**WASTEWATER
TREATMENT-V**

__SOLUTION__

1. Describe the differences in the nature of recycling
between activated sludge process and trickling filter. **(3)**

**Solution:**

·
The sludge or
biomass is recycled in the activated sludge process ·
Purpose of
recycling is to maintain a high biomass concentration in the aeration tank |
·
Treated
wastewater is recycled in the trickling filter ·
Purpose of
recycling is to maintain adequate hydraulic loading rate, without changing
organic loading rate, so that all portions of the filter may be wetted
adequately all the time. |

2. A tricking filter with the following dimensions is
available. Depth: 2 m, Surface area: 150
m^{2}. The media consists of
stones of 7-10 cm diameter. This filter
will be used to treat 0.6 MLD wastewater with BOD_{5} = 300 mg/L. The trickling filter will be operated in the
high-rate mode, i.e., OLR: 0.48 – 0.96 Kg/m^{3}/d, HLR: 10 – 40 m^{3}/m^{2}/d,
re-circulation ratio: 1-2. Based on this
information, calculate the expected BOD_{5} removal efficiency.

**Hint:**

_{}, _{}, _{}

Where, S_{o} = BOD_{5} in Raw Wastewater, mg/L

S_{e} = Total
BOD_{5} of settled effluent from the filter, mg/L

S_{a}
= Total
BOD_{5} of wastewater applied to the filter, mg/L

k = Treatability
constant, 2.36

D = Depth
of the Trickling Filter, m

Q = Total
Flow rate applied to the filter without recirculation, m^{3}/d

A = Surface Area of the Trickling Filter, m^{2}

n = 0.5

V = Volume of the Trickling Filter, m^{3} **(7)**

**Solution:**

Volume
of trickling filter = D.A = (2).150 = 300 m^{3}

Organic
Loading Rate (OLR) = _{} (within limits)

Without
recycle, Hydraulic Loading Rate (HLR) = _{}

(The
above value is inadequate and hence must be increased). So, let R = 2

Hence,
HLR = _{}. This value is
adequate.

Now, to calculate BOD_{5} removal efficiency.

It
is known that, * _{}*,
Where,

Also, _{}

or, _{} or, _{}

or, _{};

Therefore,
BOD_{5} removal efficiency: _{} or, 96.64%