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Lv 6
? asked in Science & MathematicsMathematics · 2 months ago

How to achieve 90% new transmission fluid ?

My vehicle holds a total of 11.2 quarts of automatic transmission fluid.

Removing the transmission pan and changing the filter, results in removing 5 quarts of old fluid, and adding 5 quarts of new fluid. Which was done.

So now there's about 45% new fluid in the trans

Without changing the filter, 3.5 quarts of fluid are exchanged.

So my question; with taking the pan off and replacing 3.5 quarts at a time, how many times does this take to achieve 90% new fluid in the transmission ?

The vehicle will be driven a bit between each fluid change, to mix-up the new & old fluid.

Update:

Thank you houoski & Ian H for answering . I have all but forgotten any math skills and so appreciate your efforts.

2 Answers

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  • 2 months ago
    Favorite Answer

    It's easier to look at the fraction of old fluid, since each step will multiply that by a factor of (6.2 / 11.2) on a filter change, and (7.7 / 11.2) on a fluid.  Those fractions are:

       6.2/11.2  ~~ 0.55357

       7.7/11.2  =   0.6875

    So, after the initial filter change (all fluild was old before that), you have an old-fluid concentration of 0.55357, or about 55.4%

    A regular 3.5 qt. change (after mixing) will leave a concentration of about:

        0.55356 * 0.6875  ~~ 0.3805

    About 38%.  Another regular change gets to:

       0.3805 * 0.06875  ~~ 0.2616

    You can keep that up, with a calculator (putting 0.6875 in a memory, I hope, so you don't have to retype it) or use a spreadsheel program like Excel if you're familiar with using formulas.  The high school algebra 2 approach would be to use logarithms.

    After 1 filter change and n regular changes, the concentration will be:

        x = (0.55356) * (0.6875)^n

    You want that final concentration to be x = 0.1 (10% old oil means 90% new oil).  Set x to 0.1 and take logarithms:

        log 0.1 = (log 0.55356) + n*(log 0.6875)

        n = [(log 0.1) - (log 0.55356)] / (log 0.6875)

    ...and a calculator says that's about 4.57 changes.  You'll need the filter change plus 5 regular changes to reduce the old-fluid level to 10% or less.  The last change leaves about 8.5% of the fluid that was in the transmission before the filter change.

    PS: I've used * as the "multiplication dot" symbol and x^n to mean "x raised to the nth power".

  • Ian H
    Lv 7
    2 months ago

    Note: Bad plan because each removal wastes some of the new fluid.

    Better to drain as much as possible, (probably leaving 10%) then refiil.

    However, following your instructions ........

    After first stage, transmission contained 5 new in total of 11.2, (about 45%)

    2] Removing 3.5 means resulting volume is (11/16) of 11.2, so then

    the transmission contained 55/16 of new in a volume of 7.7.

    Replacing 3.5 gives 111/16 new in volume of 11.2,

    3] Removing 3.5 results in volume of new being 111/16*11/16 = 1221/256

    Replacing 3.5 gives 2117/256 of new in volume of 11.2, (about 74%)

    4] Removing 3.5 gives volume of new as 2117/256*11/16 = 23287/4096

    Replacing 3.5 gives 37623/4096 of new in volume of 11.2, (about 82%)

    5] Removing 3.5 gives volume of new as 37623/4096*11/16 = 413853/65536

    Replacing 3.5 gives 9.8149 of new in volume of 11.2, (about 87.6%)

    You need 6 stages of removal + replace of 3.5 to exceed 90%

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