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Trains have batteries. They don't last forever, though, and in modern times, even a simple passenger car comes with a sophisticated software... which may be ill designed and shut down the system in the 'wrong' kind of emergency. (I could tell stories.)
"Catenary" is American English for "overhead wire"; though recently it is increasingly in use in rail literature this side of the pond, too (I prefer to use it, too). *Lunatic*, n. One whose delusions are out of fashion.
And you may not want to run a train across a section of broken but still possibly live wire. It shouldn't be a problem (trains are pretty good Faraday cages), but with this sort of voltages you prefer safe over sorry.
- Jake Friends come and go. Enemies accumulate.
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.
For the electric part of your calculation, I can only give slightly different numbers which don't result in changes of magnitude. For a European catenary type, let's take Re200, with a maximum 0.15 Ω/km (and as low as 0.04 Ω/km if there are parallel lines). For ormondotvos's Swedish example, the voltage is 15 kV; and we can easily assume 3 MW, giving us 200 A. That's 6 kW/km. In the extreme case of IORE locos (those double locos on the photo downthread) working at full power (10.8 MW, hence 720 A), we get 77.76 kW/km.
However, I am less sure about the first part of your calculation. I think we should be calculating heat flows rather than amounts of heat. One kilometre of 100 mm² wire has a surface of 35 m² (ignoring the carrying wire here), ice has a minimum thermal conductivity of 2.22 W/mK at 0°C. So even in the IORE case, for steady state, the temperature of the wire needs to rise above that of the air on the other side of the ice only by the numerical value of the width of the ice coat in millimetres (assuming it's uniform). To melt the ice, you really need a burst of power. *Lunatic*, n. One whose delusions are out of fashion.
Don't forget the R1 insulative effect of the still air around the conductor. I can believe the burst vs continuous argument, even if we move to a very narrow pulse, say 0.5s, that's still only about 1MW which could be provided by the existing infrastructure.
You don't want nice strong impedances, especially with AC train control systems nearby. *Lunatic*, n. One whose delusions are out of fashion.
As the cable impedance depends on frequency, increasing with frequency, even a low high frequency current results in heating the cable, without having to apply a high voltage.
I remember someone talking about that but I can't recall if it was applied on train catenaries.
However, neither system was found sufficient. At the Swiss line, the problem is the time between switching off the heating and the transit of the first train. On the high-speed line, my (German) source doesn't say what's the problem, but says that the more conventional catenary brushing cars are in use, too.
(Other conventional methods I am aware of: spraying a de-icing fluid, sending out a loco in the morning with both pantographs up so that one scraps the ice while the other draws current, and using a 'flamethrower' to de-ice the worst spots.) *Lunatic*, n. One whose delusions are out of fashion.
"Catenary" is American English for "overhead wire"; though recently it is increasingly in use in rail literature this side of the pond, too (I prefer to use it, too).
In French: Caténaire. Comes from the same Latin root:
Catenary - Wikipedia, the free encyclopedia
The word catenary is derived from the Latin word catena, which means "chain". Huygens first used the term catenaria in a letter to Leibniz in 1690. However, Thomas Jefferson is usually credited with the English word catenary.
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