Question

A city of area 20 km × 40 km receives wind perpendicular to the 20 km side at 4 m/s. Inversion layer height = 100 m. PM$_1$ emission rate = 1 kg/s. Find steady-state PM$_1$ concentration (µg/m$^3$), rounded to one decimal place, using a Box model.

Hint:

Box model: concentration equals emission rate divided by ventilation flow (UHL). Always convert the final concentration to µg/m$^3$.

Answer 1

Correct Answer: 125

Step 1: City dimensions. \[ L = 20\ \text{km} = 20000\ \text{m},\quad W = 40\ \text{km} = 40000\ \text{m} \] Wind blows across width: \[ U = 4\ \text{m/s} \] Mixing height: \[ H = 100\ \text{m} \] Step 2: Box model concentration. Steady-state: \[ C = \frac{Q}{U\, H\, L} \] \[ C = \frac{1}{4 \times 100 \times 20000} \] \[ C = \frac{1}{8 \times 10^6} = 1.25 \times 10^{-7}\ \text{kg/m}^3 \] Convert to µg/m$^3$: \[ 1\ \text{kg/m}^3 = 10^{9}\ \mu\text{g/m}^3 \] \[ C = 1.25 \times 10^{-7} \times 10^{9} = 125\ \mu\text{g/m}^3 \] Rounded: \[ \boxed{125.0\ \mu\text{g/m}^3} \]