Question

Consider the air over a city to be a box 100km on a side that reaches up to an altitude of 1.0 km. Clean air is blowing into the box along one of its sides with a speed of 4 m/s. Suppose an air pollutant with a reaction rate k=0.20/hr is emitted into the box at a total rate of 10.0kg/s. Find the steady-state concentration if the air is assumed to be completely mixed.

Answer 1

Explanation:

It is known that equation for steady state concentration is as follows.

            $C_{a} = \frac{QC}{Q + kV}$

where,   Q = flow rate

              k = rate constant

              V = volume

              C = concentration of the entering air

Formula for volume of the box is as follows.

                 V = $a^{2}h$

                    = $100 \times 100 \times 1$

                    = $10000 km^{3}$

Now, expression to determine the discharge is as follows.

                  Q = Av

                      = $100 \times 1 \times \frac{4 m}{s} \times \frac{km}{1000 m}$

                      = 0.4 $km^{3}/s$

And,    m (loading) = 10kg/s,

           k = 0.20/hr

as   1 $km^3 = 10^{12}$ L (if u want kg/L as concentration)

Now, calculate the concentration present inside as follows.

     $C_{in} = \frac{10kg/s}{0.4 km^3/s}$

                 = 25 $kg/km^3$

Now, we will calculate the concentration present outwards as follows.

       $C_{out} = {C_{in}}{(1 + k \times t)}$,

and,      t = $\frac{V}{Q}$

               = 25000 s or 6.94 hr

Hence,   $C_{out} = \frac{25}{(1 + 0.20 \times 6.94)}$

                         = 10.47 $kg/km^3$

Thus, we can conclude that the the steady-state concentration if the air is assumed to be completely mixed is $C_{out} = 10.47 kg/km^3$ and $C_{in} = 25 kg/km^3$ .