Test: Environmental Engineering- 4 - Civil Engineering (CE) MCQ

Eutrophic Lakes: These are the lakes of high productivity level, hence support high growth of algae.

Mesotrophic Lakes: These are the lakes of intermediate productivity level, hence support intermediate growth of algae. 

Oligotrophic Lakes: These are the lakes of least productivity level, hence support negligible growth of algae.

Mean cell residence time in activated sludge process in aeration tank is in range of 4 to 15 days.

The ultimate BOD value of waste is the amount of biodegradable organic matter presents in the sewage sample. At any temperature the amount of organic matter present does not change, so the ultimate BOD will remain same for any temperature.

[k_D]_T^o_C = [k_D]₂₀^o_C [1.047]^{T - 20}               

[k_D]₄₀^o_C = 0.13 [1.047]^{40 - 20}               

[k_D]₄₀^o_C = 0.13 [1.047]²⁰               

[k_D]₄₀^o_C = 0.326 per day

Autoclaving:- The method used for the disposal of biomedical waste is , in which it is brought in closed contact with steam under controlled pressure and temperature condition in order to carry out disinfection.

Pulverization and Shredding: Basically this is not a method of disposal but in these method heavier solids are broken into the lighter one either by cutting action or by grinding action.

Pulverization refers to the action of crushing and grinding, whereas shredding refers to the action of cutting and tearing.

Pyrolysis: is a method of disposal of the refuse in which burning of refuse is generally done at high temperature in the almost absence of oxygen.

Concept:

Ultimate BOD = L_o

BOD_5­ = 5day BOD, of water/waste water

BOD at any time ‘t’ is (L_t)
Lt = L_o(1 − e^−kDt)

Where L_o = ultimate BOD

K_D = Rate constant

Calculation:

L_t = 300 × (1 – e^{-0.3585 × t})

L₅ = BOD₅ = 300 × (1 – e^{-0.3585 × 5})

= 250 mg/L

Relative stability (Sr) at 20°C
S_r = 100 [1 − (0.794)^t]
= 100 [1− (0.794)²⁰]
= 99.01%

Q = 0.9 MLD

5 day BOD, Y₅ = 200 mg/L

= 2000

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

t = 0.3472 days

Oxygen deficit after time t

 

D_t = 2.35 mg/l

D.O at 36 km downstream = 12 - 2.35

= 9.65 mg/l