Section 4: Soil and Water Conservation Engineering
Hydrology
Subtopic: Hydrograph analysis, unit hydrograph theory and application
Question Statement:
Two catchments M and N are meteorologically similar. The basin length (L), length to the centroid (L_ca), and the drainage area (A) of catchment M are 36 km, 18 km, and 250 km² respectively. The L, L_ca and A for catchment N are 50 km, 30 km, and 400 km² respectively.
For the 6-h unit hydrograph in catchment M, the peak discharge of 50 m³·s⁻¹ occurs at 12 h from the beginning of the rainfall excess. Using Snyder’s method, the peak of the 6-h unit hydrograph for catchment N (in m³·s⁻¹) is ________. (Rounded off to two decimal places)
Official Answer Key:
64.20 to 64.80
Step-by-Step Solution & Explanation
Meteorologically similar catchments share the same Snyder coefficients C_t and C_p. Determine them from M, then apply to N.
Step 1 — catchment M shape factor:
(L · L_ca)⁰·³ = (36 × 18)⁰·³ = 648⁰·³ = 6.975
Step 2 — adjusted basin lag for M.
Time to peak from start of rainfall excess = t_pr + t_R/2
12 = t_pr + 6/2 ⇒ t_pr = 9 h
Step 3 — recover the standard lag t_p using Snyder’s duration correction (standard duration t_r = t_p/5.5):
t_pr = t_p + 0.25(t_R − t_r) = t_p + 0.25(6 − t_p/5.5) = 0.95455 t_p + 1.5
9 = 0.95455 t_p + 1.5 ⇒ t_p = 7.857 h
Step 4 — coefficient C_t:
t_p = C_t (L·L_ca)⁰·³ ⇒ C_t = 7.857/6.975 = 1.1266
Step 5 — coefficient C_p from the observed peak of M:
Q_p = 2.78 C_p A/t_pr
50 = 2.78 × C_p × 250/9 ⇒ C_p = 0.6475
Step 6 — apply to catchment N.
(L · L_ca)⁰·³ = (50 × 30)⁰·³ = 1500⁰·³ = 8.971
t_p = 1.1266 × 8.971 = 10.106 h
t_pr = 0.95455 × 10.106 + 1.5 = 9.646 + 1.5 = 11.147 h
Step 7 — peak discharge for N:
Q_p = 2.78 × 0.6475 × 400/11.147 = 720.0/11.147 = 64.59 m³·s⁻¹
Note: skipping the duration-correction step (Step 3) gives ≈ 62.2 m³·s⁻¹, which falls outside the official range — the correction is essential.
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