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Drainage density, form factor, channel slope

Short answer

Drainage density = 4.67 km/km²; average overland flow length = 0.107 km; form factor = 0.317; channel slope = 0.052 (5.2%).

Find the drainage density, average length of overland flow, form factor and channel slope for a basin with the following data:

Area of basin (A) = 140 km2^2 Distance between the outlet to the farthest point (L) = 21 km Elevation difference between the outlet and the farthest point (h) = 1090 m Total length of channels of all order (LsL_s) = 654 km

Data: A=140A = 140 km2^2, L=21L = 21 km, h=1090h = 1090 m, total channel length Ls=654L_s = 654 km.

Drainage density

Dd=LsA=654140=4.67 km/km2D_d = \frac{L_s}{A} = \frac{654}{140} = 4.67\ \text{km/km}^2

Average length of overland flow

Lo=12Dd=12×4.671=0.107 km=107 mL_o = \frac{1}{2D_d} = \frac{1}{2\times4.671} = 0.107\ \text{km} = 107\ \text{m}

Form factor

Rf=AL2=140212=140441=0.317R_f = \frac{A}{L^2} = \frac{140}{21^2} = \frac{140}{441} = 0.317

Channel slope

S=hL=109021 000=0.0519  (5.19 %)≈51.9 m/kmS = \frac{h}{L} = \frac{1090}{21\,000} = 0.0519 \;(5.19\ \%) \approx 51.9\ \text{m/km}

Answer: Dd=4.67D_d = 4.67 km/km2^2; Lo=0.107L_o = 0.107 km; form factor =0.317= 0.317; channel slope =0.052= 0.052 (5.2 %).

The high drainage density and steep slope show a well-drained basin with quick runoff. The low form factor shows an elongated basin, which gives lower and flatter peak flows than a circular basin.

✓ Answer checked by Claude Sonnet 5.5

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