Solution:

Estimation of Average Depth of Rainfall

The average depth of rainfall over the basin can be estimated by taking the weighted average of the individual rainfalls at the gauge stations. The weights are assigned based on the area covered by each gauge station.

Let A1, A2, A3, and A4 be the areas covered by the gauge stations and R1, R2, R3, and R4 be the corresponding annual rainfalls.

Then, the average depth of rainfall over the basin is given by:

Avg. Depth = (A1 × R1 + A2 × R2 + A3 × R3 + A4 × R4) / (A1 + A2 + A3 + A4)

For example, if A1 = 100 km2, A2 = 75 km2, A3 = 50 km2, and A4 = 25 km2, and R1 = 108 cm, R2 = 79 cm, R3 = 73 cm, and R4 = 68 cm, then the average depth of rainfall over the basin is:

Avg. Depth = (100 × 108 + 75 × 79 + 50 × 73 + 25 × 68) / (100 + 75 + 50 + 25) = 87.65 cm

Determination of Error

The error in the estimation of the average depth of rainfall over the basin can be determined using the formula:

Error = (|Actual Value - Estimated Value| / Actual Value) × 100%

where the actual value is the true average depth of rainfall over the basin, and the estimated value is the value obtained using the weighted average formula.

In this case, let's assume that the actual average depth of rainfall over the basin is 100 cm. Then, the error in the estimation using the existing four gauges is:

Error = (|100 - 87.65| / 100) × 100% = 12.35%

As we can see, the error is more than the allowed 11% error. Therefore, we need to add more rain gauges to reduce the error.

Determination of Additional Number of Gauges

The additional number of gauges needed can be determined using the formula:

n = (Zα/2 × σ / E)2

where n is the number of additional gauges needed, Zα/2 is the standard normal deviate for the desired level of confidence (e.g., 1.96 for 95% confidence), σ is the standard deviation of the rainfall data, and E is the desired error limit (11% in this case).

The standard deviation of the rainfall data can be calculated using the formula:

σ = √[Σ(Ri - R̄)2 / (n - 1)]

where Ri is the rainfall at the i-th gauge station, R̄ is the mean rainfall over all the gauges, and n is the total number of gauges (including the existing four gauges).

For example, let's assume that the standard deviation of the rainfall data is 15 cm. Then, the additional number of gauges needed for a 95% confidence level and 11% error limit is:

n = (1.96 × 15 / 11)2 = 10.37

Since we cannot add fractional gauges, we need to add 11 more