EXAMINATION-II

SOLUTION

The bonus question carries 5 marks, over and above the 30 marks for the regular questions. Marks obtained by the student in the bonus question will be added to the total marks obtained in other questions.

1. Give brief answers to the following questions with appropriate reasoning:

A. Why is recycling necessary in activated sludge process (2)

Solution:

In activated sludge process we try to increase the substrate utilization rate without increasing the specific substrate utilization rate, i.e., in the equation,

, we want to increase by lowering . However, q cannot be increased, because, , and thus increasing q will increase S, which is undesirable.

Hence X has to be increased. This is achieved by collecting the biomass leaving the aeration tank and re-circulating it back to the aeration tank.

B. What are the differences between microbial physiology in high rate and an extended aeration activated sludge processes. (2)

Solution:

Assuming Q, S_o and V_a (hence ) to be the same in both cases,

In high rate aeration, is low, hence m and q are high. Since q is high, S is high. Since m is high DX is high. X in the aeration tank is comparatively low. Oxygen requirement is low, as rate of biomass production is high. Nutrient requirement is high.

In extended aeration, is high, hence m and q are low. Since q is low, S is low. Since m is low DX is low. X in the aeration tank is comparatively high. Oxygen requirement is high, as rate of biomass production is low. Nutrient requirement is low.

C. Why a pure oxygen aeration system is often used for activated sludge treatment of high strength waste (2)

D. Why is there an upper and lower limit for detention time in a secondary settling tank used in the activated sludge process. (2)

E. Raw wastewater influent to the aeration tank of activated sludge processes has inorganic suspended solids of small size (<20 mm), due to incomplete particle removal in the primary settling tank. This material is completely inert, and is generally completely removed from wastewater during secondary settling following aeration, and ultimately expelled from the system during sludge wasting. However, sludge recycling results in maintenance of a certain steady-state inorganic suspended solid concentration (C) in the aeration tank, which is much higher than the influent inorganic suspended solid concentration (C_o). Based on the above information, derive the expression for steady-state inorganic suspended solids concentration (C) in the aeration tank of an ASP, and hence calculate C, given C_o = 20 mg/L, q = 4 hours, and q_c = 6 days. (2)

Solution:

Mass Balance for the whole system: Rate of mass Input = Rate of Mass Output,

i.e.,

Also,

Mass Balance for the secondary sedimentation tank,

Rate mass input = = Rate mass output =

or,

or, ,

or,

2. 1 MLD of 400mg/L Glucose solution (C₆H₁₂O₆) is to be treated by a completely-mixed activated sludge process (ASP). Hydraulic detention time in the aeration tank is 4 hours. The effluent from the process should have 20mg/L glucose. Biological solids retention time (BSRT) of the ASP is 6 days. Based on this information, calculate the oxygen requirement in the aeration tank in Kg/d. Assume all necessary nutrients that are required for bio-mass growth are present in excess. Also assume that BOD_u = 1.5 x BOD₅. Y_T and K_d values for microbial degradation of glucose are 0.5 mg/mg and 0.05 /d respectively (calculated based on BOD₅ values). (10)

Solution:

Thus, 180 g glucose requires 192 g oxygen for complete oxidation.

Therefore, BOD_u for 400 mg/L glucose solution is , and corresponding BOD₅ = , i.e., S_o = 284.44 mg/L

Also, BOD_u for 200 mg/L glucose solution is , and corresponding BOD₅ = , i.e., S = 14.22 mg/L

Given, , therefore, , and

Also, , So,

Also,

Therefore,

Oxygen Requirement (Kg/d) =

3. The aeration tank volume in an activated sludge plant was calculated to be 5184 m³, to be provided in four parallel units. The depth of the tanks will be 3 m. Aerators are available with power of 1, 2, 5, 10, 25, 50 KW and their areas of influence are 25 m² for 1, 2, and 5 KW, and 36 m² for 10, 25 and 50 KW aerators respectively. The expected maximum oxygen requirement for the process was calculated to be 13700 Kg/d. Manufacturers specify that the oxygen transfer capacity of these aerators is 2.0 Kg O₂/kW-h under standard conditions. Based on this information, design an adequate aerator arrangement for each aeration tank. The operating temperature of the aeration tank is expected to be 30^oC. The steady state dissolved oxygen concentration in the aeration tank should be 1 mg/L. Saturation concentration of oxygen in water at 20^oC is 9.1 mg/L and at 30^oC is 7.5 mg/L. (10)

,

,

where,

Solution:

Volume of each tank =

Surface area of each tank =

Provide 36 m length and 12 m width for each tank

If a 1 kW aerator is used under standard conditions:

Standard

Where V = 25m² or 36m² c/s area x 3 m depth

Under actual conditions,

or,

= 1.232 kg/hr

Oxygen Requirement per tank =

Power Requirement per tank =

Using 12 aerators per tank, 2 breadth-wise and 6 length-wise, power requirement per aerator =.

Hence provide 10 KW aerators @12 per tank to take care of oxygen requirement.

Bonus Question:

4. An activated sludge plant is designed for treating a flow (Q) and influent BOD₅ concentration (S_o) such that the effluent BOD₅ concentration is S. Due to some unforeseen circumstances, the value of S_o increases to 2.S_o, while Q remains the same. Explain what changes must be made to the treatment plant such that the effluent quality does not degrade. Building new tanks is not a viable option. (5)