QUESTION 2 A food processing plant has been discharging its wastewater to the lo
ID: 90985 • Letter: Q
Question
QUESTION 2 A food processing plant has been discharging its wastewater to the local sewage system after on-site primary treatment. On average, 640 m /day are discharged with a BOD of 380 mg L. However, the regional sewage treatment plant is nearing its maximum capacity and the water authority is insisting that the plant's wastewater be further treated on-site in a secondary treatment system before it is discharged to sewer. The company has thus been advised that in the near future any discharge to the sewer should have a maximum BOD level of 5 mg/L. Published pilot-plant data for a similar wastewater indicates that the biomass growth rate (r in mg VSS/L.hr) in an aerobic secondary treatment stage may be adequately described by the following equation where S is the substrate level (in mg BOD/L). Endogenous respiration effects (i.e. cell death and lysis in the pilot-plant were accounted for by using an overall observed yield (Yobserved with a reported value of 0.42 mg VSS/mg BoD. r, 0.15 100 SExplanation / Answer
Biomass production rate is given by the formula,
Yobs[(S0 - Sf)/V ]Q
where, Yobs is the observed yield, S0 - is the initial BOD, Sf is the final BOD, V is the BOD concumption rate and Q is the average discharge.
Biomass production rate
= 0.42 [(380 - 5) / 5.013] 640
= 0.42 (375/5.013) 640
= 0.42 x 74.80 x 640
= 20106.24 mg VSS/day
= 20.106 kg VSS/day
= 0.8377 kg VSS/hr