1. Three centimeters 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.

25920 ha-cm

Explanation:

Since water evaporates from the reservoir at a rate of 3 cm in 24 hours, its height changes at a rate of 3 cm/24 × 3600 s = 3 cm/86400s = 3.472 10⁻⁵ cm/s.

Now, the volume loss is dV/dt = dV/dh × -dh/dt

= dV/dt × -3.472 × 10⁻⁵ cm/s

= -3.472 × 10⁻⁵ cm/sdV/dh

The reservoir increases in volume at a rate of 3 m³/s = 3 × 10⁶ cm³/s in 24 hours.

So, the net rate of volume change per unit time of the reservoir is

3 × 10⁶ cm³/s  - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt where A = area of vertical walled reservoir and dh/dt = change in height of the reservoir with respect to time

So, 3 × 10⁶ cm³/s  - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt

Since dh/dt = 0 in 24 hours(since the water level remains the same after 24 hours, that is dh = 0)

3 × 10⁶ cm³/s  - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt

3 × 10⁶ cm³/s  - 3.472 × 10⁻⁵ cm/sdV/dh = A × 0

3 × 10⁶ cm³/s  - 3.472 × 10⁻⁵ cm/sdV/dh = 0

3.472 × 10⁻⁵ cm/sdV/dh = 3 × 10⁶ cm³/s

dV/dh = 3 × 10⁶ cm³/s ÷ 3.472 × 10⁻⁵ cm/s

dV/dh = 8.64 × 10¹¹ cm²

dV = (8.64 × 10¹¹ cm²)dh

Integrating both sides with V from 0 to V and h from h = 0 to h = 3 cm, we have

∫dV = ∫(8.64 × 10¹¹ cm²)dh

∫dV = (8.64 × 10¹¹ cm²)∫dh

V = (8.64 × 10¹¹ cm²)[h]₀³

V = (8.64 × 10¹¹ cm²)[3 cm - 0 cm]

V = (8.64 × 10¹¹ cm²)(3 cm)

V = 25.92 × 10¹¹ cm³

V = 2.592 × 10¹² cm³

V = 2.592 × 10¹² cm² × 1 cm

Since 1 ha = 10⁸ cm²,

V = 2.592 × 10¹² cm² × 1 ha/10⁸ cm² × 1 cm

V = 2.592 × 10⁴ ha-cm

V = 25920 ha-cm

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