Lbrewer34
Well-known member
So I was thinking, just how many water changes does it take to remove nitrates in a tank and the answer was MUCH more complicated than I thought. Here is how I approached the problem:
I assumed that I had a 100-gallon tank with 100ppm NO[SUB]3[/SUB]. I wanted to find how many water changes it would take to lower the nitrates to below 1ppm NO[SUB]3[/SUB]. Originally I wanted to figure for zero NO3 but realized this wasn't feasible. I also assumed that I would do 10% water changes.
Here is the shortcut formula I came up with:
[SUP](N[SUP]c[/SUP]/V[SUP](c-1)[/SUP])
[/SUP]Where:
N=(Total NO[SUB]3[/SUB]) – (NO[SUB]3[/SUB] removed based on % of water change)
V=Total Volume of Tank
c=Number of Water Changes
So to solve for how much NO[SUB]3[/SUB] is removed in a certain number of water changes is pretty easy, just plug in the numbers. So let’s say we did five 10% water changes in the tank from the example we get:
(N[SUP]c[/SUP]/V[SUP](c-1)[/SUP])
where:
N= (100ppm NO[SUB]3[/SUB]) -(10ppm NO[SUB]3 [/SUB])
V= 100 gallons
we get
(90[SUP]5[/SUP]/100[SUP](5-1)[/SUP])=59.049ppm NO3
Here is where it gets complicated. If I want to know how many water changes it takes to get the NO[SUB]3[/SUB] to a specific number. For this example, let’s just say we want to get to 25ppm NO[SUB]3[/SUB] but do 25% water changes instead what we have is this:
(N[SUP]c[/SUP]/V[SUP](c-1)[/SUP])= 25ppm NO3
where:
N=(100 ppm NO[SUB]3[/SUB]) -(25 ppm NO[SUB]3 [/SUB])
V= 100 gallons
c= Variable to solve for
so we get
(75[SUP]c[/SUP]/100[SUP](c-1)[/SUP])=25ppm NO3
=100(0.75)[SUP]c[/SUP] = 25
=0.75[SUP]c [/SUP]= (25/100)
To solve for the exponent “c” we have to take a function built into a calculator called “natural logarithm”, or “ln” on a calculator:
0.75[SUP]c [/SUP]= (25/100)
=c ln(0.75) = ln (25/100)
c ln(0.75) = ln(0.25)
c= [ln(0.25)/ln(0.75)]
c=4.82 (rounded)
Round it up to 5 water changes.
For this formula, you should be able to type your parameters in and solve. If your nitrates are high, you can use this formula to see the difference a 10%water change makes compared to a 30% water change, as well as how many water changes you need to get to below 1ppm based on the percent of water you change.
For the math enthusiasts who are curious how I came up with the formula here is what I did based on the example given at the beginning:
C[SUB]0[/SUB]=100ppm NO[SUB]3[/SUB]
C[SUB]1[/SUB]=
[(90(100ppm NO[SUB]3[/SUB])]/(100gal) = 90ppm NO[SUB]3[/SUB] remaining after 10% water change
C[SUB]2[/SUB]=
[90(90ppm NO[SUB]3[/SUB])]/(100gal) = 81ppm NO[SUB]3[/SUB] remaining after second 10% water change
C[SUB]3[/SUB]=
[90(81ppm NO[SUB]3[/SUB])]/(100gal)
=(90)[(90[SUP]2[/SUP]/100)/(100)]
=(90[SUP]3[/SUP])/(100[SUP]2[/SUP])
From this pattern I get:
(Nc/V(c-1))where:
N=(Total NO[SUB]3[/SUB]) – (NO[SUB]3[/SUB] removed based on % of water change)
V=Total Volume of Tank
c=Number of Water Changes
For those graphically inclined, I have attached a graph to the post. From the graph, the NO[SUB]3[/SUB] never really reaches zero, it just gets closer and closer (an asymptote). Of course eventually the remaining NO[SUB]3[/SUB] although present, won’t be testable.
These are just my results and thoughts, if anyone can improve on them or sees anything wrong let me know. Hopefully my math annotation is correct, it looks a lot better in word where I can insert the fractions. I can’t take full credit for all the work, when I got stuck I ran what I had by my calculus teacher for a little help.
I'm not sure if all the formula images and the graph will show, so if anyone wants the word doc just send me a pm
Lee
I assumed that I had a 100-gallon tank with 100ppm NO[SUB]3[/SUB]. I wanted to find how many water changes it would take to lower the nitrates to below 1ppm NO[SUB]3[/SUB]. Originally I wanted to figure for zero NO3 but realized this wasn't feasible. I also assumed that I would do 10% water changes.
Here is the shortcut formula I came up with:
[SUP](N[SUP]c[/SUP]/V[SUP](c-1)[/SUP])
[/SUP]Where:
N=(Total NO[SUB]3[/SUB]) – (NO[SUB]3[/SUB] removed based on % of water change)
V=Total Volume of Tank
c=Number of Water Changes
So to solve for how much NO[SUB]3[/SUB] is removed in a certain number of water changes is pretty easy, just plug in the numbers. So let’s say we did five 10% water changes in the tank from the example we get:
(N[SUP]c[/SUP]/V[SUP](c-1)[/SUP])
where:
N= (100ppm NO[SUB]3[/SUB]) -(10ppm NO[SUB]3 [/SUB])
V= 100 gallons
we get
(90[SUP]5[/SUP]/100[SUP](5-1)[/SUP])=59.049ppm NO3
Here is where it gets complicated. If I want to know how many water changes it takes to get the NO[SUB]3[/SUB] to a specific number. For this example, let’s just say we want to get to 25ppm NO[SUB]3[/SUB] but do 25% water changes instead what we have is this:
(N[SUP]c[/SUP]/V[SUP](c-1)[/SUP])= 25ppm NO3
where:
N=(100 ppm NO[SUB]3[/SUB]) -(25 ppm NO[SUB]3 [/SUB])
V= 100 gallons
c= Variable to solve for
so we get
(75[SUP]c[/SUP]/100[SUP](c-1)[/SUP])=25ppm NO3
=100(0.75)[SUP]c[/SUP] = 25
=0.75[SUP]c [/SUP]= (25/100)
To solve for the exponent “c” we have to take a function built into a calculator called “natural logarithm”, or “ln” on a calculator:
0.75[SUP]c [/SUP]= (25/100)
=c ln(0.75) = ln (25/100)
c ln(0.75) = ln(0.25)
c= [ln(0.25)/ln(0.75)]
c=4.82 (rounded)
Round it up to 5 water changes.
For this formula, you should be able to type your parameters in and solve. If your nitrates are high, you can use this formula to see the difference a 10%water change makes compared to a 30% water change, as well as how many water changes you need to get to below 1ppm based on the percent of water you change.
For the math enthusiasts who are curious how I came up with the formula here is what I did based on the example given at the beginning:
C[SUB]0[/SUB]=100ppm NO[SUB]3[/SUB]
C[SUB]1[/SUB]=
[(90(100ppm NO[SUB]3[/SUB])]/(100gal) = 90ppm NO[SUB]3[/SUB] remaining after 10% water change
C[SUB]2[/SUB]=
[90(90ppm NO[SUB]3[/SUB])]/(100gal) = 81ppm NO[SUB]3[/SUB] remaining after second 10% water change
C[SUB]3[/SUB]=
[90(81ppm NO[SUB]3[/SUB])]/(100gal)
=(90)[(90[SUP]2[/SUP]/100)/(100)]
=(90[SUP]3[/SUP])/(100[SUP]2[/SUP])
From this pattern I get:
(Nc/V(c-1))where:
N=(Total NO[SUB]3[/SUB]) – (NO[SUB]3[/SUB] removed based on % of water change)
V=Total Volume of Tank
c=Number of Water Changes
For those graphically inclined, I have attached a graph to the post. From the graph, the NO[SUB]3[/SUB] never really reaches zero, it just gets closer and closer (an asymptote). Of course eventually the remaining NO[SUB]3[/SUB] although present, won’t be testable.
These are just my results and thoughts, if anyone can improve on them or sees anything wrong let me know. Hopefully my math annotation is correct, it looks a lot better in word where I can insert the fractions. I can’t take full credit for all the work, when I got stuck I ran what I had by my calculus teacher for a little help.
I'm not sure if all the formula images and the graph will show, so if anyone wants the word doc just send me a pm
Lee
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