Discussion
A liquid flows under steady and incompressible flow conditions from station 1 to station 4 through pipe sections P, Q, R, and S as shown in figure. Consider, \(d\), \(V\), and \(h\) represent the diameter, velocity, and head loss, respectively, in each pipe section with subscripts ‘P’, ‘Q’, ‘R’, and ‘S’. \(\Delta h\) represents the head difference between the inlet (station 1) and outlet (station 4). All the pipe sections are placed on the same horizontal plane for which the figure shows the top view.
[The top view shows pipe P from station 1 to station 2, then a wider pipe Q from station 2 to station 3, after which the flow divides into two parallel branches, R and S, that rejoin at station 4 before the outlet.]
Which one of the following options is correct for the given flow loop?
Answer: C
The correct pair is \(\Delta h = h_{P} + h_{Q} + h_{R}\) and \(V_{P}d_{P}^{2} = V_{Q}d_{Q}^{2} = V_{R}d_{R}^{2} + V_{S}d_{S}^{2}\).
P and Q are in series, so their head losses add and they carry the same flow. R and S are in parallel between stations 3 and 4, so each has the same head loss, \(h_{R} = h_{S}\), and the loss across the pair is counted once. Hence \(\Delta h = h_{P} + h_{Q} + h_{R}\).
The flow rate \(Q = \frac{\pi}{4}d^{2}V\) is the same in P and Q, and it divides between R and S, so \(V_{P}d_{P}^{2} = V_{Q}d_{Q}^{2} = V_{R}d_{R}^{2} + V_{S}d_{S}^{2}\).
Adding \(h_{S}\) as well double-counts the parallel branch, and writing \(V_{R}d_{R}^{2} = V_{S}d_{S}^{2}\) equal to the total flow treats the parallel branches as if each carried all of it.
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