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: Define the conveyance of each channel.
Conveyance $K_x = A_{area}\,R^{2/3}$ carries the channel shape; since $n$ and $S$ match everywhere, $Q=(\sqrt{S}/n)\,K_x$, so conveyances add just like discharges do at the junction.

Step 2: Conveyance of A ($b=2$, $y=2$).
$A_{area}=4$ m$^2$, $P=6$ m, $R=0.667$ m, $R^{2/3}=0.763$, so $K_A=4\times0.763=3.05$.

Step 3: Conveyance of B ($b=1$, $y=2$).
$A_{area}=2$ m$^2$, $P=5$ m, $R=0.4$ m, $R^{2/3}=0.543$, so $K_B=2\times0.543=1.09$.

Step 4: Add conveyances for Channel C.
\[ K_C = K_A+K_B = 3.05+1.09 = 4.14 \]
Step 5: Trial-solve for the width $b$ that gives this conveyance at depth $2$ m.
Try $b=2.45$: $A_{area}=4.90$, $R=4.90/6.45=0.760$, $R^{2/3}=0.831$, $K=4.07$ (a bit low).
Try $b=2.50$: $A_{area}=5.00$, $R=5.00/6.50=0.769$, $R^{2/3}=0.839$, $K=4.20$ (a bit high).
Interpolating between these brackets for $K=4.14$ lands close to $b\approx2.47$ m.

Final Answer:
Matching conveyances at the confluence gives Channel C a bottom width close to $2.47$ m. \[ \boxed{b_C \approx 2.47\ \text{m}} \]