The head loss in a sudden expansion from 8 cm diameter pipe to 16 cm diameter pipe in terms of velocity V1, in the smaller pipe is

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ESE Mechanical 2015 Paper 1: Official Paper
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  1. \(\frac{1}{4}\left( {\frac{{V_1^2}}{{2g}}} \right)\)
  2. \(\frac{3}{{16}}\left( {\frac{{V_1^2}}{{2g}}} \right)\)
  3. \(\frac{1}{{64}}\left( {\frac{{V_1^2}}{{2g}}} \right)\)
  4. \(\frac{9}{{16}}\left( {\frac{{V_1^2}}{{2g}}} \right)\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{9}{{16}}\left( {\frac{{V_1^2}}{{2g}}} \right)\)
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Detailed Solution

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Concept:

Equation of continuity: A1V1 = A2V2

A = area of cross section = \(\frac{\pi }{4}{D^2}\)

Head loss due to sudden expansion = \(\frac{{{{\left( {{V_1} - \;{V_2}} \right)}^2}}}{{2g}}\)

Where, V1 = velocity before expansion

V2 = velocity after expansion

g = acceleration due to gravity

Calculation:

Given, D1 = 8 cm, D2 = 16 cm

Using continuity equation:

\(d_1^2{V_1} = d_2^2{V_2}\)

\({V_2} = {\left( {\frac{8}{{16}}} \right)^2}{V_1}\)

\({V_2} = \frac{1}{4}{V_1}\)

Therefore, head loss,

\({H_l} = \frac{{{{\left( {{V_1} - {V_2}} \right)}^2}}}{{2g}} = \frac{{V_1^2{{\left( {1 - \frac{1}{4}} \right)}^2}}}{{2g}} = \frac{{V_1^2\left( {\frac{9}{{16}}} \right)}}{{2g}}\)

\(H_l=\frac{9}{{16}}\left( {\frac{{V_1^2}}{{2g}}} \right)\)

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