I am currently studying the Hardy Cross method for water distribution networks. In the beginning of his 1936 paper, Hardy Cross states:

the distribution of flow in the network ... is controlled by two sets of conditions, both simple and obvious:

(a) The total flow reaching any junction equals the total flow leaving it (continuity of flow)

(b) The total change in potential along any closed path is zero (continuity of potential).

Condition (b) is not clear to me. Why must the total change in potential along a closed pipe circuit equal zero? Does the potential not continuously drop in the loop as the water loses hydraulic head due to friction?


It's just Kirkhoff's current laws except for molecules instead of charge.

Potential is another word for pressure.

Assume the pipes are large enough that there is negligible friction loss. Then assume somewhere in the loop there are constrictions (resistors) that resist the flow, and elsewhere in the loop are pumps (batteries) that boost the flow.

When the water flows through a constriction, its pressure drops, because it takes pressure to move the water through it. When water flows through the pump, its pressure increases, because the pump is pushing it.

If you are a little drop of water going around the loop, the pressure you feel drops as you go through the constriction, and then increases as you go through the pump, back to what you felt before - zero change.

If the pump gets weaker and gives you less pressure increase, then when you get to the constriction you have less pressure, so you flow more slowly through it, so it has less pressure drop. So again, the total pressure change is zero.

  • $\begingroup$ Sorry I just edited the original question. I meant that the water loses head due to friction, not pressure. $\endgroup$
    – user32882
    Dec 20 '16 at 11:49

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