I recently learned that the Shanghai Metro uses overhead catenary at 1500 volts DC in almost all of its system. I wasn’t aware of this when I travelled it earlier this century, and I appreciate that the Boston MBTA Blue Line uses overhead catenary for parts of its route for sea-spray corrosion resistance, but I still find it an odd thing for any underground line to use overhead lines.
What are the advantages that Shanghai Metro uses so that they are using overhead 1500 DC instead of third-rail power? Surely the overhead lines are more expensive, aren’t they?
Generally speaking a catenary is preferred to a third rail and shoe for high speed trains. For example, the Newark Light Rail which runs in a tunnel for a good part of its route used a catenary. Obviously there has to be room for the overhead wiring and pantograph.
With the Newark Light Rail in the tunnels there is not a true catenary, that is a power wire suspended from a second wire allowed to fall into the shape of a geometric catenary. Rather there is a power wire suspended from a second wire that drops a short distance from the ceiling of the tunnel. a
As far as the 1500 volts DC, for a given amount of power (number of watts) high voltage electricity requires a smaller wire than lower voltage with higher amperage. However, that is just for transmission. Today with electronic equipment it is common to change and change and change electricity so the motors may operate on AC which is produced by an electronic inverter.
I 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. And the same size wire can be used to transmit that greater amount of power. When you are building an electrically operated mass transit system and will need many miles of copper using the smallest possible wire to transmit the power you need becomes important.
PS. You may recall from high school physics Ohm’s law: I = E / R where I = current in Amperes, E = electromotive force in Volts and R = Resistance in Ohms.
As you increase the current in an electric wire you increase the heat produced according to the formula H = I squared R. Notice the heat increases according to the square of the current. 30 Amps produces 4 times as much heat as 15 Amps. Also notice the value that does not appear: E or Volts. As voltage increases heat does not increase.
For total power the formula is W = EI where W = Watts. So for a given amount of power you can increase E the Voltage and decrease I the Amperes. That allows you to use a smaller wire size without overheating.
I suspect that is why in Shanghai they are using 1500 Volts.
Can’t say about Shanghai, but there is at least one subway line in Tokyo that uses pantographs (of standard design) to contact an overhead third rail suspended from the tunnel roof on insulators.
The biggest advantage for overhead distribution is the safety factor of removing the ‘hot’ conductor from ground level - especially useful where the tunnel might be subject to minor flooding. Also a lot safer for track workers, who don’t have to kill power or worry about killing themselves. The pans and hot rail DO require slightly more tunnel clearance, and put a definite limit on vertical rolling stock dimensions.
It is becoming a lot more common in locations where clearance is a problem. Like spans on electrified drawbridges and in tunnels used for HSR. These look nothing like the ones used at ground level, rather they look like a very straight and level piece of cantenary contact wire, and typically they maintain the stagger of the adjoining contact wire.
Seattle Light Rail also uses 1500V overhead catenary, but streetcar runs at 750V trolley wire, ditto Tacoma streetcar. And some day they probably will connect!
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.
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.
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?
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.
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.
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.
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…
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.
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.
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.
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…
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.
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.