GATE 2026 ES – Question 58
An activated-sludge plant has influent flow $.150\,\mathrm{m^3/s}$,soluble influent and effluent $BOD_5$ of 84 and 11 mg/L, mean cell residence time 5 days, yield .5 kg MLVSS/kg $BOD_5$ removed and decay constant $.050\,\mathrm{day^{-1}}$. Find net waste activated sludge in kg VSS/day, to two decimals.
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Correct answer: 378 to 379
Explanation
Daily flow is $.150(86400)=12960$ m³/day and removed BOD is $(84-11)/1000=.073$ kg/m³. Thus removal is $12960(.073)=946.08$ kg/day. Net observed yield includes decay: $Y_{obs}=Y/(1+k_d\theta_c)=.5/(1+.050\cdot5)=.4$. Net sludge production is $.4(946.08)=\mathbf{378.43}$ kg VSS/day.