Calculating Pollutant Concentration After River Mixing

This problem involves calculating the resulting concentration of a pollutant when a tributary stream carrying a certain flow rate and pollutant concentration mixes completely with a larger river also carrying its own flow rate and pollutant concentration. This is a classic mass balance problem.

The principle of mass balance states that in a closed system (assuming no pollutant is gained or lost during mixing), the total mass of the pollutant entering the mixing zone per unit time must equal the total mass of the pollutant leaving the mixing zone per unit time.

Let's define the terms:

• Q_t: Flow rate of the tributary
• C_t: Concentration of pollutant 'X' in the tributary
• Q_r: Flow rate of the river
• C_r: Concentration of pollutant 'X' in the river
• Q_mix: Flow rate of the mixed stream downstream
• C_mix: Concentration of pollutant 'X' in the mixed stream downstream

Given values from the problem:

• Q_t = 4 m³/s
• C_t = 12 mg/L
• Q_r = 8 m³/s
• C_r = 30 mg/L

The total flow rate of the mixed stream downstream is the sum of the individual flow rates:

\( Q_{mix} = Q_t + Q_r \)

The total mass flow rate of the pollutant into the mixing zone is the sum of the mass flow rates from the tributary and the river. Mass flow rate is calculated as flow rate multiplied by concentration (\( \text{Mass Flow} = Q \times C \)).

\( \text{Total Mass Flow In} = (Q_t \times C_t) + (Q_r \times C_r) \)

The total mass flow rate of the pollutant out of the mixing zone is the flow rate of the mixed stream multiplied by its concentration.

\( \text{Total Mass Flow Out} = Q_{mix} \times C_{mix} \)

By the principle of mass balance:

\( \text{Total Mass Flow In} = \text{Total Mass Flow Out} \)

\( (Q_t \times C_t) + (Q_r \times C_r) = Q_{mix} \times C_{mix} \)

Substituting \( Q_{mix} = Q_t + Q_r \) into the equation:

\( (Q_t \times C_t) + (Q_r \times C_r) = (Q_t + Q_r) \times C_{mix} \)

We can rearrange this equation to solve for \( C_{mix} \), the concentration in the mixed stream:

\( C_{mix} = \frac{(Q_t \times C_t) + (Q_r \times C_r)}{Q_t + Q_r} \)

Now, let's plug in the given values:

• \( Q_t = 4 \, \text{m}^3/\text{s} \)
• \( C_t = 12 \, \text{mg/L} \)
• \( Q_r = 8 \, \text{m}^3/\text{s} \)
• \( C_r = 30 \, \text{mg/L} \)

Calculate the numerator (total mass flow in):

\( (4 \, \text{m}^3/\text{s} \times 12 \, \text{mg/L}) + (8 \, \text{m}^3/\text{s} \times 30 \, \text{mg/L}) \)

\( (48 \, (\text{m}^3/\text{s}) \times (\text{mg/L})) + (240 \, (\text{m}^3/\text{s}) \times (\text{mg/L})) \)

\( 48 + 240 = 288 \, (\text{m}^3/\text{s}) \times (\text{mg/L}) \)

Calculate the denominator (total flow rate out):

\( Q_t + Q_r = 4 \, \text{m}^3/\text{s} + 8 \, \text{m}^3/\text{s} = 12 \, \text{m}^3/\text{s} \)

Now, calculate \( C_{mix} \):

\( C_{mix} = \frac{288 \, (\text{m}^3/\text{s}) \times (\text{mg/L})}{12 \, \text{m}^3/\text{s}} \)

\( C_{mix} = 24 \, \text{mg/L} \)

The concentration of pollutant X in the downstream mixed water is 24 mg/L.

Let's compare this result with the given options:

Our calculated value of 24 mg/L matches Option 4.

Pollutant Mixing Calculation Revision

To ensure understanding, let's review the key information used in the pollutant mixing calculation:

The total flow downstream is \( Q_{mix} = Q_t + Q_r = 4 + 8 = 12 \) m³/s.

The total mass flow rate of pollutant X downstream is \( (Q_t \times C_t) + (Q_r \times C_r) = (4 \times 12) + (8 \times 30) = 48 + 240 = 288 \) (m³/s) \times (mg/L). Note that the units \( (\text{m}^3/\text{s}) \times (\text{mg/L}) \) represent mass per unit time (e.g., kg/s, since 1 m³ = 1000 L, so (m³/s)*(mg/L) = (1000 L/s)*(mg/L) = 1000 mg/s = 1 g/s = 0.001 kg/s).

The final concentration \( C_{mix} \) is the total mass flow rate divided by the total volume flow rate:

\( C_{mix} = \frac{288 \, (\text{m}^3/\text{s}) \times (\text{mg/L})}{12 \, \text{m}^3/\text{s}} = 24 \, \text{mg/L} \)

Additional Information on Pollutant Mixing and Mass Balance

The method used here is based on the principle of conservation of mass. When two streams mix, the total mass of a substance (like a pollutant) per unit time is conserved, assuming no reactions or settling occur in the mixing zone. This assumes complete mixing, meaning the pollutant is uniformly distributed across the cross-section of the downstream river shortly after the confluence.

• Complete Mixing: This is an important assumption. In reality, complete mixing might take some distance downstream from where the tributary joins the river. The length required for complete mixing depends on factors like the flow rates, velocities, channel geometry, and turbulence.
• Conservative Pollutants: The calculation assumes pollutant X is 'conservative', meaning it does not decay, react, settle out, or volatilize during the mixing process. If the pollutant were non-conservative (e.g., a decaying organic substance), a reaction term would need to be included in the mass balance equation, typically involving an exponential decay factor based on time and the decay rate constant.
• Units: It's crucial to maintain consistent units throughout the calculation. In this case, flow rates are in m³/s and concentrations in mg/L. The calculation correctly handles these units to yield a concentration in mg/L. Note that m³/s can be converted to L/s (1 m³ = 1000 L). If we used L/s for flow, the numerator would be \( (4000 \, \text{L/s} \times 12 \, \text{mg/L}) + (8000 \, \text{L/s} \times 30 \, \text{mg/L}) = 48000 \, \text{mg/s} + 240000 \, \text{mg/s} = 288000 \, \text{mg/s} \). The denominator would be \( 4000 \, \text{L/s} + 8000 \, \text{L/s} = 12000 \, \text{L/s} \). The resulting concentration would be \( \frac{288000 \, \text{mg/s}}{12000 \, \text{L/s}} = 24 \, \text{mg/L} \), which is the same result.