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?

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

"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 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.

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

"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/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.

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.

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 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/

"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.

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

"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.