Water is coming from a reservoir through a pipeline ABC which con

Water is coming from a reservoir through a pipeline ABC which con
| Water is coming from a reservoir through a pipeline ABC which consists of two pipes AB and BC connected in series. The diameter of pipe AB is 10 cm and pipe BC is 20 cm. The total head loss in terms of small pipe velocity V will be (Neglect major losses)

A. <span class="math-tex">\(\frac {9V^2}{16g}\)</span>

B. <span class="math-tex">\(\frac {9V^2}{32g}\)</span>

C. <span class="math-tex">\(\frac {17V^2}{32g}\)</span>

D. <span class="math-tex">\(\frac {10V^2}{32g}\)</span>

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Right Answer is: A

SOLUTION

Concept:

The water is flowing from reservoir to pipe AB and BC and then exit, the total losses incurred will be 

Entrance loss (From reservoir to Pipe AB)

Sudden expansion loss (AB to BC)

Exit loss (BC to exit)

The various losses are given by 

  • Sudden expansion loss:

\({\left( {{h_L}} \right)_{exp}} = \frac{{{{\left( {{v_1} - {v_2}} \right)}^2}}}{{2g}}\)

  • Exit loss:

\({\left( {{h_L}} \right)_{exit}} = \frac{{v_2^2}}{{2g}}\)

  • Entrance loss:

\({\left( {{h_L}} \right)_{ent}} = \frac{{0.5 × v_1^2}}{{2g}}\)

Calculation:

Given small pipe velocity = V ⇒ VAB = V1 = V;

From continuity, A1 V1 = A2 V2 ⇒ 102 × V = 202 × V2

⇒ V2 = 0.25 V;

Now the losses will be 

\({\left( {{h_L}} \right)_{exp}} = \frac{{{{\left( {{V} - {0.25V}} \right)}^2}}}{{2g}} = \frac {9V^2}{32g}\)

\({\left( {{h_L}} \right)_{ent}} = \frac{{0.5 × V^2}}{{2g}} = \frac {V^2}{4g}\)

\({\left( {{h_L}} \right)_{exit}} = \frac{{(0.25V)^2}}{{2g}} = \frac {V^2}{32g}\)

Total losses will be 

\(h = \frac {18V^2}{32g}=\frac {9V^2}{16g}\)