1. **Problem Statement:** Estimate the Curve Numbers (CN I, CN II, and CN III) for a catchment area with given land use percentages and soil groups B and C. 2. **Given Data:** - Land use and soil group percentages: - Cultivated land (Paddy): 30% (B), 45% (C), total 75% - Scrub Forest: 6% (B), 4% (C), total 10% - Waste land: 9% (B), 6% (C), total 15% 3. **CN II values from Table 1:** - Cultivated Paddy: CN II = 95 (for all soil groups) - Scrub Forest: CN II = 47 (soil group B), 64 (soil group C) - Waste land: CN II = 80 (soil group B), 85 (soil group C) 4. **Assumptions:** - CN I corresponds to the antecedent moisture condition I (dry), CN II is average moisture, CN III is wet condition. - CN I and CN III can be estimated from CN II using the following formulas: $$CN_I = \frac{CN_{II}}{2.281 - 0.01281 \times CN_{II}}$$ $$CN_{III} = \frac{CN_{II}}{0.427 + 0.00573 \times CN_{II}}$$ 5. **Calculate weighted CN II for soil group B:** $$CN_{II,B} = \frac{30}{45} \times 95 + \frac{6}{45} \times 47 + \frac{9}{45} \times 80$$ $$= \frac{30 \times 95 + 6 \times 47 + 9 \times 80}{45}$$ $$= \frac{2850 + 282 + 720}{45} = \frac{3852}{45} = 85.6$$ 6. **Calculate weighted CN II for soil group C:** $$CN_{II,C} = \frac{45}{55} \times 95 + \frac{4}{55} \times 64 + \frac{6}{55} \times 85$$ $$= \frac{45 \times 95 + 4 \times 64 + 6 \times 85}{55}$$ $$= \frac{4275 + 256 + 510}{55} = \frac{5041}{55} = 91.65$$ 7. **Calculate overall CN II for the catchment:** $$CN_{II,total} = \frac{45}{100} \times 85.6 + \frac{55}{100} \times 91.65 = 0.45 \times 85.6 + 0.55 \times 91.65$$ $$= 38.52 + 50.41 = 88.93$$ 8. **Calculate CN I and CN III for soil group B:** $$CN_{I,B} = \frac{85.6}{2.281 - 0.01281 \times 85.6} = \frac{85.6}{2.281 - 1.096} = \frac{85.6}{1.185} = 72.2$$ $$CN_{III,B} = \frac{85.6}{0.427 + 0.00573 \times 85.6} = \frac{85.6}{0.427 + 0.490} = \frac{85.6}{0.917} = 93.3$$ 9. **Calculate CN I and CN III for soil group C:** $$CN_{I,C} = \frac{91.65}{2.281 - 0.01281 \times 91.65} = \frac{91.65}{2.281 - 1.174} = \frac{91.65}{1.107} = 82.8$$ $$CN_{III,C} = \frac{91.65}{0.427 + 0.00573 \times 91.65} = \frac{91.65}{0.427 + 0.525} = \frac{91.65}{0.952} = 96.3$$ 10. **Calculate overall CN I and CN III for the catchment:** $$CN_{I,total} = 0.45 \times 72.2 + 0.55 \times 82.8 = 32.49 + 45.54 = 78.03$$ $$CN_{III,total} = 0.45 \times 93.3 + 0.55 \times 96.3 = 41.99 + 52.97 = 94.96$$ **Final estimated Curve Numbers:** - CN I = 78.0 - CN II = 88.9 - CN III = 95.0 --- **Summary Table:** | Land Use | Soil Group B (%) | Soil Group C (%) | CN II (B) | CN II (C) | |----------------|------------------|------------------|-----------|-----------| | Cultivated Paddy | 30 | 45 | 95 | 95 | | Scrub Forest | 6 | 4 | 47 | 64 | | Waste land | 9 | 6 | 80 | 85 | | Soil Group | Weighted CN II | CN I | CN III | |----------------|-------------------|-----------|-----------| | B | 85.6 | 72.2 | 93.3 | | C | 91.65 | 82.8 | 96.3 | | Overall Catchment | CN II | CN I | CN III | |------------------|-----------------|-----------|-----------| | | 88.9 | 78.0 | 95.0 |