Question

In a stack emission measurement at an industry, the stack cross-sectional area at 30 m height was divided into four equal sectors. The measured velocities and SO₂ concentrations through these sectors at this height are given in the table.
The mean SO₂ concentration from the stack is ______ mg/m³ (rounded off to two decimal places).

Sector Number	Velocity (m/s)	SO2 Concentration (mg/m3)
1	15	1000
2	17	1150
3	19	1250
4	21	1275

Hint:

Since the sectors have equal area, weight each concentration by its own velocity and divide by the sum of velocities, rather than taking a plain arithmetic mean.

Answer 1

Correct Answer: 1181.6

Step 1: Recognise that the mean concentration must be a flow-weighted average.
Since the stack cross-section is divided into four sectors of equal area, the volumetric flow rate through a sector is proportional to its measured velocity \(V_i\). The overall mean concentration leaving the stack is therefore the flow-weighted average \[ \bar{C} = \frac{\sum V_i C_i}{\sum V_i} \] not a simple arithmetic average of the four concentrations.

Step 2: Compute the product \(V_i C_i\) for each sector.
\[ V_1 C_1 = 15 \times 1000 = 15000 \] \[ V_2 C_2 = 17 \times 1150 = 19550 \] \[ V_3 C_3 = 19 \times 1250 = 23750 \] \[ V_4 C_4 = 21 \times 1275 = 26775 \]

Step 3: Sum the products and the velocities.
\[ \sum V_i C_i = 15000 + 19550 + 23750 + 26775 = 85075 \] \[ \sum V_i = 15 + 17 + 19 + 21 = 72 \]

Step 4: Divide to get the mean concentration.
\[ \bar{C} = \frac{85075}{72} = 1181.597\ \text{mg/m}^3 \]

Step 5: State the final answer.
Rounding to two decimal places, the mean SO2 concentration from the stack is \(1181.60\) mg/m\(^3\).