0
\$\begingroup\$

I'm trying to design a 4 Pole probe to measure electrical conductivity and I having some doubts about the theoretical calculation of its cell constant.

I know the formula, K(cell constant)=L(Distance)/A(Sectional Area) and i how to calculate in a 2-Pole Probe, in which the L is the distance of the probes and the A is the sectional area of them.

My probe design is like the one in the figure in which the figure. The current sourcing electrodes are the external and are ring shaped and the voltage measurement poles are metallic tips.

So for the cell constant calculation, in the distance part, which distance should I consider, the distance between current electrodes or the voltage measurement poles? And since the current poles are rings, what is cross sectional section, the section made by the voltage poles (rectangle) or the one made by rings (circle).

Thank you. enter image description here

\$\endgroup\$
1
\$\begingroup\$

You want to measure the conductivity of a liquid in the tube? If the current distribution inside the tube is homogenius, the resistor to be measured is a cylinder with a diameter of the inside diameter of the tube. The length of the cylinder is the distance of the two voltage measurement poles, this is the Distance L in the formula. The area A is calculated from the radius of the cylinder, the inside radius of the tube.

\$\endgroup\$
  • \$\begingroup\$ Yes. I agree with you, but the current distribution is the aspect that raises me some doubts. Because i found this probe on a Science Paper and they wrote: "To achieve the desired cell constant of 50 m^-1, the distance between the voltage terminals is approximately 20 mm ( L= K*2*A ) . Why L= K*2*A ? Could it be because only half of the current is going to inside the tube and rest is going to the liquid outside ? \$\endgroup\$ – Impe_dancer Sep 15 '16 at 10:47
  • \$\begingroup\$ I don't believe exactly half of the current would flow inside the tube. I would try with liquid only inside the tube, not outside. What about the use of insulating plugs at each end of the tube? Is it possible to hold the tube vertical to the liquids surface, the upper I pole within the liquid, but the end of the tube outside? \$\endgroup\$ – Uwe Sep 15 '16 at 11:25
  • \$\begingroup\$ No, this is supposed to be under water, according to them is very close to half. But I really don't know if that's the reason why they are multiplying by 2 the constant. The paper is this one, you can see it for yourself too. gim.lx.it.pt/dicsap/IMTC-6490%20-%20Revised%20Version.pdf do you think that it could be from there? \$\endgroup\$ – Impe_dancer Sep 15 '16 at 13:20
  • \$\begingroup\$ They assume the resistance of the liquid outside the tube is negligible. The distance from one end of the tube to the next I-pole is half as long as the distance between both I-poles. Under this condition, the resistance for both pathes are equal and one half of the current flows inside the tube between the I-poles and the other half from one I-pole to the next tube end, than outside the the tube to the other end and from the other end inside the tube to the other I-pole. Therefore only the resistances inside the tube determine the flow of the current. \$\endgroup\$ – Uwe Sep 15 '16 at 13:57
  • \$\begingroup\$ Ok, i understand what you are saying, but do you understand how does this can be related with the K=L/(2 *A) ? \$\endgroup\$ – Impe_dancer Sep 15 '16 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.