be accumulated feces that will be transformed in nitrates somewhat later and removed with later waterchanges. But there will, as a mean value, never be more waste than you put in, that's impossible. It is a rough approach and the advantage is the simplicity, though it consider as well bioload as waterchanges. But it is nothing more than an estimation due to the simplifications, I totally agree with that

I appreciate your comments, honestly, it proves your interest in the matter. I just want to be sure I'm well understood (english is not mother language, I'm from Belgium).

Your English is perfectly understood and very well put in fact.