Begin the design of a complete-mix BOD removal and nitrification stage for Tempe's new BPR wastewater treatment plant. As a first step determine the SRT and volume of tank required for nitrification using the following design data: Influent data after primary settling Design flow rate = 90,000 m³/d

Question

Begin the design of a complete-mix BOD removal and nitrification stage for Tempe's new BPR wastewater treatment plant. As a first step determine the SRT and volume of tank required for nitrification using the following design data:

Influent data after primary settling

Design flow rate = 90,000 m³/d
BOD[latex]_{5}[/latex] = 138 mg/L
bCOD = 220 mg/L
rbCOD = 80 mg/L
Total Kjeldahl nitrogen (TKN) = 35 mg/L
NH[latex]_{3}[/latex]-N = 25 mg/L
Soluble phosphorus = 8 g/m³
Total suspended solids (TSS) = 70 mg/L
Volatile suspended solids (VSS) = 60 mg/L
Biodegradable volatile suspended solids = 40 mg/L
Minimum sustained temperature = 20°C
pH = 7.0
Alkalinity = 220 mg/L as CaCO[latex]_{3}[/latex]

Effluent discharge standards

bCOD ≤ 20 mg/L
NH[latex]_{3}[/latex]-N ≤ 1.0 mg/L
NO[latex]_{3}[/latex]-N ≤ 10 mg/L
TSS ≤ 10 mg/L
Total phosphorus ≤ 2.0 mg/L

Assume DO for aeration tank nitrification = 2.0 mg/L; MLSS = 3,000 mg/L, MLVSS = (0.8) MLSS, NO = 80% of TKN, and RAS = 0.60(Q).

Accepted Answer

a. Assume that four identical process lines will be built for redundancy and flexibility. Each line will have a design flow rate of
[latex]\frac{90, 000 m^{3}/d}{4} [/latex] 22, 500 m³/d

b. Use Table 23-15 for the biokinetic constants because nitrification kinetics govern. The minimum sustained temperature is 20°C so no temperature corrections are required.

[latex]\mu _{n max}[/latex] = 0.75 g VSS/g VSS ⋅ d
[latex]K_{n} [/latex]= 0.74 g NH[latex]_{4}[/latex]-N/m³
[latex]k_{dn} [/latex] = 0.08 g VSS/g VSS ⋅ d
[latex]K_{0}[/latex] = 0.50 g/m³

c. Estimate μ using Equation 23-28. 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. The assumed value is 2.0 mg/L.

[latex]\frac{1}{\theta _{cmin}} \approx \mu _{m}-k_{d}[/latex] (23-28)

[latex]\mu _{n } = \left[(0.75 g VSS/g VSS)\left(\frac{1.0 g/m^{3}}{0.74 g NH_{4}-N/m^{3} + 1.0 g/m^{3}} \right)\left(\frac{2.0 g/m^{3}}{0.50 g/m^{3} + 2.0 g/m^{3}} \right) \right] [/latex]
−0.08 g VSS/g VSS ⋅ d
[latex]\mu _{n } = [(0.75)(0.57)(0.80)] − 0.08 = 0.262 d^{−1}[/latex]

d. Calculate [latex]θ_{c min}[/latex]

[latex]\frac{1}{θ_{c min}} = \mu _{n } = 0.262 d^{−1}[/latex]
[latex]θ_{c min} =[/latex] 3.817 d

e. Using a safety factor of 2.5 from Figure 23-18,

[latex]θ_{c }[/latex] = 2.5(3.817) d = 9.54 d

This is within the SRT guidelines for BPR and is at the desirable shorter end of the range.

f. Using Parts A, B, and C of Equation 23-40, determine biomass production. Use the biokinetic constants for heterotrophic bacteria (Table 23-14) for Parts A and B. Use the biokinetic constants for nitrification for Part C. The minimum sustained temperature is 20°C, so no temperature corrections are required. Use a conservative assumption of S = 0 as recommended by WEF (1998). Because a nitrogen balance cannot be computed withoutP[latex]_{x}[/latex] , use the assumed NO[latex]_{x}[/latex] fraction, NO[latex]_{x}[/latex] = 0.80(TKN) = 0.80(35) = 28 mg/L as a first approximation. Use the kinetic coefficients from Tables 23-14 and 23-15:

[latex]P_{x,vss} = \frac{QY (S_{o} − S)(10^{−3} g/kg)}{1 + k_{d} θ_{c}} [/latex] (Part A)
[latex] + \frac{(f_{d})(k_{d})QY (S_{o} − S)(θ_{c})(10^{−3} g/kg)}{1 + k_{d} θ_{c}} [/latex] (Part B)
[latex] + \frac{QY_{n}(NO_{x})(10^{−3} g/kg)}{1 + k_{dn} θ_{c}} [/latex] (Part C)
[latex] + Q(nb VSS)(10^{−3} g/kg)[/latex] (Part D) (23-40)

[latex]μ_{m} = 6.0 g VSS/g VSS ⋅ d [/latex] [latex]f_{d} = 0.15 g/g [/latex]
[latex]Y = 0.40 g VSS/gbCOD [/latex] [latex]Y_{n} = 0.12 g VSS/g NH_{4}−N[/latex]
[latex]k_{d} = 0.12 g VSS/g VSS ⋅ d [/latex]

[latex]P_{x} = \frac{(22, 500 m^{3}/d)(0.40 g VSS/g bCOD)(220 g/m^{3} − 0)(10^{-3} kg/g)}{[1 + (0.12 g VSS/g VSS ⋅ d)(9.54 d)]}\[0.5 cm]+ \frac{(0.15 g/g)(0.12 g VSS/g VSS ⋅ d)(22, 500 m^{3}/d)(0.40 g VSS/g bCOD)(220 g/m^{3} − 0)(9.54 d}{[1 + (0.12 g VSS/g VSS ⋅ d)(9.54 d)]} \[0.5 cm]+ \frac{(22, 500 m^{3}/d)(0.12 g VSS/g NH_{4}-N)(28 g/m^{3})(10^{-3} kg/g)}{[1 + (0.08 g VSS/g VSS ⋅ d)(9.54 d)}\[0.5 cm] P_{x}[/latex] = 923.16 + 158.53 + 42.87 = 1, 124.56 kg VSS/d

g. Determine the amount of NO[latex]_{x}[/latex] oxidized to nitrate using Equation 23-45.

[latex]NO_{x} = TKN_{0} − N_{e} − 0.12 \left(\frac{P_{x}}{Q} \right) [/latex] (23-45)
[latex]NO_{x} = 35 g/m^{3} − 1.0 g/m^{3} − 0.12 \left(\frac{1, 124.56 kg VSS/d}{22, 500 m^{3}/d} \right) (10^{3} g/kg)[/latex]
= 35.0 − 1.0 − 6.0 = 28.0 g/m³

h. Determine the mass of VSS and TSS in the aeration basin. With the exception of the nbVSS, the mass has already been calculated in step (f). From the influent data, the mass of nbVSS is 60 g/m³ − 40 g/m³ = 20 g/m³ and the

Mass = (nbVSS)(Q)
= (20 g/m³)(22, 500 m³/d)(10[latex]^{−3}[/latex] kg/g) = 450 kg/d

The total mass of VSS is then

[latex]P_{x, VSS}[/latex] = 1, 124.56 kg/d + 450 kg/d = 1, 574.56 or 1, 600 kg/d

The mass of MLVSS is equal to the MLVSS concentration times the volume of the aeration tank, that is, [latex](X_{VSS})(\sout{V})[/latex] . This is equal to mass generated each day times the SRT, that is, [latex](P_{x, VSS)(θ_{c})}[/latex]

Mass of MLVSS = (1, 124.56 kg/d)(9.54 d) = 10, 728.3 kg

Using the assumed ratio of MLVSS to MLSS, the mass of MLSS may be estimated as

Mass of MLSS = [latex]\frac{10,728.3 kg}{0.8} [/latex] 13, 410 kg

i. The concentration of MLSS times the volume of the aeration tank is equal to the mass of MLSS, that is,[latex](X_{MLSS})(\sout{V})[/latex] . At the assumed MLSS of 3,000 g/m³ , the volume of the aeration tank may be estimated as follows:

[latex](X_{MLSS})(\sout{V}) = [/latex] 13, 410 kg
[latex](\sout{V}) = \frac{13, 410 kg}{(X_{MLSS})} [/latex]
[latex]= \frac{13, 410 kg}{(3, 000 g/m^{3})(10^{−3} kg/g)} [/latex]= 4, 470 or 4,500 m³

j. Estimate the hydraulic detention time.

[latex]\theta = \frac{4, 500 m^{3}}{22, 500 m^{3}/d} = [/latex] 0.199 d or 4.78 h

This is in the range of typical HRTs for the aerobic stage of BPR processes.

Comments:
1. A volume of 4,500 m³ is too big to achieve complete mixing. A volume of about 3,000 m³ is about the maximum that should be used. For this example the flow should be divided into eight parallel process lines.

2. Because a portion of the effluent from the aeration basin is recycled to the anoxic stage, the basin may be segregated into two or more chambers with a reduced DO concentration in the downstream chamber. This reduces the DO entering the anoxic tank.

Question 23.17: Begin the design of a complete-mix BOD removal and nitrifica...

Related Answered Questions

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 ...

Verified Answer:

Verified Answer:

Check the assumed anoxic fill time used to design the SBR for the town of Quintuple using the data from Examples 23-13 and 23-14. ...

Verified Answer:

Check the assumed react plus aerated mix time used to design the SBR for the town of Quintuple using the data from Example 23-13. ...

Verified Answer:

Verified Answer:

Estimate the effluent phosphorus concentration for Tempe’s new BPR wastewater treatment plant. Use the data and assumptions from Examples 23-17 through 23-19. In addition, assume the following: 10 g rbCOD/g P is removed by BPR rbCOD/nitrate ratio is 6.6 g rbCOD/g NO3-N Phosphorus content of ...

Verified Answer:

Check the alkalinity for the BPR wastewater plant being designed for Tempe. Use the data from Examples 23-17 and 23- 18. Assume an alkalinity of 80 g/m³ is needed to maintain a pH of about 7. The amount of nitrogen oxidized is NOx = 28.0 g/m³ from step (g) in Example 23-17. ...

Verified Answer:

Verified Answer:

In continuing the design of the BPR process for Tempe’s new wastewater treatment plant, determine the volume of the anoxic tank. Use the data provided in Example 23-17. ...

Verified Answer:

Estimate the mass of sludge to be wasted each day from the extended aeration oxidation ditch at Cartouche Lake (Example 23-8). ...

Verified Answer: