GATE Civil Engineering (CE) 2020 Shift 1 Solved Paper

Question : 14 of 65

Marks: +1, -0

4 {m}

wide rectangular channel carries

6 {m}^{3} {/} {s}

of water. The Manning's 'n' of the open channel is

0.02

. Considering

{g}=9.81 {m} {/} {s}^{2}

, the critical velocity of flow (in

{m} {/} {s}

, round off to two decimal places) in the channel, is __________.

Solution:

Velocity

=\;{\;\text " Discharge "\;}/{\;\text " Area "\;}=\;{Q}/{b . y}=\;{q}/{y}

∴ V_{c}=\;{q}/{y_{c}}=\;{q}/{(\;{q^{2}}/{g})^{1{/} 3}}=q^{1 {/} 3} g^{1{/}3}

{V}_{ {c}}^{3}= {qg}

Also,

q=V_{c} × y_{c}

V_{c}^{3}=V_{c} \;y_{c} \;g

\;\; V_{c}^{2}=g y_{c}

∴ {V}_{ {c}}=√{ {gy}_{ {c}}}

Now,

y_{c}=(\;{q^{2}}/{g})^{1{/}3}=(\;{({6}/{4})^{2}}/{9.81})=0.612 {m}

\{ ∴ q=\;{Q}/{B}\}

∴ V_{c}=√{9.81 × 0.612}=2.44 {m} {/} {sec}

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