Simple but difficult formula

  • Thread starter Thread starter Richard
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Richard

=(K3-(K3/B3))/3.79
K3 = Tank Size
B3 = Actual g/l concentration
3.79 converting liter to gals
This formula works fine as long as the tank size is 100 and the g/l is
1
But when I use a different size tank and the g/l is other than 1, it
will not calculate correct. Why?
It's easy to calculate how many grams to ADD if it's below my
recommended number of grams per liter.
But if the g/l is over the recommended then I need to remove x amount
of water from tank to return it to the recommended g/l.
Thanks in advance!!!
 
Hi Richard

Richard said:
=(K3-(K3/B3))/3.79
K3 = Tank Size
B3 = Actual g/l concentration
3.79 converting liter to gals
This formula works fine as long as the tank size is 100 and the g/l is
1
But when I use a different size tank and the g/l is other than 1, it
will not calculate correct. Why?
Click to expand...

I think there may be an error on your formula. If B3=1 then it doesn't
matter what K3 is as the answer will always be 0.
It's easy to calculate how many grams to ADD if it's below my
recommended number of grams per liter.
But if the g/l is over the recommended then I need to remove x amount
of water from tank to return it to the recommended g/l.
Click to expand...

If the concentration is too high, then surely you need to ADD water to
dilute the solution?

I'm not sure what you're trying to achive here.
 
This has really been going on far too long. Consider a few more
examples along the same lines:

if you have a concentration of 3 gm/l and want to achieve 1 gm/l with
a cpacity of 100 litres, then you will need to remove 2/3 of the
mixture and add the same amount of water back;

if you have a concentration of 4 gm/l and want to achieve 1 gm/l you
will need to remove 3/4 of the mixture and add 3/4 of the capacity
back as plain water;

if you have 5 gm/l and want to achieve 1 gm/l you will need to remove
4/5 of the mixture and add 4/5 of the capacity back as pure water.

So, can you see how the numbers are related?

Then think of it as having an initial concentration of x, wanting to
achieve a concentration of y with a capacity of c, and see if you can
derive a formula (algebraic) that fits the examples, and then you can
convert the x, y and c to cell references that hold those variables.

Hope this helps.

Pete
 
This has really been going on far too long. Consider a few more
examples along the same lines:

if you have a concentration of 3 gm/l and want to achieve 1 gm/l with
a cpacity of 100 litres, then you will need to remove 2/3 of the
mixture and add the same amount of water back;

if you have a concentration of 4 gm/l and want to achieve 1 gm/l you
will need to remove 3/4 of the mixture and add 3/4 of the capacity
back as plain water;

if you have 5 gm/l and want to achieve 1 gm/l you will need to remove
4/5 of the mixture and add 4/5 of the capacity back as pure water.

So, can you see how the numbers are related?

Then think of it as having an initial concentration of x, wanting to
achieve a concentration of y with a capacity of c, and see if you can
derive a formula (algebraic) that fits the examples, and then you can
convert the x, y and c to cell references that hold those variables.

Hope this helps.

Pete




- Show quoted text -
Click to expand...

I think I got it!
=IF(E3="","",IF(ISERROR('Tank Sizes'!$F$22*E3/C3/2/3.78),"",('Tank
Sizes'!$F$22*E3/C3/2/3.78)))
Tank Size (Liters) x E3(Decrease g/l) / C3(Recommended g/l) / 2 / 3.78
Thanks for your help!!
 
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