Is it possible for a series of conductors carry massive currents(100 kA) at voltages over 100V? Assuming there is massive cooling. How can power plants carry MegaWatts of power?
\$\begingroup\$
\$\endgroup\$
9
-
2\$\begingroup\$ They can carry Gigawatts and commonly do so. \$\endgroup\$– jippieApr 11, 2014 at 6:02
-
\$\begingroup\$ at super high voltages and as low current as possible, for less loss and heat and explosions. Also, the transmission is done with "max power transfer" by having the transmitter and receiving end with matched impedences, but still as little "resistance" as possible. Imagine transferring thousands of Amperes though copper cables.. You would have a fire hazard for sure! As an understatement.. \$\endgroup\$– KyranFApr 11, 2014 at 6:03
-
\$\begingroup\$ @KyranF Not if the copper cables are fat enough. \$\endgroup\$– arneApr 11, 2014 at 6:11
-
2\$\begingroup\$ @arne, yeah, they have to be huge - that is expensive, copper cabling is ridiculously expensive, and heavy, and just physically silly. :D. Would be like twisted conduits 1-2 metres thick \$\endgroup\$– KyranFApr 11, 2014 at 6:15
-
1\$\begingroup\$ A superconductor actually can't carry infinite amps by the way.. Power plants deliver hundreds of MW by upping the voltage. For example, if the power plant delivered 20MW @ 230kV, then its only ~87A. Often though for really high power outputs the voltage is higher than that. \$\endgroup\$– Michael ChoiApr 11, 2014 at 7:13
|
Show 4 more comments
2 Answers
\$\begingroup\$
\$\endgroup\$
12
Power plants do have buses which carry large amounts of current, but they do so at as high a voltage as is practical. This is because the power losses in a conductor are equal to the square of the current times the resistance: $$P_{loss}=I^{2}R_{wire}$$
For the whole circuit, we know that: $$I=\frac{P_{total}}{V_{source}}$$
So, for constant power, the circuit current is inversely proportional to the voltage. If you sub that into the first equation and do some handwaving, we learn that $$P_{loss}\propto \frac{1}{V^{2}}$$
Therefore, losses are inversely proportional to the square of the voltage, so if you double the voltage, your copper losses get cut down to a quarter. Since in order to decrease the resistance of a wire you have to increase its cross-sectional area, this can be a huge savings in conductor cost. It might not be so big in, say, a small factory, but when you're dealing with megawatts the savings quickly add up.
There's another reason to use high voltage, too. The skin effect means that at mains frequency the majority of current will only flow in the outside 8mm or so of a conductor. Thus, very high current buses aren't actually solid chunks of copper or aluminum. They're actually hollow pipes. You might have also seen transmission lines with multiple wires per phase - this is, again, because of the skin effect.
So, as for your question - yes, power plant buses can carry a lot of power. I did an internship at a large hydroelectric station where each generator had a 12 kV bus with a typical figure of about 8 kA per phase (these were 300 MW units). As it's not practical to carry this much current very far, it was converted up to 500 kV at the outdoor substation for transmision. The buses were hollow pipes as mentioned above, which were in other larger grounded pipes to isolate their (significant) magnetic fields from other equipment. Of course, this pales in comparison to nuclear plants, where 1000 MW units are considered typical.
-
\$\begingroup\$ I was at DNV-GL/KEMA ("the world’s largest independent testing and certification lab for medium, high and ultra-high voltage electrical infrastructure components") the other day, they are not particularly impressed about 8kA currents. I believe it is at the lower end of their scale. \$\endgroup\$– jippieApr 11, 2014 at 7:19
-
\$\begingroup\$ That would be correct. This particular plant had a bunch of small units (ten of 'em) rather than, say, five 600 MW units. Different design philosophy, and a bit of maintenance nightmare. \$\endgroup\$ Apr 11, 2014 at 7:23
-
\$\begingroup\$ However, is it possible to carry high current(100 kA+) at low voltages(100V+)? Or it would never work? \$\endgroup\$– PupilApr 11, 2014 at 8:42
-
1\$\begingroup\$ @Key The 100V is the voltage of the conductor with respect to ground. This voltage does not determine the power dissipated in the conductor. For that you need the voltage drop that actually occurs across the conductor. And your question said to "assume massive cooling", so heating was not an issue. \$\endgroup\$– Joe HassApr 11, 2014 at 22:20
-
1\$\begingroup\$ @Key It's only hot if the conductor is small. If you spread that power over a very (very!) large conductor then it just warms up a bit. Your question put NO RESTRICTION on the size of the conductor and assumed MASSIVE COOLING so why are you now throwing heat and temperature into it? \$\endgroup\$– Joe HassApr 12, 2014 at 10:55
\$\begingroup\$
\$\endgroup\$
3
Yes, it is possible.
Generally high power is carried at as high a voltage as possible. This minimizes the current thru the wire to carry the same power. This is why long distance high power transmission lines are in the 500 kV to 1 MV range.
answered Apr 11, 2014 at 12:20
-
\$\begingroup\$ Yes, but what about 100 kA @ 100V? The point I'm emphasizing is having large currents with reasonable voltages like 100V and above. \$\endgroup\$– PupilApr 12, 2014 at 1:23
-
\$\begingroup\$ @Key You are confusing the voltage of the conductor with respect to ground and the voltage across the conductor. Olin is talking about high voltage with respect to ground. To calculate power loss in the conductor you need to know the voltage drop along the conductor itself. \$\endgroup\$– Joe HassApr 12, 2014 at 10:57
-
\$\begingroup\$ @Key: 100 kA would require massively thick cable that would be very expensive, not only the cable itself but also whatever has to support the weight. Your 100 kA at 100 V is 10 MW. The same 10 MW can be carried more economically at say 500 kV and 20 A. Converting to and from high voltage at each end adds some expense, but this is offset by the much cheaper cable and infrastructure over any reasonable distance that a power plant would be sending its power. \$\endgroup\$ Apr 12, 2014 at 12:17