Aquarium calculus

  • Thread starter Jean-Louis
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  • #1
15
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How to devise a usefull formula for aquarium water changes.

I have a 50g tank (not including substrate and animals). I want to do daily water changes so that at the end of 6 months I will have replaced 95% of the water present at the beginning of the water change regimen. I want to replace the same volume every day.

1. What volume should I replace each day?
2. What is a general formula I could use for larger or smaller tanks? Or for changing the amount of time over which I replace the 95%. (So I could use it if I decide to set up a X
gallon tank and have my water 'turn over' period every Y months.)
3. What is a general formula I could use that contains the variables from (2) but also allows me to change the periodicity (say I get lazy and only want to do weekly water changes.)
 

Answers and Replies

  • #2
101
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I'd wait for someone not as rusty to confirm this, but

Gallons of Old at time T = V[initial]exp-(VT/V[initial])

where V=constant removal rate per unit of time, V[iniital]=initial volume
units of removal rate have to equal units of T, e.g. per day, per second

so roughly .83 gallons/per day need to be removed over 180 days to have 5% or 2.5 gallons of old water remaining
 
  • #3
19
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so roughly .83 gallons/per day need to be removed over 180 days to have 5% or 2.5 gallons of old water remaining

This is incorrect as if you change the water 180 times and remove 0.83 gallons each time you get (0.83 gallons/day)*(180 days) = 149.4 gallons

--

Let your volume be X and months be Y (as you requested), then the amount of water you would change per day would be 0.95X/(Y*30) = dv/dt = 0.005277V gallons/day.


So, for your 50g tank, you would change 0.264 gallons / day if you want to for 6 months.

Let the 30 become U, where that is the how often you wish to change the water.

so: Y=period (one week, two weeks, 6 months) where you change out the desired % of water.
U = equal division of period (changing water everyday, once a week)

Just to check units....
0.95X gallons / ([Y time][U timeunits/time] = gallons / timeunits

Also, you can change the 0.95 to whatever % of water you wish to change.

My explanation may seem complicated, let me know if you need clarification


So in short, use this

PX/(YU) = gallons/time

P=overall percentage you want to change out over the period Y
X=Volume of tank
Y=period
U=equal partition of interval that corresponds to how many times you change the water per interval
 
Last edited:
  • #4
cristo
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chemhelper. We do not give out full solutions to homework questions, but are here to guide students to the correct answer. Also, please note that, until the original poster has shown his/her working, and attempts at the question, we cannot give any help. See the guidelines here
 
  • #5
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chemhelper. We do not give out full solutions to homework questions, but are here to guide students to the correct answer. Also, please note that, until the original poster has shown his/her working, and attempts at the question, we cannot give any help. See the guidelines here

No evidence that this is a homework question. The fact that someone moved it to the homework section isn't evidence either.

Chemhelper: Your solution is simplistic and wrong. You fail to take into account that the percentage of old water in the tank diminishes with time as it is replaced with new water...he's not emptying the tank, this is an exponential decayh
 
  • #6
cristo
Staff Emeritus
Science Advisor
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regor60 said:
No evidence that this is a homework question. The fact that someone moved it to the homework section isn't evidence either.

From past experience, this sounds like a homework question. Also, the fact that a mentor has moved it to the homework forum, shows that they agree with me!

Whether it is homework or not, the way to help a person learn, is not to solve the problem for them, and not to tell them what to do before they have shown to us that they have thought about the problem.
 
  • #7
19
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Sorry. I stand corrected.


Jean-Louis may come and correct me on this (in which case I am sorry for explaining how to do it), but I do not believe it is his desire or care to learn how to derive formulas; he states that he has an aquarium that he wants to apply this to, so why make him try to learn Calculus? Perhaps he has no idea about how to derive the formula he needs. When I posted, it was in the main forum. So unless you have evidence showing that he is a student, don't criticize me when I was following the rules.
 
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  • #8
15
0
Thanks for your feedback. This is not homework. I have an aquarium, and I wanted a mathematical approach. I think I founded. I was told that this not a calculus problem, since the water is not continuously changing. So the approach is algebraic.
Sorry if I caused any inconvenience.

(if you are interested I could post the algebraic solution.)
 
  • #9
101
0
Thanks for your feedback. This is not homework. I have an aquarium, and I wanted a mathematical approach. I think I founded. I was told that this not a calculus problem, since the water is not continuously changing. So the approach is algebraic.
Sorry if I caused any inconvenience.

(if you are interested I could post the algebraic solution.)

Well, if you're taking water out of the aquarium at some modest rate per period and replacing it with new water, and want to have a certain percentage of "original" water remaining after a certain amount of time, it really is a calculus problem.

The 0.83 gals/day=10.97% per week. Coincidentally, this calculator http://www.angelfish.net/DripSystemcalc.php

validates exactly the equation above if you plug in the 50 gals and 0.83 gal/day rate
 
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  • #10
177
1
I thought this was a Differential Equations kind of Calc. I couldn't do one now but I remember these types of questions in DE.
 

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