7 ms·
Or you can just convert that energy into heat, but do that on the surface. Have large resistor banks on the surface that you connect to the DC grid when the vol
by robryk 9y ago
Or you can just convert that energy into heat, but do that on the surface. Have large resistor banks on the surface that you connect to the DC grid when the voltage is too high.
- raverbashing 9y agoThat would be acceptable 100 years ago. Not today
- jrockway 9y agoWhy? Stopping the trains wastes energy regardless of whether you dump the extra power into the wheels with friction brakes or into a resistor bank. Dynamic brakes on diesel trains already sink the power into resistors.
- robryk 9y agoWhy is moving the place one dumps heat from below ground to above ground not acceptable?
- raverbashing 9y agoBecause you can use it for something else other than heating, like, putting it back into the grid, storing (either battery, flywheel or supercapacitor).
- hydrogen18 9y agoThat would work, but a large flywheel would also be a good solution. You could spin up the flywheel to store energy and if it reaches maximum speed then use the resistors. You'd also need to detect load on the grid and then run the flywheel system in reverse to assist vehicles that are moving.
- late2part 9y ago"run the flywheel system in reverse to assist vehicles that are moving" Just to be clear, you wouldn't actually run the flywheel in the opposite direction.. You'd take energy out of the flywheel versus putting it in?
- CydeWeys 9y agoCorrect.
- CydeWeys 9y agoHow about just batteries, or heck, large capacitor banks? How much energy is recovered from a single train braking anyway?
- semi-extrinsic 9y agoSince insect.sh is on the front page right now, let's try it: 0.5 x 29 tons x (50 mph)^2 to kWh 0.5 × 29 ton × (50 mi/ h)^2 -> kW·h = 2.01233 kW·h So a Victoria line train at top speed has 2 kWh of kinetic energy.
- CydeWeys 9y agoGiven that regenerative braking is ~60% efficient, we're talking about 1.2 kWh. That's really not that much; a typical Tesla battery pack is in the range of 65-100 kWh depending on model. Now, granted, I don't know if you can feed that much energy into a battery pack on the order of ~20 seconds, but using multiple packs would suffice. So it does seem feasible to do it either by installing battery packs connected to the tracks, or in the cars themselves. I wonder which would be better.
- zkms 9y agoThere's no need for storage nor batteries at all, though. It suffices to convert the energy from the DC rails (where the train dumps it) to a bus where every other accelerating train can draw current from -- since there's lots of trains, there'll always be one willing to accept the load. The only issue is that each third rail section is fed by its own set of independent rectifiers, and the third rail sections are not paralleled together. This means that if there's no other currently-accelerating train on the section of rail that you're on, it's just you and the rectifier substation -- and those rectifiers can't accept your regenerative braking current. Adding inverter circuitry to the substations would introduce a path for energy flow to go from the DC third rails into the AC grid (opposite of the normal direction of flow, hence the term "reversible substation"). Since the AC grid is connected to all the rectifier substations and since there's many trains on the rail network, there'll always be a source for regenerative braking current that's dumped onto the AC bus.
- MiguelHudnandez 9y agoIf you can run wires all the way to the surface, you can sink the power into the grid or traditional batteries.
- zkms 9y agoYou can do that, but you can also use it to power accelerating trains that live on every other track section / DC bus -- by inverting it back into AC and dumping that onto the AC grid that's used to supply your rectifiers.