drewhcan't remember what the Reading side uses.
Was the 'same' 11kV that the old PRR electrification used. Probably no point in increasing it (no long heavy trains coexisting with very fast trains).
If I remember correctly, back in '76 when 4800 was pulling the Reading MUs to Trenton, the 'issue' about their not going under their own power was one of signaling and control, not overhead voltage. At least one pan in the train was up for lighting and heat.
RME
The RER in Paris uses overhead electrification as does the Tyne and Wear metro in the UK. The Budapest metro line one does too, but it basically was a street car line that was put underground in the 1880's. Septa's centre city commuter tunnel in Philly has catenary, but of course it's the former Penn and Reading suburban lines. There is a phase break in the tunnel as the 2 systems used different voltage. In know the Pennsy side is the same as Amtrak at 12.5 kv, can't remember what the Reading side uses.
blue streak 1 as other posters have nooted there is somewhat more electronics with AC transmission but if AC traction is used anyway ??????
as other posters have nooted there is somewhat more electronics with AC transmission but if AC traction is used anyway ??????
Keep in mind that almost all of the drive electronics for AC traction motors run off of a DC bus. There are circuits called cycloconverters that will convert AC to any frequency from zero to one third of the line frequency, but for anything but a fractional horsepower motor, you will want to run it off a three phase line.
- Erik
The Chicago Transit Authority continues with 600VDC third-rail for distribution but has converted to AC traction with the 5000-series cars now being delivered. Subsequent orders now in the proposal stage will continue with AC traction.
When building a new line, many considerations are brought to bear on power distribution selection. If subways and tunnels predominate, catenary may require large tunnels for clearances. Third rail power with frequent substations might be expensive because of the cost of realestate. The question of frieght service and possible fowling of third rail by unusual freight equipment might be considered.
all these arguments about the power source ignores the fact almost all traction motors are now beiing buit AC. AMTRAK certainly will not be buying any more DC and the class 1s are very close. Any light rail / streetcar that is now being built north of N 40 appears to be all AC traction. I m sure there are exceptions. With AC traction the problems of snow, ice, sand grounding out DC motors is eliminated. as other posters have nooted there is somewhat more electronics with AC transmission but if AC traction is used anyway ??????
Naturally since the system will be integrated with the Orange Line, this is necessary. I wonder what sort of propulsion would be appropriate for a dedicated, heavy-rail line of 23 miles with 18 stations that didn't have to account for any old system? I say overhead AC to avoid the cost of acquiring land and utility for all of those substations. I wonder about acceleration between stations, too.
To be fair, they truck these prefabricated structures in and build walls around them. I also wonder about power consumption and loss as we go AC-DC-AC.
daveklepperDo not give up hope! We did get an Amtrak electrification to Boston, small extensions on NJT, and New Haven - Springfield, and SF - San Jose or Gilroy look pretty positive.
That occurs to me too, Dave. We do get a lot of people who talk about that terrible, awful, no good very bad system called Amtrak that never does anything right. Yet they replaced all of the track between Washington and Boston and even re surveyed the road, they have electrified it all the way and now they are in the process of replacing the older sections of the overhead wires. And that is just part of what they have accomplished.
Best regards, John
erikemFor my discussion of electrification, "long-distance" is more than 20-50 miles.
OK. I didn't see the Pennsylvania's electrification between Washington and New York and between Philadelphia and Harrisburg as long distance. I was thinking of perhaps a thousand or two miles.
Dave,
It might be better to adapt the "Green Goat" design to recharge the batteries off of the trolley - though pay a lot more attention to the batteries than was done in the original design. Modern solid state controllers are a better match to battery operation than the series-parallel plus rheostat controllers of the early days. OTOH, the old trolley/interurban locomotives have proved to be exceptionally durable and might be available for a lot less than a "Green Goat".
I wonder if such a locomotive would be useful on the Santee line of the San Diego Trolley.
Do not give up hope! We did get an Amtrak electrification to Boston, small extensions on NJT, and New Haven - Springfield, and SF - San Jose or Gilroy look pretty positive.
Look at how light rail mileage has climbed in the last 30 years! What would be interesting would be if one of the short-line freight operators that use light rail tracks for most of their work would finally feel the need to reduce noise and rehab a steeple cab from a trolley museum, possibly one with or equippable for short-time battery operation.
\
For my discussion of electrification, "long-distance" is more than 20-50 miles. The New Haven electrification is on the short end of "long-distance", and the NYC Grand Central electrification being on the long end of short distance. While the low voltage third rail used on the NYC could have been extended all the way to Chicago, it was certainly not economic to do so. The Milwaukee and PRR were about the only roads in the US where the electrified sections spanned multiple crew districts.
"Power Blocks" were used more by AC electrifications as they needed phase breaks, from my understanding the Milwaukee kept most of the line energized to allow for load sharing between substations.
What do you define as a long-distance heavy train?
Most "long distances" in the United States were not economically equipped with catenary. Railroads repeatedly looked at this, as far back as the Weed 'express parcels' railroad in the 1880s. SP, ATSF and others carefully considered electrification (most with DC at the time) and almost nobody 'bit' The one road that thought it would try was the Milwaukee extension, and the two comparatively-little parts they did electrify broke them.
Imagine running the typical GG1 train behind diesels -- how many of them would you need? Hint: your typical early E unit was down under 9000lb of TE above 110mph. This is because the drive is limited to the developed horsepower of the diesel motor. Electrics have very high comparative instantaneous and hourly ratings -- consider the effective horsepower of those little B-B AEM7 toasters. How much good would a diesel THAT size provide you?
Erik did not add that the wire of a catenary system is divided into power blocks, which are only 'on' when a train is going to be in them; power to the PRR block substations could be 132,000V (which was similar to normal overhead powerline voltage at the time). That is one place the Gibbs & Hill infrastructure showed how well designed it was.
Electrification of the Middle Division (and a cutoff tunnel under Horse Shoe that the electrification would have made possible) was in development, up to the point during the War years that the capacities of the locomotives involved (using 428-A motors, like the DD2) were specified. By the time work on this could resume... the economics of F7s and Geeps were providing some of the gains at humongously less first and maintenance cost... just don't go bragging about performance unless you brought a PASSEL of MUable units...
Meanwhile, over in Europe all the truly fast trains are electric; diesels just don't have the hp-to-weight to match what can be drawn from external power -- and turbines, well, the fuel efficiency still isn't there... even with the current experimentation with GTCC to get the cycle thermoefficiency up. Pity we can't even fund pervasive catenary even with the Big O's giveaway plans (and if that doesn't tell you something, think again!)
erikem. For long distance heavy electric operation, AC is better than DC due to the higher operating voltages historically possible with AC. For a given conductor size, the product of train power and substation spacing is proportional to the square of the voltage, so that's why AC is a big hit with long distance heavy electric operation.
Erik,
OK. But I have understood tha long distance heavy trains are almost always pulled with diesel locomotives. Am I wrong?
I do appreciate the electrolysis issue.
Thanks for the information, John
John,
For most transit applications, DC is better than AC, with the advantage increasing with decreasing car weight due to the transformer not being needed with DC. For long distance heavy electric operation, AC is better than DC due to the higher operating voltages historically possible with AC. For a given conductor size, the product of train power and substation spacing is proportional to the square of the voltage, so that's why AC is a big hit with long distance heavy electric operation.
For suburban commuter rail, the choice isn't so clear cut.
Traditionally, 3000V was the highest potential considered practical for a DC electrification, though Westinghouse did demonstrate a moderately successful 5000V electrification (motors and controllers were much more expensive than lower voltages). Progress in power electronics devices now make a high-voltage "DC-DC transformer" possible, though it's questionable whether this would have enough of an advantage over commercial frequency 25kV or 50kV electrifications to make development and deployment worthwhile.
Finally, one advantage of AC over DC is the much lesser problem with electrolysis.
erikemNow for a few comments of my own on AC vs DC.
Eric,
if you look at the history of electric traction in the US different companies have made different decisions about ac and dc. I know that today the issues are different than they were before modern electronics. But is there clearly one way that is or was better?
Stay warm and dry, John
Aegrotatio,
Overmod is concerned that I may have confused you rather than clarifying anything. I hope that didn't happen but if I did I hope that he he cleared up that confusion. He is absolutely right that total power from AC is calculated by a process called root-mean-square and I did not intend to suggest it was not.
For the record, root-mean-square is hard to clearly define. Here is one example. If you take a group of people you can measure their height, add all of the heights, divide the the total number in the group and you will get the arithmetic mean height. You can also square each individual height, calculate the arithmetic mean of the squared heights and take the square root of your result. The answer will be the same. Why do this elaborate exercise? If you want to measure the average ac voltage the answer would always be zero because have of the sine wave is a negative value. By squaring the negative values you use you produce a positive number. Just as +2 squared is +4 so also -2 squared is +4. So r-m-s enables you to measure your effective voltage.
Avoid the snow. John
I don't there is any reason for you to be sorry, kicking the hornet's est an lead to some very interesting discussion.
Now for a few comments of my own on AC vs DC.
AC: Use of transformers allows for a much higher line voltage than can be used with DC (though this is changing) - an advantage that has been known since the 1880's (e.g. Thomson-Houston versus Edison). Transformers are also heavy, so AC propulsion is heavier than DC propulsion. Running AC traction motors requires conversion from AC to DC and then back to AC, though the first step is usually quite efficient. Line voltage drop for a given current is higher with AC due to reactance and skin effect, though the effect of the drop is lower due to the much higher AC line voltages.
DC: Lighter propulsion (motors plus power conversion), with AC drive systems now capable of running directly off a 3kV source. DC is better for achieving phase balance. Higher line voltage may be possible in the near future with advancements in power electronics.
OvermodMany years ago, when you had a couple of shunts and running resistances for TMs, there was more point in matching voltage and current with what the motors could use. Those days are essentially gone, and I say good riddance...
That occurred to me when I was talking about overhead wires. Today with electronics power in switched around on board a train any way you want. Ac is picked up, switched to dc and switched back again to ac. So a lot of this discussion seems like talking about a Carnot cycle; really of historical interest.
I hope you'll forgive me if I don't sit down to calculate root-mean-square. I would have to pore over it for a long time. And when I finished, well I'm not in school any more and I don't have to prove I understand it. But if you use the simplest example, a square wave, you can really do it in your head.
There is no particular need to carry out current provision at TM voltage; you're going to have to invert it for motor control anyway, so why bother doing it at full current? Supercaps are inherently low-voltage devices, so there's some better logic in using LV where they are concerned ... but even at present, the cost of providing the massive banking to get voltage down to what they can handle is not particularly cost-effective even for charge buffering or the other things they're good for... easier to do AC-to-AC transversion to very LV, take out the spikes and noise, and not have to worry about compromising the nanostructure...
There are a number of reasons why the synthesized waveform from inversion is not a pure sine wave, but there's no particular reason to go into them here. The point to remember is that achieving comparable results with straight DC is more difficult, even before you get into issues like low-energy speed control. Many years ago, when you had a couple of shunts and running resistances for TMs, there was more point in matching voltage and current with what the motors could use. Those days are essentially gone, and I say good riddance...
With respect to the RMS business, see here:
Better RMS calculations
for some insights about why and how we use it.
P.S. DON OLTMANN, what is your take on all this?
OvermodA given wire can carry MUCH more power with AC than with DC
Well yes. But that is with high voltage transmission. I was talking about an overhead wire carrying the power a traction motor uses.
As far as average and rms are concerned, as long as you are referring to the mean they do mean the same thing. But I have generally heard the term root-mean-square used which is why I use it. I've forgotten most of my calculus but I think it refers to the calculation when there is a changing curved wave form.
And frankly, as voltage and amperage and the shape of waves changes the characteristics of electric power changes but I am not familiar with them. However, if you are using electric power the characteristics of the power you are using can be very important.
Think of it this way: RMS power is the same thing as average power for a given cycle of AC. A way to determine this is by summing or integrating the area under a half-wave, then comparing the result to the area under a "DC line" for that same time interval. The rectified half of full-wave rectified AC carries the same effective power (remember I^2R is always positive!) so for all practical intents and purposes the power of the recitified 'cycle' can be given a DC equivalent.
A given wire can carry MUCH more power with AC than with DC. The situation has changed somewhat in that effective DC-to-DC voltage conversion is possible; there are some interesting developments in HVDC for undersea feeders where the effects of AC make it less favorable than AC... but those don't apply to thin wire, or overhead wire, where most of the 'power' is carried in the space charge around the wire and not through the metal. The most important detail is that you want to minimize I at the sliding or rotating current contact -- even 3000VDC was too low to prevent massive droop and damage to the catenary in the French locomotive tests in the mid-Fifties, with only a single locomotive drawing power. This is less of a problem... indeed, short-range arcing can be less of a problem -- with 25kV potential. DC switching of that kind of voltage, and transversion to any sane voltage for use in a traction motor, is not as simple or safe. At least not up to now.
Overmod RMS rectified AC is fairly equivalent to DC of the same voltage.
I'm afraid I don't follow you. I don't know what "RMS rectified AC" means. I have never heard of an RMS rectifier. When AC is rectified one half of the power is either eliminated or it is changed from negative to positive or vice versa. As I understand it, root-mean-square is a measure of the true power of an alternating current.
My real point was that a given wire can carry more DC power than AC power.
John WRI did not intend to suggest that the total power of ac is the area under a curve connecting the peaks of the curves.
But that is exactly what you said.
In fact a considerable part of the area under such a straight line is outside of the curve and would have no power at all.
Which is what I said.
It looked to me as if you were confusing amplitude modulation (which is waveform-related) with power production (which uses not peak but RMS average). I think it's important not to confuse the non-engineers with analogic explanations that don't describe the phenomena properly. RMS rectified AC is fairly equivalent to DC of the same voltage... but then you have ripple to deal with, which in the old days would limit what you could do with "DC" motors on rectified AC unless you provided some possibly very capable inductance and capacitance...
OvermodYeah, except the math is wrong
1. As I understand it, the voltage from a wall plug oscillates going from 0 to +60 to 0 to -60 for a peak voltage of 120 volts. Am I wrong?
2. Also, as I understand it, to handle a 120 volt circuit at a given number of amperes you need a wire sized to handle the peak amount of power rather than the root-mean-square which is the total amount of power carried.
3. A line that connects the peaks of full wave rectified ac indicates indicates the dc power that you will have if the maximum dc power is equivalent to the peak ac power. In fact, this is considerable more total power than the ac would be because the root-mean-square is considerably less than the peaks.
4. A benefit of dc is that a wire of given size will carry more dc power than it will carry ac power.
Am I wrong here?
I did not intend to suggest that the total power of ac is the area under a curve connecting the peaks of the curves. In fact a considerable part of the area under such a straight line is outside of the curve and would have no power at all.
John WRI don't know how much of a mathematical background you have. If you take AC from a generator and run it into an oscilloscope it looks like a sine wave. Then if you take the bottom half of that sine wave and flip it up so both parts are above the axis the total power will be the area under the wave. The top of the wave is the maximum power and the wire must be able to handle that power. Now if you draw a straight line that touches the tops of the sine wave you have a graph of a DC current and the total power is the area under the line. You can see that the area is greater.
Yeah, except the math is wrong. AC 'voltage' is expressed in RMS; the 'peaks of the sine wave' are considerably higher than this (for instance, capacitors in a smoothing circuit for GTO speed control on house power will be found to have over 160V on them -- a nasty surprise!)
You are right that power is the area under the 'curve' of the rectified sine wave, biut drawing the equivalent DC line along the tops ignores the 'white space' between the peaks. There isn't any simple graphical way I know to derive a DC line from a sine-wave plot; you either have to do some numerical integration under the curve for a given crossing-to-crossing (or peak-to-peak but the graph is a bit wackier) interval, or do the mathematics.
There is another consideration with using very heavy conductor (aka 'third rail') in an overhead line at very high voltage, which is corona from the greater outside dimension to nearest ground. You would do this only in situations where there is no room for the suspension part -- the literal 'catenary' curve -- of a typical overhead-wire system, and therefore have very little effective room for insulators unless they can be recessed into the tunnel roof, and the walls of the tunnel made relatively nonconducting.
South Shore and the Illinois Central (Metra Electric to the youngsters) use 1500V DC catenary since 1926. CTA uses 600V DC third rail, no more overhead except in some yard and shop areas.
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