Need help with some basic math, and I'm ashamed...

mr.Green · May 3, 2016

Hi guys!

I’m doing an experiment at school and I’m having difficulty calculating my saltwater uses.

I’m ramping two different aquariums; one from 0 to 16 psu and the other from 16 to 0 psu. Each step I do is a change of 2 psu per day. So I will go from 8 to 10 psu for example.

I have freshwater (0 psu) and 16 psu saltwater to work with. No extra salt, just the water at these salinities.

Aquariums are 100 L each.

Large water changes can be a bit unpractical in our lab, so even though I in theory could change 100 L to a newly mixed salinity, at most 50 Liters per time is realistic.

How on earth do I calculate how much water to change to get to the new salinity? Math is not my strongest subject (though I’m taking biology as a major).

There has got to be some helpful experts here. If you have the time, I’d be very grateful if I could get a better understanding of this.

Cheers!
/ Leon

JimWelsh · May 3, 2016

This one is actually a nice, fun mathematical progression, if you set it up correctly.

Aquarium #1:

Assume each step is a 50 liter water change. Let's make each water change a mixture of your two water sources, 0 psu and 16 psu, such that we use x parts of 16 psu and y parts of 0 psu for a total of z parts, where x, y, and z are all integers, and z always = 16. For example, the first step below uses x = 4 parts 16 psu, and y = 12 parts 0 psu, for a total of z = x + y = 16 parts.

By making the total number of parts equal to the 16 psu of that water source, we make the salinity of each mixture work out to be x/16 times 16 psu = x psu (the two 16s cancel each other out). For example, if x = 4 parts, then the resulting mix is 4/16 made up of 16 psu water, so it is 16 psu * 4 parts / 16 parts = 4 psu water.

Since we are doing 50% water changes, and we want to go up by 2 psu each time, then we need the new water to always be 4 psu higher than what our current salininty is.

Step 1: Go from 0 to 2 psu
We are at 0, so x needs to be 0 + 4 = 4.
50 L 0 psu water + 50 L of (4 parts 16 psu water + (16-4) parts 0 psu water) = 0.5*0+0.5*16*4/16 = 0.5*0 + 0.5*4 = 2 psu.

Step 2: Go from 2 to 4 psu
We are at 2, so x needs to be 2 + 4 = 6.
50 L 2 psu water + 50 L of (6 parts 16 psu water + (16-6) parts 0 psu water) = 0.5*2+0.5*16*6/16 = 0.5*2 + 0.5*6 = 4 psu.

... and so on, where x goes up by 2 each time. The pattern fails at the last step, though, because you can't get from 14 psu water to 16 psu water using 16 psu water in a 50% water change; you end up with 15 psu water. In order to actually get to 16 psu in the last step, you would need to use 18 psu water in a 50% water change.

Aquarium #2:

The exact reverse. For each 50% water change, you would use new water that is 4 psu *lower* than the current salinity. So, starting at 16 psu, Step 1 would use x = 16 - 4 = 12 parts 16 psu water with 4 parts 0 psu water. Just like Aquarium #1, where x goes up by 2 each time, with Aquarium #2, x goes down by 2 each time. And, just like Aquarium #1, the progression fails at the last step, because you can't get from 2 psu to 0 psu using 0 psu water; you end up at 1 psu. Additionally, in this case, there is no such thing as -2 psu water, so there is no good solution for the last step for this aquarium.

KingBlingTX · May 4, 2016

I'd approach this problem a different way based on the description you gave. I've not worked with psu units, so I'm just assuming they would have direct (linear) correlation to an actual concentration unit (e.g. ppt, %, etc..) The equation to use is basic mass balance. The mass we are tracking in this problem is "salt".

Conceptually, the mass balance equation looks like this;
[What I Have Left] = [What I Started With] - [What I Took Out] + [What I Added]

For solutions, the "mass" is express by concentration multiplied by volume (e.g. C1V1). This mean you would write this equation as follows;

CfVf = CoVo - CoutVout + CinVin;
where Cf = final concentration after the water change, Vf= final volume after change, Co= starting concentration, Vo= starting volume, Cout= concentration of water you take out, Vout= volume taken out, Cin= concentration of replacment water, Vin= volume of replacement water added.

Since you are dealing with a fixed volume tank (Vf=V0= 100L) that simplifies things considerably because the volume of what you take out has to be the same as your replacement solution (either 0 psu or 16 psu). Therefore, Vout=Vin. (Let's call this Vx since it is the unknown you will solve for.) It's also obvious that Cout= Co since it is the water in the tank to be removed. Here is our more simplified version of the equation now:
Cf(100)=Co(100)-CoVx + CinVx; solving for Vx we get: Vx= 100(Cf-Co)/(Cin-Co). Now to apply this to your two tanks.

Tank 1-
Going from 0 to 16 psu in steps of 2 psu. That means Cf=(C0+2) and we would obviously be using the 16 psu replacement water (Cin=16). Plugging those into our equation yields;
Vx=100(Co+2-Co)/(16-Co) or
Vx=200/(16-C0).
So at whatever concentration you are currently, plug in that number for Co and that will tell you how much water to drain from the tank and how much of the 16 psu solution to add. For example, at the start your Co=0. Plugging that into the simplified equation, you get Vx=200/(16-0)= 12.5L. The only caveat to this is your comment about 50 L being the most practical limit to a water change. When you are at 14 psu, you will need to do a 100% water change. Simple enough, drain the tank and add two 50L additions.

Tank 2-
Going from 16 to 0 psu. That means Cf=(C0-2) and we would obviously be using the 0 psu replacement water (Cin=16). Plugging those into our equation yields;
Vx=100(Co-2-Co)/(0-Co) or
Vx=200/Co
In a similar fashion as above, at that start C0=16 so you water change volume =200/16=12.5L

I hope that helps. If something is confusing, or misunderstood your question please let me know.

mr.Green · May 4, 2016

You, gentlemen, are heroes. Thank you both very much for helping me with this ordeal. Jim - you were most accommodating to my needs, which I really appreciate, but it also got me to understand that I was over complicating things. I understood Kevins take on it better, and therefore went with his suggestion, even though it wasn't what I had in mind to begin with.

Thank you both for taking your time to help me, I truly appreciate it.

First water change done today, and it worked wonders.

Best Regards,
Leon