Darcy weisback formula head loss due to friction in pipe flow is 

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JKSSB JE Civil Jal Shakti 5 Dec 2022 Official Paper (Shift 2)
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  1. hf\({4fLV^2 \over 2gd}\)
  2. hf\({4fLV^3 \over 2gd}\)
  3. hf\({2fLV^2 \over 4gd}\)
  4. hf\({2fLV^3 \over 4gd}\)

Answer (Detailed Solution Below)

Option 1 : hf\({4fLV^2 \over 2gd}\)
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Detailed Solution

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Explanation:

Head loss due to friction in the pipe is given by Darcy’s formula:

\({h_f} = \frac{{4{\rm{fL}}{V^2}}}{{2{\rm{gd}}}}\;\)

f = Darcy’s coefficient or the coefficient of friction, l = length of pipe, v = velocity of liquid flow, d = diameter of pipe

F = Friction factor = 4f

Additional InformationDifferent type of minor losses in the pipe are as follows:

Losses due to the sudden expansion   \({h_{le}} = \frac{{{{({V_{^1}} - {V_{^2}})}^2}}}{{2g}} = \frac{{{V_1}}}{{2g}}{(1 - \frac{{{A_1}}}{{{A_2}}})^2}\)
Losses due to a sudden contraction

\({h_l} = \frac{{{{({V_c} - {V_2})}^2}}}{{2g}} = \frac{{{V_2}^2}}{{2g}}{(\frac{1}{{{C_c}}} - 1)^2}\)

If Cc is not given then, \({h_l} = \frac{{0.5{V_2}^2}}{{2g}}\)

Losses at the exit of the pipe \({h_l} = \frac{{{V^2}}}{{2g}}\)
Losses at the entrance to the pipe \({h_l} = \frac{{0.5{V^2}}}{{2g}}\)
Losses due to bends

\({h_l} = \frac{{K{V^2}}}{{2g}}\)

K = 1.2 for 90

K= 0.4 for 45

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