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GATE Civil Engineering (CE) 2019 Shift 1 Solved Paper

Question : 60 of 65

Marks: +1, -0

5000 \text{m}

0.10 \text{m}

2000 \text{m}

5 \text{m}

2.5 \text{m}

f=0.02

=10.3 \text{m}

g=9.81

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

h

\text{m}

round off to 1 decimal place) is

Solution:

Given data

d=0.1 \text{m}

Length of siphon

=5000 \text{m}

Length of siphon upto summit

=2000 \text{m}

Friction Factor,

f=0.02

Acceleration due to gravity,

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

Applying Energy equation between point 1 and 3 to get

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

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

[From Darcy Weisbas equation

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

\Rightarrow \;\; 5=\;\frac{0.02 \times 5000 \times Q^{2}}{12.1 \times (0.1)^{5}}

[\because Z_{1}-Z_{3}=5 \text{m}]

\Rightarrow Q=2.4597 \times 10^{-3} \text{m}^{3}/\text{sec}

Now applying energy equation between 1 and 2 to get

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

10.3+0+Z_{1}=P_{2}/\gamma+\frac{Q^{2}}{2ga^{2}}+Z_{2}+\;\frac{f l_{(1-2)} Q^{2}}{12.1 d^{5}}

\Rightarrow 10.3-(Z_{2}-Z_{1})=2.5+\;\frac{(2.4597 \times 10^{-3})^{2}}{2 \times 9.81 \times \frac{\pi}{4} \times 0.1^{2}}

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

\Rightarrow 10.3-h=4.5 \text{m}

\Rightarrow h=5.8 \text{m}

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