Summarized Solution

This problem involves steady uniform flow in open channels. We apply Manning's equation and the continuity principle to find the unknown bottom width of the combined channel.

1. Governing Equations

For steady uniform flow in an open channel, the discharge \( Q \) is given by Manning's equation:

$$ Q = \frac{1}{n} A R^{2/3} S_0^{1/2} $$

Where:

• \( n \) = Manning's roughness coefficient
• \( A \) = Cross-sectional area of flow (\( B \times y \) for a rectangular channel)
• \( R \) = Hydraulic radius (\( A / P \))
• \( P \) = Wetted perimeter (\( B + 2y \) for a rectangular channel)
• \( S_0 \) = Bottom slope

2. Simplifying the Proportionality

Since all three channels (A, B, and C) have the same lining material (same \( n \)) and the same bottom slope (\( S_0 \)), we can say:

$$ Q \propto A R^{2/3} = \frac{A^{5/3}}{P^{2/3}} = \frac{(B \cdot y)^{5/3}}{(B + 2y)^{2/3}} $$

The flow depth is given as \( y = 2 \text{ m} \) for every channel. Substituting this into the proportionality:

$$ Q \propto \frac{(B \cdot 2)^{5/3}}{(B + 4)^{2/3}} = 2^{5/3} \cdot \frac{B^{5/3}}{(B + 4)^{2/3}} $$

3. Applying Continuity

The total discharge in channel C is the sum of the discharges from channels A and B:

$$ Q_C = Q_A + Q_B $$

Substituting the proportional expressions (the constant terms \( 1/n \), \( S_0^{1/2} \), and \( 2^{5/3} \) cancel out from both sides):

$$ \frac{B_C^{5/3}}{(B_C + 4)^{2/3}} = \frac{B_A^{5/3}}{(B_A + 4)^{2/3}} + \frac{B_B^{5/3}}{(B_B + 4)^{2/3}} $$

4. Calculation of Width \( B_C \)

Given \( B_A = 2 \text{ m} \) and \( B_B = 1 \text{ m} \):

• Term for A: \( \frac{2^{5/3}}{(2 + 4)^{2/3}} = \frac{3.1748}{3.3019} \approx 0.9615 \)
• Term for B: \( \frac{1^{5/3}}{(1 + 4)^{2/3}} = \frac{1}{5^{2/3}} \approx 0.3420 \)

Summing these values:

$$ \frac{B_C^{5/3}}{(B_C + 4)^{2/3}} = 0.9615 + 0.3420 = 1.3035 $$

To solve for \( B_C \), we use trial and error or numerical iteration:

• If \( B_C = 2.47 \text{ m} \), LHS \( \approx 1.2997 \)
• If \( B_C = 2.48 \text{ m} \), LHS \( \approx 1.3072 \)

By linear interpolation, \( B_C \approx 2.475 \text{ m} \). Rounding to two decimal places, we get 2.48 m.

Final Answer: The bottom width of Channel C is 2.48 m.