3. A local municipality must upgrade its primary wastewater treatment plant to a secondary plant that can meet an efflue

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3. A local municipality must upgrade its primary wastewater treatment plant to a secondary plant that can meet an efflue

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3 A Local Municipality Must Upgrade Its Primary Wastewater Treatment Plant To A Secondary Plant That Can Meet An Efflue 1
3 A Local Municipality Must Upgrade Its Primary Wastewater Treatment Plant To A Secondary Plant That Can Meet An Efflue 1 (64.46 KiB) Viewed 39 times
3 A Local Municipality Must Upgrade Its Primary Wastewater Treatment Plant To A Secondary Plant That Can Meet An Efflue 2
3 A Local Municipality Must Upgrade Its Primary Wastewater Treatment Plant To A Secondary Plant That Can Meet An Efflue 2 (25.76 KiB) Viewed 39 times
3. A local municipality must upgrade its primary wastewater treatment plant to a secondary plant that can meet an effluent standard of 25.0mg/LBODs​ and 30mg/L total suspended solids. a) If they select a completely mixed-activated sludge system for the upgrade, estimate the aeration volume (m3 ) they will need given the following data and assumptions. b) What is the F/M ratio in mg/mg−d ? c) What is the hydraulic retention time (t) in the tank in h? 4. Assuming that the suspended solids in the effluent may be neglected and that wastage is from the return sludge flow, determine a) sludge age (Qc​ aka mean cell residence time aka solids retention time), b) cell wastage flow rate (Q0​) in m3/s, and c) the return sludge flow rate (Q,​)in m3/s for the activated sludge plant described in Problem 3 above. [Hint 1: to find Q. from the information provided combined with what was calculated in Problem 3, you should use the equation for X in terms of Qc​ and solve for Q. ​. Hint 2: You may assume m g/L MLSS"1.40∗mg/LMLVSS]
Q3 Sol: Solution a) To calculat the areation volume we know that γ=MLSS vx=1+kθc​YQθc​(S0​−S1​)​ where Q= waste wat y=−g fow rat rate Now θe​=Q0​x+(Q1​θ0​xe​xy​=0.05×72502990×0.6​ θc​=4.95≃5 doys V=(1+0.1×5)×(10240)0.6×0.05×5(218−30)​×3600×24 =158.625 m3​→ volums of freation  tank ​ b) fIm =VXQ30​=158.625×29900.05×218×103​=0.023 c) t0​=Qv​​=0.05158.625​=0.88 h×3​
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