Calculating final nitrate level after water change

Hello,

I realize this is probably a fairly simple equation, but I'm really
struggling with it. I'm trying to create a formula (perhaps to
eventually work into a classroom lesson) to predict the nitrate
concentration in the aquarium AFTER a partial water change. This is
complicated by the fact that I use natural ocean water, which, at least
in this area, is NOT nitrate free (low, but not free). I realize I
could just ignore this semi-insignificant factor, but the point of this
is to include all inputs.

Any ideas?
Thanks
Ryan Lenz

wrote:

I realize this is probably a fairly simple equation, but I'm really
struggling with it. I'm trying to create a formula (perhaps to
eventually work into a classroom lesson) to predict the nitrate
concentration in the aquarium AFTER a partial water change.

Yeah, it's pretty simple. Let's take the example of a 125 gallon tank with a
nitrate level of 50 ppm. We'll change 25 gallons of the water with water that
has no measurable nitrates. Twenty-five gallons is 20% of the water, so we will
be removing 1/5 of the nitrates. Put another way, 80% of the nitrates will
remain in the tank.

Given N is the nitrate level, TS is the tank size, GC is gallons changed, and RN
is remaining nitrates.

RN = N * (1 - (GC / TS)).

Plug in the numbers.
RN = 50 * (1 - (25 / 125)) = RN = 50 * .8 = RN = 40.

If the water you use for a change has, say, a nitrate concentration of 10 ppm,
then just add that back in. Call the concentration N2. The equation would be
RN = RN + (N2 * GC / TS).

Plug in the numbers.
RN = 40 + (10 * 25 / 125) = RN = 44.

If you want the entire thing as a single equation,
RN = (N * (1 - (GC / TS))) + (N2 * GC / TS).

George Patterson
No one ever says "It's only a game." when his team is winning.

George Patterson wrote:

If you want the entire thing as a single equation,
RN = (N * (1 - (GC / TS))) + (N2 * GC / TS).

By the way, "TS" in there is the actual amount of water in the tank, which is
usually less than the tank size because of the rock, substrate, etc. in the
tank. For example, my 125 gallon tank probably only has about 90 gallons of
water in it. Running this exercise with my tank, TS would be 90.

George Patterson
No one ever says "It's only a game." when his team is winning.

"George Patterson" wrote in message news:AU6th.2334$R65.1704@trnddc01...
Yeah, it's pretty simple.

You are correct!
It was pretty simple... and you did a good job with it :-)

Let me challenge you with a different mathematical problem:

Let's find out how many (x)% water changes we will need
to bring (Y) ppm nitrates down to (Z) ppm level :-)

Let's simplify the problem (for now) by assuming the tank
DOES NOT produce additional nitrates between water changes
and the water we use for partial changes is pu (0) ppm NO3.

For example: Jaime has now 100 ppm of nitrates in his tank.
How many 10% water changes does he need to do
next weekend to bring nitrates down to - let's say 5-10 ppm.

Of course it will be easy to plug the numbers into your previous
equation for couple of times, and than again and again to get
the final number, but what I am asking for is a generic solution :-)

Can you assemble an equation to solve this problem ? :-)

Pszemol wrote:

Can you assemble an equation to solve this problem ? :-)

Us a variation of compound interest calculations. Let n be the number of water
changes, N1 be the initial nitrate level, N2 be the desired nitrate level, and P
be the percentage of water changed. Then

n = log(N2 / N1) / log(1 - P)

Solving for N1 = 100, N2 = 10, and P = .1, get get 21.85434512 changes.
Solving for N2 = 20, we get 15.2755317.

George Patterson
No one ever says "It's only a game." when his team is winning.

"George Patterson" wrote in message news:Cc6uh.1262$FN1.1231@trnddc08...
Pszemol wrote:

Can you assemble an equation to solve this problem ? :-)

Us a variation of compound interest calculations. Let n be the number of water
changes, N1 be the initial nitrate level, N2 be the desired nitrate level, and P
be the percentage of water changed. Then

n = log(N2 / N1) / log(1 - P)

Solving for N1 = 100, N2 = 10, and P = .1, get get 21.85434512 changes.
Solving for N2 = 20, we get 15.2755317.

Very good! We should name you a Group Math Wizard

lol I was beginning to think I was back in high school algebra
"Pszemol" wrote in message
...
"George Patterson" wrote in message
news:Cc6uh.1262$FN1.1231@trnddc08...
Pszemol wrote:

Can you assemble an equation to solve this problem ? :-)

Us a variation of compound interest calculations. Let n be the number of
water
changes, N1 be the initial nitrate level, N2 be the desired nitrate
level, and P
be the percentage of water changed. Then

n = log(N2 / N1) / log(1 - P)

Solving for N1 = 100, N2 = 10, and P = .1, get get 21.85434512 changes.
Solving for N2 = 20, we get 15.2755317.

Very good! We should name you a Group Math Wizard

Pszemol wrote:

Very good! We should name you a Group Math Wizard

I don't think so. I'm pretty good at Algebra, but Trig is a lot of work, and I
was glad to see the last of Calculus.

George Patterson
No one ever says "It's only a game." when his team is winning.

Let us take the example of a 125 gallon tank and the nitrate content of 50 ppm. We will change 25 gallons of water and water, no measurable nitrate. Second, 20%, five gallons of water, we will cancel the 1 / 5 of the nitrate. In other words, 80% of the nitrate will remain in the tank.

Let's abridge the botheration (for now) by bold the tank DOES NOT aftermath added nitrates amid baptize changes and the baptize we use for fractional changes is pu (0) ppm NO3. For example: Jaime has now 100 ppm of nitrates in his tank. How abounding 10% baptize changes does he charge to do next weekend to accompany nitrates down to - let's say 5-10 ppm.