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Post by Deleted on Jan 27, 2016 0:14:45 GMT
Hello all,
I've read on here a number of great replies that explain how the fourth rail system of electrification works. I understand that a positive and negative Voltage with respect to the earthed running rails carry the current and that if one power rail faults to earth then the other power rail "takes over" if you like and carries the full Voltage. I also understand each of the positive and negative rails are connected to the earthed running rails by large resistors, one double the value of the other.
My questions are, what path is current taking if one power rail is out of action and how does the potential divider with two large resistors give nominal Voltages of 420V and -210V?
Also, out of curiosity, where are these large resistors located? In the substation?
Kind regards,
Aidan.
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Post by norbitonflyer on Jan 27, 2016 12:29:30 GMT
I understand that a positive and negative Voltage with respect to the earthed running rails carry the current and that if one power rail faults to earth then the other power rail "takes over" if you like and carries the full Voltage. I also understand each of the positive and negative rails are connected to the earthed running rails by large resistors, . There are people on here much more knowledgeable than I, but I understand that there is no electrical connection between the running rails and the conductor rails in a true 4-rail system. That is actually the point - to insulate the conductor rails from the running rails and stop stray currents passing through the running rails - which are earthed - into the surroundings (particularly, but not exclusively, cast iron tunnel linings) causing interference, electrolytic corrosion, etc. (One can argue that there is no difference between an insulator and a very (very!) large resistance, but in practical terms both conductor rails are insulated as well as possible - one is not deliberately made only half as good an insulator as the other. I understand that, as you say, in general the rails are held at +420V and -210V, but at some locations the "negative" rail is at zero potential and the "positive" at the full 630V. I seem to recall reading that this is done at stations as a safety measure, so that people falling off the platform are not electrocuted if they touch the "negative" centre rail (the positive rail is always on the opposite side to the platform, except (obviously), at double sided platforms. Earth-potential negative rails are also used north of Queens Park and south of Gunnersbury and Putney Bridge, so that NR 3rd-rail units can get the full voltage from just the positive rail. (Converesely, "Sarah Siddons" was adapted to work on 3rd rail by providing an electrical connection between her negative collector shoes and her running gear) I was not aware of any "failover" system designed to allow one rail to take a higher the voltage if the other is accidentally earthed - maybe someone more au fait with the electrical supply can enlighten us? The trains still see the same potential difference, whether the voltages are +630/0 or +420/-210 (or indeed +315/-315, or +840/+210) - just as there is no practical difference between falling to the ground off a 630 foot building, and falling off a 420 foot building into a 210 foot deep hole.
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Chris M
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Post by Chris M on Jan 27, 2016 14:14:43 GMT
The trains still see the same potential difference, whether the voltages are +630/0 or +420/-210 (or indeed +315/-315, or +840/+210) - just as there is no practical difference between falling to the ground off a 630 foot building, and falling off a 420 foot building into a 210 foot deep hole. That is probably the best way of describing potential difference I've ever seen. Thank you.
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Post by trt on Jan 27, 2016 14:54:01 GMT
just as there is no practical difference between falling to the ground off a 630 foot building, and falling off a 420 foot building into a 210 foot deep hole. Apart from the splash radius...
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Post by Chris M on Jan 27, 2016 15:04:16 GMT
just as there is no practical difference between falling to the ground off a 630 foot building, and falling off a 420 foot building into a 210 foot deep hole. Apart from the splash radius... Depends how big the hole is...
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class411
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Post by class411 on Jan 27, 2016 15:35:17 GMT
Apart from the splash radius... Depends how big the hole is... And if you want to be pedantic. I mean really pedantic. Stupidly, mind numbingly, 'so pedantic even the king of the pedantic people would think you were being unnecessarily pedantic', pedantic. There is a difference, because the mean value of g acting upon you during the course of your fall would be microscopically smaller if you did the 'jump into a hole' version. (I did warn you.)
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Post by norbitonflyer on Jan 27, 2016 16:14:40 GMT
There is a difference, because the mean value of g acting upon you during the course of your fall would be microscopically smaller if you did the 'jump into a hole' version. Not if the bottom of the hole next to the smaller building is the same height above sea level as the ground next to the taller building................
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class411
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Post by class411 on Jan 27, 2016 16:52:49 GMT
There is a difference, because the mean value of g acting upon you during the course of your fall would be microscopically smaller if you did the 'jump into a hole' version. Not if the bottom of the hole next to the smaller building is the same height above sea level as the ground next to the taller building................ Damn. Out pedanted! Wait. Ha! You will still get a sub microscopically smaller mean value of g in the 'hole' version, because when you are in the hole there will be mass above your head exerting a force against your body in opposition to that below. (Not sure how much more pedantry is possible. Limits are being pushed, here.)
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Post by norbitonflyer on Jan 27, 2016 17:26:09 GMT
Indeed there would. (disregarding the fact that material you have removed from the hole would even more microscopically move the centre of gravity of the planet).
But unless there is a very large electrostatically charged object hovering nearby, none of these variations in potential would apply in the voltage scenario to which this was supposed to be an analogue!
Of course if the hole was deep enough we would be in RIPAS territory, with a new tube line to the antipodes.
However, we are being discourteous to the OP, who would I expect would still like to know how the +420/-210 voltage difference is actually maintained - I'm imagining a battery of 105 6V cells connected in series, with the positive and negative rails connected to the respective ends of the row, and an earth connection between the 35th and 36th cells, (counting from the negative end)! (or perhaps more realistically, three 210V dc dynamos connected in series, with an earth connection between no 1 and no 2).
And is there any provision for the failover situation he describes?
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class411
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Post by class411 on Jan 27, 2016 18:13:46 GMT
I'd actually assumed that the voltages would be achieved by having two independent windings on the transformer in the substation yielding two distinct voltages.
Then after rectification, the positive side of lower voltage and the lower side of the higher voltage would be grounded, yielding the voltages required.
I certainly would not expect resistors to play any part in the proceedings.
I also find it extremely doubtful that they would do anything along the lines suggested to handle a failure. Apart from the technical difficulty it intuitively sounds somewhat dubious on safety grounds.
However, I'll be very interested to hear what an electrical engineer has to say on the subject as I could very easily be wrong!
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Post by nickf on Jan 27, 2016 18:37:45 GMT
I'd actually assumed that the voltages would be achieved by having two independent windings on the transformer in the substation yielding two distinct voltages. Then after rectification, the positive side of lower voltage and the lower side of the higher voltage would be grounded, yielding the voltages required. I certainly would not expect resistors to play any part in the proceedings. I also find it extremely doubtful that they would do anything along the lines suggested to handle a failure. Apart from the technical difficulty it intuitively sounds somewhat dubious on safety grounds. However, I'll be very interested to hear what an electrical engineer has to say on the subject as I could very easily be wrong! This old thread might help. Traction Current
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Post by norbitonflyer on Jan 27, 2016 20:17:25 GMT
That old thread reveals that resistors are involved after all, as part of a fault-detection system. As it says, such a fault would cause the non-faulty rail to go to +/630V. The reference earth is at the section ends, not at the generator.
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class411
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Post by class411 on Jan 28, 2016 9:58:56 GMT
I'd actually assumed that the voltages would be achieved by having two independent windings on the transformer in the substation yielding two distinct voltages. Then after rectification, the positive side of lower voltage and the lower side of the higher voltage would be grounded, yielding the voltages required. I certainly would not expect resistors to play any part in the proceedings. I also find it extremely doubtful that they would do anything along the lines suggested to handle a failure. Apart from the technical difficulty it intuitively sounds somewhat dubious on safety grounds. However, I'll be very interested to hear what an electrical engineer has to say on the subject as I could very easily be wrong! This old thread might help. Traction CurrentExcellent post by RailTechnician, there. A pity he left the site. As I suspected, the resistors are not involved in the actual voltage reduction, but their use is rather clever, providing a more fault tolerant system than the one I envisaged above.
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Post by Deleted on Jan 28, 2016 23:21:06 GMT
Thanks for posting a link to that older thread. Unfortunately it doesn't explain, electrically, what it is about the circuit that causes negative and positive rail Voltages to fall into their respective values with respect to the running rail. Also, it doesn't explain how one rail "takes on" the Voltage of the other under fault conditions.
Any insight would be fantastic.
Kind regards,
Aidan.
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Post by norbitonflyer on Jan 29, 2016 0:13:26 GMT
The generator creates a 630V potential difference which, as the item in the other thread states, is not directly referenced to earth. The big resistors he describes act as a potential divider to put earth two thirds of the way from positive to negative.
Under the fault condition described, the resistance between earth and one of the rails falls to zero - a short circuit in parallel with a big resistor (remember that the effective resistance of two resistors in parallel is given by 1/R = 1/r1 + 1/r2. However enormous r1 is, if r2 is zero then 1/r2 is infinite, so 1/R is infinite, and R is zero)
Effectively what the short circuit is doing - changing the ratio of the two resistances and thus the voltages) either side of the earthed point from 2:1 to 0:1 (or 2:0),
However, the generator is still forcing a potential difference of 630V between the two rails. If one rail is earthed, the other will therefore be at the full 630V above (or below) earth.
Using my height analogy again, think of the generator as a rigid vertical rod 630cm long connecting the two rails together. The positive rail must be 630cm above the negative rail. If you hold the rod one third of the way up, then the upper rail will be 410cm above your hand (the level of which we shall define as "Earth"), and the lower rail will be 210cm below it. Now hold one end of the rod. That puts one of the rails at the level of your hand, and the other is 630cm above (or below) it.
Remember that it is mere convention that defines "earth" as zero. (Just as we have chosen arbitrary values for zero in temperature scales - such as the freezing point of a particular chemical under certain atmospheric conditions found at a particular altitude on only one known planet) Planet Earth itself actually carries a small net electrostatic charge, and local surface variations occur e.g because of thunderstorms. Electrical equipment earthed to two bodies that are electrically isolated from each other - for example two spacecraft, or an aircraft and the ground - may actually be at different potentials. If they get too close, a current will flow to equalise the potentials. That is, according to one theory, what happened to the Hindenberg.
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class411
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Post by class411 on Jan 29, 2016 8:07:18 GMT
My questions are, what path is current taking if one power rail is out of action The other question has now been answered quite comprehensively. The answer to this one is that the current will actually follow exactly the same path until it 'finds' the short. After all, there is no other way for it to get in/out of the train. The only difference is that, once it reaches the point of the short to earth, it can travel through anything grounded. How much of the current does that depends on the relative resistance of all paths available.
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roythebus
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Post by roythebus on Jan 29, 2016 8:31:49 GMT
I often wondered how the earth fault relays on rolling stock worked. My son, Who Knows About These Things as he is fleet engineer for the Southern Railway, explained it a while ago. Whilst this applies to his trains, presumably modern tube stock has earth full detection.
I always assumed that all the electrics on Southern stock was earthed through the bodywork; indeed on older stock such a sQ and Standard stock it probably was as everything worked on line volts. but, with the advent of stock with low-voltage lighting and control, everything is separately earthed, isolated from the train body, so any fault between line volts (in his case the running rails) and the low voltage from battery or MG source) needs to be addressed. So they have an earth fault relay or MCB. It works bit like the RCD (or whatever the new name is for it) that you find in your house which detects either + or - to earth and trips out before you electrocute yourself.
I guess the large (value not physical size) resistors mentioned in Rail technician's thread do the same job, so as soon as one of the supply rails ups the voltage, this is detected by a relay and trips out the substation, preventing 650 volts going through to earth by whatever means. Which is also why there's lengthy neutral sections between LT and NR power supply systems at Putney, Gunnersbury and Queens Park.
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class411
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Post by class411 on Jan 29, 2016 9:54:34 GMT
I guess the large (value not physical size) resistors mentioned in Rail technician's thread do the same job, so as soon as one of the supply rails ups the voltage, this is detected by a relay and trips out the substation, preventing 650 volts going through to earth by whatever means. Which is also why there's lengthy neutral sections between LT and NR power supply systems at Putney, Gunnersbury and Queens Park. No, that's not quite the same thing. The arrangement on LU allows for continued operation with one conductor rail grounded. Only if both are grounded would the system 'blow'. Of course, they could put detectors across the resistors to either warn of a fault, or, if they so desired, automatically shut the system down.
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