Trapezoidal Canal Parameters

The problem asks for the critical depth ($y_c$) in a trapezoidal canal.

• Bottom width, $b = 5.0 \text{ m}$
• Side slope, $m = 1$ (since it's $1 \text{ H} : 1 \text{ V}$)
• Discharge, $Q = 20 \text{ m}^3 \text{ s}^{-1}$
• Acceleration due to gravity, $g \approx 9.81 \text{ m s}^{-2}$

Critical Depth Calculation

The condition for critical flow is given by the equation:

Where:

• $A_c$ is the cross-sectional area at critical depth $y_c$. For a trapezoidal section, $A_c = b y_c + m y_c^2$.
• $T_c$ is the top width at critical depth $y_c$. For a trapezoidal section, $T_c = b + 2 m y_c$.

Substituting the expressions for $A_c$ and $T_c$:

Calculate the left side of the equation:

Now, we test the given options for $y_c$ to see which one satisfies the equation.

Testing Option A: $y_c = 1.09 \text{ m}$

• Area, $A_c = 5.0(1.09) + 1(1.09)^2 = 5.45 + 1.1881 = 6.6381 \text{ m}^2$
• Top width, $T_c = 5.0 + 2(1)(1.09) = 5.0 + 2.18 = 7.18 \text{ m}$
• Calculate $\frac{A_c^3}{T_c}$:

This value ($40.74$) is very close to $\frac{Q^2}{g}$ ($40.77$). Therefore, $y_c = 1.09 \text{ m}$ is the critical depth.

Testing Other Options (Optional Verification)

For completeness, let's check another option, e.g., Option B ($y_c = 1.18$ m):

• Area, $A_c = 5.0(1.18) + 1(1.18)^2 = 5.9 + 1.3924 = 7.2924 \text{ m}^2$
• Top width, $T_c = 5.0 + 2(1)(1.18) = 5.0 + 2.36 = 7.36 \text{ m}$
• Calculate $\frac{A_c^3}{T_c}$:

This value ($52.64$) is significantly different from $40.77$, confirming that $1.18 \text{ m}$ is not the critical depth.

Conclusion

Based on the calculations, the critical depth $y_c$ that satisfies the condition $\frac{Q^2}{g} = \frac{A_c^3}{T_c}$ is approximately $1.09 \text{ m}$.