Environmental Engineering Test 4

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:

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:

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