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You do not need to 'short' the overhead wire.  You just need to run enough current through it to form a layer of water sufficiently large for the ice to drop off.  Let's do some calculations:

Assume ice at -20C - requires 40kJ/kg to reach 0 plus 334kJ/kg to melt - so let's say 400kJ / kg.

A catenary with a 100g of ice / m would be heavily loaded, and we only want to melt a layer 0.5mm thick on a 1cm dia wire (00 gauge), or about 1.5g of water / m - i.e. 3kg/km or about 600kJ / km

To provide this heat we might heat the wire for 20s every 5 minutes (or perhaps per section ahead of the train?), requiring 30kW of power when heating.  The resistance of 00gauge wire is about 0.25ohm and i^2r = 30kW => 350A which is well within the current rating of the OHW.  To provide this current we do not need to provide 350A*25kV, but rather v^2/r => 86V per km.

To get this 90V per km we might treat up and down wires as separate lines, running the current up one side and back down the other.  We would connect the overheat wires together at one and, and insert a 350A, 90V*km*2 transformer between them at the other end, normally shorted together, but turned on to deice.  Such a transformer might consist of four turns of catenary wire on the secondary, and for a 25kV supply, 1000/(km*2) turns on the primary.

If an electric train normally uses 1MW, on a 25kV line, it should in fact produce enough heating to melt the ice (apparently not quickly enough to prevent damage), but this approach would only need 30kW*20s/300s = 2kW average to maintain the ice free wire.

Numbers are all made up; substitute better ones if you know them.

by njh on Sat Dec 4th, 2010 at 08:54:57 PM EST
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