I have the following simple Y shape water pipe network:

Given $S_{out}$ ( in $m^3/s$), can we compute $S_1$ and $S_2$?

For the pipe, we know for each pipe their corresponding diameter $d$, length $L$, roughness coefficient $C$, we also know the height for each node relative to ground, and hence I guess we do know about pressure in this case.

All the pipe properties ( such as $d$, $L$ and $C$) are assumed constant regardless of their position in the pipe layout.

An additional remark here (this remark has nothing to do with my current question; but it serves to identify what the kinds of inputs I have with this question.) For a close loop pipe network, we can use the hazen william formula and hardy cross method to compute the head loss. This means that whatever that is required to be known about the pipe is already known, as long as they are used in hazen william formula and hardy cross method.

My question is, given the above information, is there anyway we can uniquely compute $S_1$ and $S_2$ for the Y shape network above? From what I know, we have one equation but two unknowns ( the conservation of source):


And all the pipe properties are not relevant as far as this Y Shape pipe network is concerned.

My conclusion is that above network is unsolvable. But I hear the claim that it is actually solvable as long as we take into account of the aforementioned pipe properties. But I think this is not the case.

What do you think?

Bonus: It would be great it one can provide a formulation for the solution of a general shape water pipe network, with or without loops.

  • $\begingroup$ It is solvable if you know S1 and S2 have the same pressure--- this is the most common condition when they are attached to a main. $\endgroup$
    – Ron Maimon
    Apr 18 '12 at 13:07
  • $\begingroup$ We know about the pressure for each node-- they may or may not be the same $\endgroup$
    – Graviton
    Apr 18 '12 at 14:18

As you say, the above problem stated barely is not solvable. I add that this is in the sense that there are many solutions, based on a thought experiment approach.

It is unstated in the problem what the pressures at the three nodes of the network are. If the pressures are all the same, the flow will be for practical purposes zero. If the pressures are different, there will be flow. One aspect of the pressure differences is the geometry relative to the local gravitational field, which again is unstated. If this is flat, with no other pressures acting, no flow, if it's vertical, flow.

If the sources you see claiming that this is a solvable problem are knowledgeable, I would assume that they mean "given the relative pressures and geometry". It's clear also that if the way in which the pipes are joined is nontrivially asymmetric, in a practical setting, that may modify the effective C-values of the various pipes (in other words, it may not be enough to measure the C-values of the three pipes before the whole is constructed; knowing whether the effects of the join will have to be taken into account in a given situation is a matter of engineering experience).

Thanks for the links to the hazen william formula and the hardy cross method, which were interesting. Needless to say, my answer is ab initio, I do not have experience in pipework. I hope it's helpful, but I guess I'm taking a risk, having just downvoted kleingordon.

  • $\begingroup$ We do know about the pressure for the $s_1$ and $s_2$ node as we know their height. Does this help? As for the $C$ value, we have to assume them constant regardless of the geometry and pressure $\endgroup$
    – Graviton
    Apr 18 '12 at 14:20
  • $\begingroup$ Also, I am keen to learn on how does the knowledge of the pressure at the above two nodes help to solve this problem? Any links or specific? $\endgroup$
    – Graviton
    Apr 18 '12 at 14:22

Pipe network problems of this sort are isomorphic to circuit problems. You mention in your question that you are considering closed loop pipe networks. So assuming your diagram forms a closed loop, then in addition to the junction equation expressing conservation of flow (analogous to conservation of current in a circuit junction), you've got loop equations to express that head loss is independent of path taken (analogous to Kirchoff's loop law for circuits). You can find the head loss along each path in terms of the unknown currents, using the pipe geometries to compute the hydraulic resistances.

  • $\begingroup$ I think you misunderstand my question; my remark on close loop network has nothing to do with my current question; but it serves to identify what the kinds of inputs I have with this question. I am concerning with the Y shape network I posted above and nothing more $\endgroup$
    – Graviton
    Apr 18 '12 at 9:01
  • $\begingroup$ With apologies, -1. If the Wikipedia pages cited in the question are correct, pipe network problems are clearly not isomorphic to circuit problems because the relationship of flow to pressure is significantly nonlinear (of course the relationship of electric current to electric potential is also not linear, but, unless there is qualification, it is generally taken to be linear in the regime of "circuit problems"). If the Wikipedia pages are incorrect about this, and your claim of isomorphism is correct, it is incumbent on you to cite a source. $\endgroup$ Apr 18 '12 at 11:29
  • $\begingroup$ @PeterMorgan You shouldn't downvote an answer if you are unsure. Many common "circuit problems" are nonlinear, for example any problem including a diode or a transistor. The analogy with an electric circuit is a very useful one, because it allows using a circuit simulation software such as SPICE. It's easy to implement the required nonlinear "resistors" using nonlinear dependent sources. $\endgroup$
    – mmc
    Apr 18 '12 at 23:00
  • $\begingroup$ @mmc, I only rarely downvote, and only when I can identify why I have done so in a comment, so I can be corrected. With the information in your comment, which looks as much like an answer, kleingordon's answer becomes more useful. Does SPICE allow hydrodynamic flow problems to be solved, or does the user have to work through the isomorphism with electric circuits by hand? If I only did anything when I was sure, I would never do or learn anything. Thanks for your comment. $\endgroup$ Apr 19 '12 at 3:14
  • $\begingroup$ @PeterMorgan Sorry, my comment ended up sounding harsher than I intended. You need to make the translation by hand, but it's quite simple: merely a question of renaming and multiplying all the constant factors to get a single constant. I will try to find time to write a more complete answer. $\endgroup$
    – mmc
    Apr 19 '12 at 12:01

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