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.
Thank you houoski & Ian H for answering . I have all but forgotten any math skills and so appreciate your efforts.
- husoskiLv 72 months agoFavorite 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 HLv 72 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%