Project ForgeA fraction f of the feed (from the RO problem above, CF=2 mg/L) is bypassed and mixed with the permeate to obtain treated water having a fluoride concentration of 1 mg/L. Here also the flow rate of the reject stream is 60% of the flow rate entering the reverse osmosis unit (after the bypass). The value of f, up to 2 digits after the decimal point, is _____
Explanation
Basis total feed=100: bypass=100f goes straight to product; (100−100f) enters the RO unit, producing permeate=0.4×100(1−f) at CP=0.1 mg/L. Fluoride balance on treated water: 100f×2+40(1−f)×0.1=[100f+40(1−f)]×1. 200f+4(1−f)=100f+40−40f. 196f+4=60f+40 ⟹ 136f=36 ⟹ f≈0.265≈0.26.
Source: GATE CH 2013 (2013)
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