GATE Civil Engineering (CE) 2019 Shift 1 Solved Paper

Question : 60 of 65

Marks: +1, -0

Two water reservoirs are connected by a siphon (running full) of total length

5000 {m}

and diameter of

0.10 {m}

, as shown below (figure not drawn to scale).

The inlet leg length of the siphon to its summit is

2000 {m}

. The difference in the water surface levels of the two reservoirs is

5 {m}

. Assume the permissible minimum absolute pressure at the summit of siphon to be

2.5 {m}

of water when running full. Given: friction factor

{f}=0.02

throughout, atmospheric pressure

=10.3 {m}

of water, and acceleration due to gravity

{g}=9.81

{m} \/ {s}^{2}

. Considering only major loss using Darcy- Weisbach equation the maximum height of the summit of siphon from the water level of upper reservoir,

{h}

(in

{m}

round off to 1 decimal place) is

Solution:

Given data

{d}=0.1 {m}

Length of siphon

=5000 {m}

Length of siphon upto summit

=2000 {m}

Friction Factor,

{f}=0.02

Acceleration due to gravity,

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

Applying Energy equation between point 1 and 3 to get

\;{P_{1}}/{γ}+\;{V_{1}^{2}}/{2 g}+Z_{1}=\;{P_{3}}/{γ}+\;{V_{2}^{2}}/{2 g}+Z_{3}+h_{f(1-3)}

10.3+0+Z_{1}=10.3+0+Z_{3}+\;{f l Q^{2}}/{12.1d^5}

[From Darcy Weisbas equation

h=\;{ {flQ}^{2}}/{12.1 {d}^{5}}

⇒ \;\; 5=\;{0.02 × 5000 × Q^{2}}/{12.1 ×(.1)^{5}}

[ ∵ {Z}_{1}- {Z}_{3}=5 {m}]

⇒ {Q}=2.4597 × 10^{-3} {m}^{3} {/} {sec}

Now applying energy equation between 1 and 2 to get

\;{P_{1}}/{γ}+\;{V_{1}^{2}}/{2 g}+Z_{1}=\;{P_{2}}/{y}+\;{V_{2}^{2}}/{2 g}+Z_{2}+h_{f(1-2)}

10.3+0+Z_{1}=P_2/γ+Q^2/{2ga^2}+Z_2+\;{f l_{(1-2)} Q^{2}}/{12.1d^5}

⇒ 10.3-(Z_{2}-Z_{1})=2.5+\;{(2.4597 × 10^{-3})^{2}}/{2 × 9.81 × \;{π}/{4} × 0.1^{2}}

+\;{0.02 × 2000 ×(2.4597 × 10^{-3})^{2}}/{12.1 ×(0.1)^{5}}

⇒ 10.3-h=4.5 {m}

⇒h=5.8 {m}

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