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Reservoir level change, water balance, 1600 ha

Short answer

Runoff volume = 55% of 16 mm × 1600 ha = 140,800 m³; net storage change after 6 h of canal, evaporation, and seepage is 116,176 m³; level rise = 116,176 / (45 ha) = 0.258 m.

The catchment area of a reservoir is 1600 ha. A uniform precipitation of 8 mm/hr for 2 hour was observed on a particular day. 55% run off reached the reservoir. A canal carrying a flow of 1 m3^3/s is taken from the reservoir. The rate of evaporation was 0.8 mm/h/m2^2. Assuming seepage loss is 40% of evaporation loss, find the change in the reservoir level for 6 hours, if the water spread of the reservoir was 45 ha.

Given: catchment area A=1600A = 1600 ha =16×106= 16 \times 10^6 m²; rainfall 8 mm/h for 2 h; runoff coefficient 0.55; canal flow 1 m³/s; evaporation 0.8 mm/h/m² of water spread; seepage = 40% of evaporation; water spread =45= 45 ha =4.5×105= 4.5 \times 10^5 m²; period T=6T = 6 h.

Water balance: ΔS=Inflow−Canal outflow−Evaporation−Seepage\Delta S = \text{Inflow} - \text{Canal outflow} - \text{Evaporation} - \text{Seepage}

Step 1: Inflow

Rainfall depth =8×2=16= 8 \times 2 = 16 mm =0.016= 0.016 m

Vin=0.55×0.016×16×106=140 800 m3V_{in} = 0.55 \times 0.016 \times 16\times10^6 = 140\,800\ \text{m}^3

Step 2: Outflows over 6 h

  • Canal: 1×6×3600=21 6001 \times 6 \times 3600 = 21\,600 m³
  • Evaporation: 0.0008×4.5×105×6=2 1600.0008 \times 4.5\times10^5 \times 6 = 2\,160 m³
  • Seepage: 0.4×2160=8640.4 \times 2160 = 864 m³

Step 3: Change in storage

ΔS=140 800−21 600−2 160−864=116 176 m3\Delta S = 140\,800 - 21\,600 - 2\,160 - 864 = 116\,176\ \text{m}^3

Step 4: Change in level

Δh=116 1764.5×105=0.258 m\Delta h = \frac{116\,176}{4.5\times10^5} = 0.258\ \text{m}

Answer: The reservoir level rises by about 0.258 m (25.8 cm) in 6 hours.

✓ Answer checked by Claude Sonnet 5.5

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