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Streaming current

Does pumping water through the tube itself produce a current — and how much?

The formula
I=−εr⁢ε0⁢ζ⁢A⁢Δpη0⁢L,Vs=−εr⁢ε0⁢ζ⁢Δpη0⁢σ
I
Streaming current, A
εr
Relative permittivity of the liquid
ζ
Zeta potential of the wall, V
A
Bore cross-section, πD² ÷ 4, m²
Δp
Pressure drop, Pa
η0
Liquid viscosity, Pa·s
L
Section length, m
Vs
Streaming potential, V
σ
Liquid conductivity, S/m
LaTeX 
I = -\frac{εr \cdot \varepsilon_0 \cdot ζ \cdot A \cdot Δp}{η0 \cdot L}, \qquad Vs = -\frac{εr \cdot \varepsilon_0 \cdot ζ \cdot Δp}{η0 \cdot σ}

Work it out

The liquid in the tube. Its permittivity sets the double layer's charge and its viscosity how fast the flow drags it. A gas has no double layer.

mV

The potential at the shear plane of the wall's double layer. Glass and most plastics against tap water: −20 to −60 mV.

mm

The current scales with the cross-section; the voltage does not depend on it at all.

m

The length of tube over which the pressure drop is measured, between the electrodes.

kPa

The pressure driving the flow through the section. Both current and voltage are proportional to it.

µS/cm

The liquid's conductivity. Sets how easily the streaming charge leaks back, and so the open-circuit voltage. Distilled water is about 1 µS/cm; tap water, hundreds.

or see a worked example with typical numbers

Method

  1. Look up the liquid's relative permittivity and viscosity — 80 and 1 mPa·s for water — and multiply the permittivity by ε₀ for the absolute value, 7.1 × 10⁻¹⁰ F/m.
  2. Find the bore's cross-section, πD² ÷ 4.
  3. Helmholtz–Smoluchowski: the streaming current is −ε·ζ·A·Δp ÷ η·L. The minus sign is the convention that a negative wall gives a positive current in the direction of flow, because the flow carries the positive counter-ions of the diffuse layer.
  4. With the ends open the charge builds until conduction back through the liquid cancels the streaming current: the streaming potential is −ε·ζ·Δp ÷ η·σ. It does not depend on the bore or the length, only on the pressure drop and the liquid.
  5. A source of open-circuit voltage V and short-circuit current I delivers at most I·V ÷ 4 into a matched load.

Assumptions

  • A thin double layer: the Debye length — under a micron in any water with dissolved ions — is far smaller than the bore, which for 25 mm it is by four orders of magnitude. The Helmholtz–Smoluchowski form is then independent of the tube's shape.
  • No surface conduction along the wall, which matters only for very low-conductivity liquids in very fine capillaries, and not here.
  • The pressure drop entered is the one across the section, and the flow is what that drop drives — laminar or turbulent, the double layer sees only the wall shear, and the result holds for both.
  • The zeta potential is the wall's against this liquid, and it depends on the wall material, the pH and the dissolved ions far more than on anything else on this page. Glass, PVC and most plastics run −20 to −60 mV against tap water; it is the least certain input by a factor of three.
  • This is an electrokinetic effect with nothing magnetic in it. It moves charge along the tube, and it would be measured between electrodes at the ends of the section; a coil round the tube sees a steady current of ten nanoamperes as a steady field of nothing, and a pulsed one as a signal a thousand million times below its noise.
  • Nothing here is Meyer's. The archive is asked whether water flowing through the tube is itself a source of current, and it is — this is the arithmetic of how much, which is the part the question needs.