Question

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

Answer 1

the effluent concentration is 0.06236 mg/L

Step-by-step explanation:

Given that;

treatment capacity
`V_(0)` = 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
`CA_(0)` = 0.86 mg/L

-
`r_(A)` =
`KC_(A)`

Now, performance equation for CSTR is expressed as follows;

`(V)/(V_(0) )` =
`\frac{CA_(0) -CA _{} }{-r_(A) }`

`(V)/(V_(0) )` =
`\frac{CA_(0) -CA _{} }{KC_(A) }`

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