Leak resistance across a tubular cell
Through what resistance does the charge on this tube leak away, given the water's conductivity?
- R
- Leak resistance, Ω
- a
- Inner diameter, in
- b
- Outer diameter, in
- l
- Length, in
- σ
- Conductivity, µS/cm
LaTeX
R = \frac{\ln(b / a)}{2\pi \cdot σ \cdot l}Method
- Convert the conductivity to siemens per metre (1 µS/cm is 10⁻⁴ S/m) and the length to metres.
- The resistance between coaxial cylinders through a conducting medium is ln(b/a) ÷ (2πσl) — the same logarithm as the capacitance, inverted, because the current spreads through the same annulus the field does.
- Divide by the length in inches for the per-inch figure.
Assumptions
- The water's conductivity is what the meter says and is uniform. Under a high field it is not: ions crowd the electrodes, gas forms at them, and the effective resistance during a pulse differs from the meter reading, usually upward.
- DC resistance. At the pulse frequency the water's loss is partly dielectric, which the Cole-Cole calculation covers; this is the ionic conduction alone.
- No electrode effects. The electrochemical double layer at each tube adds a series capacitance and a polarisation resistance that this ignores.