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My question comes from exercises with parallel pipes such as this.

enter image description here
I know that to analyse internal flow, we use general energy equation which origins from the 1st law so each time we apply it we are using a Control Volume. However for this type of question, if I apply a CV that goes from 1 to 2, part of the mass flows toward C and so starting from the 1st law the equation should become like this:

enter image description here enter image description here

enter image description here

However the solution tells me that it is:

enter image description here

I dont understand why you can procede as the solution. It ignores the mass flow toward 3, almost as if there is only mass flow from 1 to 2 which permits it to cancel out completely the mass flow.

EDIT: Equations from 1 to J and from J to 2. The velocity heads still remains. Have I done anything wrong? enter image description here

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  • $\begingroup$ There is pressure drop in the piping and in the valve going to tank 3 that is not accounted for in your Bernoulli equation approach. $\endgroup$ – David White Dec 13 '18 at 1:58
  • $\begingroup$ Yeah i didnt account for that because my CV only goes from 1 to 2. My confusion is basically that the solution treats the energy equation as if it is bernoulli by following the same streamline. But we take a CV, no matter how,for example from 1 to 2 or the covering the whole fluid, it is inevitable to consider the energy that carries the mass in 3, which isnt in the solution. $\endgroup$ – Richard Dec 13 '18 at 16:11
  • $\begingroup$ Questions: 1) Are the levels in the 3 tanks constant? 2) Are you open to a somewhat different approach to solving this problem? 3) Do you have a way to estimate pressure drop in piping and the valve as a function of flow rate? 4) Are there elevation differences between the tanks? $\endgroup$ – David White Dec 13 '18 at 16:16
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The term involving $V_B$ takes into account the fact that it doesn't involve the full flow. You also need to include the continuity equation which helps to establish the flow rate split between branches B and C. Also, the pressure at junction J has to be the same for the entrance to branches B and C, and this too helps to constrain the split.

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  • $\begingroup$ The term that you are referring is f(L/d)Vb^2/g right? So does it mean that we are just considering the flow rate of B along the way from 1 to 2? If that is the case how can we still consider the head loss through branch A with the velocity of A? $\endgroup$ – Richard Dec 13 '18 at 16:00
  • $\begingroup$ No. The frictional term from 1 to J includes the total flow (the sum of the flows in B and C). The frictional term from J to 2 includes only the flow in B. Write the equations for 1 to J, J to 2, and J to 3 separately and see what you get. $\endgroup$ – Chet Miller Dec 13 '18 at 16:16
  • $\begingroup$ I have updated my post. Could you check it? $\endgroup$ – Richard Dec 13 '18 at 16:34
  • $\begingroup$ It looks OK to me. $\endgroup$ – Chet Miller Dec 13 '18 at 16:52
  • $\begingroup$ I did disregard them both. From 1 to J V1 is 0 but Velocity in J is Va. From J to 2, the velocity in J is Vb. If I have considered it to be Va, that would mean my CV will be before the junction, and so I would need to consider energy of the mass that leaves toward 3. $\endgroup$ – Richard Dec 13 '18 at 16:53

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