One for al of the math whizzes

[LEROY]

So, I just purchased a refractometer and my sg is not 1.024 as it read on my hydrometer but actually 1.028 Now I am wondering what equation I can use to bring the sg to where I want it 1.024. Is it safe to bring down the sg more than .002 per waterchange? So if I do a 20% waterchange what should the sg of the refill water be if I want to lower the sg to say 1.026 or 1.024? I used to be good at math but I would like an experienced reefer's input on this one. I have a 10 gallon with 2 false percs in it right now.

-Thanks in advance, Steve.

[Jacobnano]

Did you calibrate the refractometer?

[LEROY]

Yes sir I did, as soon as I got it.

[circusordie16]

i say just do water changes with water tahts at about 1.023. if you do a 20% water change that will bring it to about 1.027. (1.028- [(1.028-1.023)*.2]) do that a few times and you should be ok. unless youre experiencing problems with the high salinity then do it a little faster. but id think 20% water changes at 1.023 until youre done would be ok.

[LEROY]

Cool, thanks for the replies, I'll give that a go and get the sg on it's way down.

[nematoad]

Why not just top off with RO till you get to 1.024 and then do a WC as normal?

[LEROY]

I was also considering that option, I'll figure it out.

[MitchReef]

Actually the concentration for a 20% water change to drop you to 1.026 needs to be a 20% change at 1.018.

 

Here is the math:

 

1.026=(1.028 * 0.8) + (X * 0.2) this takes into account the concentrations spread over the percentages of total water present. Therefore:

 

1.026 - (1.028 * 0.8) = (X * 0.2) Therefore:

 

(1.026 - (1.028 * 0.8) ) / 0.2 = X

 

Plug it into a calculator and you will get 1.018.....

[dylanserbin]

Actually the concentration for a 20% water change to drop you to 1.026 needs to be a 20% change at 1.018.

 

Here is the math:

 

1.026=(1.028 * 0.8) + (X * 0.2) this takes into account the concentrations spread over the percentages of total water present. Therefore:

 

1.026 - (1.028 * 0.8) = (X * 0.2) Therefore:

 

(1.026 - (1.028 * 0.8) ) / 0.2 = X

 

Plug it into a calculator and you will get 1.018.....

 

 

+1 1337 pwnage

[illuminano]

So, I just purchased a refractometer and my sg is not 1.024 as it read on my hydrometer but actually 1.028 Now I am wondering what equation I can use to bring the sg to where I want it 1.024. Is it safe to bring down the sg more than .002 per waterchange? So if I do a 20% waterchange what should the sg of the refill water be if I want to lower the sg to say 1.026 or 1.024? I used to be good at math but I would like an experienced reefer's input on this one. I have a 10 gallon with 2 false percs in it right now.

-Thanks in advance, Steve.

 

What salt are you using? I can tell you exactly

[MitchReef]

I'm not sure just what the type of salt has to do with a simple concentration calculation.....V*C + V*C = V*C, no matter what brand of "C" you use.....

[lakshwadeep]

However, specific gravity is not concentration (there's no unit). +1 to just adding water until your specific gravity is what you want it to be. It's easier to dilute the saltwater than to do water changes with water at a lower specific gravity.

 

Just because I like chemistry, here's a method to calculate the specific gravity of a 2 gallon water change for a 10 gallon tank, changing the 10 gallon tank's SG from 1.028 to 1.024.

 

SG = ρ / ρH2O

where

SG = specific gravity

ρ = density of fluid or substance (kg/m3)

ρH2O = density of water (kg/m3)

 

The density of water can be found in many tables, such as this one:

http://www.simetric.co.uk/si_water.htm

 

I used a temperature of 80˚F, which is 26.7˚C (technically I can't know the .7˚C, but we'll assume 80.0˚F). Temperature is important for specific gravity but not for salinity (which your refractometer should show).

 

Using the above table ("Density of Water (g/cm3) at Temperatures from 0C (liquid state) to 30.9C by 0.1C inc"), the density of water at 26.7˚C is 0.996594 g/cm^3. We don't have to convert to kg/m^3 since s.g. is unitless, so we can convert the density of water into a concentration of g/mL (note g/cm^3=g/mL because a cubic centimeter ~ one milliliter). We'll convert gallons to milliliters so the concentrations can be used without being converted, using 1 gallon is 3785.4 mL.

 

Ultimately we want SG of the 10 gallon tank to be 1.024. Using the SG formula, we would have 1.024 = ρ/0.996594. Solving for ρ gives 1.021 g/cm^3 = 1.021 g/mL. The original SG of the tank was 1.028, and using the same steps, we get a concentration of 1.024 g/mL. Now we can use the dilution formula:

 

C1V1+C2V2=C3V3

 

C1=initial concentration in 10 gallon tank (1.024 g/mL)

V1=volume in tank, which is 8 gallons because you took out 2 gallons to do the water change (8 gal=30283 mL)

C2=concentration in the water change water (unknown)

V2=volume of water change water (2 gal=7570.8 mL)

C3=total concentration in the 10 gallon tank after the water change (1.021 g/mL)

V3=total volume of tank with water change water (10 gal=37854 mL)

 

C2=(C3V3-C1V1)/V2 =(1.021*37854-1.024*30283)/7570.8 = 1.0090 g/mL

 

1.0090 g/mL = 1.0090 g/cm^3 = ρ

 

Now go back to the original SG formula to calculate SG of the water change water

 

SG = ρ/ρH20 = 1.0090/.996594 = 1.012

 

 

However, the major assumptions are that the water is pure enough (should be good with ro/di), that the temperature is held constant, and that you can accurately know how much total water is in the tank. These last two are the most problematic, which is why I think adding freshwater (not all at once) is the easiest method to follow. Reef tanks handle accelerated lowering specific gravity (think of it like rain in a natural reef) better than accelerated raising.

[LEROY]

You guys are awesome, my initial idea was to just add freshwater to the tank to achieve the lower specific gravity but was not sure about lowering too fast. I also got a nifty math/chemistry lesson out of it too, good think you like chemistry lakshwadeep.

-Thanks again, Steve.