Concept:

Productivity of Lake: Productivity is a measure of the ability to support the food chain. It is a measure of algal growth. Higher algal growth leads to decreased water quality. Depending upon the increasing level of productivity, the lakes can be classified as:

(a) Oligotrophic lake

(b) Mesotrophic lake

(c) Eutrophic lake

(d) Senescent lake

Oligotrophic: Oligotrophic lakes have a low level of productivity due to a severely limited supply of nutrients to support algal growth. Euphotic zones extend to hypolimnion which in this case becomes aerobic. Algal growth is negligible.

Mesotrophic: It has medium algal growth i.e. medium productivity and hypolimnion remain aerobic although substantial depletion of oxygen occurs.

Eutrophic: It has fairly high productivity and algal growth. The euphotic zone will partially extend to epilimnion. Hypolimnion will be anaerobic.

Concept:

Trickling Filter:

• The rate of oxidation in a conventional bed is at a peak at the surface and just below the surface, where oxygen may possibly be limiting, and declines down through the bed as the food concentration decreases.

• In a conventional filter bed the microorganisms tend to be stratified with large numbers of bacteria and fungi carrying out carbonaceous oxidation in the upper layers. Their numbers decline with depth into the bed as the carbon content of the feed decreases and nitrifying bacteria become established in the lower layers.
• As the film thickness increases, food and O₂ cannot penetrate deep inside the film layer and endogenous metabolism starts at the film-medium interface. Thus the bond weakens and sloughing starts.

Hence the aerobic layer will be at the surface whereas the anaerobic layer will be deeper inside the film where O₂ cannot penetrate. Facultative bacteria are the predominant organisms in any trickling filter.

Oxidation pond:

• In an aerobic pond, the depth should be 0.5 m at the max so that sunlight can penetrate the whole depth. For totally aerobic pond PST is necessary so that turbidity reduces and sunlight can penetrate the whole depth.
• The actual oxidation pond used for domestic sewage is facultative in which three zones exist throughout the pond depth, viz, aerobic zone at the surface, anaerobic zone at the bottom, and facultative zone at the mid-depth of the pond.

Imhoff Tank:

An Imhoff tank is an improvement over the septic tank, in which the incoming sewage is not allowed to get mixed up with the sludge produced, and the outgoing effluent is not allowed to carry with it large amount of organic load, as in the case of the septic tank. They are sometimes also known as Two-storey Digestion Tanks. It removes 60 to 65% solids and 30 to 40% BOD.

The primary bacteria carrying out the digestion is anaerobic.

Concept:

As per the Indian standard for the maximum permissible limit of different parameters in the effluent disposal in a different type of water is as follows:

Concept:

Sludge Volume Index (SVI): SVI is defined as volume occupied in ml by 1 gm of solids in the mixed liquor after settling for 30 minutes. Its unit is ml/gm. Sludge recirculation and settleability are determined by SVI. SVI indicates the physical state of sludge in the biological aeration system.

\(SVI = \frac{{Volume\;of\;Settled\;Sludge\;\left( {m\ell /L} \right)}}{{MLSS\;Concentration\;\left( {gm/L} \right)}}\)

Calculation:

Volume of settled sludge = 200 ml

MLSS concentration = 2000 mg/L

∴ SVI = \(\frac{{200\;ml/L}}{{2000\; \times \;{{10}^{ - 3\;}}gm/\;L}}\) = 100

Time required for certain amount of D.O at 36 km downstream is given by

t = \(\frac{{Distance\;downstream}}{{Velocity\;of\;flow\;in\;stream}}\)

t = \(\frac{{36\; \times \;{{10}^3}}}{{1.2}}\) = 30000 sec

t = 0.3472 days

Oxygen deficit after time t

D_t = \(\frac{{{{\rm{k}}_{\rm{D}}}{\rm{L}}}}{{{{\rm{k}}_{\rm{R}}}{\rm{\;}}---{\rm{\;}}{{\rm{k}}_{\rm{D}}}}}\left[ {{{exp}^{ - {{\rm{k}}_{\rm{D}}}{\rm{t}}}} - {{exp}^{ - {{\rm{k}}_{\rm{R}}}{\rm{t}}}}} \right]\)\(+ {\rm{\;}}\left( {{{\rm{D}}_o}{\rm{\;}} \times {{exp}^{ - {{\rm{k}}_{\rm{R}}}{\rm{t}}}}} \right)\)

D_t = \(\frac{{0.2\; \times \;19}}{{0.45{\rm{\;}} - 0.20}}\left[ {{{exp}^{ - 0.20{\rm{\;}} \times {\rm{\;}}0.3472}} - {{exp}^{ - 0.45{\rm{\;}} \times {\rm{\;}}0.3472}}} \right]\)\(+ \left[ {0\; \times \;{{exp}^{ - 0.45{\rm{\;}} \times {\rm{\;}}0.3472}}} \right]\)

D_t = 1.17895 mg/l

D.O at 36 km downstream = 12 - 1.17895

Total BOD produced per day \(= \frac{{10000 \times 250 \times 320}}{{{{10}^6}}} = 800\;kg/day\)

Organic loading rate = 200 kg/day/ha

\(\begin{array}{l} \therefore Area\;requried = \frac{{Total\;BOD\;produced}}{{Organic\;loading\;rate}}\\ = \frac{{800}}{{200}} = 4\;ha \end{array}\)

Concept:

i) If the progress of sludge digestion is assumed to be linear:

Volume of the digestor (V) is given by:

\({\rm{V}} = \left( {\frac{{{{\rm{V}}_1} + {{\rm{V}}_2}}}{2}} \right) \times {\rm{t}}\)

ii) For parabolic changes:

\({\rm{V}} = \left( {{{\rm{V}}_1} - \frac{2}{3}\left( {{{\rm{V}}_1} - {{\rm{V}}_2}} \right)} \right) \times {\rm{t}}\)

Where,

V₁ = Raw sludge added per day (m³/d)

V₂ = Equivalent digested sludge produced per day on completion of digestion (m³/day)

t = Digestion period (day)

Calculation:

For parabolic change:

\({{\rm{V}}_1} = \frac{{40000}}{{{{10}^3}}} = 40\;{{\rm{m}}^3}{\rm{\;per\;day}}\)

\({{\rm{V}}_2} = 0.3 \times 40 = 12 \;{{\rm{m}}^3}{\rm{\;per\;day}}\)

Concept:

Biochemical Oxygen demand (BOD): It is the total amount of oxygen required to oxidize the biodegradable organic matter present in wastewater through microbial utilization of organics.

Dilution Ratio \(\frac{{{Q_{Total}}}}{{{Q_{sewage}}}}\)

Calculation:

Q_R = 200 ℓ/s, BOD_UR = 20 mg/ℓ, BOD_US = 400 mg/ℓ , BOD_{U mix} = 70 mg/L

Let the sewage discharge be Q_s

\(BO{D_{U\;mix}} = \frac{{{Q_R}\: \times \:BO{D_{UR}}\: + \:{Q_s}\: \times \:BO{D_{US}}}}{{{Q_R}\: + \:{Q_{s}}}}\)

\(\Rightarrow 70 = \frac{{200\; \times \;20 \;+ \;{Q_s}\; \times \;400}}{{200\; + \;{Q_s}}}\)  ⇒ Q_s = 30.3 ℓ/sec

∴ Dilution Ratio \(= \frac{{{Q_{Total}}}}{{{Q_{s}}}} = \frac{{{Q_R}\; + \;{Q_s}}}{{{Q_s}}} = \frac{{200\; + \;30.3}}{{30.3}} = 7.60\)

Concept:

\({\rm{Sludge\;age}} = {{\rm{θ }}_{\rm{C}}} = \frac{{{\rm{V}} \times {{\rm{X}}_{\rm{T}}}}}{{{{\rm{Q}}_{\rm{W}}} \times {{\rm{X}}_{\rm{R}}} + \left( {{\rm{Q}} - {{\rm{Q}}_{\rm{W}}}} \right) \times {{\rm{X}}_{\rm{E}}}}}\)

Where

X_T = Concentration of solids in the influent of the Aeration Tank, called the MLSS, i.e. Mixed Liquor Suspended Solids (in mg/l).

V = Volume of Aerator

Q_W = Volume of wasted sludge per day.

X_R = Concentration of solids in the returned sludge or in the wasted sludge (both being equal) (in mg/l).

Q = Sewage inflow per day

X_E = Concentration of solids in the effluent in mg/l

Calculation:

Q = 30000 m³/day, V = 11000 m³, X_T = 2500 mg/l, X_R = 9500 mg/l, Q_W = 220 m3/day, X_E = 30 mg/L, 

\({{\rm{θ}}_{\rm{C}}} = \frac{{11000 \times 2500}}{{220 \times 9500 + \left( {30000 - 220} \right) \times 30}}\)

Concept:

Standard Rate trickling Filter removes BOD upto 90% and the efficiency of standard rate trickling filter is given as

\(\eta = \frac{{100}}{{1\; +\; 0.44\;\sqrt {\frac{{{Q_0}{S_0}}}{V}} }}\)

Where,

Q₀ = Discharge in L passing through the filter per day.

S₀ = BOD applied in kg.

V = Volume of the trickling filter media (in m³).

The diameter of the trickling filter is limited to a maximum of 60 m.

Calculation:

Q₀ = 150 m³/day = 150 × 10³ L/day, S₀ = 180 mg/L

\(\therefore \eta = \frac{{100}}{{1 + 0.44\;\sqrt {\frac{{150\; \times\; {{10}^3}\; \times \;180\; \times \;{{10}^{ - 6}}}}{V}} }}\)

\(\Rightarrow 90 = \frac{{100}}{{1\; + \;0.44\;\sqrt {\frac{{27}}{V}} }} \Rightarrow V = 423.40\;{m^3}\)

Let the diameter of the filter be D.

\(\therefore \;\frac{\pi }{4}{D^2} \times h = 423.40 \Rightarrow \frac{\pi }{4} \times {D^2} \times 2 = 423.40\)

⇒ D = 16.42 m < 60 m