GATE Civil Engineering (CE) 2020 Shift 1 Solved Paper

Question : 57 of 65

Marks: +1, -0

Water flows at the rate of

12 {m}^{3} {/} {s}

in a

6 {m}

wide rectangular channel. A hydraulic jump is formed in the channel at a point where the upstream depth is

30 {cm}

(just before the jump).
Considering acceleration due to gravity as

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

and density of water as

1000 {kg} {/} {m}^{3}

, the energy loss in the jump is

Solution:

Head loss in a hydraulic jump

= {h}_{1}=\;{( {y}_{2}- {y}_{1})^{3}}/{4 {y}_{1} {y}_{2}}

Here,

y_{1}=0.3 {m}

We know that,

Y_{2}=\;{y_{1}}/{2}(-1+√{1+8 F_{r_{1}}^{2}})

So,

F_{r_{1}}=(\;{q^{2}}/{q y_{1}^{3}})^{1 \/ 2}=(\;{2^{2}}/{(9.81 × 0.3^{3})})^{1 {/} 2}=3.88

∴ y_{2}=\;{0.3}/{2}[-1+√{1+8 × 3.88^{2}}]

⇒ y_{2}=1.505 {m}

Now,

h_{2}=\;{(1.505-0.3)^{2}}/{4 × 1.505 × 0.3}

=0.968 {m}

Power lost

= {r}_{ {w}} {Qh}_{2}=9.81 × 12 × 0.968 {kW}

=114.04 {kW}

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