Any (electric) railway engineers on here? I have a question.
Discussion
I have recently started taking the train to work and as I'm interested in all things technical, I got to thinking...
The train I usually catch appears to only have a single Pantograph to pick up power from the overhead line.
My question is - how does this apparently small point of contact manage to transfer so much current?
I'm guessing that an electric train must consume 100's (if not 1000's?) of amps, particularly during the acceleration phase.
I realise that the voltage is very high, which I presume means less current - but is it AC or DC?
TIA for any explanations!
MODS: Please don't move this, as this is a general technical kind of question, which may also be of interest to others.
The train I usually catch appears to only have a single Pantograph to pick up power from the overhead line.
My question is - how does this apparently small point of contact manage to transfer so much current?
I'm guessing that an electric train must consume 100's (if not 1000's?) of amps, particularly during the acceleration phase.
I realise that the voltage is very high, which I presume means less current - but is it AC or DC?
TIA for any explanations!
MODS: Please don't move this, as this is a general technical kind of question, which may also be of interest to others.
Lines are high-voltage AC (low voltage would incurr huge transmission losses, DC is less useful for efficient electric motors). I don't know what the power requirements are, but I think 100 amps at 25kV is a decent ballpark for train energy consumption (I would say an order of magnitude accuracy).
That would let an 800t train accelerate to 60mph in approximately 100s (assuming ~30% efficiency), a bit slow, but that's ~the heaviest TGV listed on Wiki.
Ross: You're doing a BEng in Railway Tech, yet you haven't had electricity 101 and mechanics 101?
That would let an 800t train accelerate to 60mph in approximately 100s (assuming ~30% efficiency), a bit slow, but that's ~the heaviest TGV listed on Wiki.
Ross: You're doing a BEng in Railway Tech, yet you haven't had electricity 101 and mechanics 101?
Edited by Krikkit on Thursday 28th June 20:04
There are at least 2 x 1.5" (ish) carbon strips in the pantograph head, this is sufficient contact area to transfer enough power to the transformer via the VCB. The voltage is then tapped at various voltages depending on the class of train.
Power is voltage x current so if your motor circuits are ( for example) 650v @ 200a then you draw 5.2a through the pantograph carbons.(in simple terms).
Power is voltage x current so if your motor circuits are ( for example) 650v @ 200a then you draw 5.2a through the pantograph carbons.(in simple terms).
Harrytsg said:
There are at least 2 x 1.5" (ish) carbon strips in the pantograph head, this is sufficient contact area to transfer enough power to the transformer via the VCB. The voltage is then tapped at various voltages depending on the class of train.
Power is voltage x current so if your motor circuits are ( for example) 650v @ 200a then you draw 5.2a through the pantograph carbons.(in simple terms).
And because carbon is such a wonderful conductor, the amount of heat generated by conduction is tiny, so it could carry huge amounts of energy without problem.Power is voltage x current so if your motor circuits are ( for example) 650v @ 200a then you draw 5.2a through the pantograph carbons.(in simple terms).
Krikkit said:
Lines are high-voltage AC (low voltage would incurr huge transmission losses, DC is less useful for efficient electric motors). I don't know what the power requirements are, but I think 100 amps at 25kV is a decent ballpark for train energy consumption (I would say an order of magnitude accuracy).
That would let an 800t train accelerate to 60mph in approximately 100s (assuming ~30% efficiency), a bit slow, but that's ~the heaviest TGV listed on Wiki.
Ross: You're doing a BEng in Railway Tech, yet you haven't had electricity 101 and mechanics 101?
Yes we did, nearly 3 years ago, and I went down a different route! More of telecommunications and signalling that anything! That would let an 800t train accelerate to 60mph in approximately 100s (assuming ~30% efficiency), a bit slow, but that's ~the heaviest TGV listed on Wiki.
Ross: You're doing a BEng in Railway Tech, yet you haven't had electricity 101 and mechanics 101?
Edited by Krikkit on Thursday 28th June 20:04
Krikkit said:
Lines are high-voltage AC (low voltage would incurr huge transmission losses, DC is less useful for efficient electric motors). I don't know what the power requirements are, but I think 100 amps at 25kV is a decent ballpark for train energy consumption (I would say an order of magnitude accuracy).
That would let an 800t train accelerate to 60mph in approximately 100s (assuming ~30% efficiency), a bit slow, but that's ~the heaviest TGV listed on Wiki.
Ross: You're doing a BEng in Railway Tech, yet you haven't had electricity 101 and mechanics 101?
I wouldn't even say its 101, its GCSE physics That would let an 800t train accelerate to 60mph in approximately 100s (assuming ~30% efficiency), a bit slow, but that's ~the heaviest TGV listed on Wiki.
Ross: You're doing a BEng in Railway Tech, yet you haven't had electricity 101 and mechanics 101?
Edited by Krikkit on Thursday 28th June 20:04
Harrytsg said:
There are at least 2 x 1.5" (ish) carbon strips in the pantograph head, this is sufficient contact area to transfer enough power to the transformer via the VCB. The voltage is then tapped at various voltages depending on the class of train.
Power is voltage x current so if your motor circuits are ( for example) 650v @ 200a then you draw 5.2a through the pantograph carbons.(in simple terms).
Can you clarify the figures you quoted?Power is voltage x current so if your motor circuits are ( for example) 650v @ 200a then you draw 5.2a through the pantograph carbons.(in simple terms).
650v at 200A equates to 130000W by my calculations!
I can't see how you draw 5.2A to run a motor at 650v / 200A?
P = V * A
So, 650V at 200A = 130kW - ten times a domestic socket.
Up the voltage to 25000V and you need a much lower current to supply the same power, P = V*A again (although slightly modified due to the waveyness of AC current, but that's getting a bit technical). Net result is significantly less current, but with the same power transferred.
Think of it this way - the voltage is how much energy each electron in the wire carries, the current is how many electrons are moving through it (it doesn't really work this way, but it's a very good model for all intents and purposes here), give each electron more energy and you need fewer of them to come through to turn your motor.
So, 650V at 200A = 130kW - ten times a domestic socket.
Up the voltage to 25000V and you need a much lower current to supply the same power, P = V*A again (although slightly modified due to the waveyness of AC current, but that's getting a bit technical). Net result is significantly less current, but with the same power transferred.
Think of it this way - the voltage is how much energy each electron in the wire carries, the current is how many electrons are moving through it (it doesn't really work this way, but it's a very good model for all intents and purposes here), give each electron more energy and you need fewer of them to come through to turn your motor.
Max_Torque said:
Yeah, but a 130kW train isn't going to be much use now is it?
No, a typical 4 car DMU (for example a class 321) works on a horsepower figure of 1328hp, 996kw. This would translate to approx 39A, but as it wasn't a specific question (or answer) about a class of unit the figures were only for demonstrating the principle. Gassing Station | General Gassing | Top of Page | What's New | My Stuff