From a Survey of India map: the overland path length (to the critical point) is 20 km and the elevation drop is 193 m. Estimate the overland flow time to the outlet using an empirical travel-time relation.

Introduction:Overland flow time is the travel time for surface runoff from the hydraulically most distant point to the channel/outlet. It is a key component of the time of concentration, which controls peak flow estimation in hydrologic design.

Given Data / Assumptions:

• Length L = 20 km = 20000 m.
• Elevation difference ΔH = 193 m ⇒ average slope S = 193/20000 = 0.00965.
• Use a standard empirical relation suitable for surface flow on natural slopes.

Concept / Approach:A commonly used relation for time of travel on natural surfaces is the Kirpich-type equation (metric form):t = 0.0195 * L^0.77 * S^-0.385where t is in minutes, L in metres, and S is slope (dimensionless). This is appropriate as a first estimate when detailed microtopography and roughness data are unavailable.

Step-by-Step Solution:Compute S = 193/20000 = 0.00965.Evaluate L^0.77 ≈ 20000^0.77 ≈ 3.90 * 10^3.Evaluate S^-0.385 ≈ (0.00965)^-0.385 ≈ 3.28 (approx.).t ≈ 0.0195 * 3900 * 3.28 ≈ 239 minutes.Convert to hours: 239 min ≈ 3.98 hours ≈ 4 hours.

Verification / Alternative check:Magnitude check: a 20 km path on a gentle slope yields several hours of travel time—consistent with comparable basin studies and with the computed 4 hours.

Why Other Options Are Wrong:

• 2 h, 2 h 30 m, 3 h, 3 h 30 m: Each underestimates the empirical travel time given L and S; rounding the computation points to ~4 hours.

Common Pitfalls:

• Using slope in percent rather than fraction (would under/overestimate).
• Forgetting unit conversions for L.

Final Answer:4 hours.