Determining The Water Volume Of Your Whole Reef System

Discussion in 'Water Chemistry' started by ReefSparky, Aug 12, 2008.

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  1. WhiskyTango

    WhiskyTango Eyelash Blennie

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    Why didn't I think of that.. Strong Work.;)
     
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  3. Matt Rogers

    Matt Rogers Kingfish Staff Member

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    I think it has to be more than that. It is a 36 gallon tank, a sump with about 15 gallons in it (I pulled 10 out of it yesterday) and a refugium that is about 4 gallons.

    I can't use his updated equation:
    http://www.3reef.com/forums/water-c...our-whole-reef-system-51209-3.html#post919063

    ... because I didn't add ro/di water, I added diluted saltwater. I was just curious if you could still pull volume from the numbers I posted last post.
     
  4. Reeron

    Reeron Blue Ringed Angel

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    Here's what I used for a formula:

    x number of gallons tank volume at 40.2ppt + 10 gallons at 24.6ppt =
    New salinity of tank at 35ppt.

    So: 40.2x + 246 = 35(x +10 gallons) where x is the unknown quantity.

    Solving for x I got 20 gallons. 20 +10 = 30 gallons total.

    Maybe my formula is wrong?

    Using the formula posted by ReekSparky (as it sits) can't be done as you took out 10 gallons of initial water volume. His formula is based on adding to the initial volume and getting a new volume (old volume + additional volume = new volume).

    The only way to solve using ReefSparky's posted formula is to add 1 gallon of RO/DI water to the current tank and test the new SG. Plugged into his formula, you will get the initial water volume in your tank.
     
  5. ReefSparky

    ReefSparky Super Moderator Staff Member

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    I attempted it Matt, but can't make the equasion work when the added water isn't of 0 SG. I'll dig around and see if I can get the mother equasion from what I posted above. It shouldn't be too difficult, but I need to first find an equasion that accounts for any SG added to the initial SG.

    I performed the test on the small volume of water to confirm the equasion was right, and it worked. Of course, as the system volume increases, the accuracy of this method will decline; given a small degree of dilution is more difficult to discern than a drastic one. Suffice it to say that if your system is more than 150 gallons, the more RO/DI water one adds to dilute, the more accurate the outcome.
     
  6. Eddie

    Eddie Flamingo Tongue

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    The dilution of solution equation could work too :C1V1=C2V2 | C1C2 initial and final concentrations of "something you test for. | V1V2 are initial and final volumes.

    If you have a gallon or 5 gallon jug at a know concentration of "x", that would be C1V1 in the dilution equation. Then measure C2 (new concentration of "x") after adding to tank.
    Then your final volume, total volume, would be given by V2(vTOT)=C1V1/C2. Just make sure "x" isn't copper.
     
  7. Eddie

    Eddie Flamingo Tongue

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    No? Is that too easy?
     
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  9. Eddie

    Eddie Flamingo Tongue

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    One last thought:
    If the "x" you choose to measure is already present in your tank, then C2 can be replaced by dx/dV, or the change in concentration of x. given by [x]f-[x]i or (Conc. "x" final -Conc. "x" initial), or delta-[x], as the C2 divisor. Right?
     
  10. ReefSparky

    ReefSparky Super Moderator Staff Member

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    Can't say I follow, Eddie. Wanna run that by me once more? :)
     
  11. Eddie

    Eddie Flamingo Tongue

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    When diluting one liquid into another, it is true that the initial concentration times the initial volume is equal to the final concentration times the final volume: C1 x V1 = C2 x V2 .
    So it must also be true that V2 = (V1 x C1)/C2.
    ex; 5 ml x 10,000 ppm = x ml times 10 ppm. x ml = (5ml x 10,000ppm)/10ppm or x ml = 5,000 ml or 5 Liters. Had C2 only changed by 1 ppm, the answer would have been 50,000 ml or 50 liters. Right?
     
  12. Eddie

    Eddie Flamingo Tongue

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    (C1 x V1) is in your hand, and of known volume and concentration.
    C2 will be measured, and V2 is the only unknown.

    To be clear let's use Copper because C2=C2, (not C_final-C_initial) since there is no Cu initially.

    So, we add one gallon of 1000 ppm Cu to the tank.
    We then wait and measure the resulting [Cu] in the big tank (V2).
    Keeping all units the same, the equation would yield:

    1gal x 1000ppm Cu = V2gal x measured [Cu] in ppm.

    then, in gallons, V2 is equal to (1gal x 1000ppm)/measured [Cu].

    If the new copper reads say, 25ppm. That would mean your final volume, V2, would be (1000ppm x 1gal)/25ppm = 40 gal.

    a reading of 5ppm would mean (1000ppm x 1gal) / 5ppm = 200 gal

    If it were Strontium, since it is already present, C2 wou;ld be replace by the final-initial [Strontium], or "change in Sr concentration". Then it would read (1000ppm Sr x 1 gal)/ ([Sr]f-[Sr]i)= V2

    And so forth. Does this make sense?