Determining The Water Volume Of Your Whole Reef System

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

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  1. ReefSparky

    ReefSparky Super Moderator Staff Member

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    Makes sense, yes. I tried your equasion with Matt's scenario, and I'm getting 7.14 gallons total volume, which wouldn't be accurate for a 34 or so gallon tank and a sump too. Could you work that out here?
     
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  3. Eddie

    Eddie Flamingo Tongue

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    Nope, no sir not a good result at all. I am assuming that you kept compatible units throughout the calculations, thus avoiding any NASA type errors.

    Another variable here I suppose would be turn-over rate times time. I am not sure at what point R x t= enough turnovers. I guess the sump and main tank should BOTH be measured/tested and they must give the same value for [x] to be sure of homogeneity.

    Another variable might be the element itself you are testing for. If it is being consumed at all before you can accurately test it, that would give a wrong answer. But it would be too high, not too low...hmmm
    If your units and calculations are good, and the dilution is complete and homogeneous, I can't explain your answer. :confused:
     
    Last edited: Sep 2, 2010
  4. Reeron

    Reeron Blue Ringed Angel

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    That's what I attempted to do with my formula.

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

    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.

    Mat seems to think that 30 gallons is too little. But unless I missed something (which nobody has mentioned as of yet), the answer IS 30 gallons.
     
  5. ReefSparky

    ReefSparky Super Moderator Staff Member

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    Hmm.. Wish I could say yes. Your examples are a bit esoteric. If you could take Matt's numbers from his "flub" post, and write out the equasion, I'd be indebted.

    Thanks. :)
     
  6. Eddie

    Eddie Flamingo Tongue

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    Esoteric? Ah, I see my error. I was fixed on inferring volume by the dilution of a known high elemental concentration from a small volume, into a much larger volume of lesser concentration. In your work here, you are dealing with a property, specific gravity, of water which is caused by the dilution of many "salts" of elements in water.
    Also, I see that we are not diluting our "known here", but rather increasing its concentration. Adding a 24.6 ppt solution into a 40.2 ppt solution is not a dilution, but an increase in the control concentration. The only thing being "diluted" here is the unknown volume from 1.030 to 1.0x. I think I went a stray here? Should have paid more attention to the opening thread I guess, sorry. This might explain why the dilution equation falters here, there is more going on than a simple dilution. In this case C2 > C1.
     
    Last edited: Sep 3, 2010
  7. Seano Hermano

    Seano Hermano Giant Squid

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    Eddie, I wish I could understand your & Reefsparky's conversation. But your use of big words & complicated mathematical problems is just too much for me.
     
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  9. Eddie

    Eddie Flamingo Tongue

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    What I was saying was that in a simple dilution, the equation C1*V1=C2*V2 can be very useful for finding V2, total volume when rearranged as (C1*V1)/C2 = V2. I would add an element like Sr, Ca, Mo, Mn, or just about anything safe that can be accurately tested for, at a high concentration and then calculate the decrease in concentration, giving a solid number for V2. This is known as the "dilution equation", and C2 is always less that C1. if C2 is greater than C1, this approach is the incorrect one. These folks were calculating volume using salinity by adding large volumes to a larger volume with Cx increasing. It doesn't work in this situation. A larger C2 will of course lead to a wrong conclusion, too small V2 and meaningless.
     
  10. ReefSparky

    ReefSparky Super Moderator Staff Member

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    Unless I'm missing something, your statement that C2 > C1 is false. You're saying the salinity of the system is higher after addition of lower salinity water. That's not the case. In Matt's example, the SG of his tank was 1.030. Then he added 10 gallons of water with an SG of 1.018. In the end, the SG of his tank dropped to 1.026.

    This would still be a dilution equasion. You are diluting the system's salt water by adding salt water with a lower SG. The final result is a system of lower salinty.

    I'm asking you to plug Matt's numbers into the equasion.
     
  11. Eddie

    Eddie Flamingo Tongue

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    Not at all Sparky. What I am saying is that what I call the "known" is in hand and what I am calling C1*V1, and am solving for C2. It is not the "system" concentration or dilution that concerns me in the dilution equation(DE), but rather that the system's effect on the known. since C1*V1 are known and C2 will be measured as affected by V2. The system is being diluted rather than the "known" source.


    This probably still isn't making any sense is it? lemme try to rephrase this somehow...Why the conditions are not right for the DE using yours or Matts numbers.

    I know it seems strange that it should make a difference, but it does. From my point of view, C1, the initial concentration, is less than C2, the final system concentration. I used this equation daily as a chemist at an electroplating shop where the metal concentrations of things like nickel, copper, gold, silver, etc..in the "system" were measured. Adds were made based on c1*v1=c2*V2. In this case, I would solve for V1, since V2, C2 and C1 were known, and C2< C1 ( the system was 400 gallons and at say 80ppm, I had some 10,000 ppm nickel liquid. The system needed to be at 90ppm. So I would tell the electro-plater to add (10,000)*X= 400 * 90 => X= (400*90)/10,000 = 3.6 gallons of say, NiCL2 (nickel chloride) to the "system, the change in volume being negligible (the NiCl2 was 10,000ppm Ni).
     
  12. blackraven1425

    blackraven1425 Giant Squid

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    Now, for a better formula for finding SG, when the salinity of the added water is a factor (when Sparky's equation fails, due to lack of an added salinity variable), just use this and do the real math, or toss it into your TI-89 and tell it to solve for you:

    (salttank x volumetank) + (saltadded x volumeadded) = (saltfinal x volumetotal)

    In Matt's case:

    volumetotal = volumetank + volumeadded, just as a clarification.

    1.030 * Y + (1.018 x 10) = 1.026 * (10 + Y)

    1.030 * Y + 10.18 = 1.026 * (10 + Y)

    1.03Y + 10.18 = 10.26 + 1.026Y

    1.03Y = .08 + 1.026Y

    At this point, I threw it into my trusty TI-89 (which I could, and would, have done to begin with, but I needed to show steps for this post), and got Y=20. Matt, seemingly, had 20 gallons in the tank before he added his water.

    Now, I realize this doesn't sound right. You're probably thinking, "He has more than 20 gallons in his system, for sure!".

    BUT, what happens is that Y, or tank volume, is only a portion of system volume. You have to add the added volume, as I showed above:

    volumetotal = volumetank + volumeadded

    That means the total volume, in Matt's case, is 30 gallons. It may not seem like it's enough, but that also assumes the 2 water masses have had sufficient time to thoroughly mix, and the tank is at a perfectly even distribution of salt. It also assumes that 10 gallons is an accurate figure for the amount of water added. Still yet, it assumes SG and PPT are converted via a linear equation; if not, you need to use PPT. The last thing it assumes is that I know how to do basic math. I'm much better at creating equations than solving them, thanks to my TI-89's influence.
     
    Last edited: Sep 5, 2010