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October 1st, 2018, 11:59 PM  #1 
Newbie Joined: Oct 2018 From: USA Posts: 1 Thanks: 0  Engine coolant ratio
My engine's cooling system has a capacity of 10.95L of coolant which is a 50:50 mix of antifreeze + water. My radiator was low so I added 1.44L of water, to bring it back up to 10.95L of antifreeze + water; the resulting ratio is now approximately 43:57 antifreeze to water in my radiator. How much (in liters) of this 43:57 mix will I have to remove from the radiator, so that I can replace it with same amount of 100% antifreeze, to attain a ratio of 50:50 again? Many thanks. 
October 2nd, 2018, 07:47 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,713 Thanks: 1806 
You need to avoid the 43:57 approximation to get an accurate result. Prior to adding 1.44L of water, the radiator held 4.755L of antifreeze and 4.755L of water, so the result was 4.755L of antifreeze and 6.195L of water. To restore 50:50, you need to reduce the water content to 5.475L by removing 0.72L of water, which implies removing 0.72/6.195 of the 10.95L of mixture, i.e. (2628/2065)L = 1.2726392...L of mixture, and then add that amount of antifreeze only. It's unlikely you could remove exactly the right amount, but as I don't know how accurate your figures are or how accurately you can measure volume, I've not rounded the answer. If, when the water level in the radiator was low, the mixture wasn't exactly 50:50, you are stuck with needing more information in order to get an accurate answer. 
October 2nd, 2018, 08:35 AM  #3  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,302 Thanks: 935  Quote:
Add 1.44L of antifreeze to the large container. Pour back into your 10.95L cooling system. You'll be wasting very little antifreeze, but no calculating headaches  

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coolant, engine, ratio 
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