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  •  oh boy. Ohm's law? (25+ / 0-)

    This is dusting off some very cobwebby memories, but if V = IR, given a constant resistance, doesn't the current increase linearly with voltage?

    (ouch.  I think I may have strained something in my brain)

    Hay hombres que luchan un dia, y son buenos Hay otros que luchan un año, y son mejores Hay quienes luchan muchos años, y son muy buenos. Pero hay los que luchan toda la vida. Esos son los imprescendibles.

    by Mindful Nature on Tue Jan 22, 2013 at 10:00:50 AM PST

    [ Parent ]

    •  only if total power consumption goes up... (47+ / 0-)

      In your equation, if voltage went up, current would actually go down to make the equation stay in balance.

      The equation you need to consult is P=I*V

      So, if the total power requirement goes up, the current will go up (regardless of the voltage). This is why power lines can sag on hot days when there is a lot of power being consumed.. a lot of current running through the lines, and heating up the lines...

      Freedom isn't free. So quit whining and pay your taxes.

      by walk2live on Tue Jan 22, 2013 at 10:12:07 AM PST

      [ Parent ]

      •  And more directly (14+ / 0-)

        P = R* I^2

        or

        P = R / V ^2

        which is what gives you the 6x decrease in power loss for a 2.5x increase in voltage (resistance of the cable is the same).

        •  Actually, the second equation is: (6+ / 0-)

          P = V^2 / R

          and I'm not sure how you draw the conclusion you do.

          The man who moves a mountain begins by moving away small stones. -Confucius

          by Malachite on Tue Jan 22, 2013 at 01:20:28 PM PST

          [ Parent ]

          •  You're right (5+ / 0-)

            I screwed up the 2nd equation, but it's important to realize which V you're talking about. When it comes to losses in the transmission line, the V is the voltage drop across the line (which will be lower at lower I), not the input voltage.

            So, at higher voltage at constant delivered power, the current will be lower and the reduced power loss will go like the square of the decreased current.

            •  Here's a link to a handy ohms law calculator (1+ / 0-)
              Recommended by:
              kyril

              usually the square chart is shown as a circle, but all the formulas are correct.  There are also some 3 phase AC formulas that one might find useful.

              http://www.angelfire.com/...

              Google Ohm's Law Calculator to find more.

              Republicans are like alligators. All mouth and no ears.

              by Ohiodem1 on Tue Jan 22, 2013 at 09:03:51 PM PST

              [ Parent ]

              •  You also need to change the transformer at each (2+ / 0-)
                Recommended by:
                goodpractice, kyril

                end of the powerline to both boost the voltage on the line, and to step it down to distribution useful voltages, like 13,800 line to line, 7200 line to neutral distribution lines and such.

                The transformers in the switchyard also have to have sufficient distances between phases to handle the magnetic forces that are generated, so it isn't as easy as it may seem, just to boost voltage.  Before it can be done, there is also an arduous and fairly long term process to modify standards for cross country powerlines and towers before this kind of change can be implemented.  So the regulatory environment must also change as well as the crossbars.  Much study for wind and ice loading, capacitance, effects on power factor correction and a lot more needs to be done before changes can be made.

                But I like the idea, and agree with the higher efficiency of high voltage long distance power transmission.

                Republicans are like alligators. All mouth and no ears.

                by Ohiodem1 on Tue Jan 22, 2013 at 09:11:40 PM PST

                [ Parent ]

                •  In general (2+ / 0-)
                  Recommended by:
                  kyril, Ohiodem1

                  our national grid is long overdue for an upgrade, and boosting long-haul interconnections is going to be a major part of any strategy that relies on localized renewable sources.

                  Seems like a good time to revist the 'standards' for high voltage transmission and make the needed changes to reduce transmission losses, especially if it can be done relatively cheaply.

      •  This is not true!!! And I think a bit confusing... (18+ / 0-)
        In your equation, if voltage went up, current would actually go down to make the equation stay in balance.
        Actually, current would increase to meet Ohm's Law.

        What Mindful Nature is confusing is the placement of the voltage and load (resistance) when applying ohm's law. Each household represents a load. The voltage at the house is still going to be the same! That's the key! People are most familiar with 120V lines in their house. Their house is still going to draw the same average current at the same 120V voltage. So even their power consumed will also be the same. And that makes sense! So what are we changing?

        We are changing (boosting) the transmission voltage going to the house and inserting new transformers to step that larger transmission voltage back down to 120V at the load (home). So it's a more complex circuit than a single voltage, current and resistance. You can't apply V=I*R to anything more than a single, lumped-parameter resistance.

        The transmission line itself can be modeled as a very, very small resistance. That's why it is used as a conductor! And any current flowing through the line will generate a very small voltage across that line segment. That is where V=I*R does apply directly to the transmission line. But the voltage across the transmission line that we are talking about is very small relative to the voltage being carried by the line in relation to overall circuit ground.

        What is saved/reduced here is the amount of current required at the load end of distribution, specifically at the transformer. A higher voltage on the primary/transmission-line side of the transformer means less current is needed on that same side of the transformer for a fixed power load. The power load is determined by the secondary/household side. This is where P=V*I applies. So if P is fixed for the house (which it is), a larger voltage allows a proportionally smaller current.

        So if the voltage on the transmission lines can be higher, the current needed in those same transmission lines to deliver the original amount of power to the homes is smaller. And with less current in the transmission lines, there is less loss. And with less loss, there is more overall efficiency.

        I think. ;)

        The man who moves a mountain begins by moving away small stones. -Confucius

        by Malachite on Tue Jan 22, 2013 at 12:18:51 PM PST

        [ Parent ]

        •  Yeesh, I'll just come right out and say, (21+ / 0-)

          I don't understand a lot of what you guys are talking about, but man do I love running across conversations like this on dkos. Cheers!

          And my baby's my common sense, so don't feed me planned obsolescence.

          by vadasz on Tue Jan 22, 2013 at 12:34:20 PM PST

          [ Parent ]

        •  I think the source of the confusion is (5+ / 0-)

          that "voltage" in the familiar V=IR is a different quantity than "voltage" in "high voltage power lines."  V=IR would apply to the voltage lost over the length of the cable, with appropriate caveats as the equation really applies to direct current.  "High voltage" would be the amplitude of the alternating voltage in the transmission line, akin to the "120 V" of household supply.

          Sorry, just had to chip in to work some of the cobwebs out of that section of my brain.  Your post was informative and, as far as I can tell, correct.

        •  Huh? (2+ / 0-)
          Recommended by:
          KenBee, Ohiodem1

          You said what I wrote was "not true", then came to the same conclusion:

          if the voltage on the transmission lines can be higher, the current needed in those same transmission lines to deliver the original amount of power to the homes is smaller.
          That's exactly what I was saying... just using a simple equation. Actually all these equations are very powerful, and mostly universal.

          Freedom isn't free. So quit whining and pay your taxes.

          by walk2live on Tue Jan 22, 2013 at 01:31:46 PM PST

          [ Parent ]

          •  The portion... (1+ / 0-)
            Recommended by:
            kyril

            I highlighted and immediately corrected is at the top of my comment. About ohm's law. It has nothing immediately to do to with your conclusion. Here it is again:

            In your equation, if voltage went up, current would actually go down to make the equation stay in balance.
            This is not true.

            The man who moves a mountain begins by moving away small stones. -Confucius

            by Malachite on Tue Jan 22, 2013 at 02:32:13 PM PST

            [ Parent ]

            •  sure it is... (1+ / 0-)
              Recommended by:
              Ohiodem1

              example:

              P=I*V

              P=200
              I=10
              V=20
              200 = 10 * 20

              Now, increase the voltage, but keep power the same:

              P=200
              I=x
              V=40

              solve for I...

              I=5

              voltage went up... current went down.

              Freedom isn't free. So quit whining and pay your taxes.

              by walk2live on Tue Jan 22, 2013 at 05:01:55 PM PST

              [ Parent ]

              •  First, let me apologize. I have a relative in the (1+ / 0-)
                Recommended by:
                kyril

                hospital. It's a bit stressful here. So my tone is quick, short, poor and not like me normally. Really. I am sorry. Normally I would elaborate at length and be very subtle about my feelings.

                Second, your comment seems very strongly to be responding to Mindful Nature's comment. I mean it is a reply to that comment after all. So the expression 'your equation' seems to refer to the equation he cites, namely ohm's law. You even go on then to say that the equation he needs is a different equation, namely the power equation.

                Do you see the confusion there? Again, sorry that I am a bit short and thus less subtle/polite (ergo, dickish) tonight. Things are nasty here and I am using dKos as an escape. I felt like more perspective and clarity would reduce confusion.

                The man who moves a mountain begins by moving away small stones. -Confucius

                by Malachite on Tue Jan 22, 2013 at 07:24:32 PM PST

                [ Parent ]

                •  Oh, I see.. (0+ / 0-)

                  no problem... happens all the time, these threads are hard to keep straight. I hope your relative gets well soon. There are certainly  more important things in life than discussing power equations on the internet.

                  Freedom isn't free. So quit whining and pay your taxes.

                  by walk2live on Wed Jan 23, 2013 at 09:10:10 AM PST

                  [ Parent ]

              •  I agree with this. nt (1+ / 0-)
                Recommended by:
                kyril

                Republicans are like alligators. All mouth and no ears.

                by Ohiodem1 on Tue Jan 22, 2013 at 09:13:12 PM PST

                [ Parent ]

    •  Limited by the transformer's impedance (2+ / 0-)
      Recommended by:
      flitedocnm, kyril

      What comes out of the high-voltage transformer is high voltage and (relatively) low current. You can increase the voltage in an AC system without increasing the current.

    •  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 ]

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