Consider the following data recorded at a three rainguage stations- Stations A B C- Normal annual rainfall mm 300 400 500- Annual rainfall for year 2010 130 X 240- -The value of -X- is

Since the value of normal precipitation of station 'A' and 'C' is not within ± 10% of normal precipitation i.e (0.9 × 400 = 360 , 1.1 × 400 = nbsp; 440 ). So, ...

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Hydrology & Irrigation

Normal Ratio Method

Consider the ...

Question

Consider the following data recorded at a three rainguage stations

$\begin{matrix} Stations & A & B & C Normal annual rainfall (mm) & 300 & 400 & 500 Annual rainfall for year 2010 & 130 & X & 240 \end{matrix}$

The value of 'X' is

178

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195

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183

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190

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Solution

The correct option is C 183
Since the value of normal precipitation of station 'A' and 'C' is not within $\pm 10 %$ of normal precipitation i.e
$(0.9 \times 400 = 360, 1.1 \times 400 = 440)$ .
So, missing value of X will be found out using normal ratio method

$P_{B} = \frac{N_{B}}{n} (\frac{P_{A}}{N_{A}} + \frac{P_{c}}{N_{c}})$

$= \frac{400}{2} (\frac{130}{300} + \frac{240}{500}) = 182.67 ≃ 183 m m$

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Consider the following data recorded at a three rainguage stations StationsABCNormal annual rainfall (mm)300400500Annual rainfall for year 2010130X240 The value of 'X' is

The correct option is C 183Since the value of normal precipitation of station 'A' and 'C' is not within ±10% of normal precipitation i.e (0.9×400=360,1.1×400=440). So, missing value of X will be found out using normal ratio method PB=NBn(PANA+PcNc) =4002(130300+240500)=182.67≃183 mm

Consider the following data recorded at a three rainguage stations

$\begin{matrix} Stations & A & B & C Normal annual rainfall (mm) & 300 & 400 & 500 Annual rainfall for year 2010 & 130 & X & 240 \end{matrix}$

The value of 'X' is