Water change math

[Bill Stock]

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?

[Altum]

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.

--
Did you read the FAQ?
http://faq.thekrib.com

[carlrs]

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/

[IDzine01]

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

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

[Altum]

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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[carlrs]

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

[Bill Stock]

"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.

[Bill Stock]

"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. :-)

[Bill Stock]

"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.

[Altum]

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/AllBrowsers/3401/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.

--
Did you read the FAQ?
http://faq.thekrib.com

[carlrs]

Bill Stock wrote:

>
> 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.

That sounds like a great way of installing this unit, I never used one
this way.

Carl

[Bill Stock]

"carlrs" > wrote in message
oups.com...
>
> Bill Stock wrote:
>
>>
>> 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.
>
> That sounds like a great way of installing this unit, I never used one
> this way.
>
> Carl

We'll see what I've forgotten in a few weeks. I'm going to install a couple
of float valves, one to shut of the water if it gets too high and one to
shut off the pump if the water gets too low.

I was going to add the water to the bottom and take the effluent off the
top. But NetMax suggested plumbing my overflow to the bottom and adding the
water to the top. I should get more crap sucked out the overflow that way.
I'm not sure how much it will help, but it sounds good in theory.

[Bill Stock]

"Altum" > wrote in message
. com...
> 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/AllBrowsers/3401/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.

The Nitrates in the existing tank (75g) range between 40 and 80+, even with
weekly 50% water changes, reduced feeding and gravel vaccing. The big guy
(girl) must be close to 9" now and has a fair girth. I'm hoping the
increased water volume/flow in the new system will let me keep it below 40.

> 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.


Thanks Altum, I'm a little surprised that it's 21. I had expected higher.

[Altum]

Bill Stock wrote:

> The Nitrates in the existing tank (75g) range between 40 and 80+,
even with
> weekly 50% water changes, reduced feeding and gravel vaccing. The big guy
> (girl) must be close to 9" now and has a fair girth. I'm hoping the
> increased water volume/flow in the new system will let me keep it below 40.

21 grams of crap in 140 gallons comes in right around 40 ppm. Looks
like you'll meet your goal with a lot less work.

> Thanks Altum, I'm a little surprised that it's 21. I had expected higher.

Hope I solved the diffeq right! I'll be really curious to hear how the
actual measurements turn out.

Have you considered putting a cheap flourescent shop light over your
sump and growing duckweed or anacharis to lower the nitrates? You could
even harvest duckweed and feed it to the GF (mine love duckweed).

--
Did you read the FAQ?
http://faq.thekrib.com

[Bill Stock]

"Altum" > wrote in message
. net...
> Bill Stock wrote:
>
> Hope I solved the diffeq right! I'll be really curious to hear how the
> actual measurements turn out.

Yeah, I'm pretty sure you did. I ran the numbers through Excel and I came
very close to the numbers produced by your equation. I assumed the reduction
per hour for the additional Crap (.15) would be close to the average
addition per hour. I got 25.15, you had 25.22. LOL.

> Have you considered putting a cheap flourescent shop light over your
> sump and growing duckweed or anacharis to lower the nitrates? You could
> even harvest duckweed and feed it to the GF (mine love duckweed).

I considered some WH (Water Hyacinths) for the existing setup, but I would
have had to use a Sump Pump and hang-on overflow for the plumbing. A little
too much chance of a leak for my taste. The new sump is'nt a very good shape
for growing plants, but should make a good settling tank.

[swarvegorilla]

"Bill Stock" > wrote in message
...
> "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. :-)
>
what is this 'usenet' you speak of?
heh heh
Always nice to know your an uber fish geek.
Even nicer to have someone work it all out for ya.
Da international brotherhood of fish keepers (long haired brothers included)
w000t