Dupit’s equation is expressed for series-connected pipes as:

(where L1, L2, and L3 are lengths of pipe 1, 2 and 3, d1, d2, d3 are diameter of pipe 1, 2 and 3, L is equivalent length of pipe are D is equivalent diameter of pipe)

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  1. L/d5 = L1/d15 - L2/d25 - L3/d35
  2. L/d5 = L1/d15 + L2/d25 + L3/d35
  3. L/d5 = L1/d15 - L2/d25
  4. L/d6 = L1/d16 + L2/d26 + L3/d36

Answer (Detailed Solution Below)

Option 2 : L/d5 = L1/d15 + L2/d25 + L3/d35
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Detailed Solution

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Concept:

In series: If n numbers of pipes are arranged in series then the discharge in all the pipes are equal, from the conservation of mass principle.

F1 S.S Madhu 13.01.20 D3

i.e. Q1 = Q2 = Q3 = Q4 = Q5 = … = Q

The local losses are all the minor losses that occur because of sudden expansion or sudden contraction, entry and exit losses, and due to bends.

And the major losses due to friction i.e.

\({h_l} = \frac{{fl{V^2}}}{{2gd}},{h_1} = \frac{{f{L_1}V_1^2}}{{2g{D_1}}},{h_2} = \frac{{f{L_2}V_2^2}}{{2g{D_3}}}\)

\({h_{eq}} = \frac{{f{L_{eq}}V_{eq}^2}}{{2g{D_{eq}}}} = \frac{{f{L_{eq}}Q_{eq}^2}}{{12D_{eq}^5}} = \left( {minor\;losses + major\;losses} \right)\;in\;each\;pipe\)

\(\frac{{{L_{eq}}}}{{D_{eq}^5}} = \frac{{{L_1}}}{{D_1^5}} + \frac{{{L_2}}}{{D_2^5}} + \frac{{{L_3}}}{{D_3^5}} + \frac{{{L_4}}}{{D_4^5}} + \frac{{{L_5}}}{{D_5^5}} + \ldots .\)

In parallel: if n number of pipes are arranged in parallel then the discharge in all the pipes are different and the discharge is equal to sum discharge of individual pipe also the head loss in all the pipes are equal.

i.e. \(Q = {Q_1} + {Q_2} + {Q_3} + {Q_4} + {Q_5} + \ldots \)

and h = h1 = h2 = h3 = h4 = h5 = …

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