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Old 11-27-2002, 04:56 PM   #1
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Reef math question


Say your tank holds 250 gallons and you have 250 gallons or water made up in a storage tank for a water change.

You pump in clean water at exactly the same rate that you are siphoning water out of the tank and the tank has a lot of circulation in it (the new water and old water are being mixed).

When the storage tank is empty, what percentage of water have you changed in the tank?


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Old 11-27-2002, 05:01 PM   #2
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A LOT OF WATER... I dont know ... maybe some one does...
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Old 11-27-2002, 05:09 PM   #3
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I'm guessing around 60%.
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Old 11-27-2002, 05:31 PM   #4
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Depends on the density (temparture/salinity/force being added with) ratios of the water and how quickly one displaces/intermixes with the other.
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Old 11-27-2002, 05:35 PM   #5
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Oh Lordy

I should have know LOL Chris stop over thinking this.

all factors are the same -

Let's say we're adding/siphoning slowly with massive circulation.
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Old 11-27-2002, 05:37 PM   #6
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I'm with galleon.

I have read somewhere though about someone that made their new water warmer (I think) than their tank water temp so it stayed on top and pulled old water out from the bottom. The final result being the amount of water pumped in was very close to the amount of water changed.

Sounded interesting, but I'm not going to chance it.
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Old 11-27-2002, 06:13 PM   #7
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STP, eh Jerel?

Hmm. Well, in that case, I still don't think you can know without knowing how the tank size/dimensions are affecting your mixing. Also, with no stratification, and with any kind of forced (plus any kind of thermal kinetic energy the water already has) will instantly mix your pollutants. Ya know. You could always do 5 replicates of trying this, measuring how much the concentration of DIN changes. Then do statistical analysis (maybe just a Student's t-test) to see whether any standard deviation can be attributed to random experimental error, or you actually are getting variable results with this method. If it turns out any variance in the data can be stuck to random error, then you can do a mean and calculate your rate of change and thus determine the rate/ratio pollutants are exchanged.

...you asked.
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Old 11-27-2002, 06:30 PM   #8
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This is a MATH question.

I need a loan officer in a bank. This is no different than figuring out compound interest!
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Old 11-27-2002, 06:48 PM   #9
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...but it is. There are so many more variables than just doing a logarithmic equation. What would be interesting is if you did find the calculations, then do what I described, and see how close you came. Hmmmmm.... honestly, my malfunctioning intuition says 50%, especially with high mixing.

Besides, what is the mantra of science? When you don't know/can't remember how to calculate it, reinvent the wheel.
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Old 11-27-2002, 06:52 PM   #10
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LOL
Quote:
I'll look to see if I can find the equation if you really want it, its around here somewhere in old notes.
Please do, that's all I need - just a general idea. I'm too old for science! Besides, I'm a biologist - I don't believe in science.

Is it about/around/maybe/possibly/could be/more or less 60%?
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Old 11-27-2002, 07:02 PM   #11
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Jerel, see my edited post above.
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Old 11-27-2002, 07:10 PM   #12
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I dunno, I think it's somewhere between 60-75%.

I think if you had 75% of volume, you'd get closer to 50%
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Old 11-27-2002, 07:29 PM   #13
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...just something about diluting at the same rate as emptying with such high mixing. With the new water being mixed so fast, you could be removing just as much new water as old with the siphon, thats where the 50% comes from.
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Old 11-27-2002, 08:53 PM   #14
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Wink

"I'm too old for science! Besides, I'm a biologist - I don't believe in science."

Spanky = Scrooge

Right time of the year.
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Old 11-27-2002, 09:11 PM   #15
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Jerel, FWIW , and believe me its nothing, but given the scenario described = in and out with reasonble currents , blah, blah, blah my gut instinct tells me 50% is prolly best case scenario. Couldn't one measure a known level in the main that is absent in the replacement water , before and after and arrive at a usable answer?
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