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Sodium Carbonate Dilution

Sodium Carbonate Dilution

Sodium Carbonate Dilution

Hello, I have a tank(s)/pipe system with the following set up: I have one tank with 250 gallons of 5% sodium carbonate in solution. This tank is fed into another tank with water added to achieve a final sodium carbonate solution of  0.10% in soultion (by wieght).

My questions are:
1) How much water do I need to add to the second tank to cut the 5% solution to 0.10 % solution?
2) If the flow rate from the second tank is 90 gpm, how much water gpm would I need to add to 5% solution to reduce the sodium carbonate to 0.10 % by weight?

Thank you for your help.  

RE: Sodium Carbonate Dilution

Using g (gram) as mass unit:

Solution #1: salt (5) + water (95) = 100  
Solution #2: salt (5) + water (4,995) = 5000

Assuming density of solution #1 is 1.05 g/cc, its volume would be 95.24 cc.

Add (4995-95)= 4900 g = 4900 cc water to 100 g (= 95.24 cc of solution #1) to get 4900+95.24 = 4995.24 cc of  solution #2. The density of this solution would be about 1.0 g/cc.  

Therefore for any volume flow rate of solution #1 add water in the ratio of 4900÷95.24 = 51.45 volumes of water, totalling 52.45 volumes of solution #2.

For 90 gpm the water addition would be 90×51.45÷52.45 = 88.284 gpm water to 1.716 gpm solution #1. In round figures 88.3 gpm water for every 90 gpm solution #2.

I truly hope the above is correct. Please check my assumptions and logic.


RE: Sodium Carbonate Dilution

Hi 25362,

Thank you for your detailed and accurate response.

Yes, your logic is correct but I want to validate my understanding with your concurrence.

1) So if I had a 250 gallon tank of 5% sodium carbonate in solution, I would need to add 250 * (51.45/52.45) = 245.23 gallons of water. The final volume would be 250 gallons + 245.23 gallons = 495.23 total gallons of 0.1 sodium carbonate in solution.

2) Does your final statement infer that I need to add 88.3 gpm of water to 90 gpm of 5% sodium carbonate in solution to arrive at 88.3+90 = 178.3 gpm of 0.1% sodium carbonate?

3) Last, why did you subtract 4995-95=4900 cc? Where did 52.45 come from?

Thanks again for your help.

RE: Sodium Carbonate Dilution

Dracula, let's start from the last question backwards:

3) A 5% solution already contains 95 mass units of water and 5 mass units of salt. In a 0.1% solution 5 mass units of salt would need 4995 mass units of water to reach a total of 5000 mass units. Since the original solution (100 units) already contains 95 units of water you need to add the difference (=4900 units) to reach a toral of 5000 units.

Now, while water mass units = volume units, the original 100 mass units of the 5% solution correspond only to 95.24 volume units, the ratio between 4900 volume units of water to 95.24 volume units of the original 5% solution is 51.45.

This means that to 1 volume of the original solution you must add 51.45 volumes of water, reaching a total of 52.45 volume units.

2) No, it means that to reach 90 gpm you need to take about 1.7 gpm of the 5% solution plus about 88.3 gpm of water to keep the 51.45 ratio found above. The rounding of the figures gives a ratio of 51.9 instead of 51.45.

1) So, if you have 250 gal of a 5% solution you must add 250 × 51.45 volumes of water, namely 12862.5 gal of water to drop the concentration to 0.1%.

To make a rough check. To reduce a solurion from a concentration of 5% to 0.1% one needs to dilute it about 50 times (5÷0.1), right? Thus, we are on the right track.

I sincerely hope I made myself clear.

RE: Sodium Carbonate Dilution

Hi 25362,

Your rough check of 50:1 is exactly what I calculated when I first looked at the flows. This seemed very high and thats why I made my post.

3)Item is understood and correct.
2)Item is understood and correct.
1)Yes, again you have validated my first assumption.

Great work and thanks again for all you help!

RE: Sodium Carbonate Dilution

Why does this sound like homework?


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