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A treatment plant uses a CMFR as the reactor for the removal of manganese via an oxidation reaction by the addition of potassium permanganate. If the influent manganese concentration is 0.86 mg/L, the plant has a treatment capacity of 3,800 m3/d, the reactor has a volume of 45 m3, and manganese reacts with potassium permanganate in a first order reaction with a reaction rate constant of 0.0125 s-1, what is the effluent concentration

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Answer:

the effluent concentration is 0.06236 mg/L

Explanation:

Given that;

treatment capacity [tex]V_{0}[/tex] = 3,800 m³/d =  ( 3,800 × 86.4) =  43.98 L/sec

reactor's volume V = 45 m³ = (45 × 1000) = 45,000 L

reaction rate constant K =  0.0125 s⁻¹

influent manganese concentration [tex]CA_{0}[/tex] = 0.86 mg/L

-[tex]r_{A}[/tex] = [tex]KC_{A}[/tex]

Now, performance equation for CSTR is expressed as follows;

[tex]\frac{V}{V_{0} }[/tex] = [tex]\frac{CA_{0} -CA _{} }{-r_{A} }[/tex]

[tex]\frac{V}{V_{0} }[/tex] = [tex]\frac{CA_{0} -CA _{} }{KC_{A} }[/tex]

So we substitute

45000L / 43.98 L/sec = ( 0.86 mg/L - CA) / 0.0125 CA

we cross multiply

562.5CA = 37.8228 - 43.98CA

562.5CA + 43.98CA = 37.8228

606.48CA = 37.8228

CA = 37.8228 / 606.48

CA = 0.06236 mg/L

Therefore, the effluent concentration is 0.06236 mg/L

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