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.

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