Runoff volume using Horton's equation, 850 km2

Given: basin area $= 850$ km² $= 850\times10^6$ m²; $f_0 = 6.5$ mm/h, $f_c = 1.5$ mm/h, $k = 0.15$ h⁻¹; hourly mass curve (mm): 0, 10, 22, 30, 39, 45.5, 50, 55.5, 60, 64, 68.

Step 1: Hourly rainfall

10, 12, 8, 9, 6.5, 4.5, 5.5, 4.5, 4, 4 mm (total 68 mm in 10 h).

Step 2: Infiltration capacity in each hour

Horton's cumulative capacity: $F(t) = f_c t + \dfrac{f_0 - f_c}{k}(1 - e^{-kt})$ = $1.5t + 33.33(1 - e^{-0.15t})$.

The rainfall exceeds the infiltration capacity in every hour, so infiltration takes place at capacity throughout:

$$F(10) = 1.5\times10 + 33.33\,(1 - e^{-1.5}) = 15 + 25.90 = 40.90\ \text{mm}$$

Step 3: Runoff

$$R = 68 - 40.90 = 27.10\ \text{mm} = 0.0271\ \text{m}$$ $$V = 0.027104\times850\times10^6 = 2.30\times10^7\ \text{m}^3$$

Answer: Runoff depth = 27.1 mm; runoff volume = 2.30 × 10⁷ m³ (23.0 million m³).