1. feedstream calculation

It’s late and I’m brain dead…
I need to dilute a “feedstream” so that the resultant discharge concentration is <= 400 mg/L.

Example:
If I have a rate of 450 gpm for 30 minutes at a concentration of 600 mg/L. The second “dilutant” stream has a concentration of 100 mg/L and is added for the full 30 minutes. At what rate do I have to add the second stream so that the resultant discharge concentration is 400 mg/L?

I need to do this for several different rates and concentrations of the feedstream, with the “dilutant” always at 100 mg/L and running the same length of time.

It shouldn’t be hard, but I can’t wrap my brain around it. Suggestions?

Thanks!

2. A rough (though probably practical) calculation. The conversion from gallons to Liters is not needed as it cancels. The 30 min time is also immaterial since it is the same for both streams. Thus the mimimum rate of the "dilutant stream" can be calculated:
=450 *(600-400) / (400 - 100)

If the times may be different, then you should use:
=450 *(600-400) * Time1 / (400 - 100) / Time2

The number is not exact. It presumes firstly that the 2 streams will be the same temperature and pressures. If not the densities of the material will be different before they are mixed and after, which will change the volumetric concentrations. Also since the 2 streams have different compositions, the volumes may not be additive. When the volumes of 2 substances are added together, the total new volume may be the same, but it is often larger or smaller depending on the interactions of the materials and the molecular forces between them (they sometimes attract and sometimes repel). [That is why it is better to use mass flows, since mass is additive and unaffected by temperatures and pressures]. I don't think the differences will matter in the practical sense of what you are doing.

Steve

3. The Following User Says Thank You to sdckapr For This Useful Post:

jepalmer (2012-08-30)

4. thanks for the help!

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