Municipal Solid Waste Management Practice Questions

We are given: OFMSW bulk density = $315\ \text{kg/m}^3$ , water content = $30$ Sludge bulk density = $700\ \text{kg/m}^3$ , water content = $70$ Final mixture water content = $40$ Let $1\ \text{kg}$ of OFMSW be mixed with $x\ \text{kg}$ of sludge Step 1: Mass of water and solids in OFMSW Water in OFMSW = $1 \times 0.30 = 0.30\ \text{kg}$ Solids in OFMSW = $1 - 0.30 = 0.70\ \text{kg}$ Step 2: Mass of water and solids in sludge Water in sludge = $x \times 0.70 = 0.70x\ \text{kg}$ Solids in sludge = $x - 0.70x = 0.30x\ \text{kg}$ Step 3: Total water and solids in the mixture Total water = $0.30 + 0.70x$ Total solids = $0.70 + 0.30x$ Total mass = $1 + x$ Water content of mixture = $\frac{0.30 + 0.70x}{1 + x} = 0.40$ Step 4: Solve for x $\frac{0.30 + 0.70x}{1 + x} = 0.40$ $0.30 + 0.70x = 0.40(1 + x)$ $0.30 + 0.70x = 0.40 + 0.40x$ $0.30x = 0.10$ $x = \frac{0.10}{0.30} = 0.33$ Step 5: Density of mixture Total mass = $1 + 0.33 = 1.33\ \text{kg}$ Volume of OFMSW = $\frac{1}{315} = 0.00317\ \text{m}^3$ Volume of sludge = $\frac{0.33}{700} = 0.0004714\ \text{m}^3$ Total volume = $0.00317 + 0.0004714 = 0.0036414\ \text{m}^3$ Mixture density = $\frac{1.33}{0.0036414} \approx 365\ \text{kg/m}^3$

I. Various MSW separation techniques (a) Screening process (b) Size reduction process Reduction in size of MSW can be done using following mechanical devices: (a) Shredders (b) Flail mill hammer (c) Glass cruster (c) Density separation process It is the process of separation of light materials such as paper and plastic from heavy materials such as metals. It is carried out using air density separators. (d) Magnetic separation process It is the process of separating ferrous metals from MSW It is carried out using magnetic separators. (e) Current separation process It is the process of separating aluminium and copper from MSW. It is carried out using Eddy current separator. II. Various MSW treatment technique (a) Sanitary landfills (b) Composting (c) Incineration (d) Pyrolysis Hence, correct option is (B).

Organic waste $=\mathrm{C}_{240} \mathrm{O}_{200} \mathrm{H}_{180} \mathrm{~N}_{5} \mathrm{~S}$ Weight of organic waste $=1 \mathrm{~kg}$ Burning of organic waste converts sulphur into sulphur dioxide as follows: $\mathrm{C}_{240} \mathrm{O}_{200} \mathrm{H}_{180} \mathrm{~N}_{5} \mathrm{~S}+\mathrm{O}_{2} \rightarrow \mathrm{xCO}_{2}+\mathrm{yH}_{2} \mathrm{O}+\mathrm{zH}_{2} \mathrm{O}+\mathrm{SO}_{2}$ i.e. 6362 of organic waste produces $64 \mathrm{~g}$ of sulphur dioxide. So, $1 \mathrm{~kg}$ of organic waste produces $\frac{64}{6362} \times 1000 \mathrm{gm}$ of $\mathrm{SO}_{2}$ So, Amount of $\mathrm{SO}_{2}$ produced $=10 \; \mathrm{grams} / \mathrm{kg}$

Assuming $1 \mathrm{Kg}$ of solid waste

Energy produce from $1 \mathrm{Kg}$ work $=13.6 \mathrm{~mJ}$ Moisture content $=26 \%$ Solid content $=100-26=74 \%$ Therefore $1 \mathrm{Kg}$ of waste consist of $=0.74 \mathrm{Kg}$ dry mass $\therefore \quad$ Energy content based on dry weight

(A) is false. The Higher Heating Value (HHV) of Refuse-Derived Fuel (RDF) processed from Municipal Solid Waste (MSW) is typically higher than the HHV of unprocessed MSW. This is because RDF processing removes non-combustible materials from MSW and concentrates the combustible fraction. (B) is true. RDF can be produced in various forms including pellets, fluff, and powdered form. The form of RDF depends on the specific process and intended end use. (C) is partially true. RDF processing typically removes a significant portion of the inorganic fraction of MSW, but not all of it. The remaining inorganic fraction is typically mixed with the RDF product. (D) is false. RDF can be used as a fuel in a variety of combustion systems, including those that burn oil. In fact, RDF is often used as a supplemental fuel in cement kilns and other industrial processes that also use oil as a fuel.

In hauled container system, 'Haul' time includes. The time required to reach the disposal site, starting after the container whose contents are to be emptied has been loaded on the track, plus the time after leaving the disposal site until the truck arrives at the location where the empty container is to be redeposited. Time spent at the disposal site is not included.

Wt. of water initially present $=frac{10}{100} \times 30=3mT$ Wt. of solid $=30-3=27mT$ In ideal condition, Wt. of solid $=27mT$ $M/C=50$ Wt. of water $=27mT$ Amount of water needed to get ideal condition $= 27 - 3 = 24 mT$

Typical speicifiv weight and moisture content data for residential, commercial, industraial, and agricultural wastes. Moisture content Mcf > Mcp > Mcg Typical values for short residue and energy conent of residentail MSW. Energy content ECpp > ECf > ECg

$C_{100}H_{250}O_{80}N+xO_2\rightarrow 100CO_2+\frac{247}{2}H_2O+NH_3$

1 mole of MSW required 121.75 mole of oxygen No. of moles of MSW $=\frac{1 \; ton}{(100 \times 12+250+80 \times 16+14)g}$ No. of moles of O2 required $=121.75 \times \frac{1 \; tonne}{2744g}$

Some heuristic guidelines that should be taken into consideration when laying out routes are as follows : 1. Existing policies and regulations related to such items as the point of collection and frequency of collection must be identified. 2. Existing system characteristics such as crew size and vehicle types must be coordinated. 3. Wherever possible, routes should be laid out so that they begin and end near arterial streets, using topographical and physical barriers as route boundaries. 4. In hilly area, routes should start at the top of the grade and proceed downhill as the vehicle becomes loaded. 5. Routes should be laid out so that the last container to be collected on the route is located nearest to the disposal site. 6. Waste generated at traffic-congested locations should be collected as early in the day as possible. 7. Sources at which extremely large quantities of waste are generated should be serviced during the first part of the days. 8. Scattered pickup points (where small quantities of solid waste are generated) that receive the same collection frequency should, if possible, be serviced during one trip or on the same day.

Solid waste generated by the use of raw material, processing and finished goods

Let 1 kg of row municipal solid waste (MSW) Decomposable material = 70% of 1 kg = 0.7 kg Water content = 35% Weight of solids in MSW $=\left(\frac{100-35}{100}\right)\times0.7=0.455\;\mathrm{kg}=455\;\mathrm{gms}$ Given: C:H:O:N: other = 44 : 6: 43: 0.8:6.2

Total no. of kms. Travelled by all the vehicles $=20000\times 12000\: km=24\times 10^{7}\: km$ Total $NO_{x}$ emission $=2\: g/km\times 24\times 10^{7}\: km$ $=48\times 10^{7}\: g$ $=48\times 10^{4}\: kg$

Coal burned in one year

Sulfur content= $\frac{2}{100}\times 1.5768\times 10^{7}\times 10^{-6}$ $=315.36\; tonnes/year$ Sulphur content in ash

Sulphur converted to $SO_{2}=315.36-18.92$ $=296.44\; tonnes/year$ $\; \; \; S\, +\, O_{2}\rightarrow SO_{2}$ 1 mole of S is present in 1 mole of $SO_{2}$ 32 gm of S is present in 64 gm of $SO_{2}$ $\therefore$ Rate of $SO_{2}$ emission $=\frac{64}{32}\times 296.44=592.88\: tonnes/year$

Currently no explanation available

Energy on dry basis: It will be the total energy when whole mass is dry In current situation only 60% mass is dry energy corresponding to 60% dry mass=5805 kJ/kg Energy corresponding to 100% dry mass= $\frac{5805}{60}\times 100=9675\: kJ/kg$ So, energy as on dry basis=9675 kJ/kg Energy based on as discharged basis=5805 kJ/kg So, difference=9675-5805=3870 kJ/kg

Let density of sludge is $\rho$ , $\therefore \; \; \frac{c_{1}+c_{2}}{\rho }=\frac{c_{1}}{\rho _{1}}+\frac{c_{2}}{\rho _{2}}$ Assuming $c_{1}+c_{2}=100$ $\therefore \; \; \rho =\frac{100}{\frac{c_{1}}{\rho _{1}}+\frac{c_{2}}{\rho _{2}}}$

Solid waste generated per day $=2\times 200000=400000\: kg$ Total solid waste generated in 25 years $=400000\times 365\times 25=3.65\times 10^{9}\: kg$ Compaction ratio, $4=\frac{Volume\; before\; compaction}{Volume\; after\; compaction}$ Volume of waste before compaction, $V=\frac{3.65\times 10^{9}}{100}=3.65\times 10^{7}\: m^{3}$ Volume of waste after compaction, $V'=\frac{3.65\times 10^{7}}{4}=9.125\times 10^{6}\: m^{3}$ $\frac{SW+cover}{SW}=\frac{SW}{SW}+\frac{cover}{SW}=1.5$ $\frac{Cover}{SW}=0.5$ $Cover=0.5\times 9.125\times 10^{6}$ $\; \; \; =4.5625\times 10^{6}\: m^{3}$

Incineration can be adopted when the calorific value of the MSW is high. Land fill can be adopted when the density of MSW is high. Since in the given MSW the quality of inorganic material is low, its density is less. Composting can be adopted when the MSW contains high organic content. Barging the MSW into the sea is now a days generaly not used and has become obsolete.