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Basic Electronics » High voltage lines

March 11, 2010
by aluck
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One thing I couldn't understand no matter how hard I've tried is the reason behind high voltage lines. Several sources suggest that high voltage, low current line has less losses that low voltage, high current. Why is it so? Isn't Watt's (or it sometimes called Joule's) law applicable here? P=IV, so according to it the rate of voltage to current doesn't matters! Am I missing smth?

March 11, 2010
by BobaMosfet
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Yes, you are missing some things. Those transmission lines don't carry voltage. Voltage is a measure of how much potential current can be made to flow (safely) along that conductor. In the US, that's approximately 22KV (3-phase) which drops to about 7KV (at a substation) and then lower (at the pole transformer), when then drops it further to your breaker panel (usually 240 into the house and then breakers drop that to 120/240 depending on need).

How much current drawn across those lines depends on what YOU have plugged in at your location. The generator will generate as much potential as it takes to feed your current demand (until a safety trips-- usually a breaker in your house or at a feeder or substation).

High voltage, low current means less loss- because less current is flowing (and thus less energy is lost in heat, EMI, and impedance). Watt's Law describes a relationship between 3 elements of energy in accordance with the Law of Conservation of Energy.

It's a little bit like comparing a 30-Watt Lightbulb, to an LED. The LED uses a whole lot less current-- but it won't light a room.

BM

March 11, 2010
by mongo
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Another thing to consider is economics. The cross-country lines push even higher voltages, in the order of 200KV. The higher voltages require smaller conductors and keeps it all down in weight as well as cost. The higher voltages also experience less loss over distance and being AC as opposed to DC, much longer transmission distances are achievable.

March 11, 2010
by aluck
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High voltage, low current means less loss- because less current is flowing (and thus less energy is lost in heat, EMI, and impedance). What is a dependency here? And according to what law? I cannot see why losses in heat, EMI and impedance depend on current, and not voltage. :(

March 11, 2010
by aluck
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The higher voltages require smaller conductors and keeps it all down in weight as well as cost.

According to what law/equation?

being AC as opposed to DC, much longer transmission distances are achievable

I've seen quite the contrary statement: http://en.wikipedia.org/wiki/High_voltage_line#HVDC

Although HVDC has at long lines less loss than AC, at short lines the losses of the rectifier and inverter can be higher than that of the line.

I'm confused.

March 11, 2010
by n3ueaEMTP
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The simplest answer is voltage drop (loss). The generating station supplies outrageously high voltage that nobody can use. Since all conductors have some amount of resistance, 22KV or 200KV can be measured at the generator but at the substation, due to voltage drop, you may only see 20KV which depends on multiple factors (conductor size, distance from the generating station, condition of the conductors, etc.). The substation drops the voltage to somewhere around 1KV. From there, it's shipped to those grey "cans" on the telephone poles that reduce the voltage further to something useable 120/240VAC.

Chris B.
n3ueaEMTP

March 11, 2010
by aluck
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I understand this system. What I do NOT understand is exactly WHY energy losses are minimized by trading current for voltage?

March 11, 2010
by mongo
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The higher voltages require smaller conductors and keeps it all down in weight as well as cost.

According to what law/equation?

Ohm's law, for the most part. Heavier current requires heavier conductors. Higher voltages require lower currents for the same net work. Prices also are a factor. Conductors to handle higher current cost more.

being AC as opposed to DC, much longer transmission distances are achievable

I've seen quite the contrary statement: http://en.wikipedia.org/wiki/High_voltage_line#HVDC

Although HVDC has at long lines less loss than AC, at short lines the losses of the rectifier and inverter can be higher than that of the line.

The lower loss is due to lower coronal loss. AC loses a little bit in flipping the polarity 50 or 60 times a second. It's not quite contradictory, just says it in a different way. It's in the conversion to make it useful that causes most of the loss.

I maintain three substations that add up to about 8 MW. They bring in 13.2KVAC and transform down to 480VAC. If this were a DC input, instead of three #8 wires coming in, there would be two #2 wires and the converter to make it useful would be massive and expensive to operate.

I could go into analogies with water, tanks and pipe sizes to visualize it but not right now. It really does get interesting but I am tired...

March 12, 2010
by aluck
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Ohm's law, for the most part. Heavier current requires heavier conductors.

I still don't get it. R=V/I, as soon as I could understand R is constant for a given wire. Dissipated power P=IV, so applying double the voltage (2V) according to Ohm's law we will get double the current I=2V/R. And dissipated power P=IV will turn into P=2I2*V = 4IV.

I don't get it. :(

March 12, 2010
by BobaMosfet
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aluck-

Quote:
>by aluck   
>I understand this system. What I do NOT understand is exactly WHY
>energy losses are minimized by trading current for voltage?

This is because you are NOT TRADING CURRENT FOR VOLTAGE. You don't understand the application of the equation. High Voltage Drop = High Current Movement. That IS what Ohm's Law tells you. Watt's law is an equation that describes how much WORK that energy can (usually referenced over some arbitrary period of time) perform.

Only if you use a transformer do you transform that ratio in other ways- using an electromagnetic field.

Let's apply PIE and EIR to real world examples so everyone can follow (I use the correct nomenclature for the equation -- E = "Electromotive force", I = "Intensitie" (French), and R = "Resistance" -- If you prefer to use "V" in place of E, do so):

Example 1: 9 Volt Battery

E = 9V
R = 1 Ohm
I = E/R

Therefore (No voltage or current is traded):

I = 9/1
I = 9 Amps

In other words, A voltage drop of 9 Volts = 9 Amps, and P = I * E, so P = 18 Watts.

Example 2: Wire Resistance

R = .1003 Ohms per 1000 feet
I = 220 Amps
E = Voltage Dropped (lost) to get current that length required

In other words, if we use AWG 0 conductor for a mile, to
transmit 220 Amps of current, it will require 'E' volts to get
the current that far.  That is voltage lost, attraction
fulfilled.

Whenever you force elecrons to move through a conductor, they
generate an electron field-- induction.  This is because (simply
put) they don't want to move.  They are happy where they are.
That field is the means by which they resist change.  The larger
that field (meaning the more current flowing), the more voltage
(aka attraction) is required to cause electrons to move.

So, to move 220 Amps of current an entire mile across a 0 AWG
conductor (about .37" inches across as I recall), We drop 'E'
volts as follows:

E = 220 * (5.280 * .1003)
E = 116.508 V

The reason transmission lines have such high voltage, is because it takes tremendous attraction to pull the required amount of electrons back & forth through the conductor, 60 times every second. A LOT of current.

R = Loss -- due to friction (electrons moving through a conductor generate heat), due to interference causing inductive or capacitive resistance (called impedance), and many other aspects.

You ask for a Law: The Law of Conservation of Energy. Energy can neither be truly lost, or gained- but may only change form. In power, you aren't exchanging voltage and current, you are working with current and resistance. Voltage is related only as a means to describe the amount of attractive force required to move a sufficient number of electrons to satisfy demand (by your refrigerator, alarm clock, lamp, toaster oven-- what have you).

Quote
>so applying double the voltage (2*V) according to Ohm's law we
>will get double the current I=2V/R.

Yes, you get double the current because you input double the voltage. What you're failing to understand is that there is a DIRECT PROPORTION RELATIONSHIP between Voltage and current. You don't get one without the other.

You can't sit down and say:

R = 5
I = 7
E = 12

...and expect that to work. You must use the DIRECTION PROPORTION relationship between 2, to derive the third.

Does this help at all?

BM

March 13, 2010
by aluck
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First of all, thank you for the thorough explanation. Unfortunately, I still don't get it. Let me comment on your post:

This is because you are NOT TRADING CURRENT FOR VOLTAGE.

I am! Having 550W power supply, according to Watt's law, I have two options: either to transfer 110VAC 5A or use a transformer and trade it for 550VAC 1A before transfer. I do not understand benefits of second case - why should I have less voltage drop?

You don't understand the application of the equation. High Voltage Drop = 
High Current Movement. That IS what Ohm's Law tells you

I don't see how. Ohm's Law according to Ohm's Law explanation doesn't tell us ANYTHING about voltage DROP - only about applied voltage.

As soon as I understand, in Ohm's Law R=E/I (using your notation) to the exact battery (E=const) and exact wire of exact length (R=const) we have given E and R. And Ohm's law allows us to calculate current I by equation I = E/R.

For the very same battery E is constant. For the very same piece of conductor (wire) R is constant. How should I calculate voltage drop across the wire? Maybe if you kindly explain this - I would get somewhere...

March 13, 2010
by mongo
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Ohm's law is what would be a purist law. Ideal conditions only, and does not take into consideration certain factors like temperature changes, impurities and other natural variations that happen in the real world.

If you had a transformer that would take 550 VAC at 1A on the primary, the secondary can be the same as if it were a 110 VAC primary. Same trade-off for the same application. Ideally, it would be 550Watts but there is loss in the coil windings and there is also heat generated in the core due to eddy currents and internal conductivity and resistance.

March 13, 2010
by aluck
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Ok, it seems that I finally figured it out by myself (with my dad's help - he is an electrical engineer, actually). My transition in power dissipation equation from P=I^2 * R to P=E^2 / R was incorrect. In fact, it should be P=dE^2 / R, where dE - voltage drop into the conductor, not total voltage applied to the wire. Am I right?

March 13, 2010
by BobaMosfet
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aluck-

Your transformer example works the way it does, because the winding ratio from the primary to the secondary transformer is 5:1, not 1:1. As such, only 20% of the VOLTAGE attraction of the primary winding is even seen by the secondary winding (stepping down from more turns of wire to 1/5th as many on the secondary). Since the Law of Conservation of Energy applies, this means the secondary winding must still accept all the current from the primary-- so current increases proportionately.

This can be described the the standard (ideal) equation by:

Vp / Is = Vs / Ip

Where Vp = Voltage Primary, Vs = Voltage Secondary, Ip = Current Primary, and Is = Current Secondary.

Really straightforward. Before you worry about voltage versus current, I think it would behoove you to simply try to understand what voltage is, and what current is. Unless you understand those two things by themselves, all the rest of this is meaningless.

As for the wiki article, bear in mind that since your understanding is flawed, you are not understanding the larger concepts the article is presenting.

But we have faith, eventually you can get this!

BM

March 13, 2010
by aluck
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BobaMosfet, am I right or wrong in my last guess about dE?

March 13, 2010
by bretm
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Ohm's law is what would be a purist law.

No, it's a mathematical relation follows from the definition of resistance. Resistance is defined as voltage divided by current, so it's impossible for Ohm's law to be an approximation.

Ideal conditions only, and does not take into consideration certain factors like temperature changes, impurities and other natural variations that happen in the real world.

Sure it does, unless you expect resistance to be constant. Impurities can increase or decrease resistance. Temperature changes also affect resistance. Etc. The law doesn't change, just the values do. Sometimes resistance even changes depending on current flow, but always V=IR, because the definition of R is V/I. In these cases of varying resistance it makes more sense to talk about the differential resistance R=dV/dI. But the three values are always related in that way.

there is loss in the coil windings and there is also heat generated in the core due to eddy currents and internal conductivity and resistance.

That's not a deficiency of Ohm's law, it's a failure to correctly model the components in question by ignoring those factors. If you use an idealized model of a transformer, then of course the math isn't going to come out right because you're calculating V=IR when you haven't really accounted for all the factors that make up R.

March 13, 2010
by mongo
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Nothing in the real world can exactly be reflected by a mathematical equation. Ohm's law is based on cut and dried numbers and is an exact set of formulas. Reality is the approximation, not the math.

The losses in the core may not be directly related to Ohm's law, but yes, it will have some effect on the resistance where in the purist sense, Ohm's law can be applied. In this context however, it really is inconsequential.

The whole thing here is being over-thought and more information is brought into knowledge via empirical data. Formulations are then figured out to correspond with what was actually observed. From that data, new figures can be obtained as known variables that can be plugged into existing math to better describe reality, but nothing is ever absolute and perfect as is mathematics.

Best to just learn it for what it's worth and not cause too much brain damage by over-analyzing.

March 13, 2010
by bretm
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I agree. It doesn't help that there are so many "fuzzy" statements out there about stuff like this. I'm guilty of it to. I was talking about Ohm's law as if it was V=IR. That's not actually Ohm's law. It's only Ohm's law if R doesn't depend on I or V. All it really is is a definition of R. Ohm's law is just V=kI where k is a constant. It happens to match V=IR for ohmic materials, so people loosely call V=IR Ohm's law.

March 13, 2010
by BobaMosfet
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aluck-

Sorry to take so long on this.

All the rest--

Let's stick with idealized examples please until he understands some things?!?!? Only in idealized examples does the simpler math work out, because we are not talking about all the complex equational issues related to losses/interferences, and all that jazz of every kind.

Yes, R is normally referred to as a constant, but that (at this stage) will simply confuse him more.


aluck-

You're kinda of right and kinda of not right. Forget what you've read elsewhere for a moment and consider this (because this is how it actually works at the electron level:

CURRENT: A measurement of the flow of electrons over a period of time. VOLTAGE: This is an attractive force- it does not push, it only pulls. It does not exist at any single point in your circuit. You cannot take any instrument and touch a single probe to any point (anywhere in the universe) and get a voltage reading. This is a fact of the physical universe and cannot be changed. Your dad should be able to confirm this. One point has no polarity. It is neutral.

The ONLY way you can read voltage-- whether this is from any power source (battery, generator, lightning bolt-- anything) is to route the current THROUGH a measuring device. This would be your voltmeter.

Think about a lightning bolt. Lighting travels from the Ground (Earth Ground) to the Clouds (Positive Terminal). It does this whenever there is a difference in polarity between 2 points that is great enough (over 50,000V) to overcome the resistor (air) in it's path. And in fact, the true and only real definition of voltage is that it is an indicator (a number, a measurement if you will) of the difference in electron charge between two points. Always between two points. If there is no difference, there is no voltage, because neither side is trying to attract electrons from the other side.

Whenever there is a difference in charge between two points, given the opportunity to do so, electrons will travel from the side with too many, to the side with too few, until they are equal.

Therefore, returning to our lightning bolt, once the electron current has transferred a sufficient number electrons to equalized the charge-- the lightning bolt ends. It happens so quickly, that it seems instantaneous. But if we could measure both the current (electron quantity traveling) and the voltage attraction (charge difference) simultaneously during the lighting strike, we would see voltage start at maximum and begin a steady drop to 0, and conversely, we would see current start at 0 and rise to maximum.

As such, voltage as given in the E=IR circle formula (and anywhere in electronics or electrical engineering) refers to POTENTIAL ATTRACTION that is LOST/DROPPED as current flows.

When you use your voltmeter and place it's probes across two points in a circuit or power-supply, your meter becomes PART OF the circuit. In parallel with whatever is between it's probes. As such, inside the meter, they have a very high (MEGAOHM) resistor that resists virtually ALL the voltage attraction across the meter. This means that only very few-- a specific and precise number of-- electrons flow through the delicate instruments of the meter and are counted. It then scales that number up, based on the setting of the meter, and can tell you how many volts were consumed (aka DROPPED) driving current through the meter.

In all of electronics and electrical engineering, there is but one thing that moves-- electrons. Everything else is a means by which we measure either the attraction at the positive terminal, or the resistance between two bipolar polarity points.

So, whether you use 'dE', 'E', 'V', or POTENTIAL- these are the same thing.

I hope that helps. I understand where you are coming from I do. Too many people in this field toss voltage and current around without truly understanding the nature of the two-- and worse, they are very lax in the words they use with them. This is one of the reasons there is such confusion. Voltage (drop) is not AT anywhere. It is only ACROSS.

I believe you can do this, and I really want you to gain a clear understanding of voltage and current. Once you do, everything else gets easier. But you have to build a really solid foundation so that as you learn, you can correctly formulate expectations and if they are not met in testing, you will be able to more quickly deduce why or develop accurate means to test and determine why-- and then fix it.

Please feel free to ask more questions.

BM

March 13, 2010
by aluck
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Thanks, Boba! Spectacular explanation. Think now I've got it.

One more thing: why there is no voltage between + of one battery and - of the other?

March 13, 2010
by BobaMosfet
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aluck-

Quote: >One more thing: why there is no voltage between + of one >battery and - of the other?

A single battery: There is voltage/potential. Voltage is highest between the two terminals when they are not connected. Your meter can measure this current, because you make a complete circuit when you put the meter across the terminals-- but because the meter's resistance is so high- your meter can measure all of that potential (as explained previously). Remember, the meter's ability to measure this depends upon electron movement.

In case you meant two different batteries, the answer is the same (essentially): There is voltage/potential between the two opposing terminals, but if the other two terminals are not connected, no electrons can move, because the circuit is incomplete. Since no electrons move, the meter cannot register a drop.

BM

March 13, 2010
by aluck
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In case you meant two different batteries, the answer is the same (essentially)

Yes! I meant two different batteries. Then I do not understand what a "potential" is. I thought that at - terminal of a battery there is an excess of electrons, and on + terminal is a shortage. Am I wrong?

March 14, 2010
by BobaMosfet
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aluck-

No, there IS a potential-- the problem is that the electrons cannot move. Think of a tube with marbles in it. The two batteries, with a meter between the positive terminal of one battery and the negative terminal of the other battery form an incomplete circuit.

The marbles at the negative end of the tube have no where to go-- and they cannot cross the air gap (giant resistor) because the battery is (for our example, let's say 9 volts) only a small voltage, whereas the resistor (air) requires 50,000+ volts to overcome/pass-thru it.

Likewise, there is no room for marbles to come in to the positive end of the tube (because those marbles cannot move towards the negative end of the tube-- it's like an electron traffic jam-- and due (again) to the "air resistor" there aren't any electrons trying to come in.

So YES, the potential exists-- but because the electrons cannot move-- you can't register it-- but you know it's there (the math tells us so).


You originally wondered about High voltage .v. low current resulting in low losses. The very simplest answer to your question (if you can understand it based on what I've explained) is a battery.

The battery, with nothing across it's terminals, represents high voltage, zero current. In effect, there is high voltage and no loss. This is an extreme example of your test case in a tangible form.

R = 9/0 R = INFINITY (for all intents and purposes)

The above equation works, because zero is not really zero-- but it's CLOSE. It is a number so small, we can't measure it easily. Certainly not with a regular voltmeter.

Does that help?

BM

March 16, 2010
by aluck
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Yes! Thanks, BM

February 02, 2011
by Jalex
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There is another one here that I didn't see. The higher the AC frequency is the farther from the center of the wire the electrons flow until they almost want to leave the wire completly. Eventually a pipe will work better.

February 02, 2011
by mongo
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Right. AC has what is called the "skin effect". As the frequencies increase, the more prevalent it becomes. Lightning, though basically DC (static?) acts a lot like HF AC. It travels along the surface of a conductor, rather than through it. (a little does go through). If you see how lightning arrestors are connected, there are no sharp turns in the wire or cable. Lightning will actually jump off them instead of being dissipated in the ground, so smooth round curves are made in place of going around corners.

I kind of wonder though, as lightning strikes generally from the cloud down for the most part, there are return strokes back and forth. Since they happen in rapid succession, wouldn't that actually act like an AC discharge, as the imbalance becomes equalized?

High tension lines in the grid system are steel. It is not a very good conductor by comparison but it usually has a cladding of copper or another better conductor on the outside. The steel is just there to carry the physical load.

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