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.

--
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

"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.