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I Water model for electric circuits: Pipe diameter?

  1. Nov 1, 2016 #1

    There's this nice water circuit model for electric circuits where pressure corresponds to electric potential and the (mass or volume) flow rate to electric current.

    In the water model, we can vary the pipe diameter along the circuit. Since water is practically incompressible, the flow rate remains constant, i.e. the flow velocity must increase when the pipe diameter decreases.

    Is there an electrical analog to the pipe diameter? It certainly does not correspond to a resistance, since no energy is dissipated if we assume frictionless flow.

    Also, which pressure is the closest analog to electric potential? I guess it must be total pressure, since it should be the same at two points of different pipe diameter (if there's no dissipative element between them) to correctly model zero voltage drop along perfect conductors, and static and dynamic pressure would not be constant individually when the flow velocity changes.
  2. jcsd
  3. Nov 2, 2016 #2

    Jonathan Scott

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    Gold Member

    The obvious equivalent is the diameter of the conductor, which does effectively change the resistance, especially if it is too small relative to the amount of current. Assuming frictionless flow is equivalent to assuming a superconductor.
  4. Nov 3, 2016 #3


    Staff: Mentor

    In the analogy the diameter would be one of the factors that determines resistance. Specifically, the resistance of the pipe varies as 1/r^4.


    Note, the pipe-circuit analogy is actually fairly poor for a lot of reasons. The biggest problem is that fluid flow in a pipe is horrendously complicated, far more complicated than the electrical circuit laws. So the analogy should only be taken very qualitatively. Asking for quantitative details like that quickly exposes this prime weakness in the analogy.
  5. Nov 3, 2016 #4
    A more accurate quantitative analog between water flow and electrical current flow would be flow through a porous medium. In such a case, for the porous medium, the equation would be $$\vec{v}=-\frac{k}{\mu}\nabla P$$where ##\vec{v}## is the superficial water velocity, k is the permeability of the porous medium, and ##\mu## is the viscosity of the water. The corresponding equation for current flow would be $$\vec{j}=-\frac{1}{r}\nabla V$$where V is the electrical potential, r is the resistivity of the conductor, and ##\vec{j}## is the current density.
  6. Nov 6, 2016 #5

    David Lewis

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    Gold Member

    The electrical analog would be resistance. If you double the cross sectional area of a conductor, its resistance is cut in half.
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