Problem 1 10 points

A sample of wastewater was incubated for seven days at 20oC and showed a BOD of 208 mg/L. (a) Calculate the 5-day BOD. (b) Calculate the 10-day BOD. (c) Calculate the ultimate BOD (Assume the BOD rate constant k equals 0.15 d-1, base e).

Answer to Problem 1

For these solutions we use the BOD equation, BODt = �� (1 − ��−����)

(a) To calculate the 5 day BOD, we first use the information given at 7 days to obtain the ultimate BOD. Then we can solve for the 5-d BOD

208 = �� (1 − ��−(0.15 ��−1)(7 ��)) L = BODu = 320 mg/L

BODt = 320 mgL(1 − ��−(0.15 d−1)(5 d)) BOD5 = 169 mg/L

(b) BODt = 320 ������(1 − ��−(0.15 ��−1)(10 ��)) BOD5 = 249 mg/L

(c) See part (a) for the ultimate BOD calculation. L = BODu = 320 mg/L

Problem 2 10 points

What is the ThOD of grease, assuming that grease has the composition of C51H99O6? Answer to Problem 2

To solve this, we need to determine the stoichiometric coefficients. For oxidation of grease, with no biomass production (since it is theoretical)

C51H99O6 + aO2 = bCO2 +cH2O

O: 6 + 2a = 2b + c

C: 51=b

H: 99=2c, So: c = 49.5, a = 72.8, b=51

The molecular weigth of grease is calculated from the given formula as 807 g/mol. The moecular weight of O2 is 32 g/mol.

ThOD =g−O2 needed

g−substrate =72.8 mol O2

mol−grease

Problem 3 20 points

mol−grease 807 g−grease

32 g−O2

mol−O2= 2.89 ����

An activated sludge process has the following operational data:

• Influent flow to the aeration tank (Q) = 10,000 m3/d

• Influent BOD –180 mg/L

• Aeration Tank Volume – 2,500 m3

• MLSS in the aeration tank – 2000 mg/L

• Yield coefficient – 0.5

• Decay coefficient – 0.06 d-1

• Solids in return sludge – 10,000 mg/L

• Solids in effluent – 30 mg/L

• BOD in effluent – 20 mg/L

Determine (a) the sludge age (θc) and (b) the rate of waste sludge flow (Qw), if the solids concentration in effluent (Xe) cannot be ignored.

Problem 4 20 points

Given the following data, determine the Kg/day SS to be wasted:

∙ Aeration Tank Volume – 4164 m3

∙ MLSS – 2100 mg/L

∙ Influent Flow – 462 m3/h

∙ % VS – 67%

∙ BOD –108 mg/L

∙ Desired F/M – 0.3 g BOD/day/g VSS

Problem 5 25 points

A complete-mix aerobic reactor without solids recycle is used to treat a wastewater containing 100 mg/L phenol (C6H6O) at 20 °C. using the following kinetic coefficients determine (a) the minimal hydraulic retention time t in days at which the biomass can be washed out faster than they can grow, (b) the effluent phenol and biomass concentration at a θ value of 4.0 d at 20 °C, and (c) the amount of oxygen required in kg/d for a θ value of 4.0 d assuming a flowrate of 100 m3/d. Plot the phenol and biomass concentration and the amount of oxygen required versus θ in days, for θ from 3.3 to 15 d at 20 °C.

Coefficient

Unit

Values

K

g phenol/g VSS •d

0.90

Ks

mg phenol/L

0.20

Y

g VSS/g phenol

0.45

b

g VSS/g VSS• d

0.10

Problem 6 15 points

What is the predicted effluent BOD concentration from a trickling filter with the following conditions: wastewater flowrate of 1 mgd; TF diameter of 30 ft and 20 ft height; 30 ft2/ft3cross flow media; 22oC; influent sBOD=100 mg/L; R = 0.2. (Hint: consider using the modified Velz Equation, note the units!)