Is there a formula to calculate drop in salinity?

[bleedingthought]

Is there a way to calculate how much the water salinity will drop when adding a certain percentage of fresh water to it?

For example, if you had 100 gallons of 1.025 specific gravity water in a tank, and added 10% of it (10 in this scenario) in fresh water, what would the salinity drop to?

I figured someone here knew about these things!

 

[PezJunkie]

Salinity Calculator for evaporation:
http://www.saltyzoo.com/SaltyCalcs/EvapSalinityCalc.php

Since you're doing the opposite of evaporation, enter a negative number in the "Gallons evaporated:" field.

Looks like 1.025 SG for 100 gallons = 1.023 for 110 gallons

(Now if you're taking out 10 gallons first & then replacing it with freshwater, then you're back at 100 gallons, with a SG of 1.022)

 

[Len]

The way I do it is this:
1. Convert S.G. to salinity (1.025 = 33 PPT).
2. Subtract percentage of introduced freshwater from 100% (e.g. 100% - 10% = 90%)
3. Multiply this value by the current salinity (33 PPT x 0.90 = 29.7 PPT)
4. Convert salinity back to S.G. (29.7 PPT = 1.022).

 

[bleedingthought]

Ah, thanks guys!

Makes perfect sense.

 

[Len]

Oops. I read it as replacing 10 gallons of the 100 gallons, not adding 10 gallons on top of it.

Instead of redoing my instructions (I'm lazy ) bealsob has a equation that works .... and without converting to salinity:
100/110*1.025+ 10/110*1.000=1.0227

 

[pcardone]

I would add 10 gallons and then test it

 

[Anonymous]

By far the easiest way to do this is C1*V1 = C2*V2

Where C1 is the initial concentration (ie, salinity), and V1 is the initial volume. You also know V2, which is volume after adding the 10 gallons (110 gallons). Rearrange the equation to find C2 or the final concentration/salinity:

C2 = (C1*V1)/V2.

C2 = (35*100)/110

C2 = 31.8

 

[Norvix]

Say your salinity is 1.023, I think you could just take off the 1 because that is the SG of one mL water. So you could say .023 times 100 (gal) / whatever #gallons are left and add the one back. So, .023 times 100 / 80 plus 1. (salinity 1.0287) Right?

I guess measuring conductivity is more accurate than seatest meter, well if you don't know how much salt is in tank to begin with...

 

[Anonymous]

The way I do it is this:
1. Convert S.G. to salinity (1.025 = 33 PPT).
2. Subtract percentage of introduced freshwater from 100% (e.g. 100% - 10% = 90%)
3. Multiply this value by the current salinity (33 PPT x 0.90 = 29.7 PPT)
4. Convert salinity back to S.G. (29.7 PPT = 1.022).

Close, but not exact. It works for small values but as the freshwater input becomes larger it becomes more skewed. Consider if your topoff was 100 gallons in this scenario; plug in the numbers and you get 0 ppt.

In step 2 you need to divide the starting gallonage (100) by the ending gallonage (110), which gives you 0.90909090..., not 0.90, and the end result is a salinity of 30.0 ppt, not 29.7 ppt.

 

[Anonymous]

Or what Dolph Lundgren said.

I think pcardone's advice is good too. For one thing, it's damn near impossible to say for certain that you have exactly 100 gallons in your system. Pipes, rock, sand, skimmer, etc. The best way to figure it out is to add it slowly and measure the change.

How are you measuring your salinity?

 

[bleedingthought]

Or what Dolph Lundgren said.

I think pcardone's advice is good too. For one thing, it's damn near impossible to say for certain that you have exactly 100 gallons in your system. Pipes, rock, sand, skimmer, etc. The best way to figure it out is to add it slowly and measure the change.

How are you measuring your salinity?

Well, 2 things I wanted to use this for were when I mix new salt water and the salinity ends up being too high, about how much water to add to get it closer to what I need. Second, I'm trying to calculate how much water the new system will be able to handle (physically and "salinity-lly" ) in case something goes haywire with the top-off system. So, it does have to be 100% accurate, just close enough.

I'm using a refractometer.