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

It helps to see why a simple average of the four concentrations would be wrong, and the fastest way to do that is to bring in an actual sector area $A$ and work out real mass and volume flow rates, even though $A$ will cancel out in the end. Let each of the four equal sectors have cross-sectional area $A$. The volumetric flow through sector $i$ is $q_i = A V_i$, and the mass flow rate of SO2 through that sector is $m_i = q_i C_i = A V_i C_i$. The total volumetric flow leaving the stack is

\[ Q_{total} = A(V_1+V_2+V_3+V_4) = A(15+17+19+21) = 72A \]

and the total mass flow rate of SO2 is

\[ M_{total} = A(V_1C_1+V_2C_2+V_3C_3+V_4C_4) \]

Computing each term: $V_1C_1 = 15(1000) = 15000$, $V_2C_2 = 17(1150) = 19550$, $V_3C_3 = 19(1250) = 23750$, $V_4C_4 = 21(1275) = 26775$. Summing, $M_{total} = A(15000+19550+23750+26775) = 85075A$. The true mean concentration leaving the stack is, by definition, the total mass flow divided by the total volumetric flow,

\[ \bar{C} = \frac{M_{total}}{Q_{total}} = \frac{85075A}{72A} = \frac{85075}{72} \]

The sector area $A$ cancels exactly, confirming that only the velocity-weighted ratio matters. Evaluating, $\bar{C} = 1181.597\ \text{mg/m}^3$. Rounding to two decimal places,

\[\boxed{\bar{C} = 1181.60\ \text{mg/m}^3}\]