Lecture 6: Understanding 2x25kV ac traction system (Auto Transformer Feeding System)
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- Опубліковано 18 жов 2024
- This video provides the basic understanding of 2x25 kV AC traction system (Auto Transformer Feeding System). It explains its advantages (e.g. reduced voltage drop, reduced EMI effect) using basic circuit diagrams. Furthermore, it provides the voltage level required as per EN 50163/EN 50388 and different types of connection techniques such as Scott connected, V-connected/center tapped traction transformers.
I was hunting for an explanation of the 2x25kV system (as I work in electricity transmission but not railways) and I have stumbled across this video. Great explanation! I'm very grateful - thumbs up from me.
Many thanks
Actually It's amazing that you were so kind to share all this knowledge. Thank you very much!
sir, your teaching skill is fabulous. so simple to digest and understand what you teach. thank u sir again.
Many thanks
Hi there. This presentation was helpful in improving my understanding. Thank you for taking the time to share with us.
your efforts are tremendous and Scott-T calculation you did so good!! Good Job Sir..
Many thanks
Excellent
Very clear explanation
Great video. Thanks a lot!
Great explanation! Thanks
Many thanks
Wonderful lecture-thumbs up from me son
Many thanks
You made it so simple
Many Thanks.
Excellent sir
Thanks
8:16 I have seen other diagrams of 2 x 25kV systems but not explained in this understandable detail. Thank you.
In your experience is it still necessary to add a separate feeder in parallel with the catenary and contact wire like in DC systems to increase current capacity of the circuit or is the contact wire in parallel with the +25kV feeder adequate for the higher current flowing in the subsections between ATs.
Can you please share a single line diagram of whole system showing isolators, circuit breakers and track switches.
What the the clearance between the +25kV and ~25kV feeders on mast poles and in substation. Is it the same as for 66kV as the next higher standard voltage
Excellent😊
Many thanks!
It’s really a marvellous system. We recently energised an Autotransforner Substation on an existing 2 x 25kV railway , and despite an electrical engineering degree I was still a little confused during the energisation testing / section proving, how we managed to get both catenary and feeder busbars energised at 2 x 25kV by energising the substation via single feeder (in this particular instance it was via the -25kV track feeder). However, by sketching out an SLD, it becomes obvious that applying 25kV between any pole of the AT and earth, the flux developed in the AT core will then induce the opposite 25kV pole. Then when it comes to introducing all feeders from the track (and ultimately the adjacent feeder station) it become obvious why this method is so efficient.
Good information
Many thanks
Sir, can you pl explain the necessity for running AEC & BEC? Why can't we use only rail as return path?
AEC and BEC are required for multiple reasons. Some of them are listed below
1. Cross bonding and other measures are not sufficient to reduce rail potential so that it falls within the limit.
2. To ground OCS poles and Wayside equipment with every time drilling or cadwelding with rails.
3. AEC also helps in reducing EMI.
Running rails along with AEC and BEC form the part of return circuit. You can avoid them completely by increasing cross bonding between tracks, design for headway mentioned in contract and not overdesign, evaluate rail potential as per annex C provided in EN 50122-1, providing earth pits along the route. All this require detailed analysis.
Great video
Thanks
Hello, thank you for the video - very informative. I am looking to model an ATF system using LT spice, however I am having trouble setting up a reliable system: I have modelled the centre tapped transformer as two 25 kV (peak) 50 Hz AC Voltage sources with a 180 degree phase shift; the Inductors (used to represent the Auto transformers) as 0.2 H; The train as a resistive load of 80 Ohm. The issue is when I check the current in the Voltage sources, I am getting 800 A peak. This is far greater than the current in the resistive load, which reads 300 A.
1. I would like to know whether anyone is able to give any advice as to whether I can make any improvements with my model to yield more realistic results.
2. I am also aware that Lt Spice is typically used for lower voltage electronics modelling. Is anyone able to recommend a similar free (and new user friendly) application that may be more suited to high voltage circuit modelling?
Many thanks.
Is it necessary to install harmonic filters in traction substations 2x25kV?
thank you Sir
Sir, why can't we centre tapped the secondary of Scott connected transformer as we are doing for V-connected transformer?
Why we have to use Auto Transformer after Scott connected transformer?
What do the notations SSP and SP mean? And what type of connections are there and why they are required?
Very well presented, thank you!
Would be interesting to see a comparison of RC vs BT vs 2x25kV in terms of losses in transformers, transformer kVA (capital cost) and short-circuit MVA.
Many thanks. Thanks for your suggestion. Would try to make a video on it.
Hi there... The main advantage of 2x25 kV system over 1x25 kV system is to reduce the voltage drop by a factor of 2 for same load and at same distance. But this can be achieved by increasing the effective cross section of the catenary by running a feeder wire at the same potential as the catenary and then connecting it with the catenary at various locations. Why was this approach not taken in place of the 2x25 kV system explained in your video.
Hi, you are correct on the part to provide additional contact wire in parallel to existing and this practice is done. Either we increase the cross-section or add an additional conductor (also called reinforcement feeder if cable is used instead of conductor). This can help reduce to resistance and to some extent self inductance, however, effect of mutual inductance will be still prevalent. Thus, such practices are mainly popluar in case of DC traction.
Moreover, 2x25kV has many other advantages which 1x25 kV cannot offer and the main being less electromagnetic interference and lower transmission losses.
Please tell me what is the difference between the Booster Transformer System and Auto Transformer System?
Kindly refer to lecture 3 of traction seriers
Kindly post video about boundary post
When current ,I flows in +ve feeder & I flows in -ve feeder,I2R loss will be 2 times ,so same as 1ph * 25kV system,so how loss will be less,Can you please clarify ??
Let us suppose resistance of all conductors whether in 2x25kV or in 1x25kV is same and is R. Furthermore, neglecting losses due to other conductors that form return circuit and only considering +ve & -ve feeders in 2x25kV and contact in 1x25kV.
In 2x25kV as highlighted by you = (I^2)*R (+ve feeder) + (I^2)*R (-ve feeder) = 2(I^2)*R
In 1x25kV = (2I^2)*R = 4(I^2)*R
Hi, I was rewatching your video again and wanted to ask what is the difference and benefits of using center-tap for secondary of supply transformer (So far I haven’t seen, with the systems where I work, center-tapped secondary). Thanks!
Hi, it basically a V-connection. Scott is more beneficial than this in terms of voltage unbalance. However, both fair nearly the same in case of short circuit MVA is very high at source end.
Do you knwo Static Frequency Converters (SFC) ?
Yes but not done much work related to it. Mainly handled projects till now with other traction equipment.
Hello.
This was helpful but there is one thing I still don't understand: Looking at the 9:11 video time.... I don't see how the -25kV line negates EMI for lineside equipment between the two autotransformers where the train is.... since the autotransformers supply current for the train through the +25kV rail and it returns to them through the running-rails to the auto-transformer centre tap.
Since we have higher current flowing through the running rails.... would this not expose lineside equipment to EMI?
Thanks.
When there is single train in the section then there will not be any reduction in interference as rightly pointed out by you. However, the current will be low enough to not cause interference. This is typically useful when headway is less and number of trains in sections are more which leads to more flow of current that causes EMI issues.
@@railwayengineeringinsights6463 Thank you for taking the time to reply. I really appreciate it.
Hi, this video illustration the current flow is based on the voltage level of +25kV for catenary, 0V for rail, and -25kV for negative feeder. But since this is an AC system, so the voltage waveform of both catenary and negative feeder fluctuate between -25kV to 25kV am I correct?
Because your video explain very nicely, but I am attempting to translate it into a single line simulation model. In simulation, the catenary and negative feeder is both connected to a 25kV busbar, am I correct?
Hi, yes you are correct it will fluctuate.
Yes catenary and feeder are both connected to negative 25kV busbar which are feed from traction transformer.
@@railwayengineeringinsights6463 Thank you sir
Hello Sir, Could you please share En standards you are referring to while explanation..that would help users a lot .
Nice sir
Many thanks.
Where is complete ckt diagram which is show on the video first before started
Hi will we have separate SSP and AT locations? If we do how physically we connect +25kv to catenary system? I mean should we connect through an overlap or directly connect to catenary by means of a 2 pole switch?
Yes in some location you can have separate AT. This might sometimes be required to maintain the voltage range as defined by EN 50163. Yes it can be directly connected by means of switching devices without overlap if section is not required at that particular location.
Wonderful video & explanation. Can you please describe SSP and other abbreviations used.
Hello Sahil,
If I have railway system with 2x25kV,50Hz
with autotransformer 50/25kV, how I calculate the impedance of the autotransformer with the general equation : Zat=(Uk%/100)(Un^2/Sn)
,so in this case,what is the value of Un, 50kV? or 25kV?
Where: Uk% is the short circuit autotransformer impedance and Sn is the nominal apparent power in [VA]
Best regards,Dennis
Hi Dennis,
Typically we calculate impedance on secondary side, i.e. 27.5kV side. For example, for AT rating of 15MVA and Uk of 1%, the impedance of AT on secondary side will be 0.504 ohms [(0.01) *(27.5^2)/15].
Kind regards,
Sahil Bhagat
@@railwayengineeringinsights6463 THANK YOU Sahil!
Wonderful video. I have been working with 2*25 kV system but never got this type of clarity.
Thanks for making this video.
Do post this type of video.
If possible plz share your number.
Many thanks. Sure will upload more such.
Hi sir please do a video on metro train power supply distribution system i.e from panto to wheel
Sure. But upcoming videos will be on EMC
@@railwayengineeringinsights6463 When will be uploaded? Anxiously being waited
WoW!
Sir, why don't we call here Booster Transformers, but as Auto Transformers
Hi, beacuse booster are connected in series with contact-catenary system and have a different connection whereas auto transformer have different connection as explained
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Send me your WhatsApp number !
Great explanation! But can you elaborate more on the current distribution depending on the train position between two substations.
As I understood, if train is in the middle then current distribution will be even (half and half).
So, the question is, when train is closer to one of the substations, voltage drop decreases and current increases on the one side and opposite on the side of RR? And if yes, does it impact train load current distribution on the catenary (contact wire)?
Example:
Train load is 2I. Since the train іs closer to the substation on the left (as a reference to your circuit), let’s say that on the RR current splits 1.5I to the left and 0.5I to the right. On the auto transformers windings they gonna split evenly (1.5I/2 = 0.75I, 0.5I/2 = 0.25I). On the negative feeder side 0.75I+0.25I=1I, which will be added with 0.75I (auto transformer’s trolley winding), so train is drawing 1.75I from supply and 0.25I - return current from auto substation.
Correct me if I wrong.
Many thanks. Your current distribution in example is correct. As you go closer to a substation the impedance seen by the load is less one side compared to the other. Thus current on closer side increases as shown by you in example compared to the other👍
@@railwayengineeringinsights6463 Thank you, usually in the papers there is always an example with 3/4 and 1/4 currents and it might get confused (as it was for me initially).
Also, can you explain how current will be distributed when there are two trains on the same track (what is the RR current direction in this case - on one side they will flow in direction to autos, but on the common side between trains current will be 180 degrees to each other, so should they cancel/subtract each other)?
Let us suppose we have two trains, T1 and T2 and two AT, AT1 and AT2. T1 is closer to AT1 and current distributes along RR as 4I/3 towards AT1 and 2I/3 towards AT2 considering train takes 2I. Similarly for T2 the distribution will be 4I/3 towards AT2 and 2I/3 towards AT1. Therefore, in between two trains at RR we have 2I/3 in phase opposition. But we shall not cancel it in order to complete the flow. Thus, 4I/3 + 2I/3 will become 6I/3 (2I) at the center point of each AT and will split I in each winding.
The net voltage drop due to current (2I/3) on running rails will have less effect at that instant. However, rail potential at that will be governed by the other two currents towards AT only for that instant.
@@railwayengineeringinsights6463 Thanks for the answer. I have been stuck with this one for a while. Looking forward for your new videos👍
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