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Thread: Explain why this equation works

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    Explain why this equation works

    Hi, I was looking for an equation (outside of the pool math calculator) for adding chlorine to my pool and found this:

    Initial chl solution * x = desired solution * pool volume

    I tried it out on my pool numbers and checked it against the pool math calculator and it appears to work (at least, I should say the results agree). I was trying to think how to explain this, but I can't really seem to articulate to myself WHY it works. Can someone help me out with this? I can't really figure out what the value is that you achieve when multiplying the desired solution by the water volume.

    Thanks (trying to improve my long abandoned math skills).

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    pabeader's Avatar
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    Re: Explain why this equation works

    What are the numbers you are using?
    Bob - Palm Beach by San Juan Pools. approx 5000 gals Pentair 320 cartridge filter (all new guts installed by me) Goldline SWG Intermec 104 TImer Test kit: TF-100 w/Speed Stir

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    Re: Explain why this equation works

    I don't usually go down this road, but I had to test myself. So here's what I got using "5" as initial chl (FC) and "8"as desired FC in a 33K pool:
    Initial chl solution * x = desired solution * pool volume .... so ...... we need to solve or find "X":

    5 * X = 8 * 33K ..... so ... first complete the right side ... 8 * 33K = 264,000
    Then .... 264,000 / 5 = 52,800 ... so "X" = 52,800

    To check my work ... 5 * 52,800 = 264,000

    Smell that? That's my brain on fire. Not sure how this would be helpful, except showing my kids how to solve for "X", so I'll stick to the Poolmath Calculator.
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    Re: Explain why this equation works

    Quote Originally Posted by pabeader View Post
    What are the numbers you are using?
    12.5% chl solution = 125,000ppm

    Assuming 0 chl in pool and wanting to attain 5ppm.

    Pool volume of 13,500 = 1,728,000oz

    So,

    125,000ppm * x = 5ppm * 1,728,000oz

    x = 69.12oz to add to pool

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    Re: Explain why this equation works

    It works because both sides are solving for the same thing - essentially the total amount of chlorine.

    On the left side is the total amount of chlorine in your bleach

    On the right is the total amount of chlorine in your pool

    It only works when the measurements are the same (ppm and oz). Honestly I think your own explanation is much more eloquent than mine.
    16K Gal Plaster | Compupool SWG | Intelliflow VF | TF-100

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    Teald024's Avatar
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    Re: Explain why this equation works

    Quote Originally Posted by Tammy S View Post
    Hi, I was looking for an equation (outside of the pool math calculator) for adding chlorine to my pool and found this:

    Initial chl solution * x = desired solution * pool volume

    I tried it out on my pool numbers and checked it against the pool math calculator and it appears to work (at least, I should say the results agree). I was trying to think how to explain this, but I can't really seem to articulate to myself WHY it works. Can someone help me out with this? I can't really figure out what the value is that you achieve when multiplying the desired solution by the water volume.
    This equation is basically finding the ratio of actual chlorine to volume of liquid. Since the units have to be the same on both sides, then x has to be a volume of liquid (chlorine solution). This formula basically will tell you the amount of bleach to add to increase the pool chlorine by a desired ppm (at reasonable ppm levels). It does not give you the final ppm of the pool water, unless you started with 0ppm of chlorine in the pool.


    my opinion is that it is a cumbersome way to figure it out ppm needed.

    if you simplify the equation it gets easier though:

    Reduced to basic terms:

    x (oz of bleach to add) = 0.0128* Pool volume (in gallons) * desired chlorine to add (in ppm)
    Initial bleach concentration (in %)

    work listed below(lol)


    My actual method uses this formula above to make a home made measure cup out of an old used juice container. I have a reliable supply of 10% bleach and I always use the same pool volume in my calculation. This allows me to make the personal measuring cup of sorts. I know that a 1ppm addition to pool chlorine takes 15.4 oz of 10% bleach in my 12,000 gallon pool.

    I just took the half gallon juice container (because it was clear) and poured in 15.4 oz of water, and marked a line. This represented the equivalent chlorine to add to increase my pool by 1ppm. I added another 15.4 oz of water and marked another line. I did this 4 times. I then went back and labeled each line with 10%, 1ppm / 2ppm... and actual volume each line represented.
    When it comes time to dose the pool, I just fill from the bleach bottle into my personal jug to the appropriate line for the ppm I need. No calculations and easy to measure.

    *Changing the bleach concentration changes this method (5.25%, 8%, 10%, 12.5%)
    *You can use any size jug you have
    *Obviously make sure you label the container so you don't use it for other uses. Rinse well after each use.

    similar to this:
    http://www.troublefreepool.com/threa...l=1#post535690

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~

    work shown below (lol):

    Initial chl solution * x = desired solution * pool volume

    x (gallons of bleach to add) = Pool volume (in gallons) * desired chlorine to add (in ppm)
    ---------------------------------------------Initial bleach concentration (in ppm)


    since it is much easier to measure in oz than gallons:
    1 gallon = 128 lq oz.

    x (oz of bleach to add) = 128* Pool volume (in gallons) * desired chlorine to add (in ppm)
    ---------------------------------------------Initial bleach concentration (in ppm)


    since bleach is labeled in % Sod Hypo and not ppm Sod Hypo:
    1 % = 10,000 ppm

    x (oz of bleach to add) = 128* Pool volume (in gallons) * desired chlorine to add (in ppm)
    ---------------------------------------------10,000 * Initial bleach concentration (in %)
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