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Determine the Safe height of the Chimney required for the safe Dispersion of the Pollutants

A factory uses 1.5 ML of fuel oil per month. The exhaust gases from the factory contain the following quantities of pollutants per ML per year.

(i)Particulate matter = 4 t/year

(ii) $SO_{2}$ : 20 t/year

(iii) NOx: 5 t/year

(iv) HC, CO and other : 3 t/year

Determine the safe height of the chimney required for the safe dispersion of the pollutants.

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The concentrations of $\mathrm{NO} x, \mathrm{HC}, \mathrm{CO}$ and others are generally very much less than the concentration of $\mathrm{SO}_{2}$ in various industries. Hence the Board has made only $\mathrm{SO}_{2}$ as the criterion for design, along with the particulate matter. (a) Height of chimney on the basis of particulate matter. $$ h=74\left(Q_{p}\right)^{0.27} $$ Particulate matter emission $=4(1.5 \times 12)=72 \mathrm{t} /$ year Assuming 300 working days in a year, and 24 hours working a day. $$ \begin{aligned} Q_{p} &=\frac{72}{300 \times 24}=0.01 \mathrm{t} / \mathrm{hour} \\ h &=74(0.1)^{0.27}=21.34 \mathrm{~m} \end{aligned} $$ (b) Height of chimney on the basis of $\mathrm{SO}_{2}$ Now, $$ \mathrm{SO}_{2} \text { emission }=20(1.5 \times 12)=360 \mathrm{t} / \text { year } $$ $\therefore$ $$ \begin{aligned} Q &=\frac{360}{300 \times 24}=0.05 \mathrm{t} / \text { year }=50 \mathrm{~kg} / \mathrm{hr} \\ h &=14(50)^{1 / 3}=51.58 \mathrm{~m} \end{aligned} $$ $$ \therefore \quad h=14(50)^{1 / 3}=51.58 \mathrm{~m} $$ $\therefore$ Required height of chimney $=52 \mathrm{~m}$ (maximum of $21.34 \mathrm{~m}$ and $51.58 \mathrm{~m}$ )

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