Continuous Dilution of CYA - Math Help

[del]

I think that I saw an equation by chem geek, or someone else maybe, when I was looking at old posts. Now, I cannot find. I am looking for the equation for figuring out the concentration of a pool chemical, CYA in my case, as a function of the volume of pure water added to a pool. This might be called something like continuous dilution. I am assuming that the pool is always full and perfectly mixed. A rule of thumb I learned years ago was that for every 2.3 volumes of fluid added to a fixed volume of solution, the solute concentration would go down 90%. This helps some, but it not really what I need now. Of course, I can check the equation against that rule of thumb. In this scenario, a garden hose fills the pool, somehow the new water is 100% mixed instantly and perfectly with the pool water, and the excess just drains out at an equal that of the incoming garden hose water. The excess overflows and goes to drain. The level of the pool never changes. I know that there are some differential equations to figure this out, but there must be a simper equation that is derived from the differential equation. If someone even knows that name of the equation, that would be helpful, as I am sure that I can do a web search to find it. Myi old chemical engineering text books provide the information to set up the differential equation, but are useless from a practical perspective.

As a side note, I am not really going to spill out a lot of expensive water, but the equation will help me calculate a few things with respect to alternate means to reduce CYA in a pool. Thanks.

del

 

[Richard320]

Are you aware that there is a company in Escondido that performs Reverse Osmosis on swimming pools? It'll cost more than the water, but it strips out everything, CYA, Calcium, Salt... everything.

 

[JoyfulNoise]

I don't think the math question ever got answered. We all know that, when you want to reduce the concentration of some solute in the pool water, it is better to do larger exchanges of water (eg, 50% drain and refill) then to do smaller successive exchanges (eg., five 10% drains). But why?

Assume that:

C_i is the initial concentration​

C_i+1 is the concentration after the next incremental dilution​

V_r is the volume of water removed​

V_tot is the total pool volume​

Then, for the first dilution, you'd get -

C_i+1 = (1 - (V_r/V_tot)) x C_i

The second dilution would be -

C_i+2 = (1 - (V_r/V_tot)) x C_i+1

OR

C_i+2 = (1 - (V_r/V_tot)) x (1 - (V_r/V_tot)) x C_i

OR

C_i+2 = (1 - (V_r/V_tot))² x C_i

For the third dilution you'd get -

C_i+3 = (1 - (V_r/V_tot)) x C_i+2 = (1 - (V_r/V_tot))³ x C_i

So on and so forth, such that you can write the compact form as -

C_final = (1 - (V_r/V_tot))ⁿ x C_initial, where n is the number of successive dilutions.

As an example, say you started out with 100ppm CYA and you did five (5) 5% water removals. At the end of the five removals the CYA would be -

[CYA]_final = (1 - (0.05))⁵ x 100ppm = (0.7738) x 100ppm = 77ppm

If you had removed 25% of the water initially as opposed to five 5% removals, then your new [CYA] would be 75ppm, not 77ppm. For small numbers of dilutions, the reduction in concentration is approximately close enough to the total dilution but when you try to do a lot of little dilutions, the differences are more significant. The amount you remove also affects the magnitude of the reduction -

One 30% removal = 30% reduction

Three 10% removals = 27.1% reduction

Six 5% removals = 25.6% reduction

In all cases, you removed 30% of your water but the concentration change is larger for largest amount of water removed in the fewest cycles. So, in the end, the more water you can remove and replace in the fewer number of cycles, the better off you will be in lowering concentrations.

Hope that helps.

Matt

 

[Patrick_B]

Matt!!

That just gave me some horrific flashbacks of a grouchy old Math Teacher. :blah::blah: :jocolor:

 

[JoyfulNoise]

Matt!!

That just gave me some horrific flashbacks of a grouchy old Math Teacher. :blah::blah: :jocolor:

Yeah, if I had it to do over again, I would have studied math in college and grad school.

 

[del]

Thank you joyfulnoise. Yes, this is helpful. You pretty much found a "numerical methods" approximation to the calculus problem. I did find a write up of asimilar problem online since I posted, which is also helpful, but not as simple as yours. Your method is certainly good enough for what I need at this point.

Thank you, again.

del