**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 (C_{6}H_{12}O_{6})
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 BOD_{5}.
Y_{T} and K_{d} values for microbial
degradation of glucose are 0.5 mg/mg and 0.05 /d respectively (calculated based
on BOD_{5} 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 BOD_{5} = _{}, i.e., S_{o}
= 284.44 mg/L

Also, BOD_{u}
for 200 mg/L glucose solution is _{}, and corresponding BOD_{5} = _{}, 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 m^{3}, 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 m^{2} for 1,
2, and 5 KW, and 36 m^{2} 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 O_{2}/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^{o}C. The steady state dissolved oxygen
concentration in the aeration tank should be 1 mg/L. Saturation concentration of oxygen in water
at 20^{o}C is 9.1 mg/L and at 30^{o}C 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 = 25m^{2 }or 36m^{2} 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 BOD_{5} concentration (S_{o}) such that
the effluent BOD_{5} 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)**