To solve this problem, we'll use the given water balance equation for the reservoir and compute the required volumes step by step. Let's list the variables provided and the steps we'll follow.

Given dataTo solve this problem, we'll use the given water balance equation for the reservoir and compute the required volumes step by step. Let's list the variables provided and the steps we'll follow.

Given

• $A = 380 \, \text{km}^2$ (Catchment area around the reservoir)
• $B = 34 \, \text{km}^2$ (Plan area of the reservoir)
• $R = 31 \, \text{mm}$ (Monthly rainfall)
• $E = 200 \, \text{mm}$ (Measured evaporation)
• $X = 186.1 \, \text{m}$ (Initial water level)
• $Y = 185.7 \, \text{m}$ (Final water level)
• $Q = 460 \times 10^6 \, \text{litres/day} = 0.46 \, \text{MCM/day}$ (Abstraction rate from the reservoir)
• Time duration: $t = 30 \, \text{days}$

Step-by-Step Solution:

1. Volume change in the reservoir: The change in water level in the reservoir over 30 days can be calculated as: $\Delta h = X - Y = 186.1 - 185.7 = 0.4 \, \text{m}$ The corresponding change in volume for the reservoir is the plan area $B$ multiplied by this height difference $\Delta h$: $V_{\text{change}} = B \times \Delta h = 34 \, \text{km}^2 \times 0.4 \, \text{m} = 34 \times 10^6 \, \text{m}^2 \times 0.4 \, \text{m} = 13.6 \, \text{MCM}$ (Note: 1 km² = $10^6 \, \text{m}^2$, and MCM = million cubic meters.)

2. Volume of rainfall directly onto the reservoir: The volume of rainfall on the reservoir is calculated as the plan area of the reservoir $B$ multiplied by the rainfall depth $R$ (converted to meters): $V_{\text{rainfall}} = B \times \left( \frac{R}{1000} \right) = 34 \, \text{km}^2 \times \left( \frac{31}{1000} \, \text{m} \right) = 34 \times 10^6 \, \text{m}^2 \times 0.031 \, \text{m} = 1.054 \, \text{MCM}$

3. Volume of water abstracted from the reservoir: The total volume of water abstracted over the 30-day period is: [ V_{\text{abstraction}} = Q