coolant ratios...

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Okay, here is the issue:

I have a solution of 50/50 antifreeze & water. I also have a solution of 100% antifreeze.

I need to know how much coolant to add to the 50/50 mixture to change it to 60 antifreeze / 40 water.

Given 2 cups of the 50/50 mixture I figure will need to add 1/2 cup of antifreeze to change it to a 60/40 ratio.

Anyone see anything wrong with my thinking? ...it would not surprise me if there is!
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You are correct.

Originally Posted By: boxcartommie22
its the water that does the cooling not anti freeze..i use 60%water and 40% g-05 coolant in all my fords and lincolns

Both mixtures will remove the same amount of heat.
 
Ok, great! Thank you for the confirmation.

The reason I am interested in a higher concentration of antifreeze is that I am often in an area of country where it has the potential to be and very often is quite cold during the winter. Check the map below:

http://preview.tinyurl.com/5edlf6
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The summers? Well, 90 degrees Fahrenheit is considered a major heat wave.
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A = 50% antifreeze
B = 100% antifreeze

You want x amount of A + y amount of B = 60% antifreeze.

xA + yB = 0.6
x(50%) + y(100%) = 0.6
0.5x + y = 0.6

x+y has to add up to 100%, so x+y=1.

x+y=1
x=1-y

Plug that in up top...

0.5(1-y) + y = 0.6
0.5 - 0.5y + y = 0.6
-0.5y + y = 0.1
0.5y = 0.1
y=0.2

So you want 20% B and 80% A. 2 cups A and 1/2 cup B will do just what you want.

Who says algebra is useless in daily life?
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Fantastic... there must be other algebraic formulas that can be applied to coolant/antifreeze.

I wasn't paying attention during Algebra so I ended up in accounting.
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