GATE Civil Engineering (CE) 2019 Shift 2 Solved Paper

Question : 42 of 65

Marks: +1, -0

At the foot of a spillyway, water flows at a depth of

23 {cm}

with a velocity of

8.1 {m} {/} {s}

, as shown in the figure.

The flow enters as an

{M}-3

profile in the long wide rectangular channel with bed slope

=\;{1}/{1800}

and Manning's

{n}=0

.A hydraulic jump is formed at a certain distance from slope the foot of the spillway. Assume the acceleration due to gravity,

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

. Just before the hydraulic jump, the depth of flow

{y}_{1}

(in

{m}

, round off to 2 decimal places) is __________-

Solution:

Given data:

y = 0.23 m

V = 8.1 m{\/}sec.

Slope

{s}=\;{1}/{1800}

{h}=0.015

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

{q}= {yV}=8.1 × 0.23

\;\; =1.863 {m}^{3}{/} {sec} {/} {m}

It is given that flow enters as an

{M}-3

profile in the long wide rectangular channel with bed slope

=\;{1}/{1800}

From manning equation at

{M}_{3}

profile section.

Q=\;{1}/{n} R^{\;{2}/{3}} √{s} A

q ⋅ B=\;{1}/{n} y^{2{/} 3} √{s} ⋅ y ⋅ B

[for wide channel

{R}= {y}]

1.863=\;{1}/{0.015} √{\;{1}/{1800}} {y}^{3 {/} 3}

⇒ \;\; y_{n}=1.11 {m}

The flow profile will follow normal depth

( {yn}_{ {n}})

after jump.
For wide rectangular channel,

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

∴ \;\; y_{c}=(\;{1.863^{2}}/{9.81})^{1 {/}3}=0.707 {m}

Here,

{y}_{ {n}}= {y}_{2}=1.11 {m}

{F}_{2}^{2}=\;{ {q}^{2}}/{ {gy}_{2}^{3}}

⇒ \;\; F_{2}=√{\;{1.863^{2}}/{9.81 × .11^{3}}}=0.508

Again,

\;{y_{1}}/{y_{2}}=\;{1}/{2}[√{1+8 F_{2}^{2}}-1]

⇒ \;\; \;{y_{1}}/{y_{2}}=\;{1}/{2}[√{1+8 × 0.508^{2}}-1]

⇒ {y}_{1}=0.417 {m}< {yc}[

supercritical

]

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