Water change math
 

I'll be setting up an automatic water changer in a few weeks and I've been
having trouble with the math.
The reading I've done suggests that a 20% daily water change is still going
to leave the fish swimming in 82% crap (r = e ^ -.2) as opposed to 50% to
100% crap with large weekly water changes. BUT, the detailed formula I've
seen shows the crap remaining as r=(1-a/n)^n, where a is the percentage of
water changed and n is the number of changes. Now assuming I change 1 gph
and a drop is 1/3600 gallons, then I'm left with 99.99% crap because the
changes are so small. (One drop at a time) Am I missing something here or do
I need to remove some water first to make this work?

Water change math
 

Bill Stock wrote:
I'll be setting up an automatic water changer in a few weeks and I've been
having trouble with the math.
The reading I've done suggests that a 20% daily water change is still going
to leave the fish swimming in 82% crap (r = e ^ -.2) as opposed to 50% to
100% crap with large weekly water changes. BUT, the detailed formula I've
seen shows the crap remaining as r=(1-a/n)^n, where a is the percentage of
water changed and n is the number of changes. Now assuming I change 1 gph
and a drop is 1/3600 gallons, then I'm left with 99.99% crap because the
changes are so small. (One drop at a time) Am I missing something here or do
I need to remove some water first to make this work?

This kind of math problem is known as a mixing problem, and it's solved
with differential equations. The algebraic model you're trying to use
assumes that a is constant during the water change, but it's not.

If you assume perfect mixing, constant water volume during the change,
and no crap in the tap water, here's the basic system of differential
equations where Crap(t) is the amount of crap in the water at t hours
and gph is the rate you're changing water.

dCrap/dt = crap fish add/hour - gph * (Crap(t)/tank volume in gallons)
Crap(0) = initial amount of crap in the water

From there, you integrate, plug in the initial condition and solve for
Crap(t). However, I don't think you can assume perfect mixing. The
equations get messy with imperfect mixing and you need information about
the mixing rates to solve them. It's probably easiest to set the
changer up for 20% a day and measure nitrates at the start and end of a
week.

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Water change math
 

Bill Stock wrote:
I'll be setting up an automatic water changer in a few weeks and I've been
having trouble with the math.
The reading I've done suggests that a 20% daily water change is still going
to leave the fish swimming in 82% crap (r = e ^ -.2) as opposed to 50% to
100% crap with large weekly water changes. BUT, the detailed formula I've
seen shows the crap remaining as r=(1-a/n)^n, where a is the percentage of
water changed and n is the number of changes. Now assuming I change 1 gph
and a drop is 1/3600 gallons, then I'm left with 99.99% crap because the
changes are so small. (One drop at a time) Am I missing something here or do
I need to remove some water first to make this work?

I believe you are using a static equation, as there are so many
variables, such as cross circulation vs vertical circulation. I like to
have both types of circulation in an aquarium, except in some breeding
enviroments.
I have only used this type of product a few times (it has been many
years ago, and they may have improved) and found that did not do a good
job of removing a lot of the organic muck that ended up in the nitrogen
cycle.

Carl
http://aquarium-nitrogen-cycle.blogspot.com/

Water change math
 

Seriously, you guys need to get out more... go on a date... have a
beer... something.
-------------------------

I'm kidding... I meant that light spiritedly.

Water change math
 

IDzine01 wrote:
Seriously, you guys need to get out more... go on a date... have a
beer... something.
-------------------------

I'm kidding... I meant that light spiritedly.

*chuckle* But I NEVER got to write dCrap/dt in my math class. :-p

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Water change math
 

IDzine01 wrote:
Seriously, you guys need to get out more... go on a date... have a
beer... something.
-------------------------

I'm kidding... I meant that light spiritedly.

Hey, I have kids, it is a common subject.

Carl

Water change math
 

"Altum" wrote in message
.. .

This kind of math problem is known as a mixing problem, and it's solved
with differential equations. The algebraic model you're trying to use
assumes that a is constant during the water change, but it's not.

If you assume perfect mixing, constant water volume during the change, and
no crap in the tap water, here's the basic system of differential
equations where Crap(t) is the amount of crap in the water at t hours and
gph is the rate you're changing water.

dCrap/dt = crap fish add/hour - gph * (Crap(t)/tank volume in gallons)
Crap(0) = initial amount of crap in the water

From there, you integrate, plug in the initial condition an solve for
Crap(t). However, I don't think you can assume perfect mixing. The
equations get messy with imperfect mixing and you need information about
the mixing rates to solve them. It's probably easiest to set the changer
up for 20% a day and measure nitrates at the start and end of a week.

Thanks Altum, but my Differential Equations class was more years ago then I
care to remember. Sadly the dreaded Beer has erased all those brain cells.
I'm assuming 35g of initial Crap (Nitrates), .15 g/hr of additional
Nitrates, 140 gallons of water and 1 gallon per hour of additional
water/outflow. I'm also assuming perfect mixing, although hoping to do
better by adding the fresh water to the bottom of the sump (tall plastic
barrel) and taking the old water from the top of the "sump".

A math refresher would be appreciated if you have time.

Water change math
 

"IDzine01" wrote in message
ups.com...
Seriously, you guys need to get out more... go on a date... have a
beer... something.
-------------------------

I'm kidding... I meant that light spiritedly.

LOL, you should be the looks I get when discussing this stuff at work. You
can just feel the eyes rolling in the heads. It can be amusing to lay it on
a little thick. :-)

Water change math
 

"carlrs" wrote in message
ups.com...

I believe you are using a static equation, as there are so many
variables, such as cross circulation vs vertical circulation. I like to
have both types of circulation in an aquarium, except in some breeding
enviroments.
I have only used this type of product a few times (it has been many
years ago, and they may have improved) and found that did not do a good
job of removing a lot of the organic muck that ended up in the nitrogen
cycle.

Yes, I'm assuming perfect mixing to make my life easy. The overflow will
actually happen in the sump (large plastic barrel). I'm hoping it might act
as a settling chamber as well, since the filter will be after the sump to
allow as much Crap to fall out of suspension before it reaches the filter.

Altum also mentioned the complexity of the variables and in the end I'll
just have to try it. But I thought it made an interesting problem.

Water change math
 

Bill Stock wrote:
"Altum" wrote in message
.. .

This kind of math problem is known as a mixing problem, and it's solved
with differential equations. The algebraic model you're trying to use
assumes that a is constant during the water change, but it's not.

If you assume perfect mixing, constant water volume during the change, and
no crap in the tap water, here's the basic system of differential
equations where Crap(t) is the amount of crap in the water at t hours and
gph is the rate you're changing water.

dCrap/dt = crap fish add/hour - gph * (Crap(t)/tank volume in gallons)
Crap(0) = initial amount of crap in the water

From there, you integrate, plug in the initial condition an solve for
Crap(t). However, I don't think you can assume perfect mixing. The
equations get messy with imperfect mixing and you need information about
the mixing rates to solve them. It's probably easiest to set the changer
up for 20% a day and measure nitrates at the start and end of a week.

Thanks Altum, but my Differential Equations class was more years ago then I
care to remember. Sadly the dreaded Beer has erased all those brain cells.
I'm assuming 35g of initial Crap (Nitrates), .15 g/hr of additional
Nitrates, 140 gallons of water and 1 gallon per hour of additional
water/outflow. I'm also assuming perfect mixing, although hoping to do
better by adding the fresh water to the bottom of the sump (tall plastic
barrel) and taking the old water from the top of the "sump".

A math refresher would be appreciated if you have time.

LOL! Setting these suckers up is a lot easier than solving them. My
diffeq class was a few years ago, so I recognized the problem but it's
been a while since I solved one. I used this web page to set the
equations up.
http://tutorial.math.lamar.edu/AllBr...1/Modeling.asp
There's a section on that website on solving first order linear
equations too.

I attempted a solution with your conditions and got:
Crap(t) = 21 + 14 * e ^ -t/140

If I solved it right, with a continual 1 gph drip and perfect mixing,
running the changer for 24 hours will drop the crap to from 35g to
32.8g. After a week, it will be at 25 grams. After three weeks the
tank will stabilize at 21 grams of crap.

For a continual drip, the most useful part of the solution is the
constant, since as t gets large e ^ -t/140 approaches zero. The general
equation for the constant is crap/hour * (140 gal/gph water change). It
derives from the integrating factor used to solve the equation. You can
use this if you want to change your crap/hour estimate (almost 50
ppm/week seems awfully high for a normal fish load) or want to estimate
the crap left with different constant flow rates.

If anyone finds an error, could you post the correction? This is useful
enough that I may write it up for my website at
http://fish.turquoisewave.com.

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http://faq.thekrib.com