3 cm of water evaporated from a 200-hectare vertical walled reservoir during 24 hours. Storm water was added to the reservoir at a constant rate of 3 m3/s during this period. Determine the volume in ha-cm of water released during the period (through the bottom of the reservoir) if the water level was the same at the beginning and the end of the day.

water released through the bottom of the reservoir in 24 hrs is 1992 ha-cm

Explanation:

Given the data in the question;

A = 200 hectare =  2 × 10⁶ m²

water evaporated Ve = 2 × 10⁶ m² × 3 × 10⁻² = 60000 m³ { in 24 hrs }

Water added by storm in 24hrs Vi = 3 × 24 × 3600 = 259200 m³

now let water released be Vr

ΔV = V_ini - V_final = 0

Vi - Ve - Vr = 0

Vr = Vi - Ve

Vr = 259200 m³ - 60000 m³

Vr = 199200 m³ = 19920000 m² - cm

Vr = 1992 ha-cm

Therefore, water released through the bottom of the reservoir in 24 hrs is 1992 ha-cm

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