Calculating Downstream Concentration Using Mass Balance

This problem involves mixing two streams with different flow rates and chloride concentrations. Since chloride is treated as a conservative substance and complete mixing is assumed, the principle of mass balance can be applied.

The total mass of chloride entering the mixing point must equal the total mass of chloride leaving the mixing point per unit time.

Mass Balance Calculation

Let:

• $Q_1$ = Flow rate of the first stream = 10.0 m$^3$/s
• $C_1$ = Chloride concentration of the first stream = 20.0 mg/L
• $Q_2$ = Flow rate of the tributary = 5.0 m$^3$/s
• $C_2$ = Chloride concentration of the tributary = 40 mg/L
• $Q_{downstream}$ = Downstream flow rate = $Q_1 + Q_2$
• $C_{downstream}$ = Downstream chloride concentration (to be calculated)

The mass balance equation is:

$ Q_1 C_1 + Q_2 C_2 = (Q_1 + Q_2) C_{downstream} $

To find the downstream concentration, we rearrange the equation:

$ C_{downstream} = \frac{Q_1 C_1 + Q_2 C_2}{Q_1 + Q_2} $

Now, substitute the given values:

$ C_{downstream} = \frac{(10.0 \, \text{m}^3/\text{s} \times 20.0 \, \text{mg/L}) + (5.0 \, \text{m}^3/\text{s} \times 40 \, \text{mg/L})}{10.0 \, \text{m}^3/\text{s} + 5.0 \, \text{m}^3/\text{s}} $

Calculate the numerator:

$ \text{Numerator} = (200.0 \, \text{m}^3/\text{s} \cdot \text{mg/L}) + (200.0 \, \text{m}^3/\text{s} \cdot \text{mg/L}) = 400.0 \, \text{m}^3/\text{s} \cdot \text{mg/L} $

Calculate the denominator:

$ \text{Denominator} = 15.0 \, \text{m}^3/\text{s} $

Finally, calculate the downstream concentration:

$ C_{downstream} = \frac{400.0 \, \text{m}^3/\text{s} \cdot \text{mg/L}}{15.0 \, \text{m}^3/\text{s}} $

$ C_{downstream} \approx 26.67 \, \text{mg/L} $

Rounding to one decimal place, the downstream concentration is 26.7 mg/L.