After a flood had passed an observation station on a river, an engineer visited the site. By locating flood marks, performing appropriate surveying and doing necessary computations determine that cross-sectional area, wetted perimeter and water surface slope at the time of peak flooding were 2860 $m^{2}$ , 320 m and 7.2 $\times 10^{- 4}$ respectively. The channel bottom roughness coefficient is estimated to be 0.027. The peak flood discharge at the observation station is (in cumecs).
12240.8

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Solution

The correct option is A 12240.8
Given,
$A = 2863 m^{2}, P = 320 m, S = 7.2 \times 10^{- 4}, n = 0.027$

$V e l o c i t y o f f l o w, v = \frac{1}{n} R^{2 / 3} S^{1 / 2}$

$= \frac{1}{0.027} \times {(\frac{2860}{320})}^{2 / 3} (7.2 \times 10^{- 4})^{1 / 2}$

= 4.28 m/sec

Peak flood discharge, Q = AV

$= 2860 \times 4.28$

= 12240.8 cumecs

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The correct option is A 12240.8Given, A=2863m2, P=320m, S=7.2×10−4, n=0.027 Velocity of flow,v=1nR2/3S1/2 =10.027×(2860320)2/3(7.2×10−4)1/2 = 4.28 m/sec Peak flood discharge, Q = AV =2860×4.28 = 12240.8 cumecs