Skip to main content

View Diary: The simple innovation that could make wind power a big player (230 comments)

Comment Preferences

  •  Ohm's Law in this case (6+ / 0-)

    applies to the voltage drop along the wire.  The wire is of fixed resistance (resistivity, technically, but the particular wire is of fixed cross section, material, and length, so the resistance is fixed).  By V=IR, with R fixed, V is proportional to I.

    (Just a word of warning here.  I'm a VI-3 -- computer science -- guy, and this is VI-1 -- electrical engineering.  Back at that little trade school on the north bank of the Charles, they combine electrical engineering and computer science into one department, which they, and we, fondly or otherwise refer to as Course VI.  We VI-3's had to get our hands dirty with actual circuits, though; they wouldn't let us become full Ancient and Honorable Nerds of the Infinite Corridor without it.  So that's where my knowledge, such as it is, comes from.  You can imagine the various puns on "Course Six".  And you can Google 'em.  But I won't go there, beyond noting that there's now a VI-2, which is a middle ground between VI-3 and VI-1.  Think of VI-2 as the Very Serious People inside the Beltway who are constantly advocating compromise.  The "VI" in this paragraph has nothing to do with V and I elsewhere in this screed, although I'm sure there are VI-1's who would beg to differ.)

    Again, V in this case refers to the voltage drop along the length of the wire.  It's independent of the voltage difference between the two wires in the circuit, but it's why the efficiency of transmission improves so dramatically with higher voltage.

    The other thing to remember is that P=IV (power is the product of current and voltage).  If we step up the voltage, we require proportionately less current for the same amount of power.  But remember in this case the voltage is the generated voltage. not the voltage drop along the wire.

    So let's try an example.  Suppose we have a power line that we want to carry 1 gigawatt of power (I don't know what the actual numbers are; they're probably much lower for any given wire, but whatever).  If we transmit that power at 400 kilovolts (KV), it will require 2500 amperes.  Now let's suppose the total resistance of the power line is 40 ohms (again, I don't know what the resistance of real world power lines is).  To sustain a current of 2500 amperes through 40 ohms requires 100 KV (at the other end, you'll see only 300 KV).  100 KV * 2500 ohms is 250 megawatts.  So fully 1/4 of the power is wasted.

    Now, let's transmit that power at 1 megavolt (MV).  This requires only 1000 amperes.  With a 40 ohm resistance, that will be 40 KV voltage drop along the power line; the far end will see 960 KV.  So instead of 25% loss, there's 4% loss.  25 divided by 4 is...let's see...6.25.  The factor of 6.

    There are other problems with the higher voltage.  Air has finite resistance, so there will be some losses by conduction through the air.  But the efficiency should still increase by a lot.

    •  Ah, but what about power factor? (6+ / 0-)

      Dusting off my old EE degree, and remembering all those months in power class... (some of the lab sessions could get kind of exciting), when AC is involved you can never neglect the power factor.  This is a bad thing in electrical distribution systems, and it happens when the sine waves of the current flow are out of phase with the sine waves of the voltage.

      Numerous parts of the generation and distribution network become less efficient when the power factor is large.  Utilities go to great expense to counteract the effects of power-factor-boosting things.   Impedance of power lines, transformers, and the loads those annoying customers insist on connecting, all contribute to an increase in the out-of-phaseness.   The motor in your refrigerator, for instance.

      If I remember correctly (it has been some decades) power factor arises from the current phase being retarded behind the voltage phase, and the retardation effect goes up with the amount of current in the lines.  So increasing the transmission voltage and thus decreasing the current, could have the additional benefit of decreasing the power factor effect of the lines.

      I can't remember how much this amounts to in the total schema of things - my summer jobs in the electrical industry were limited to the Distribution Department (from the sub-station to your house), not the long distance Transmission Department, where all these high voltage thingies happen.   We are talking upwards of a million volts, here.   Quite spectacular when one of those insulators fails.   We had a lab where such things were made to happen on purpose.  Like a scene out of Frankenstein.

      A really interesting case is Japan, where half of the country runs on 50 Hz power and the other half on 60 Hz.  You can't easily convert one to the other at these power levels, and if you change the frequency, you also change our old friend the power factor.  So to enable power sharing around the country, Japan has had to build some really big motor-inverter sets, and also really big vacuum tubes to do this.

      It works out that actually using DC for long distance transmission has some benefits.  No power factor losses at all.  But again, big converters at both ends.

      Though ones hates to think that Edison might have been right about something.

      •  Poor power factory simply means (5+ / 0-)

        that you need to increase the current to deliver the same amount of usable power to the load(s), because the voltage and current are out of phase. The usable (real) power delivered is P = IR cos p where p is the phase angle (how many degrees out of phase the voltage and current are), and cos p X 100%  is the power factor - bigger is better, perfect is p = 0, cos p = 1, PF = 100%.

        So if you need 120 watts at 120v, in phase you need 1 Amp. With a phase angle of 30 degrees, PF = 87% and you now need 1.15 amps (15% more current) to service the same 120 watt load. If the wire's resistance in 1 ohm, you'd lose 1 watt in transmission with 100% PF, but 1.3 watts (30% worse) with 87% PF. (There are also 18 watts of "imaginary power", or the nameplate rating on a single device with 87% PF would 138 VA instead of 120 watts - VA is Volt-Amps).

        In places that use a lot induction motors (like factories), they install large banks of capacitors to bring the voltage and current closer to in phase. They do this not because they want to save energy, but because the electric utility charges them for three things: the power they actually use, the power factor they use it at, and demand (sort of the peak power they draw).

        Computer power supplies also have terrible power factor, but residential and commercial users usually don't pay for it, directly anyway. You can buy power factor corrected computer power supplies to replace the original equipment. If decent power factor were mandated, given the huge number of computers and other electronic equipment, there would be a large energy savings.

        Frequency affects power factor only because it affects impedance - the imaginary or reactive part of impedance varies with frequency, so that affects the phase angle between V and I.

        Edison was still wrong, because the technology to step up or step down DC efficiently has only existed in the last 30-40 years or so. But DC transmission lines don't suffer from reactive losses like AC lines do. They still have to deal with ohmic losses.

        Modern revolutions have succeeded because of solidarity, not force.

        by badger on Tue Jan 22, 2013 at 01:24:37 PM PST

        [ Parent ]

    •  I'm not even a VI-1 (0+ / 0-)

      But I do remember cramming all those equations into my head many years ago, the night before a physics 1 exam. But how can the equation V = IR suggest anything other that at a given resistance, if you increase V then you increase I?

      •  If you have a perfect voltage source (0+ / 0-)

        then increasing the voltage across a particular resistance will increase the current proportionately.  But in practice, any voltage source (like a battery) has an internal resistance, which sums with the load (and transmission) resistance.

        But what's actually going on is that we have a power source -- something that generates a constant (at least at a given point in time) IV.  If we increase the voltage of the power source, we correspondingly decrease the current.

    •  It's not Resistance that costs you (2+ / 0-)
      Recommended by:
      Nebraskablue, PrahaPartizan

      It's hysteresis, the alternating current.  Current flow lags voltage and that creates big losses in long transmission lines.  The best solution is high voltage direct current.  Upwards of 500,000 volts DC.  Current is relatively low, which limit losses due to resistance.  You convert 3-phase alternating current to direct current, transport it as DC, then convert ot back to AC at the receiving end.

    •  Bravo (0+ / 0-)

      Nicely done.

      -5.38, -2.97
      The NRA doesn't represent the interests of gun owners. So why are you still a member?

      by ChuckInReno on Tue Jan 22, 2013 at 04:30:04 PM PST

      [ Parent ]

Subscribe or Donate to support Daily Kos.

Click here for the mobile view of the site