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Doesn't the current to run the trains heat the wires?

When trains aren't running enough to keep the wires warm, wouldn't it be cheaper to warm them than to repair them?

I see that the trains are dead with no overhead wirefeed.

Shouldn't there be a small generator onboard for emergencies, to at least keep the toilets running? Perhaps a tank of natgas for heat?

I presume catenaries refers to the shape the feed wires take between supports.

Align culture with our nature. Ot else!

by ormondotvos (ormond.otvosnospamgmialcon) on Sat Dec 4th, 2010 at 06:01:52 PM EST
[ Parent ]
Doesn't the current to run the trains heat the wires?

I hope not. Miles of cabling putting out enough heat to melt ice might be less than entirely efficient.

Separate de-icing systems are available, but I'm not sure how widely they're used.

Shouldn't there be a small generator onboard for emergencies, to at least keep the toilets running? Perhaps a tank of natgas for heat?

Trains have very limited space, and weight is an issue.

In the UK failed electrics are hauled to safety by troubleshooter diesels. But if a long stretch fails, the diesels have to do multiple runs to rescue multiple trains. Which takes a while.

There were plans to create a hybrid diesel/electric next generation Intercity train in the UK, but weight and space requirements made it impractical solution.

I expect when nuclear fusion becomes small enough to fit into an engine car these problems will be solved. But until then catenaries, with all of their faults, remain the best choice.

by ThatBritGuy (thatbritguy (at) googlemail.com) on Sat Dec 4th, 2010 at 06:19:42 PM EST
[ Parent ]
To heat the overhead wires with electricity, you have to short-circuit them, which is both a waste and dangerous. (I know of someone who did it in a cold spell, though...) But serious icing on overhead wires is a rare occurrence and there are some devices against it. However, when overhead wires snap, it usually has to do with some defect. Most often it's some failure of the system to compensate thermal contraction (like a stuck roller), sometimes material failure, and sometimes something falling on or getting stuck in the wires.

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.

by DoDo on Sat Dec 4th, 2010 at 06:41:32 PM EST
[ Parent ]
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.)

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.

by JakeS (JangoSierra 'at' gmail 'dot' com) on Sat Dec 4th, 2010 at 06:51:38 PM EST
[ Parent ]
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
[ Parent ]
First a nitpick: the 350 A heating system you describe is shorting the catenary, just not by earthing the normal 'working' voltage. Your system is certainly practical, especially if it is DC; I suspect some Nordic countries use something like that.

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.

by DoDo on Sun Dec 5th, 2010 at 04:50:03 AM EST
[ Parent ]
Why does DC/AC matter?  The heating loop is galvanically isolated from the supply by the use of the transformer.

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.

by njh on Sun Dec 5th, 2010 at 11:13:31 PM EST
[ Parent ]
Why does DC/AC matter?

You don't want nice strong impedances, especially with AC train control systems nearby.

*Lunatic*, n.
One whose delusions are out of fashion.

by DoDo on Mon Dec 6th, 2010 at 03:58:59 AM EST
[ Parent ]
Isn't there a system that allow for ice melting by making pass a high frequency current through the cable?

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.

by Xavier in Paris on Mon Dec 6th, 2010 at 11:55:55 AM EST
[ Parent ]
I do not understand this statement.
by njh on Mon Dec 6th, 2010 at 06:23:04 PM EST
[ Parent ]
I searched around a little; turns out that the system you propose is already in use on some lines in Europe. One is the Cologne-Frankfurt high-speed line (another one electrified with the 15 kV/16.7 Hz system), where the heating current is applied in bursts in the night hours; unfortunately, I couldn't find details (voltage, frequency if any, length of sections, method of shorting). Another is a rack railway in Switzerland, which runs on DC, but the heating current applied during the night is 50 Hz AC, and applied continuously.

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.

by DoDo on Mon Dec 6th, 2010 at 02:01:25 PM EST
[ Parent ]
The scary guy I mentioned who shorted the high voltage did so when, faced with a real bad icing, he judged that freeing the catenary with the flamethrower would take all day. (BTW, I know him well, but I heard this story from others.)

*Lunatic*, n.
One whose delusions are out of fashion.
by DoDo on Mon Dec 6th, 2010 at 02:13:31 PM EST
[ Parent ]
DoDo:
"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.
by Bernard (bernard) on Sun Dec 5th, 2010 at 09:20:13 AM EST
[ Parent ]

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