Question Bank • QB_SWCE_ADV_117 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Hydrology

Subtopic: Analysis of precipitation data

Question Statement:

The normal annual precipitations at rain gauge stations X, A, B, and C are $N_X = 900.0\text{ mm}$, $N_A = 800.0\text{ mm}$, $N_B = 1000.0\text{ mm}$, and $N_C = 950.0\text{ mm}$. During a storm event, stations A, B, and C recorded storm rainfalls of $P_A = 65.0\text{ mm}$, $P_B = 85.0\text{ mm}$, and $P_C = 78.0\text{ mm}$, while gauge X was inoperative. Using the Normal Ratio Method, the estimated storm rainfall $P_X$ at station X in $\text{mm}$ is ________ (round off to one decimal place).

Answer Key:

74.5

Step-by-Step Solution & Explanation

**Method 1: Normal Ratio Formula** When normal annual rainfalls differ by more than $10\%$, the Normal Ratio Method is prescribed: $P_X = \frac{N_X}{m} \sum_{i=1}^m \frac{P_i}{N_i}$ Here $m = 3$ surrounding index stations: $P_X = \frac{900.0}{3} \left[ \frac{65.0}{800.0} + \frac{85.0}{1000.0} + \frac{78.0}{950.0} \right]$ 1. Individual ratios: - $\frac{65.0}{800.0} = 0.08125$ - $\frac{85.0}{1000.0} = 0.08500$ - $\frac{78.0}{950.0} = 0.082105$ 2. Sum of ratios: $\text{Sum} = 0.08125 + 0.08500 + 0.082105 = 0.248355$ 3. Estimated precipitation: $P_X = 300.0 \times 0.248355 = 74.5065\text{ mm} \approx 74.5\text{ mm}$ **Method 2: Weighted Normalized Anomaly** $P_X = N_X \times \text{Average Weight} = 900.0 \times \frac{0.248355}{3} = 74.51\text{ mm}$
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