Question

Two rectangular channels, A and B, join to form a large rectangular channel, C. Each rectangular channel is lined with the same material and has the same bottom slope. The bottom widths of Channels A and B are \(2\) m and \(1\) m, respectively. If the flow depth in each channel is \(2\) m, the bottom width of Channel C (in m) is ________. (Rounded off to two decimal places)

Hint:

Add the individual channel conveyances and solve for the width of Channel C by trial.

Answer 1

Correct Answer: 2.47

Step 1: Write Manning's equation with the common terms grouped.
Since \(n\) and slope \(S\) are the same for A, B and C, write \(Q = K \times A_{area} \times R^{2/3}\), where \(K = \sqrt{S}/n\) is common to all three channels.

Step 2: Find \(Q_A\) (width \(2\) m, depth \(2\) m).
Area \(=2\times2=4\) m\(^2\), wetted perimeter \(=2+2(2)=6\) m, \(R=4/6=0.667\) m.
\(R^{2/3}\approx0.763\), so \(Q_A = K\times4\times0.763 = 3.05K\).

Step 3: Find \(Q_B\) (width \(1\) m, depth \(2\) m).
Area \(=1\times2=2\) m\(^2\), perimeter \(=1+4=5\) m, \(R=2/5=0.4\) m.
\(R^{2/3}\approx0.543\), so \(Q_B = K\times2\times0.543 = 1.09K\).

Step 4: Apply continuity at the junction.
\[ Q_C = Q_A+Q_B = 4.14K \]
Step 5: Solve for the width \(b\) of Channel C, taking its depth as \(2\) m too.
Area \(=2b\), perimeter \(=b+4\), \(R=2b/(b+4)\); solve \(2b[2b/(b+4)]^{2/3}=4.14\) by trial.
At \(b=2.47\): \(2b=4.94\), \(R=4.94/6.47=0.764\), \(R^{2/3}=0.835\), product \(\approx4.13\), just under the target.
At \(b=2.48\): product \(\approx4.15\), just over the target, so \(b\) lies close to \(2.47\) m.

Final Answer:
Balancing the conveyance at the junction gives a combined bottom width of about \(2.47\) m. \[ \boxed{b_C \approx 2.47\ \text{m}} \]