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Reservoir level change, water balance, 15.5 km2

Short answer

Inflow is 65% of 0.01 m rainfall over 15.5 km² = 100,750 m³. Outflows: canal 39,600 m³, evaporation 3,600 m³, seepage 1,620 m³. Net increase 55,930 m³ over 0.6 km² water spread gives a level rise of 0.093 m (9.3 cm).

The catchment area of a reservoir is 15.5 km2^2. A uniform precipitation of 0.5 cm/h for 2 hr was observed on a particular day. 65% of precipitation reached into the reservoir. A canal carrying a flow of 1.1 m3^3/s is taken from the reservoir. The rate of evaporation was 0.6 mm/hr/m2^2. Assuming seepage loss to be 45% of evaporation loss, find the change in reservoir level for 10 hrs, if the water spread of the reservoir was 0.6 km2^2.

Given: catchment area A=15.5A = 15.5 km² =15.5×106= 15.5\times10^6 m²; rainfall 0.5 cm/h for 2 h =1= 1 cm =0.01= 0.01 m; runoff reaching reservoir 65%; canal 1.1 m³/s; evaporation 0.6 mm/h per m²; seepage 45% of evaporation; water spread =0.6= 0.6 km² =6×105= 6\times10^5 m²; period T=10T = 10 h.

ΔS=Vin−Vcanal−Vevap−Vseep\Delta S = V_{in} - V_{canal} - V_{evap} - V_{seep}

Step 1: Inflow

Vin=0.65×0.01×15.5×106=100 750 m3V_{in} = 0.65 \times 0.01 \times 15.5\times10^6 = 100\,750\ \text{m}^3

Step 2: Outflows in 10 h

  • Canal: 1.1×10×3600=39 6001.1 \times 10 \times 3600 = 39\,600 m³
  • Evaporation: 0.0006×6×105×10=3 6000.0006 \times 6\times10^5 \times 10 = 3\,600 m³
  • Seepage: 0.45×3600=1 6200.45 \times 3600 = 1\,620 m³

Step 3: Storage change

ΔS=100 750−39 600−3 600−1 620=55 930 m3\Delta S = 100\,750 - 39\,600 - 3\,600 - 1\,620 = 55\,930\ \text{m}^3

Step 4: Level change

Δh=55 9306×105=0.0932 m\Delta h = \frac{55\,930}{6\times10^5} = 0.0932\ \text{m}

Answer: The reservoir level rises by about 0.093 m (9.3 cm) in 10 hours.

✓ Answer checked by Claude Sonnet 5.5

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