if raw sewage contains 30 mg/L phosphate and a secondary sewage treatment plant removes 25% of the phosphate would a secondary treatment plant provide portable water if 0.3 mg/L is the maximum phosphate concentration allowable in drinking water?
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30 mg/L
If 25% is removed, that leaves 75%. Thus 0.75 x 30mg/L = 22.5mg/L left
If you removed another 25%, thus leaving 75%, you would have 22.5mg/L x 0.75 = 16.8 mg/L
This is still very much above acceptable limits.
If 25% is removed, that leaves 75%. Thus 0.75 x 30mg/L = 22.5mg/L left
If you removed another 25%, thus leaving 75%, you would have 22.5mg/L x 0.75 = 16.8 mg/L
This is still very much above acceptable limits.