Determine the effective height of a stack, with the following given data:

(a) Physical stack is 180 m tall with 0.95 m inside diameter.

(b) Wind velocity is 2.75 m/sec

(c) Air temperature is 20°C

(d) Barometric pressure is 1000 millibars

(e) Stack gas velocity is 11.12 m/sec

(f) Stack gas temperature is 160°C

The given data is symbolised as below: $\begin{array}{lll}h=180 \mathrm{~m} & D=0.95 \mathrm{~m} & u=2.75 \mathrm{~m} / \mathrm{sec} \\ T_{a}=20^{\circ} \mathrm{C}=20+273=293 \mathrm{~K} & P=1000 \text { millibars } & \mathrm{v}_{\mathrm{s}}=11.12 \mathrm{~m} / \mathrm{sec} \\ T_{s}=160^{\circ} \mathrm{C}=160+273=433 \mathrm{~K} & & \\ \text { Using equation, we have } & & \end{array}$ $$ \begin{aligned} \Delta h &=\frac{v_{s} \cdot D}{u}\left[1.5+2.68 \times 10^{-3} P D\left(\frac{T_{s}-T_{a}}{T_{s}}\right)\right] \\ &=\frac{11.12 \times 0.95}{2.75}\left[1.5+2.68 \times 10^{-3} \times 1000 \times 0.95 \times \frac{433-293}{433}\right] \\ &=\frac{11.12 \times 0.95}{2.75}\left[1.5+\frac{2.68 \times 0.95 \times 140}{433}\right]=8.92 \mathrm{~m} \\ \Delta h &=8.92 \mathrm{~m} \\ H &=\text { Effective height of stack } \\ &=h+\Delta h=180+8.92=188.92 \mathrm{~m} \end{aligned} $$