My Math Forum Engine coolant ratio

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 October 2nd, 2018, 12:59 AM #1 Newbie   Joined: Oct 2018 From: USA Posts: 2 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, 08:47 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,974 Thanks: 1850 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. Thanks from Gearhead
October 2nd, 2018, 09:35 AM   #3
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Quote:
 Originally Posted by Gearhead 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?
Empty the whole 10.95L into a larger container.
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

October 3rd, 2018, 01:33 AM   #4
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Quote:
 Originally Posted by skipjack 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.
Thank you skipjack, that is the answer I was looking for. About 1.25L should do the trick. I had actually added 6 cups of water and converted that to the 1.44L. I appreciate the explanation, and was just interested in the mathematical solution as I strained my feeble mind. It's pretty awesome of you and the other mathletes here to help out strangers with their math problems. Thanks again and all the best.

Quote:
 Originally Posted by Denis Empty the whole 10.95L into a larger container. 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
Nice try Denis, but I wasn't looking for a practical solution, I was looking for a mathematical one, eh? Your solutions are like square roots. Mostly irrational. Keep yer stick on the ice, bud.

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