To determine the Biochemical Oxygen Demand (BOD) of the river water 10 km downstream of the mixing point, we need to follow several steps:

1. Calculate the ultimate BOD of the mixed stream immediately after mixing.
2. Calculate the velocity of the river after mixing.
3. Calculate the time it takes for the water to travel 10 km downstream.
4. Apply the de-oxygenation formula to find the BOD remaining at that downstream location.

Ultimate BOD Calculation at Mixing Point

When two water streams mix, the ultimate BOD of the combined stream can be calculated using a mass balance approach. This ensures that the total BOD load from both sources is conserved in the mixed stream.

• Waste water flow (\(Q_w\)) = 2 m\textsuperscript{3}/sec
• Waste water ultimate BOD (\(L_w\)) = 90 mg/litre
• River flow (\(Q_r\)) = 12 m\textsuperscript{3}/s
• River ultimate BOD (\(L_r\)) = 5 mg/litre

The total flow of the mixed stream (\(Q_m\)) is the sum of the waste water flow and the river flow:

\(Q_m = Q_w + Q_r = 2 \text{ m\textsuperscript{3}/s} + 12 \text{ m\textsuperscript{3}/s} = 14 \text{ m\textsuperscript{3}/s}\)

The initial ultimate BOD of the mixed stream (\(L_0\)) is calculated as a weighted average:

\[L_0 = \frac{(Q_w \times L_w) + (Q_r \times L_r)}{Q_w + Q_r}\]

Substituting the given values:

\[L_0 = \frac{(2 \text{ m\textsuperscript{3}/s} \times 90 \text{ mg/litre}) + (12 \text{ m\textsuperscript{3}/s} \times 5 \text{ mg/litre})}{2 \text{ m\textsuperscript{3}/s} + 12 \text{ m\textsuperscript{3}/s}}\]

\[L_0 = \frac{180 \text{ mg/s} + 60 \text{ mg/s}}{14 \text{ m\textsuperscript{3}/s}}\]

\[L_0 = \frac{240 \text{ mg/s}}{14 \text{ m\textsuperscript{3}/s}} \approx 17.14286 \text{ mg/litre}\]

So, the initial ultimate BOD of the mixed river water is approximately 17.14 mg/litre.

River Velocity Calculation

To determine how long it takes for the water to travel downstream, we need to calculate the velocity of the river. The velocity can be found by dividing the total flow by the cross-sectional area of the river.

• Total flow (\(Q_m\)) = 14 m\textsuperscript{3}/s
• Cross-sectional area of the river (A) = 50 m\textsuperscript{2}

The velocity (V) is given by:

\[V = \frac{Q_m}{A}\]

Substituting the values:

\[V = \frac{14 \text{ m\textsuperscript{3}/s}}{50 \text{ m\textsuperscript{2}}} = 0.28 \text{ m/s}\]

The river water flows at a velocity of 0.28 m/s after mixing.

Time to Travel Downstream

Next, we calculate the time (t) it takes for the river water to travel 10 km downstream from the mixing point. We need to convert the distance to meters and time to days to match the units of the de-oxygenation rate constant.

• Distance (D) = 10 km = 10,000 m
• Velocity (V) = 0.28 m/s

The time (t) is calculated as:

\[t = \frac{D}{V}\]

Substituting the values:

\[t_{\text{seconds}} = \frac{10000 \text{ m}}{0.28 \text{ m/s}} \approx 35714.286 \text{ seconds}\]

Now, convert this time from seconds to days (1 day = 24 hours × 3600 seconds/hour = 86400 seconds):

\[t_{\text{days}} = \frac{35714.286 \text{ seconds}}{86400 \text{ seconds/day}} \approx 0.41336 \text{ days}\]

The water takes approximately 0.41336 days to reach 10 km downstream.

BOD Downstream Calculation

The ultimate BOD remaining in the water stream at a certain time 't' downstream from the mixing point can be calculated using the first-order de-oxygenation kinetics. The formula is:

\[L_t = L_0 e^{-Kt}\]

Where:

• \(L_t\) = Ultimate BOD at time t (mg/litre)
• \(L_0\) = Initial ultimate BOD of the mixed stream (mg/litre) = 17.14286 mg/litre
• \(K\) = De-oxygenation rate constant = 0.25/day
• \(t\) = Time taken to reach downstream point = 0.41336 days

Substituting the values into the formula:

\[L_t = 17.14286 \times e^{(-0.25 \text{/day} \times 0.41336 \text{ days})}\]

\[L_t = 17.14286 \times e^{(-0.10334)}\]

Calculating the exponential term:

\[e^{(-0.10334)} \approx 0.90186\]

Now, calculate \(L_t\):

\[L_t = 17.14286 \times 0.90186 \approx 15.460 \text{ mg/litre}\]

Thus, the BOD of the river water 10 km downstream of the mixing point is approximately 15.46 mg/litre.

The final answer is 15.46 mg/litre.