Environmental Engineering

Concentration of SO_2 in ppm = 0.04
Let's equivalent concentration in =\mu g/m^3 is x.
x\mu g of SO_2 present in 1m^3 of air.
We know, 1 mole of SO_2 (at 0^{\circ}C, 1 atm) has volume of 22.4 L.
at 25^{\circ}C and 1 atm, volume of SO_2
=\frac{22.4}{273+0}\times (273+25)=24.45 lit
\frac{x \times 10^{-6}}{64} mole of SO_2 has volume
\frac{24.45 \times x \times 10^{-6}}{64} lit \; in \; 1m^3 of air.
0.382x \times 10^{-3}m^3 of SO_2 present in 10^{6}m^3 of air.
0.382x \times 10^{-3}=0.04
x=104.7 \approx105
0.04 ppm of SO_2=105\mu g/m^3 of SO_2

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

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

From figure s(x,t)
at x=0\Rightarrow s(0,t)=s_0=5
at x=0.4L\Rightarrow s(0.4L,t)=s_0=5
at x=L\Rightarrow s(L,t)=0
From x = 0\; to \;x = 0.4 L
Concentration of pollutant 5\mu \; mol/m \times (0.4 \times 10m) =20 \mu \; mol
x = 0.4L\; to \;x = L
Concentration of pollutant = 0
Total concentration of pollutant in 10 m =20 \mu \; mol
In infinite time this concentration will be diluted so concentration of pollutant per m
=\frac{20}{10} \mu \; mol/m=2 \mu \; mol/m
Under steady state condition, concentration of pollutant will be uniformly distributed.
Steady state concentration at x=\frac{L}{2} =2\mu \; mol/m

C_{100}H_{250}O_{80}N+xO_2\rightarrow 100CO_2+\frac{247}{2}H_2O+NH_3
\begin{aligned} 80+x \times 2&=200+\frac{247}{2}\\ x&=121.75 \end{aligned}
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}
\begin{aligned} \text{wt. of }O_2&= \frac{121.75}{2744}\times 32 \; tonne\\ \text{Mass of air }&= \frac{1.42}{0.23}\times 10^3 Kg\\ &=6.174 \times 10^3kg\\ \text{Density of air }&= 1.3 Kg/m^3\\ \text{Volume of air }&= \frac{6.174}{1.3}\times 10^3 m^3\\ =4749.2 m^3 &\text{per tonne of MSW} \end{aligned}

HNO_3\rightarrow H^++NO_3^{-}
\begin{aligned} \text{1 mole of }HNO_3&=\text{1 mole of }H^+\\ \text{No. of mole of }HNO_3&=\frac{100}{(1+14+16 \times 3)}\\ &=1.587 m\; mol\\ [H^+] \text{ in 1 lit. of water }&=1.587 m\;mol/Lit\\ &=1.587 \times 10^{-3}\;mol/Lit\\ pH&=-\log[H^+]\\ &=\log[1.587 \times 10^{-3}]\\ &=2.8 \end{aligned}

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.

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

VOC emission = 5 Kg/h
Capturing Efficiency =95%
Fugitive emission = (5/100)*5 Kg/h=0.25 Kg/h

\begin{aligned} V_o&=\frac{500}{5 \times 40} =2.5m/h \\ \eta &=\frac{V_s}{V_o}\times 100=\frac{1}{2.5}=0.4 \; or \; 40% \end{aligned}