Question

You are provided the following information about a municipal wastewater treatment plant. This plant uses the traditional activated-sludge process. Assume the microorganisms are 55 percent efficient at converting food to biomass, the organisms have a first-order death rate constant of 0.05/day, have a maximum specific growth rate of 0.1/day, and the microbes reach half of their maximum growth rate when the BOD5 concentration is 10 mg/L. There are 150,000 people in the community (their wastewater production is 225 L/day-capita, 0.1 kg BOD5/capita-day). The effluent standard is BOD5 = 20 mg/L and TSS = 20 mg/L. Suspended solids were measured as 4,300 mg-MLSS/L in a wastewater sample obtained from the biological reactor, 15,000 mg-MLSS /L in the secondary sludge, 200 mg-MLSS /L in the plant influent, and 100 mg-MLSS/L in the primary clarifier effluent. SRT is equal to 4 days.

a. (0.5 points) What is the design volume of the aeration basin (m3 )?

b. (0.25 points) What is the plant’s aeration period (days)?

c. (0.25 points) How many kg of secondary dry solids need to be processed daily from the treatment plants?

d. (0.25 points) If the sludge wastage rate (Qw) is increased in the plant, will the solids retention time go up, go down, or remain the same?

e. (0.5 points) Determine the F/M ratio in units of kg BOD5/kg MLVSS-day (assume 0.6 g MLVSS per g MLSS)?

f. (0.25 points) Determine the critical SRT value.

Answer 1

Solution:

(a)

Given,

kd = first-order death rate constant =  0.05/day

SRT = 4 days

flow rate = Qo = 150000 x 225 L/day-capita = 33.75 x 10^6 L/day

V=microorganisms are 55 percent efficient at converting food to biomass = 0.55 gm SS/gm BOD5

Y = Suspended solids were measured = 4,300 mg-MLSS/L

S= 20 mg/L

Assuming that 30% o f the plant influent BOD5 is removed during primary sedimentation, this means that So = 444 mg/L × 0.70 = 310 mg/L.

1/SRT = [(Qo × Y)/(V × X)) × (So-S)] – kd

1 / 4 days=[33.75x10^6 L/day (0.55 gm SS/gm BOD5) /V 4,300 mg SS/L (310 mg/L – 20 mg/L)] – 0.05/day

V = 5,000,000 Liters = 5000 m3

(b)

=> plant’s aeration period hours = Wastewater is aerated during the activated sludge process

Hydraulic detention time of the biological reactor HRT= V/Q = 5,000,000 L / 33.75 x 106 L/day = 0.15 days

(c)

Weight of solids processed from the primary sedimentation tanks equals to  the difference in suspended solids concentrations multiplied by the plant flow rate.

33.75x106 L/day (200 mg SS/L – 100 mgSS/L) × kg/1,00,000 mg = 2,275 kg primary solids per day

Determine the weight of secondary solids produced daily in the same manner that used for primary solids.

Solids retention time = Qw Xw  

X = Suspended solids were measured as 4,300 mg-MLSS/L

V = 5,000,000 Liter

SRT = V × X / Qw × Xw

4 days = V × X / Qw × Xw = 5,000,000 L (4,300 mg SS/L) / Qw × Xw

Qw × Xw = 5375000000 mg = 5375 kg

(d) =>

If the sludge wastage rate (Qw) is increased in the plant,

The solids retention time is go down.

Solids retention time is inversly proposonal to the sludge wastage rate (Qw)

SRT = V × X / Qw × Xw

(e) =>

Q = 33.75x10^6 L/day

So= 310 mg/L

X = 4,300 mg SS/L

V = 5,000,000 L

F/M = Q So / X V = [33.75x10^6 L/day x 310 mg/L) / [4,300 mg SS/L × 5,000,000 L] = 0.4866 lbs BOD5/lb MLSS-day

(f)

Mean cell retention time of the solids = SRT =4 days

Sludge is processed on the belt filter press every 5 days

Critical SRT = 5 days.