Water is coming from a reservoir through a pipeline ABC which con
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>
Please scroll down to see the correct answer and solution guide.
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
\(h = \frac {18V^2}{32g}=\frac {9V^2}{16g}\)