1. **Problem Statement:** Determine the flow in a trapezoidal canal using Manning’s and Kutter’s formulas. Given: - Base width $b = 2$ m - Left slope $m_1 = \frac{1.5}{1} = 1.5$ (horizontal per vertical) - Right slope $m_2 = \frac{1}{1} = 1$ (horizontal per vertical) - Water height $h = 1.78$ m - Slope of canal bed $S = \frac{0.6}{100} = 0.006$ - Roughness coefficient $n = 0.015$ 2. **Formulas and Important Rules:** - Cross-sectional area $A$ of trapezoidal channel: $$A = h \times \left(b + h(m_1 + m_2)\right)$$ - Wetted perimeter $P$: $$P = b + h \sqrt{1 + m_1^2} + h \sqrt{1 + m_2^2}$$ - Hydraulic radius $R$: $$R = \frac{A}{P}$$ - Manning’s formula for flow $Q$: $$Q = \frac{1}{n} A R^{2/3} S^{1/2}$$ - Kutter’s formula for flow $Q$: $$Q = \frac{1}{n} A R^{2/3} S^{1/2}$$ (Note: Kutter’s formula is similar to Manning’s but $n$ is computed differently; here $n$ is given, so formulas coincide.) 3. **Calculate Cross-sectional Area $A$:** $$A = 1.78 \times \left(2 + 1.78(1.5 + 1)\right) = 1.78 \times \left(2 + 1.78 \times 2.5\right)$$ $$= 1.78 \times (2 + 4.45) = 1.78 \times 6.45 = 11.481$$ m$^2$ 4. **Calculate Wetted Perimeter $P$:** Calculate each sloped side length: $$h \sqrt{1 + m_1^2} = 1.78 \times \sqrt{1 + (1.5)^2} = 1.78 \times \sqrt{1 + 2.25} = 1.78 \times \sqrt{3.25} = 1.78 \times 1.803 = 3.209$$ $$h \sqrt{1 + m_2^2} = 1.78 \times \sqrt{1 + 1^2} = 1.78 \times \sqrt{2} = 1.78 \times 1.414 = 2.518$$ So, $$P = 2 + 3.209 + 2.518 = 7.727$$ m 5. **Calculate Hydraulic Radius $R$:** $$R = \frac{A}{P} = \frac{11.481}{7.727} = 1.486$$ m 6. **Calculate Flow $Q$ using Manning’s formula:** $$Q = \frac{1}{0.015} \times 11.481 \times (1.486)^{2/3} \times (0.006)^{1/2}$$ Calculate each term: $$R^{2/3} = (1.486)^{2/3} = e^{\frac{2}{3} \ln(1.486)} \approx e^{0.268} = 1.307$$ $$S^{1/2} = \sqrt{0.006} = 0.07746$$ Therefore, $$Q = 66.67 \times 11.481 \times 1.307 \times 0.07746 = 66.67 \times 11.481 \times 0.1013 = 66.67 \times 1.163 = 77.53$$ m$^3$/s 7. **Calculate Flow $Q$ using Kutter’s formula:** Since $n$ is given and no other parameters for Kutter’s formula are provided, the flow is the same as Manning’s: $$Q = 77.53$$ m$^3$/s **Final Answer:** The flow in the canal is approximately **77.53 cubic meters per second** using both Manning’s and Kutter’s formulas.