4 ms·
Power has to be stored as DC. If google wants to "own the grid" and we know they do. Then power will need to be efficiently converted from DC to AC. If you do
by drakaal 12y ago
Power has to be stored as DC.
If google wants to "own the grid" and we know they do. Then power will need to be efficiently converted from DC to AC.
If you do this at the outlet, then LED's and other DC favoring electronics could be powered by wires that are DC, and AC would only be used when necessary.
Think of all the devices you have with a power brick to take AC to DC. All of those are losing electrons along the way... (most power supplies are less than 75% efficient).
- nsrango 12y agoAdditionally AC transmission is inefficient compared to DC transmission; the skin effect[1] really kills you [1] http://en.wikipedia.org/wiki/Skin_effect http://en.wikipedia.org/wiki/Skin_effect
- abstrakraft 12y agoThe efficiency gained by stepping up AC to higher voltage (and correspondingly lower current) is far greater than that lost to skin effect. Since there is no way to efficiently step up DC (modern DC-DC converters are still horribly lossy), AC transmission ends up being more efficient, by a long shot.
- eigenvector 12y agoThere are scenarios where HVDC can be more suitable than AC, in particular, very long distance links (e.g. to a remote hydraulic station). All of the substation equipment is much more expensive in HVDC but the benefit of eliminating line reactance can outweigh this if you're doing 1000+ km without any interconnections. Also, interconnection between neighbouring AC grids is an important HVDC application since we don't have to worry about transient stability. Neither AC or DC are superior. Different technologies for different applications.
- aaren 12y agoCan you explain like I'm a physics major why HVDC becomes more efficient than AC over ~1000km? A scale analysis would be great - where does the distance D come into the equation? I get why AC is better than DC on ~100km, but I don't understand how it changes again at larger scales.
- eigenvector 12y agoIt's a cost vs. energy carried trade-off. Both systems have resistive loses proportional to the square of the current. However: 1. Total power transferred in a DC system is proportional to the voltage, whereas power transferred in an AC system is proportional to the RMS voltage (which is roughly 0.7 of nominal for a sine wave), so more energy is transmitted at the same current level in HVDC. 2. AC systems manifest impedance which has a resistive (aka DC) component as above as well as a reactive (aka AC) component, i.e. Z = R + jX. In DC systems X = 0. In a theoretical transmission line no energy is absorbed or supplied from line reactance, but in practice we have to transmit a certain amount of reactive power (VARs) to charge the line capacitance/inductance each AC cycle. This reduces the amount of our current capacity (limited by thermal constraints) that actually carries current that can be delivered to the load as active power (watts). This effect is somewhat although not directly proportional to distance (characteristic impedance has no dependence on line length, but voltage drops along the line due to resistive effects meaning the variation from the optimal reactive power-minimizing voltage level increases). The effect of (1) and (2) is that for any given conductor, at a given voltage level, more usable energy can be transmitted with DC than AC, and that differential increases with distance. That being said, building DC converter and switching stations is much more expensive than AC. So for a shorter line, or one that has many switching stations, I could counter the above by simply generating 5-8% more power at the generating station and still come out ahead (because in real engineering everything is about $). Therefore, DC is only more cost-effective ($/MVA of energy delivered) at long distances.
- aaren 12y agoReactive power is the control then. With AC, we have to increase reactive power to shove more real power along the line, but reactive power doesn't transmit very well. Follow on: in a national grid, could we just distribute the production of reactive power with capacitor banks in each town / neighbourhood? Heavy flywheels spinning at 50hz? > building DC converter and switching stations is much more expensive than AC Is this intrinsic to the technology or is it more because we have economies of scale from building infrastructure around AC for 100 years? Thank you for this. Really helpful.
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- onnoonno 12y agoAccording to Wikipedia, this is not true anymore: http://en.wikipedia.org/wiki/Hvdc http://en.wikipedia.org/wiki/Hvdc 'Depending on voltage level and construction details, HVDC transmission losses are quoted as about 3.5% per 1,000 km, which is less than typical losses in an AC transmission system.[16]'
- dfox 12y agoAt line-frequency skin effect is not significant for almost any practical conductor (ie. reasonably conductive and with reasonable cross-section). What gets significant for long distance transmission is that few thousand kilometer long power transmission line starts behaving like, well, transmission line even on frequencies as low as 50/60Hz. Also HVDC elegantly sidesteps problem of grid synchronization.
- davepage 12y agoOh, that is completely not true for commercial AC power transmission -- just look up at a big power pylon and see the multiple power conductors per phase. See the pretty blue picture: http://en.wikipedia.org/wiki/Skin_effect http://en.wikipedia.org/wiki/Skin_effect Skin effect in copper is about 9mm at 60Hz
- coryrc 12y agoThat's part of the reason those power lines are steel-cored with copper jackets; the other reason being strength, of course.
- votingprawn 12y agoThey're Aluminium over a steel core, not copper. A copper over steel version would be significantly heavier.
- coryrc 12y agoheadsmack yes, of course.
- eurleif 12y agoWhere is the DC ultimately coming from in this scenario? Is the whole grid high-voltage DC, or what?
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- dannypgh 12y agoPower does not need to be stored as DC. Flywheels store rotational energy, and starting with a spinning flywheel it's pretty simple to get AC out of it.
- eigenvector 12y agoTrue, but your electrical output would be variable-frequency AC at the drive terminal. This isn't particularly useful unless you use a back-to-back converter (i.e. DC link) to get constant-frequency AC.
- derekp7 12y agoWould it have to be variable? If it goes through a CVT, then you can keep the output whatever speed you want.
- eigenvector 12y agoI'm no mechanical engineer but at first glance the conversion efficiency of using a CVT with a synchronous AC drive (vs. direct coupling the motor rotor and the flywheel rotor and using a variable-frequency induction motor + back-to-back converter) would seem to be lower. Modern VFDs have very high part-load efficiency and we can easily maintain a constant power characteristic through the full speed range by operating in the field weakening mode. One way or another you're losing energy in frequency conversion, it's just a question of whether you do that mechanically (with CVT) or electrically (with back-to-back converters). Also keep in mind that for grid storage devices, we're usually talking about 500+ kW on each flywheel which, at low speed, is A LOT of torque.
- drakaal 12y agoThat would not be efficient. Yes you could also store the energy by pumping water up hill and letter it turn a turbine later, but you'd lose a LOT of power. You could store the energy in springs ala the da Vinci cart, but again not efficient. All of these methods require a conversion, where as a Battery, Capacitor, or Leyden Jar will store DC as DC.