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Discussion Starter #1
Your M5 is running a bit hot, or your doing your twice yearly check, so you test the antifreeze in the radiator and find it is not at the prefered 50% mixture, rather its more like 40%,

Here is what to do.

Your coolant capacity is 10 gallons (just an example). That means you need 5 gallons pure water and 5 gallons antifreeze for a 50% solution.

The tank actually has 10 gal x .4 (40% solution) = 4 gal of antifreeze and 6 gallons of pure water, not quite where you want it to be.

Now comes the fun part. How much do you need to drain from the radiator and replace with pure antifreeze to get you to your 50% prefered mix?

Well you have 4 gallons of antifreeze in your tank and you need to have 5 gallons. Let y = the amount of pure antifreeze you need to add, so.....

4 + y = 5

but for every gallon of pure antifreeze you add to the tank you also must remove an equal quantity of that 40% antifreeze, therefore....

4 - .4y + y = 5 , now solve for y,
4 + ( y - .4y) = 5
4 + .6y = 5
.6y = 1
y = 1.6666

You can easily check your results by replacing y with your calculated value for it.

4 - .4(1.666) + 1.666 = 5
4 - .6664 + 1.666 = 5
3.3336 + 1.666 = 5
4.9996 = 5 , plenty close for antifreeze

So you have to remove 1.666 gallons of the 40% antifreeze solution thats in the radiator, and repace it with an equal amount (1.666 gallons) of pure antifreeze to get to that 50% solution.

Why not just draw off 1 gallon of the 40% solution and add 1 gallon of pure antifreeze? Because you still have to replace the antifreeze that was in that 40% solution you drew off for you to achieve the 50% required.

I know, old hat for most of you, but it just might be new info for some.
 

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I like that formula, that's pretty cool...good info :cheers:


Ed
 
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