I heard recently that the Chinese are transmitting power using dc rather than ac. I don't have details about this but I wondered what advantages there would be. No inductances to worry about for instance at dc. Is there less losses with dc transmission? Of course there can be no transformers. I also thought that 3-phase was one of the most efficient methods in terms of copper of transmitting power. Also there could be no power factor problems if factories used dc. dc on the otherhand for motors makes a motor less reliable due to brushes etc. So, was Tesla wrong and Edison right after all?
Lots of places are using DC power transmission in certain spots, usually for two reasons of acute economic necessity:
#1 Reason to use DC at a couple of points in your grid: You're connecting two or more AC subgrids that are not synchronized and cannot be synchronized.
#2 Reason is that you have an existing, un-build-outable transmission line segment that you need to move more power over. Using DC lets you not worry about peak components but only the DC component in terms of insulation and current-carrying capacity. Remember, heat losses go like I2R but power only goes like IE...
If you can build a single segment for both reasons #1 and #2, then it can be economically feasible, but the converters on both ends are massively expensive.
In fact while DC is used for the long-hauls and choke-points they invariably convert to/from AC on both ends except in some very specialized applications.
China is under a lot of political pressure to look like they're doing something about urban air polution, and while power plants near large cities are certainly a source they probably aren't the biggest one. If it results in something useful actually being done, so much the better.
Edison didn't have a good way of going up and down in DC voltage on any scale. Motor-generator sets dominated for the production of DC from AC for a long, long time.
You forgot few 'insignificant' factors like overall efficiency in distributing form source to user(s), distances and terrain involved, weather, keeping it up and costs. BTW the three phases are mathematical solution to an equation which has many of those subjects as variants. Simple if you use the right math. The DC solution is with us for many years over 'very' long distances as the DC creation on one end and AC recreation on the other balances favourable against losses in such long conduction line.
Tesla/Edison each has its technical uses. It is for the designer to come with solution to local problem.
You're having a problem with reading comprehension. The long distance transmission at DC does NOT imply that the load end distribution is also DC. The HV DC is stepped down and converted to AC at the destination so it can plug into the existing infrastructure.
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There is also the real possibility of radiation loss (line acts like an antenna) from 50 and 60 Hz transmission lines. For instance, 1/4 wavelength for either will be less than 1000 miles. A high SWR will further increase losses.
For fun, go figure the radiation losses from a line even at an exact 1/4 wavelength, taking into consideration the fact that the return path is RIGHT THERE alongside (virtually along the same path in space, at these wavelengths) the "forward" path.
Ah, but if the line is balanced (and they do take some pains to keep it balanced, by rotating the wires occasionally and other tricks), what _is_ the expected radiation? Does it depend on length of the line? What sort of matched-line attenuation do you get in 1000 miles of line? And what sort of SWR do the lines run at when operating at rated power? Are long lines ever operated at high SWR, or is that considered a fault condition? The answers to some of that may surprise you.
Another consideration for DC versus AC transmission is that at DC, the current density in a cross-section of a homogeneous conductor is practically uniform, but at 60Hz, the skin depth in copper is only about 1/3 inch, and only a little more for aluminum; thus the resistance of the wire is lower for DC, significantly so for large conductors.
If you assume a two-wire line made from highly conductive round wires two inches in diameter and spaced 200 inches apart, well away from any ground and essentially air-insulated, the differential mode impedance will be pretty nearly 636 ohms resistive. To get to 1000 ohms, they'd need to be spaced 4160 inches apart, and kept much further than that from ground. It's tough to make a really high impedance RF transmission line.
If you're operating a line at a frequency where the series resistance of the conductors is an appreciable fraction of the series inductive reactance, or the shunt conductance is an appreciable fraction of the parallel capacitive reactance (in other words, at low frequencies), the line characteristic impedance may be quite reactive. The calc I just did for this hypothetical line at 60Hz and assuming room temperature copper conductivity suggests about 636-j10 ohms; assuming I didn't mess it up, you can probably forget about the reactive part for most calcs.
It's not true that there is _no_ radiation, but it's pretty low. Consider it as a linear system where each phase consists of a current in one direction in one wire and a current of the same magnitude and opposite direction in another of the three wires. You can find the radiation from that pair and vectorially add it to the radiation of the other two phases from two other wire pairs. If you're careful about it, you'll discover that there is radiation, but that the total radiated power is not so dependent on the length of the line as you may have guessed it would be. Even if you added the magnitudes of the radiations from each phase (instead of adding them vectorially), you'd be dealing with a very small number. Talk with people who actually run power through such lines and ask them what they worry about for losses. I think you'll find that radiation is pretty far down on the list.
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