Disposal Of Sewage Effluents Practice Questions

As glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ does not contains sulphur hence anaerobic decomposition of glucose does not release $\mathrm{H}_{2} \mathrm{~S}$ . Anaerobic decomposition of glucose gives $\mathrm{CO}_{2}$ , $\mathrm{CH}_{4}, \mathrm{H}_{2} \mathrm{O}$ as its byproducts.

Green House Gas allow short wave radiation to pass through to the earth surface and absorbs the long wavelength radiation coming from the earth surface.

A bulking agent (wood ships shredded leaves or types) is intended to support the structure of sludge by increasing its porosity to encourage effective aeration. In-vessel composting systems can be divided into plug flow and agitated bed modes. In plug flow mode, the relationship between particles in the composting mass stays the same through the process and the system operates on first-in, first-out principle. In an agitated bed system, composting material is mixed mechanically during the processing. The heat produced in aquatic composting cause any surplus moisture to be extracted as water vapour and hence no leachate is produced. Actinomycetes are aerobic, spore forming positive bacteria. These are thermophilic species and their development depends on aerobic conditions, temperature and water content. The heat produced in aerobic composting helps support the growth of these species. Leachate can generated from vegetable wastes.

Wt. of TSS at inlet of PST $=5000 \times 10^3\frac{L}{d} \times 200 \times 10^{-6}Kg/L=1000Kg/d$ Wt. of solid in sludge from PST $=0.6 \times 1000=600 Kg/d$ Wt. of sludge before thikeninig $=\frac{600}{\frac{2}{100}}=30000 Kg/d$ Wt. of sludge before thikeninig $=\frac{600}{\frac{6}{100}}=10000 Kg/d$ Daily volume of thickened sludge $=frac{10000}{1000}=10m^3$

Total weight generated by city $=40\times 10^{6}\: kg/year$ Weight of MSW going into landfill $=0.9\times 40\times 10^{6}\: kg/year$ $=36\times 10^{6}\: kg/year$ Compacted density $=550\: kg/m^{3}$ Compacted volume of MSW $=\frac{36\times 10^{6}\: kg/year}{550\: kg/m^{3}}$ $=65454.5454\: m^{3}/year$ Total landfill volume = Vol. of MSD + Vol. of cover Given, Volume of MSW $=0.8\times$ Total landfill volume $\therefore$ Vol. of cover $=0.2\times$ Total landfill volume $\therefore$ Total landfill volume $=\frac{65454.5454}{0.8}m^{3}/year =81818.18175m^{3}/year$ Height of landfill $=3\: m$ $\therefore$ Area of lanfill $=\frac{81818.18175}{3} =27272.7\: m^{2}/year$

As given rate of aeration is less than rate of degradation which decreases with time/distance, minimum DO is observed downstream of point of disposal 'S' where both rate of a reaction and degradation becomes equals.

Currently no explanation available