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Current IssueApril 24, 2014
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Ask The Expert: Nitrate removal with nanofiltration
08/01/2007 
 
Question:

I read the article in WC&P regarding RO treatment removing 82 percent nitrate. What is the reject rate of nitrate if a nanofiltration membrane is used? Also please clarify the following: my understanding from the NSF guidelines is that the MCL for nitrate (measured as nitrogen) is 10 mg/L (IBWA and FDA) and 50 mg/L (WHO and EC), but some other organizations are measuring the nitrate (as nitrate). In that case, what is the MCL in mg/L? Are there any changes in the above levels of 10 and 50 mg/L?

With regards, Thomas Patu K, Process Engineer Makkah Water Co., Saudi Arabia

 

Answer: 

There was a study done in Spain on this very subject, 'Performance of commercial nanofiltration membranes in the removal of nitrate ions', Desalination, Volume 185, Issues 1-3, 1 November 2005, pages 281-287 by A. Santafó-Moros, J.M. Gozlvez-Zafrilla and J. Lora-Garca. (Find it online at http://www.science direct.com)

Andrew Warnes, GE Water & Process Technologies

Though I have no data from testing specific products, generally one should not expect much reduction at all of nitrate ion utilizing nanofiltration. However, the 82 percent reduction you cite for RO seems reasonable. Nanofiltration is tailored for multivalent ion reduction; e.g., Ca++, Mg++, SO4=, at about 60-80 percent removal; not for monovalent ion removal (Na+, K+, Cl-). Monovalent ion removal would probably be expected to be less than 50 percent. Perhaps some of the other Tech Reviewers can give you specific examples from empirical data that they might have. You could try conducting a Google search to see if you can find such data.

Regarding the 10 mg/L US EPA MCL for nitrate, this is for nitrate nitrogen as "N. To calculate or convert this to mg/L as nitrate (NO3-), you have to calculate the full mass or molecular weight including the oxygen atoms. The molecular wt. of nitrate is 62 (N = 14 and O = 16 x 3 = 48). The N represents only 22.58 percent of the weight of NO3. So to convert 10 mg N to mg NO3, divide 10 by 0.2258 = 44.3. Therefore, 10 mg nitrate-N/L = 44.3 mg nitrate as NO3/L. The WHO regulation rounds that up to 50 mg nitrate as NO3/L, which to converts back to N...50 x 0.2248 = 11.3 mg nitrate-N/L.

Gary Hatch, Ph.D. Pentair Filtration, Inc.

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