# Worked example for Pt wire

If you purchase 33cm of wire of 2mm diameter, this will have a usable surface area of 3.142 X 0.2 X 30 = 18.85 square cm.
That’s a 5.6 amp Anode (300mA/cm squared say) (30 cm active surface in electrolyte)
This wire will weigh 3.142 X 0.1 X 0.1 X 33 X 21.4 = 22.2 grams which will cost 22.2 X \$40 = \$888

A 5.6 Amp Anode made by purchasing 1 mm diameter wire will cost half this much.
The length of the 1mm diameter wire will need to be 63cm (do the calculation).
The weigh of this 63cm of 1mm wire will be 3.142 X 0.05 X 0.05 X 63 X 21.4 = 10.6 grams. Half the price approx.

Should we go for wire that is 0.5mm diameter and 123cm long?
The wire you use must carry 5.6 amps at it's top end without heating too much remember. The resistance of the wire will start to become more and more the limiting factor.
The resistivity of Pt. metal is 0.0000106 Ohm Centimetres.
To calculate the resistance of a piece of wire we use it's dimensions and the resistivity.
The resistance of a wire is:

(Resistivity X Length)/Cross sectional area. (We will use cm throughout)

The resistance of a piece 3 cm long (assume 3 cm to get from connection to electrolyte) by 2mm diameter will be:
(10.6 X 10^-6 X 3)/ (3.142 X 0.1 X 0.1) = .000,404,84 Ohms.

This resistance will have to dissipate a wattage of: (current squared) X Resistance = Watts in a length of wire that is 3cm long.

That’s 5.6 X 5.6 X 0.000,404,84 = 0.127 Watts.

It will be difficult to ascertain if this will melt the lid or not but will give you some indication of the problem.
Hot Pt may also corrode inside the cell at the top.
The wire will also drop Volts = current X resistance. This may be a problem if your power supply is a bit short in the Volts department. It may stop you from pumping as much current into your cell as you had wished.
With high wire resistance the current distribution on the Anode may also suffer because of the high resistance down the length of the Anode, in this case 0.000,404,84 Ohms every 3 cm. (no problem with this particular anode)
The current will preferentially leave the Anode and travel into the electrolyte at the upper end of the Anode. The current density will be higher at the top of the Anode than at the bottom which may have erosion consequences if Anode is being driven at a current density that is close to the point of erosion.
Since Pt can be driven at ferocious current densities this will probably not be too much of a problem in most cases. In the case above (where resistance of 3cm is only 0.000,404,84 Ohms) there will be no problems with the Anode as far as current distribution is concerned.

If we keep cutting down the diameter of the wire to save cash there will come a point when the local heating at the top of the anode will be a problem. A guesstimation of problems occurring would be when you expect 3 cm of wire to dissipate 1/4 Watt or more.

If cash is short and you must scrimp on the wire you can still use very thin wire and simply have a number of Anodes to get up to the surface area you want.
Assume we use 0.25mm diameter wire. We now need 243cm to give us our surface area we want for a 5.6amp Anode. This will cost \$100. The 0.25mm Dia. will not carry 5.6 amps, as 2 Watts of power will be dissipated in the wire leading into the cell. We cut the wire into four pieces and install four Anodes into the cell connecting them at the top using a Copper busbar. Each 0.25 diameter wire now has to only carry 5.6/4 = 1.4 amps (same Anode current density in electrolyte as above). The current will distribute to each Anode if all connections are good and each Anode has similarish distance from a Cathode. It should be noted that to be exact we now need 253cm + 9cm extra as we have to go from lid to electrolyte four times.
You do not have to cut the wire if you so wish. Simply 'snake' the wire into the electrolyte four or five or six times and join at the top with a busbar. You will need extra cm's to do this.

There is a simple spread sheet (ziped) for resistance calculation here.

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