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Pascal's Law

In a reach of...

Question

In a reach of open channel flow, with $q = 2 m^{2} / s$ , the depth of flow at section A is 1.5m and at section B, 1.6 m. The difference in specific energies at the two-section can be correspondingly taken as 0.09 m. The bed slope is 1 in 2000. The mean energy slope between the sections is given as 0.003. What is the length of reach AB?

A

54 m with B on the upstream

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B

36 m with A on the upstream

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C

36 m with B on the upstream

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D

54 m with A on the upstream

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Solution

The correct option is C 36 m with B on the upstream
Critical depth,
$y_{c} = {(\frac{q^{2}}{g})}^{1 / 3} = {(\frac{2^{2}}{9.81})}^{1 / 3} = 0.74 m$

$A s y_{A} = 1.5 m a n d y_{B} = 1.6 m a r e m o r e t h a n y_{c} s o t h e f l o w i s s u b c r i t i c a l .$

$∴ E_{A} < E_{B}$

Length of reach AB assuming B on the upstream will be

$L_{A B} = \frac{E_{B} - E_{A}}{S_{0} - S_{1}} = \frac{0.09}{\frac{1}{2000} - 0.003} = - 36 m$
$S i n c e L_{A B} is negative so B is on the upstream .$

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