Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

a) Primary achieves 65% suspended solids and 35% removal. What are the average B

ID: 107736 • Letter: A

Question


a) Primary achieves 65% suspended solids and 35% removal. What are the average BOD and SS (1kg/day) entering the secondary treatment b) A complete mix activated sludge process is to be used for biological treatment. Assume average flow conditions and the primary performance as in part 2 above. Assume the following in Table 3 for the activated sludge process. Make calculations based on a single train, i.e., total plant flow for the following: i) Determine the aeration tank volume in cubic meters. ii) Determine the design hydraulic detention time, in hours. iii) Determine the mass (dry) and volumetric rates of secondary sludge wasted (kg/day, m^3/day).

Explanation / Answer

(a)At wastewater treatment especially in municipal treatment plants in US, raw municipal wastewater undergoes preliminary, primary, secondary, and in some cases, additional treatment to yield treated effluent and a concentrated stream of solids in liquid, called sludge. The sludge is treated as required for utilization or disposal, and additional treatment of effluent may be needed to accommodate specific water reuse opportunities. States must comply with the federal regulation of a monthly average of 30 milligrams per liter of biochemical oxygen demand or BOD (a measure of the amount of biodegradable organic material remaining in the treated wastewater) and 30 mg/liter of suspended solids (particles removable by filtration).

The volume of municipal wastewater sludge produced by wastewater treatment facilities is an elusive quantity because it varies as a result of typical sludge treatment. Since the mass of dry solids is conserved during most treatment processes, dry weight is a more useful basis for expressing the amount of sludge from municipal wastewater treatment. Typical primary and secondary wastewater treatment produces a total of about 1.95 lbs (0.94 kg) of dry solids per 1,000 gal (3.78 m3) of wastewater treated.

Chemical addition to sludge during conditioning and stabilization processes can appreciably increase the mass of solids in sludge. Biological stabilization acts to reduce the mass of suspended solids through oxidation of some of the volatile organic solids in sludge. For example, if sludge contains 80 percent volatile suspended solids and 50 percent of them are destroyed through oxidation, the stabilized mass of sludge solids would be reduced to 60 percent of the initial mass. Primary treatment typically removes about one-third of the BOD and one-half of the suspended solids in domestic wastewaters. Combined primary and secondary treatment is required to achieve 85 percent reduction in both BOD and suspended solids concentration to meet the regulatory definition of secondary treatment. Preliminary treatment serves to prepare wastewater for subsequent treatment, but it effects little change in wastewater quality.

The residues from preliminary wastewater treatment, screenings and grit, are not ordinarily incorporated with sludge, and they are not considered further in this report. Although the goal of primary wastewater treatment is to separate readily-removable suspended solids and BOD, wastewater constituents that exist as settle-able solids or are sorbed to settle-able wastewater solids may also be removed. Thus, primary treatment effects some reduction in the effluent concentration of nutrients, pathogenic organisms, trace elements, and potentially toxic organic compounds. The constituents that are removed are contained in primary sludge.

(b) (i) Using the below model we would derive the aeration tank volume:

1 / thetac = [YQ(S0 – S) / VX] - kd

1/ 8d = {[0.55 (1-*106 gal /d)* 3.785 / gal * (138-8) mg/l] / V3000 mg/l } – 0.04 d-1

V = [(8d) * 0.55 * ( 1*106 gal /d)* 3.785 / gal * (138-8) mg/l] / V3000 mg/l*(1+8d (0.04d-1))

= 546800 L = 547 m3

(ii) The design hydraulic detention time in hours = thetha

Theta = V/Q = 564800 L / (1*16gal / d * 3.785 L/gal ) = 0.144 d = 3.46 hours.

(iii) Knowing that theta = XV / ( Q – Qw) Xe + QWXr = XV / QwXr

So, rearranging, we can isolate the mass loading rate (QwXr)

= QwXr = XV / thetac = 3 Kg/m3 *(547m3) / 8d = 205 Kg/d

Volumetric flow (Qw) = QWXr / xr = (205kg/d) / (12Kg/m3 ) = 17m3 / d.