Estimate the volume of an extended-aeration oxidation ditch for carbonaceous BOD oxidation and nitrification for the city of Cartouche Lake using the following design data: Influent data Design flow rate = 14,200 m³/d bCOD = 205 mg/L NH3-N = 16.8 mg/L Total suspended solids (TSS) = 160 mg/L

Question

Estimate the volume of an extended-aeration oxidation ditch for carbonaceous BOD oxidation and nitrification for the city of Cartouche Lake using the following design data:
Influent data
Design flow rate = 14,200 m³/d
bCOD = 205 mg/L
NH[latex]_{3}[/latex]-N = 16.8 mg/L
Total suspended solids (TSS) = 160 mg/L
Biodegradable volatile suspended solids = 95 mg/L
Minimum sustained temperature = 12°C
pH = 7.0
Alkalinity = 90 mg/L as CaCO[latex]_{3}[/latex]
Effluent discharge standards
bCOD ≤ 20 mg/L
NH[latex]_{3}[/latex]-N ≤ 1.0
TSS ≤ 10 mg/L
Assume MLSS = 3,000 mg/L, MLVSS = (0.7) MLSS, and DO = 3.0 mg/L.

Accepted Answer

a. Assume nitrification governs, and select the following kinetic coefficients from Table 23-15:

b. Correct kinetic coefficients to 12°C because this temperature limits the growth rate of nitrifying organisms. Use the correction equations in Table 23-15.

[latex]\mu _{n max} = (0.75 g VSS/g VSS ⋅ d)(1.07)^{12 − 20} = [/latex] 0.44 g VSS/g VSS ⋅ d
[latex]K_{n} = (0.74 g NH_{4}-N/m^{3})(1.053)^{12 − 20} = 0.49 g NH_{4}-N/m^{3}[/latex]
[latex]k_{dn} = (0.08 g VSS/g VSS ⋅ d)(1.04)^{12−20}[/latex] = 0.06 g VSS/g VSS ⋅ d

c. Estimate [latex]\mu_{n}[/latex] using Equation 22-28. The value selected for N is the design effluent concentration because that is limiting. The DO is the design DO that must be achieved by the design of the aeration system.

[latex]\mu _{\eta } = \left(\frac{\mu _{nm}N}{K_{n}+N} ight) \left(\frac{DO}{K_{o}+DO} ight) -k_{dn}[/latex] (22-28)

[latex]\mu_{n} = \left[(0.44 g VSS/g VSS) \left(\frac{1.0 g/m^{3}}{0.49 g NH_{4}-N/m^{3}+ 1.0 g/m^{3}} ight)\left(\frac{3.0 g/m^{3}}{0.50 g/m^{3}+ 3.0 g/m^{3}} ight) ight] [/latex] −0.06 g VSS/g VSS ⋅ d
[latex]\mu_{n} = [(0.44)(0.67)(0.857)] − 0.06 = 0.19 d^{−1}[/latex]

d. Using Equation 23-17 calculate θ[latex]_{c min}[/latex]
[latex]\frac{1}{ heta _{c}}=\mu [/latex] (23-17)
[latex]\frac{1}{ heta _{c min}} = \mu _{n}= 0.19 d^{-1}[/latex]
[latex] heta_{c min} =[/latex] 5.26 d

e. Using a safety factor of 2.5 from Figure 23-18,
[latex] heta _{c}[/latex] = 2.5(5.26 d) = 13.15 d

f. Estimate U for BOD oxidation using Equation 23-19 and the kinetic coefficients from Table 23-14 corrected for temperature. Solve Equation 23-19 for U.
[latex]\frac{1}{ heta _{c}}= YU - k_{d} [/latex] (23-19)
[latex]U= \left(\frac{1}{ heta _{c}}+k_{d} ight) \left(\frac{1}{Y} ight) [/latex]
[latex]U = \left(\frac{1}{13.15 d} +0.09 d^{-1} ight) \left(\frac{1}{0.40} ight) = 0.415 d^{-1}[/latex]

g. Solve Equation 23-18 for θ. Call this θ[latex]_{BOD}[/latex]. Use this to estimate θ[latex]_{BOD}[/latex] for the assumed MLSS = 3,000 mg/L and a conservative assumption of S = 0 as recommended by WEF (1998):
[latex]U = -\frac{r_{su}}{X} =\frac{Q(S_{o}-S)}{\sout{V}X} =\frac{S_{o}-S}{ heta X} [/latex] (23-18)
[latex] heta _{BOD}=\frac{205 g/m^{3} − 0 g/m^{3}}{(0.415 d^{−1})(2, 100 g/m^{3})} [/latex]
= 0.235 d or 5.64 h

h. As in step (f), estimate U for nitrification using Equation 23-19 and the kinetic coefficients from Table 23-15 corrected for temperature.
[latex]\frac{1}{ heta _{c}}= YU - k_{d} [/latex] (23-19)
[latex]U = \left(\frac{1}{13.15} +0.06 ight) \left(\frac{1}{0.12} ight) =(0.136)(8.33)= 0.13 d^{-1}[/latex]

i. An estimate of the fraction of MLVSS that is nitrifying organisms is required to calculate θ for nitrification. Using Equation 22-29:
[latex]f_{N} = \frac{0.16(NH_{3}removed)}{0.6(BOD_{5}removed) + 0.16(NH_{3}removed)} [/latex] (22-29)
[latex]f_{N}= \frac{(0.16)(16.8 g/m^{3} − 1 g/m^{3})}{(0.6)(205 g/m^{3} − 0) + (0.16)(16.8 g/m^{3} − 1 g/m^{3})} [/latex]
[latex]= \frac{2.53}{123+2.53}= [/latex] 0.02

Note that the NH[latex]_{3}[/latex] concentrations were given in the problem statement. The MLVSS of nitrifying organisms is
MLVSS = (0.02)(2, 100 mg/L) = 42 mg/L

j. As in step (g), estimate θ[latex] _{nitrification}[/latex]

[latex] heta _{nitrification}=\frac{16.8 g/m^{3} − 1.0 g/m^{3}}{(1.13 d^{−1})(42 g/m^{3})} [/latex] = 0.33 d or 7.99 h

The hydraulic detention time for nitrification governs.

k. The volume of the oxidation ditch is estimated to be

[latex] \sout{V}= (Q)(θ) =[/latex](14, 200 m³/d)(0.33 d) = 4, 686 or 4, 700 m³

l. To provide redundancy and flexibility of operation, two or three ditches should be provided. Assuming two ditches, the volume of each ditch is estimated to be 12,700 m³ .

Comment. Alkalinity is required to buffer the nitrification reaction. The estimation of the alkalinity required and the chemical dose to adjust it are explained in the next example.

Question 23-8: Estimate the volume of an extended-aeration oxidation ditch ...

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