It Turkey Day and my mind is blank

asterof · Nov 27, 2014

problem

I have a solution of 25% sodium and 75% H20
During transfer I want to reduce the 25% to a ratio of
10% sodium and 90% water

If the total transfer amount is 80 gallons the
the original targets would be 20 gallons of sodium and
60 gallons of water based on the 25% solution.

So here is were I am lost
I would need to increase the water amount by ?? to maintain the
80 gallons at 10%.

For some reason I just am not getting it

jrwb4gbm · Nov 27, 2014

You would use 32 gallons of the original 25% solution and add 48 gallons of water to get your total batch of 80 gallons at 10%.

cjd1965 · Nov 27, 2014

Hi
So you have a mixture of 25% Sodium and 75% water in a tank/vessel and you want to pump some of that mix plus more water into a new tank/vessel to end up with 80 Gallons at 10% Sodium and 90% water?

Is that the question?

cjd1965 · Nov 27, 2014

aI think it looks like this at first glance

32 gallons of mix plus top the rest with water to 80 gallons

Based on 0.25 of a gallon sodium in mix 1. so 4 gallons gives 1 gallon of sodium

10% of total is sodium so 8 gallons so 8x4 is 32 gallons required plus the top up of water

asterof · Nov 27, 2014

Yea but

cjd1965 said:
aI think it looks like this at first glance

32 gallons of mix plus top the rest with water to 80 gallons

Based on 0.25 of a gallon sodium in mix 1. so 4 gallons gives 1 gallon of sodium

10% of total is sodium so 8 gallons so 8x4 is 32 gallons required plus the top up of water

Again

If the total transfer amount is 80 gallons the
the original targets would be 20 gallons of sodium and
60 gallons of water based on the 25% solution.

But we want to reduce this to 10%
So if we stay with 80 gallons total at 10% then the ratio
needs to be 25/10 or 2.5
So the 20 gallons of Sodium becomes 8 gallons of sodium
and 80 gallons of water metered at a 10% ratio
I think

so at 1 gallon of sodium we would have 10 gallons of water

1 > 10
2 > 20
3 > 30
4 > 40
5 > 50
6 > 60
7 > 70
8 > 80

I will toss some code and post it see if it works

asterof · Nov 27, 2014

code

Here is the code I cam up with
any way to do it better, cleaner is welcome

allscott · Nov 27, 2014

That must be a big turkey you are brining.

asterof · Nov 27, 2014

nah

allscott said:
That must be a big turkey you are brining.

Working on roof, cooking turkey and trying to solve a problem
that keep me awake last-night

jrwb4gbm · Nov 27, 2014

After re-thinking my first answer and reading some of the other posts, I believe I misunderstood the end point target. 29.09 gallons of the 25% solution would contain 7.2725 gallons of sodium and 21.8175 gallons of water. If you add 50.91 gallons of water you would now have a solution with 72.7275 gallons of water and 7.2725 gallons of sodium. That brings the total amount of solution to 80 gallons at 10% strength.

Tom Jenkins · Nov 28, 2014

Just to throw a monkey wrench into the works, are you actually concerned with concentration by volume or concentration by weight? Different answers are required for each.

danw · Nov 28, 2014

I hope it's really a salt, sodium chloride, not elemental sodium, a metal.

Elemental sodium reacts violently with the hydronium ions in tap water releasing enough heat to self ignite the hydrogen created.

The graphic shows the typical chemistry class experiment when someone drops a bit of sodium into a beaker of water:

asterof · Nov 28, 2014

concentration by weight

by weight but needs introduced in a ratio and diluted format

rootboy · Nov 29, 2014

danw said:
I hope it's really a salt, sodium chloride, not elemental sodium, a metal.

I was thinking the same thing.

It Turkey Day and my mind is blank

asterof

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