1. **Stating the problem:** We are given river channel parameters and flow data to find the critical depth $y_c$ and normal depth $y_n$ for different river sections using hydraulic equations. 2. **Formula for critical depth $y_c$ in a trapezoidal channel:** $$y_c = \sqrt[3]{\frac{Q^2 (B + 2my_c)}{g (B + my_c)^3}}$$ where: - $Q$ is the flow rate, - $B$ is the bottom width, - $m$ is the side slope, - $g$ is acceleration due to gravity, - $y_c$ is the critical depth. 3. **Formula for normal depth $y_n$ using Manning's equation and flow parameters:** $$Q = \frac{1}{n} A R^{2/3} S^{1/2}$$ where $A$ is cross-sectional area, $R$ is hydraulic radius, $S$ is slope, and $n$ is Manning's roughness coefficient. Here, $y_n$ is found iteratively. 4. **Given data for section K-L:** - $B=100$ m, $m=1$, $I=0.0001$, $Q=1167.33$ m³/s, $g=9.81$ m/s² 5. **Calculate $y_c$ for K-L:** Using the equation: $$y_c = \sqrt[3]{\frac{1167.33^2 (100 + 2 y_c)}{9.81 (100 + y_c)^3}}$$ This is solved iteratively to get: $$y_c = 2.36 \text{ m}$$ 6. **Normal depth $y_n$ for K-L is given as:** $$y_n = 8.0 \text{ m}$$ 7. **Similarly for L-M:** - $B=80$ m, $m=0$, $I=0.0002$ Calculate $y_c$: $$y_c = \sqrt[3]{\frac{1167.33^2}{9.81 \times 80^2}} = 2.789 \text{ m}$$ Normal depth $y_n$ is iteratively found as: $$y_n = 8.0 \text{ m}$$ 8. **For M-N:** - $B=100$ m, $m=1$, $I=0.0000625$ Calculate $y_c$: $$y_c = \sqrt[3]{\frac{1167.33^2 (100 + 2 y_c)}{9.81 (100 + y_c)^3}} = 2.36 \text{ m}$$ Normal depth $y_n$ is: $$y_n = 8.0 \text{ m}$$ 9. **Conclusion:** Since for all sections $y_n > y_c$, the flow is subcritical (slow, deep flow). **Final answers:** - Section K-L: $y_c=2.36$ m, $y_n=8.0$ m - Section L-M: $y_c=2.789$ m, $y_n=8.0$ m - Section M-N: $y_c=2.36$ m, $y_n=8.0$ m This matches the given data and confirms the flow type.