Re: daily water change/continuous percent

[NetMax]

"Donald Kerns" > wrote in message
...
> Greetings,
>
> I've (for the moment) the means and the motivation to put a daily or
> continuous water change system on my outside tub garden.
>
> Why? Well... just because... I wanna play...
>
> (I've got a spare drip irrigation valve contained by a reasonably smart
> controller and access to a couple of spare peristaltic (sp?) pumps.)
>
> So... I _know_ my water company uses Chlorine (and not huge amounts of
> it). The couple of times I've measured it out of the tap I don't read
> ANYthing.
>
> The NH3 / NO2 / NO3 are all zero, zero, zero, but the water itself gets
> green pretty easily. I've got two 3" SL goldfish, anachris, duckweed
> and water hyacinth. The fish eat the duckweed faster than it can grow.
> The water hyacinths are just about breaking even. (yea...)
>
> What sort of percent per day guys/gals?
>
> Continuous really slow flow or once/twice/thrice daily batch?
>
> -Donald

Currently thrice, adding up to 10% a day. It all depends IMHO on fish
load, parameter similarity, temperature, CO2 charge and chlorine
concentration.

Continuous would be better as activated carbon removes chlorine more
efficiently with a slow water movement. Is you negative measurement of
chlorine really indicative of an absence of chlorine, or a failure of
your test kit due to the level being below it's threshold?

NetMax

[NetMax]

"Donald Kerns" > wrote in message
...
> Anna Hayward wrote:
<snip>
>
> Or I could just overdose with AmQuel/NovAqua once a week and skip the
> math... But that's no fun. ;-)

So then your next question would be; what is the half-life of sodium
thiosulfate.

...and then, what concentration of chlorine does a standard dosage of
de-chlor treat.

....now that would be useful information

NetMax

> -Donald
> --
> "When you've lost your ability to laugh, you've lost your ability to
> think straight." -To Inherit the Wind

[Donald Kerns]

NetMax wrote:

>
> "Donald Kerns" > wrote in message
> ...
>> Anna Hayward wrote:
> <snip>
>>
>> Or I could just overdose with AmQuel/NovAqua once a week and skip the
>> math... But that's no fun. ;-)
>
> So then your next question would be; what is the half-life of sodium
> thiosulfate.
>
> ..and then, what concentration of chlorine does a standard dosage of
> de-chlor treat.
>
> ...now that would be useful information

Ask and ye shall receive.

http://www.kordon.com/kpds.htm

Kordon does real science...

-D
--
"When you've lost your ability to laugh, you've lost your ability to
think straight." -To Inherit the Wind

[Bob K.]

I've had numerous discusions about water changin over the years,
and there's a lot of YMMV involved.

I change 50% or much more on some cichlid tanks each week.
I've sometimes done about 80% .. & yes, I use Amquel when
I do 50% or more.

I have other tanks on a water overflow system that get 15-20%
several times a week.

I've had any chlorine or chloromines when I measured my water but
use the Amquel anyway to be safe. I saw what happend years
ago with a massive water change and chloromines present..
Immediate shock and death in less than 20 minutes.

As to holw much and how often,
10% change every day for 5 days is the same amount of water
change as a 50% change every 5 days,
BUT the wastes will AVERAGE less in the 50% change every 5.
as the max will eventually vary between 90 to 100% in the every
day 10% change, and will be only 50 to 100% in the once every
5 day 50% change.
If you don't quite see it, consider a 1% every day or 100% every
100 days, the fish in the 1% change will be living in some very
bad water! nearly a no-change situation. At least the others will
take 100 days to get that bad. Work the numbers back and it
is obvious you want to do larger changes more frequently for
best results. My own 2 cents? 10% isn't enough.

And finally. I was talking to an old time guppy breeder last
week, and he did 90% or better changes EVERY DAY to
grow the champion sized fish he raised. It was before chloromine,
and he 'aged' the water several days before using it.. Several
hundred gallons a day.

Bob

[Jim Seidman]

Donald Kerns > wrote in message >...
> Point taken, but the math doesn't work out that way. 10% for five days
> is actually only a 41% change.
>
> (And actually, in this application it's even worst than that because it
> is a mixing rather than a replacement. Still haven't found the mixing
> formulas... HELP?!? 1 order diff-eq probably.)

Fortunately, it's much simpler than that.

n = number of changes
r = fraction of water remaining after n changes
a = fraction of water added (and removed) in all n changes together

For a finite number of changes:
r = (1 - a/n) ^ n
So, for example, if you change 20% of the water in two batches:
(1 - .2/2) ^ 2 = 0.81 (meaning 81% of the original water remains)

Taking the natural log of this formula gives:
log r = n log(1 - a/n)

What you need is to solve this formula as n approaches infinity.
Luckily, for numbers very close to 0, log(1+x)=x. And as n approaches
infinity in the above equation, a/n will be *very* close to zero. So
it becomes:
log r = n * (-a/n)
which simplifies to
log r = -a

Now we can take the inverse log of each side to get the more useful:
r = e ^ -a

So, if you add 20% of the water through a continuous drip:
r = e ^ -.2
r = .818 (meaning 81.8% of the original water remains)

Hope this helped!

- Jim

[Donald Kerns]

Jim Seidman wrote:

> Donald Kerns > wrote in message
> >...

>> (And actually, in this application it's even worst than that because
>> it is a mixing rather than a replacement. Still haven't found the
>> mixing formulas... HELP?!? 1 order diff-eq probably.)
>
> Now we can take the inverse log of each side to get the more useful:
> r = e ^ -a
>
> So, if you add 20% of the water through a continuous drip:
> r = e ^ -.2
> r = .818 (meaning 81.8% of the original water remains)
>
> Hope this helped!

Excellent, yes it did.

I'm a bit worried that there isn't a time unit in there somewhere...

Going with r = exp(-at) yields the behavior I intuitively expect (and is
also the canonical solution to a 1o linear de.)

Whenever you get an equation with an e to the something, there is almost
always a 1o de lurking around somewhere...

But anyway, given those numbers, I'm comparing a 20% weekly change
against 10% per day. So that's 80% vs 50% after a week (assuming the
r=exp(-at)). And *I* don't have to do it.

(PS I bought the parts for the drip irrigation/charcoal refrigerator
combo. If only I could find a good low range Cl test kit...)

-Donald
--
"When you've lost your ability to laugh, you've lost your ability to
think straight." -To Inherit the Wind

[Bob K.]

>> I'm a bit worried that there isn't a time unit in there somewhere...
>
>You could certainly add a time unit by changing definitions:
>
>c = fraction of water added per hour
>t = number of hours that water gets changed
>Then:
>r = e ^ (-ct)
>
>Looking at it, you're right, with a time in there it does look like a
>DE. With my approach (limit as n approaches infinity) the time makes
>it more complicated.
>
>And the time doesn't really matter, since what we're calculating is
>the amount of "original" water remaining after the water changes.
>Suppose you compare changing 10% a day versus 10% a week. Either way,
>after 5 changes, there'll still be 59% of the original water left. Of
>course, it'll be much older original water in the latter case. :-)

Time isn't the main key. You want to measure the amount of waste over
a given period that the fish are in. They're pumping it out at a steady
rate. Any amount of water changes will decrease the amount of waste.
A careful program of fixed changes will reach a steady state of maximum
waste in the water over time.

10% daily, when it reaches a steady state, varies from 90% to 100%
of whatever that amount of waste for the rate those fish excrete.
50% change every 5 days will vary the waste from 50% to 100%.

In other words, the maximum waste will be identical in both, but
with larger changes, less frequently - the fish have a respite from
high waste for several days.

Try the extreme of 1% daily change (huge waste accumulation in time)
vs 100% change every 100 days. Maximum waste will be the same,
but they'll have many days of low waste in the water.

Obviously, you want an optimal situation and most of us do either
10% daily or 50% weekly or something in between (except when we
get lazy)

Bob,
Nice job with the math, but like my math professor taught,
thinking of the situation reveals more than the raw numbers
may show. I.E. don't get lost in the formulas

[Donald Kerns]

Bob K. wrote:

> 10% daily, when it reaches a steady state, varies from 90% to 100%
> of whatever that amount of waste for the rate those fish excrete.
> 50% change every 5 days will vary the waste from 50% to 100%.

I guess I'm making the assumption that the accumulation of fish waste is
small compared to the 3 gal/day I'll be replacing...

> In other words, the maximum waste will be identical in both, but
> with larger changes, less frequently - the fish have a respite from
> high waste for several days.

My second guess is that fish like to be "continuous" critters and
generally don't like "discontinuities" in their environment.
(Particularly my unstable pH coming out of the tap...)

-D
--
"When you've lost your ability to laugh, you've lost your ability to
think straight." -To Inherit the Wind