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The specific energy for 5m wide rectangular channel is 4Nm/N. If rate of flow of water is through the channel is $20m^3/s.$ Determine the alternate depth of flow.
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Given,

$b = 5m$

$E = 4Nm/N = 4m$

$Q = 20 m^3/s$

$E= 4$

$E = h + \frac{V^2}{2g}$

$V= Discharge/area =Q/b\ \times\ h= 20 / 5 \times\ h$


$E = h + \frac{V^2}{2g} = h + (4/h)^2 \times 1/2 g$

$E = h + \frac{8}{g \times h^2}$

$4 = h + \frac{8}{9.81 \times h^2}$

$4 = h + \frac{0.8155}{ h^2}$

$4h^2 = h^3 + 0.8155$

$h^3 - 4h^2 + 0.8155 = 0$

Solving this cubic equation we get,

h = 3.93 m or 0.48. Ans.

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