Power Loss in Transmission Lines: AC vs DC Comparison Explained with a Model

[John3509]

Intuitively I fell like AC would lose more power long a long power line than DC. I will use a model to explain my perspective.

Let me model a wire as a long straight, non-frictionless tube, about the diameter of a marble, filled with a row of marbles that are each connected to the one next to it by springs. This models the electrons and their repulsive nature.
Voltage in this model is pressure or force applied to the marbles from one end.

To simulate AC, i take a marble from one and and vibrate it back and forth. Because of the friction in the tube and all the springs a lot of that motion would not reach the marble on the other end, it would vibrate at a lower amplitude, and would also transmit less force to anything placed at its end. If you ever played with a slinky on a table you know how much and can pull and push one end before the other budges, think of something like that.

To simulate DC, just push and one marble at one end and keep on pushing. Since you are not drawing back the motion but continuing to push you will eventually compress all the springs, up until this point the force you were applying to a marble at one end is higher than the force the marble at the other end fells and transmits. But after this point, it will equalize, the marbles will move wit the same forward motion, like a rigid rod. Imagine the stretched out slinky is being compressed and then pushed across the table.

In other words, Id imagine that with DC, after the initial phase where you do not transmit much power, eventually you would reach the point where the voltage at receiving end of the transmission line is the same as provided at the source. But with AC, the "slinky" effect over long distances, would absorb a lot more voltage that would reach the end of the transmission line.

Or a more simpler model, if you have a block of wood with a spring on the table and you push on the spring eventually the compression will overcome the static friction of the block and it will move but if you pull back on the spring before this point it will not move. I am imagining this type of principle at play in a long wire.

 

[mfb]

Intuitively I fell like AC would lose more power long a long power line than DC.

Correct.
I'm not sure if the marble analogy helps, however. AC has inductive losses which don't have an equivalent for marbles, and it has the skin effect which also doesn't have an analogy. There is also the difference between peak and effective voltage for AC, that doesn't transfer well to marbles either. Synchronizing AC over a long distance is an issue, too, there I could see how this is modeled with marbles.
We have a series of Insights articles about AC power grids.

AC is used because it is easier to transform between different voltages, something necessary to transmit electricity over relevant distances. DC is used when electricity is transmitted over very long distances - above ~1000 km the lower losses per length are more important than the slightly higher losses from transformation. As power electronics get cheaper and better there are more and more DC transmission lines.

 

[anorlunda]

To simulate AC, i take a marble

That was a worthy try, but the reality of conduction in a wire is too complicated for such a simple model.

For one thing, the moving electrons have many collisions with atoms in the solid.
See https://en.wikipedia.org/wiki/Drude_model To make an analogy with marbles, the wire is not a hollow tube, it is like a Japanese pachinko game or a pinball machine.

Also electrons repel other electrons. That means all the electrons in the wire have forces that are like springs that connects every electron to every other electron. Marbles don't have that.

There are even more complexities that I won't explain. Bottom line, there is no simple analogy of electrons behaving like marbles or water that comes close to the reality of current in a wire. It is one those annoying cases where simplifications don't work.

But there's an easier way to think about your AC versus DC question, that does not mention electrons. The key relationship, P=VI, power equals voltage times current. it works for AC and DC, and it works instantaneously. So think of an instant that is so brief, that even a sine wave varying AC current is approximately constant for that instant. Constant current for an instant is the same for AC and DC. Therefore, in an instantaneous view there is not difference between AC and DC. That is the simple answer to your question about AC versus DC power losses.

This picture helps illustrate this point. A series of steps approximates a sine. The more steps, the better the approximation. During the flat part of each step, it is like DC.

It is only when we consider averages over long times, that the AC/DC differences arise. Actually averages of a number of complete AC cycle, that the complications come it. Complications like peak versus RMS measurements, imaginary power. But when you return to the instantaneous view where V and I are approximately constant for the instant, all the differences between AC and DC power losses vanish.

There is also a middle view, where V and I may vary with time but not necessarily as a sin wave. We can write differential equations including terms like IR, LdI/dt, and CdV/dt and solve those for voltages and currents versus time including inductance and capacitance. In that approach, there is also no such thing as AC, but only time varying quantities.

Let me repeat for clarity. The whole topic of AC electricity applies only when we use averages over one or more complete cycles. When we model electricity instantaneously, we do not use averages and there is no such thing as AC.

 

[sophiecentaur]

Or a more simpler model, if you have a block of wood with a spring on the table and you push on the spring eventually the compression will overcome the static friction of the block and it will move but if you pull back on the spring before this point it will not move. I am imagining this type of principle at play in a long wire.

This model of yours goes quite a long way to describing the propagation of Sound ( pressure) waves through a solid but is really doesn't remotely describe the process of electrical conduction. The only mechanical analogy for electricity that takes us very far is the flow of a fluid through a pipe (the Water Model) but even that is extremely limited and doesn't begin to deal with the electromagnetic fields associated with electricity.

 

[metastable]

Intuitively I fell like AC would lose more power long a long power line than DC.

You can divide the peak voltage of the AC by the square root of 2 to get the root mean square voltage, which is the equivalent DC voltage that dissipates the same power.

https://en.m.wikipedia.org/wiki/Root_mean_square#Definition

 

[sophiecentaur]

You can divide the peak voltage of the AC by the square root of 2 to get the root mean square voltage, which is the equivalent DC voltage that dissipates the same power.

https://en.m.wikipedia.org/wiki/Root_mean_square#Definition

Note - this only works for a true sine wave AC. If the wave shape is a square wave (or even just an approximation) the RMS value is the same for AC and DC.
The (better) DMMs that you can buy these days, actually calculate the true RMS value of low frequency AC waveforms.

 

[metastable]

In a permanent magnet motor, one can choose between FOC control mode (sine wave) or BLDC control mode (6-step).

In the latter case BLDC 6-step, which during any given step the current through the motor resembles DC, only 2/3rds of the copper and 2 out of 3 of the phases are used at the same time (assuming a WYE winding).

In the former case FOC sine wave, a 3 phase AC sine wave is generated in the motor, which uses all 3 phases and 3/3rds of the copper simultaneously. Using more of the copper results in lower losses for a given amount of electrical power at a given rpm, resulting in greater torque for the same electrical power with FOC and sine wave AC. FOC stands for Field Oriented Control and is a highly complex topic on its own.

 

[DaveE]

Bottom line, there is no simple analogy of electrons behaving like marbles or water that comes close to the reality of current in a wire. It is one those annoying cases where simplifications don't work.

Yes. This.
Many of the losses in AC power distribution are due to field effects away from the actual transmission wire, like induced currents in lossy materials (like the ground). Marbles and water don't do that.

 

[rcgldr]

The main advantage of AC for power distribution is the ability to step up to high voltages which does reduce power losses (I=P/V) (P= I^2 R).

 

[sophiecentaur]

The main advantage of AC for power distribution is the ability to step up to high voltages which does reduce power losses (P= I^2 R).

Great for intermediate distances and where you can synchronise power sources. For very long distances, DC losses are less than AC losses and nowadays, there are DC equivalent devices to the Transformer which are pretty low loss. You can transfer power between areas with different frequency standards by using a DC link.

 

[rcgldr]

Great for intermediate distances and where you can synchronize power sources. For very long distances, DC losses are less than AC losses and nowadays, there are DC equivalent devices to the Transformer which are pretty low loss. You can transfer power between areas with different frequency standards by using a DC link.

How long is long enough? USA has two major power grids (and yes they are synchronized). There are generators at Hoover Dam powering parts of southern California. The high voltage lines run from 500,000 to 700,000+ volts with what I thought was relatively low power losses. There's also the issue of USA infrastructure based on 3 phase AC power generation, but that's a separate issue from power losses.

https://en.wikipedia.org/wiki/North_American_power_transmission_grid
However parts of Europe are using high power dc distribution:

https://en.wikipedia.org/wiki/High-voltage_direct_current

 

[John3509]

Correct.
I'm not sure if the marble analogy helps, however. AC has inductive losses which don't have an equivalent for marbles, and it has the skin effect which also doesn't have an analogy. There is also the difference between peak and effective voltage for AC, that doesn't transfer well to marbles either. Synchronizing AC over a long distance is an issue, too, there I could see how this is modeled with marbles.
We have a series of Insights articles about AC power grids.

AC is used because it is easier to transform between different voltages, something necessary to transmit electricity over relevant distances. DC is used when electricity is transmitted over very long distances - above ~1000 km the lower losses per length are more important than the slightly higher losses from transformation. As power electronics get cheaper and better there are more and more DC transmission lines.

Well, this looks to be much more complex then I imagined and that there are other things at play here. Until you mentioned it I totally forgot the skin effect even existed.

But regardless of the other effects that may or may not be more influential, am I correct in assuming that specifically the mechanical properties I described would be amongst these effects that would more negatively effect AC in voltage drop over DC ? I want to look at each effects, effect individually. Is what I described even an actual effect? You mentioned Synchronizing, is that it? Can you elaborate on that please.

Also, a follow up question, why is the ability to step up or down voltages so important?

 

[John3509]

This model of yours goes quite a long way to describing the propagation of Sound ( pressure) waves through a solid but is really doesn't remotely describe the process of electrical conduction. The only mechanical analogy for electricity that takes us very far is the flow of a fluid through a pipe (the Water Model) but even that is extremely limited and doesn't begin to deal with the electromagnetic fields associated with electricity.

I now realize its much more complicated than I originally pictured and that I ignored other effects. But given that let me reformulate my question. In addition to the electro magnetic interactions, there has to at least some mechanical component to them as well right?
What I want to know is if the effect I described, which I think would arise from their mechanical interactions, would actually cause more voltage drop, and as a result power loss, in AC than DC? I want to look at how each different effect individually effects the voltage. I now know there are others like skin effect, and field effect.

 

[John3509]

The main advantage of AC for power distribution is the ability to step up to high voltages which does reduce power losses (I=P/V) (P= I^2 R).

Im glad you brought this up here because this is something someone else told me before and I did not understand it and its been in my head ever since.
How does increasing voltage decrease power loss?

 

[Nugatory]

Im glad you brought this up here because this is something someone else told me before and I did not understand it and its been in my head ever since.
How does increasing voltage decrease power loss?

All else being the same, there is more power loss with higher current. Increasing the voltage allows the same amount of power to be carried with less current, which decreases the loss.

 

[sophiecentaur]

How long is long enough?

I can't find too much about HISTORY in this thread but History is always responsible for the direction of new projects or expansion in Engineering. Compatibility Rules unless there are very good reasons to change direction. The present AC Power technology does pretty well and there are many established companies that will supply AC equipment for incremental expansion of networks.
One day, I expect nearly all electrical power distribution to use DC. One good reason for using AC at present is that Induction Motors are driven with AC and an AC distribution system has allowed big static equipment to be driven straight off the mains. Big AC motors are Everywhere! But Power Electronics is advancing all the time and it won't be long before all induction motors can be fed from AC. Even better - speed control on Induction Motors is achieved by varying the supply frequency so local DC-AC conversion has this facility built in.
Poor old Tesla must be turning in his grave (at 60Hz, I imagine).

 

[mfb]

How long is long enough?

~700 km on land, ~50 km under the sea, as the HVDC article you linked describes.
Sometimes it is used even for "zero distance" if it connects unsynchronized networks. North America has five unsynchronized regions according to the article and a while ago I saw a news article about a connection station using HVDC to transfer power between two or three of them.

But regardless of the other effects that may or may not be more influential, am I correct in assuming that specifically the mechanical properties I described would be amongst these effects that would more negatively effect AC in voltage drop over DC ? I want to look at each effects, effect individually. Is what I described even an actual effect? You mentioned Synchronizing, is that it? Can you elaborate on that please.

I don't even understand where you see some effect in the marble analogy. It is just too far away from electricity to be useful here.

 

[sophiecentaur]

am I correct in assuming that specifically the mechanical properties I described would be amongst these effects that would more negatively effect AC in voltage drop over DC ?

I don't even understand where you see some effect in the marble analogy. It is just too far away from electricity to be useful here.

I can never understand why people insist that there is merit in the models they just thought up in their heads, with no evidence or with no serious provenance. Advances in Science thinking are few and far between and they always come from those who are very well informed. Remember Newton's statement about 'standing on the shoulders of others"? He was never the most humble of men but even he acknowledged that ok.

 

[anorlunda]

there has to at least some mechanical component to them as well right?

Sorry but very wrong. Your attempt to understand electricity mechanically is so seriously wrong that you should forget it completely. Pretend that you never heard of the word electron.

 

[fluidistic]

I do not want to distract the main topic of this thread. However, the Drude's model is completely wrong and has nothing to do with reality. It is worse than the Bohr's model of an atom, and can reach some quantities correctly, most quantities completely wrong and some quantities correctly only because gross errors cancel each other out.

I personally find its teaching in solid state physics to be a disgrace, as it can lead to build a faulty intuition of the electron's behaviors in a conductor. Don't do that. Don't rely on that model, not even for intuition.
This is not equivalent to Newtonian mechanics vs GR, where Newtonian mechanics can be reached in some limits. Here we're talking of a known-to-be-wrong model that has absolutely nothing to do with reality vs, say, the free (and/or nearly free) electron model, which would be the equivalent of a good model where we have taken some limits to make the description much simpler, losing many details of course in the process (and predictive power).

 

[anorlunda]

I don't like the Drude model much either. But is it still taught for students unable to cope with the free electron model. But I think you are overstating the case for deprecation. The free electron model you mentioned is still called the Drude-Sommerfield model.

The model was extended in 1905 by Hendrik Antoon Lorentz (and hence is also known as the Drude–Lorentz model) and is a classical model. Later it was supplemented with the results of quantum theory in 1933 by Arnold Sommerfeld and Hans Bethe, leading to the Drude–Sommerfeld model.

My personal preference for teaching electric conduction is to restrict it to three levels with solid foundations.

1. Circuit analysis (no fields, no charges, no propagation)
2. Maxwell's equations.
3. Q.E.D.

Intermediate levels (we could call the free electron model level 2.6) are most often only of interest to specialized researchers.

But B level students yearn for a level 1.5. They learn Ohm's law, then they want to jump directly into atomic physics. (See the lament #18) This is fostered by many (perhaps most) teaching in basic electricity that begins with the atomic structure. If I ruled the world, it would be a crime for a teacher to mention the word electron to B level students.

 

[cjl]

How long is long enough? USA has two major power grids (and yes they are synchronized). There are generators at Hoover Dam powering parts of southern California. The high voltage lines run from 500,000 to 700,000+ volts with what I thought was relatively low power losses. There's also the issue of USA infrastructure based on 3 phase AC power generation, but that's a separate issue from power losses.

https://en.wikipedia.org/wiki/North_American_power_transmission_grid
However parts of Europe are using high power dc distribution:

https://en.wikipedia.org/wiki/High-voltage_direct_current

The US has 3 major grids (East, West, Texas), and they aren't synchronized with each other. DC is used to tie the grids together, and also for very long distance, such as the line that takes power from the pacific northwest to the LA area (https://en.wikipedia.org/wiki/Pacific_DC_Intertie).

 

[John3509]

All else being the same, there is more power loss with higher current. Increasing the voltage allows the same amount of power to be carried with less current, which decreases the loss.

How though? I know power drained by a resistor is P = I*I *R so lower I means lower power drained but how can you deliver higher voltage with less current when voltage and current are proportional? How do you wind up with the same amount of power being carried?
And, isn't the load at the end of the cable, like a vacuum cleaner for instance, just another resistance in the circuit, so aren't you just lowering the power the vacuum clearer gets?

 

[John3509]

.I don't even understand where you see some effect in the marble analogy. It is just too far away from electricity to be useful here.

I can never understand why people insist that there is merit in the models they just thought up in their heads, with no evidence or with no serious provenance.

Sorry but very wrong. Your attempt to understand electricity mechanically is so seriously wrong that you should forget it completely. Pretend that you never heard of the word electron.

I understand my model is bad but I am not insisting on using it despite all reason. I'm not trying to use it as an alternative to the water in pipe analogy given to new students or a general model for electricy flow but adress in particular how voltage travels. Like if you had a wire from here to the moon and osilated the voltage, how it would reach the other end. What I am trying is to understand a specific aspect of current, how voltage propogates trough a wire differently with AC and DC and how this effects voltage drop. I guess how I should have started this tread is to just ask "does something similar to how sound propogates trough air or waves trough water ever happen in wires" mfb mentioned synchronizing AC over long distances, i feel like the answer I am looking for will have something to do with this. For context.

I know that

1. Electrons repel and can be compressed to be more densely packed or less, like in compactor plates, so there is springiness to their motion
2. electrons can behave both as particle and wave,
3. In thermodynamics you model matter as a grid of points connected by springs in 3D
4. Current is not instantaneous throughout the wire so it must travel along the wire

This is why I assumed that in addition to the electromagnetic and quantum stuff going on there would at least in some aspect be mechanical motion.

But something i read from someone very recently tells me that current is actually pushed by the electric field that goes along the whole wire and that electrons don't actually push on each other at all like megnetic freight carts on a track. Is this true?

 

[DaveE]

how I should have started this tread is to just ask "does something similar to how sound propogates trough air or waves trough water ever happen in wires"

No. Sound and water waves are not very similar at all to electric current flow in a wire.

 

[DaveE]

But something i read from someone very recently tells me that current is actually pushed by the electric field that goes along the whole wire and that electrons don't actually push on each other at all like megnetic freight carts on a track. Is this true?

Really, if you want to try to understand electricity, you need to give up on analogies like water, marbles and magnetic freight carts.

 

[mfb]

How though? I know power drained by a resistor is P = I*I *R so lower I means lower power drained but how can you deliver higher voltage with less current when voltage and current are proportional? How do you wind up with the same amount of power being carried?
And, isn't the load at the end of the cable, like a vacuum cleaner for instance, just another resistance in the circuit, so aren't you just lowering the power the vacuum clearer gets?

You want to deliver a specific power P to the consumer. P_consumer=U_grid I. A higher voltage means you need a lower current to deliver the same power.

Losses in the cable are P_cable = I² R_cable. A lower current means lower losses in the cable.

We can also plug the first equation into the second: P_cable = R_cable P²_consumer/(U²_grid). We can't change the power the consumer wants. We can lower the cable resistance, and we can increase the cable voltage to reduce the power lost in the cable.

 

[Nugatory]

How though? I know power drained by a resistor is P = I*I *R so lower I means lower power drained but how can you deliver higher voltage with less current when voltage and current are proportional? How do you wind up with the same amount of power being carried?
And, isn't the load at the end of the cable, like a vacuum cleaner for instance, just another resistance in the circuit, so aren't you just lowering the power the vacuum clearer gets?

You can think of the load as another resistance but you must remember that the load is not the vacuum cleaner itself, but rather the step-down transformer at the end of the transmission line. Its resistance will vary with the power demand so the voltage and the current in the transmission line are not proportional. Instead, we have V=IR with R varying and V held constant by the generator so that I and R are inversely proportional to one another.

 

[anorlunda]

Maxwell's Equations tell us that electromagnetic effects propagate at a finite speed. That speed is c in a vacuum. In other media, such as a wire the propagation speed is in the range 0.6-0.8 c.

At those speeds, AC versus DC is irrelevant. Ordinary circuit analysis breaks down and you must treat wires as waveguides, and antennas.

If you are trying to teach students about that, then using circuits is the wrong approach. You need to consider Maxwell's Equations. Unfortunately, the math is difficult so it is not usually taught at the high school level, but rather in field theory courses to college seniors or post graduate students.

 

[vanhees71]

This math will also tell you that for dispersive media it is very important to clearly define which "speed of propagation" you are talking about... (phase velocity, group velocity, front velocity,...).

 

[John3509]

Really, if you want to try to understand electricity, you need to give up on analogies like water, marbles and magnetic freight carts.

Why? When I use the analogy of magnetic fright cars to describe electrons pushing each other down the wire in case someone does does not know what I mean by electrons pushing each other down the wire. I now know electrons may not literally push each other down the wire but what's wrong with using a an anology to just convey the idea of electrons pushing each other?

 

[DaveE]

Why? When I use the analogy of magnetic fright cars to describe electrons pushing each other down the wire in case someone does does not know what I mean by electrons pushing each other down the wire. I now know electrons may not literally push each other down the wire but what's wrong with using a an anology to just convey the idea of electrons pushing each other?

Analogies are ok in the proper context, but analogies are always wrong. That's why they're called analogies. If you really want to understand a subject, you need to get past that and study the subject for what it is, not what it is like.

 

[John3509]

You can think of the load as another resistance but you must remember that the load is not the vacuum cleaner itself, but rather the step-down transformer at the end of the transmission line. Its resistance will vary with the power demand so the voltage and the current in the transmission line are not proportional. Instead, we have V=IR with R varying and V held constant by the generator so that I and R are inversely proportional to one another.

But if I and R are inversely proportional does that still means V and I are proportional?

 

[John3509]

You want to deliver a specific power P to the consumer. P_consumer=U_grid I. A higher voltage means you need a lower current to deliver the same power.

Losses in the cable are P_cable = I² R_cable. A lower current means lower losses in the cable.

We can also plug the first equation into the second: P_cable = R_cable P²_consumer/(U²_grid). We can't change the power the consumer wants. We can lower the cable resistance, and we can increase the cable voltage to reduce the power lost in the cable.

Here is what confuses me about this, the load of the consumer or transformer to me exact, is another resistor in the circuit, you are minimizing the poser consumed by one resistence but maximising it for the other? How can that work? I am guessing it has something to do with the fact that you have 2 different P's in your equation, just not sure how to put the pieces together in my mind.

and V = IR, how to you achieve delivering the current with high voltage but low current?

 

[mfb]

The power you have to deliver to the consumer is determined by the consumer - you can't change that if you want to keep the grid healthy.

With a given power arriving at the customer you want to minimize the power loss in the grid. Reducing the resistance of the cables is an obvious way to do so, increasing the transmission voltage is another one.

If high voltage safety and insulation wouldn't be an issue we could deliver this high voltage directly to the customer, the customer would use a very large resistance (as P=V²/R), this resistance can be larger for high voltages, and it can be much larger than the resistance of the cable. You can't do that in practice, of course, so the voltage is transformed down near the customer.

 

[jbriggs444]

But if I and R are inversely proportional does that still means V and I are proportional?

I and R are inversely proportional if you hold V constant. Which means that this inverse proportionality, by itself, gives you exactly zero information on what happens if V is allowed to vary.

V and I are directly proportional if you hold R constant. Which means that this direct proportionality, by itself, gives you exactly zero information on what happens if R is allowed to vary.

In the case at hand, R is not held constant while you vary V. It varies, for one thing, because you will use different step down transformers if you choose to run your transmission lines at 20,000 volts versus at 40,000 volts.

 

[Nugatory]

But if I and R are inversely proportional does that still means V and I are proportional?

No. If I and R are inversely proportional with V constant, then V and I cannot be proportional - one of them is fixed and the other is not.

 

[John3509]

I and R are inversely proportional if you hold V constant. Which means that this inverse proportionality, by itself, gives you exactly zero information on what happens if V is allowed to vary.

V and I are directly proportional if you hold R constant. Which means that this direct proportionality, by itself, gives you exactly zero information on what happens if R is allowed to vary.

In the case at hand, R is not held constant while you vary V. It varies, for one thing, because you will use different step down transformers if you choose to run your transmission lines at 20,000 volts versus at 40,000 volts.

Oh right, one of them always has to be a constant of proportionality, now I remember.

 

[John3509]

So I had some time to think about it, but its just not making sense to me for some reason.
Here is how I see it
As electrons pass a resistor they loose potential, that is the voltage drop, this happening over time is the power the resistor drains
You can't control current directly, only indirectly by controlling the things that effect it, resistance and voltage.

What I don't get is:
How are you controlling the current in power distribution systems?
How can you achieve a high voltage but low current? Wouldn't increasing the transmission voltage increase the current?
Also the wording "power transmitted" is throwing me off, I know I used it too but now that I think about it it makes no sense to me, how I understand it is power used up by a resistor or load.
And finally, I am imagining a system with two resisters, one representing the total resistance in the wiring and one for the consumer load. If you minimize the amount of power consumed by the wire resistor by lowering current, aren't you also minimizing the power used by the consumer? Ideally you can have no power lost by setting current to 0, but then the consumer appliances will have no power to consume eighter. So the power both resistances consume has to be proportional right? isn't it in series?

 

[vanhees71]

The "power drain" in a resistor is due to dissipation. In the most simple classical picture you have conduction electrons in the wire which can move quasi freely, but there's friction. If you have a DC voltage after some short time you have a constant current density, i.e., the electrons are not accelerated anymore due to the electric field. That's the case when the friction force is as large as the electric force on that electrons.

The power transmission is, BTW, not through electron transport (that wouldn't make sense in the AC case at all, because there the electrons stay more or less where they are) but through the electromagnetic field. It is very illuminating to analyse the coaxial cable for DC as well as AC in detail. The DC case if masterfully discussed in

A. Sommerfeld, Lectures on Theoretical Physics, Vol. 3

 

[Nugatory]

What I don't get is:
How are you controlling the current in power distribution systems?

You don’t. You control the voltage (holding it fixed) and the current varies with the load. For example, the power company supplies power at 240 volts to my house. When I’m running a device using 240 W the current in the supply wires to my house is .1 A; turn on a second such device and the effective resistance is halved and the current doubles to .2 A so that twice as much power is being transmitted at the same voltage.

How can you achieve a high voltage but low current? Wouldn't increasing the transmission voltage increase the current?

You are forgetting the stepdown transformer. If the power company is using a 2400 V transmission line to get power to my house (where everything runs on 240 V) they will install a 10:1 stepdown transformer at the end of line and connect my house supply line to that. If they decide that a 9600 V transmission line makes more sense, they will replace the 10:1 stepdown transformer with a 40:1 one as part of the change. Nothing will change for me, I’m still getting 240 V in my house but now the power company transmission line can deliver the same amount of power with 1/4 the current.

And finally, I am imagining a system with two resisters, one representing the total resistance in the wiring and one for the consumer load. If you minimize the amount of power consumed by the wire resistor by lowering current, aren't you also minimizing the power used by the consumer? ... So the power both resistances consume has to be proportional right? isn't it in series?

You are forgetting the stepdown transformer again. The consumer appliances are not in series with the transmission line resistance; they are in series with my household wiring which is connected to the output of the stepdown transformer. It is the stepdown transformer that is in series with the transmission line resistance. The power “consumed” by it is of course the power that’s running the appliances in my house; it’s resistance varies with the load so the current in the line also varies while the voltage is fixed.

 

[sophiecentaur]

Im imagining a system with two resisters, one representing the total resistance in the wiring and one for the consumer load. If you minimize the amount of power consumed by the wire resistor by lowering current, aren't you also minimizing the power used by the consumer?

There is a logic in the way you need to approach this. You start with a SUPPLY VOLTAGE at the generator. The appliance has a particular resistance (based on the Power it is designed for) and that determines the amount of Current that flows.
A practical point to make here is that the supply cables and generator are made with as LOW a resistance as is practical (only a few Ohms total). The current running through your appliance also flows through the cables and they will get a bit warm and waste Power. If you were using 100V mains voltage then the current for a given power (say 1kW) will be 10A. Change the system so that you are using 200V supply and a suitable 1kW appliance (you need a different one) will use 5A.

The resistance of the supply cables has very little effect on the 1kW figure (we ignore any change for simplicity to start with) so you have half the current going through the 200V system, which means 1/4 of the power dissipated. (P=I²R).

The only possible problem is that the risk of Shock has gone up a bit. But the number of accidents in Europe are not significant compared with US so a higher operating voltage can be very good value.

 

[mfb]

But the number of accidents in Europe are not significant compared with US so a higher operating voltage can be very good value.

This could also be a result of the plug types. Europlugs have no metal you can touch when there is an electric connection as parts of their pins are isolated. The US plugs don't have that isolation - you can easily touch a live metal part there.

 

[sophiecentaur]

This could also be a result of the plug types. Europlugs have no metal you can touch when there is an electric connection as parts of their pins are isolated. The US plugs don't have that isolation - you can easily touch a live metal part there.

Undoubtedly. It's something that is easily dealt with.

 

[John3509]

You don’t. You control the voltage (holding it fixed) and the current varies with the load. For example, the power company supplies power at 240 volts to my house. When I’m running a device using 240 W the current in the supply wires to my house is .1 A; turn on a second such device and the effective resistance is halved and the current doubles to .2 A so that twice as much power is being transmitted at the same voltage.
You are forgetting the stepdown transformer. If the power company is using a 2400 V transmission line to get power to my house (where everything runs on 240 V) they will install a 10:1 stepdown transformer at the end of line and connect my house supply line to that. If they decide that a 9600 V transmission line makes more sense, they will replace the 10:1 stepdown transformer with a 40:1 one as part of the change. Nothing will change for me, I’m still getting 240 V in my house but now the power company transmission line can deliver the same amount of power with 1/4 the current.

You are forgetting the stepdown transformer again. The consumer appliances are not in series with the transmission line resistance; they are in series with my household wiring which is connected to the output of the stepdown transformer. It is the stepdown transformer that is in series with the transmission line resistance. The power “consumed” by it is of course the power that’s running the appliances in my house; it’s resistance varies with the load so the current in the line also varies while the voltage is fixed.

So when you increase the voltage but also use a steeper step down transformer you can lower the current? I did not know transformers could do that, in that case that answers my question spot on. Thanks.Back to the original topic, about the propagation of voltage through a long wire, someone at the beginning of the thread mentioned synchronization of the voltage is a problem on a long wire, can someone elaborate on what that means?

 

[sophiecentaur]

use a steeper step down transformer you can lower the current?

This is not the way it goes; there is a right way and a wrong way to think of a problem like this. Your domestic (250V) supply will provide a 1kW appliance with the current will pass (4A) because its resistance will be designed to be about 65Ω. The transformer will only need to take 1/100 of that current from its 25,000V supply to give 1kW (VI is the same). So the 25000 V supply will 'see' a load of 650,000Ω, when the load has been 'transformed' by the transformer.
Best to get used to that before trying to 'understand' how the currents flowing in the windings of a transformer and the magnetic fields in the Iron are related. Transformers are seldom totally understood by most of the people who use them and who design circuits with transformers in them. So no need to worry. Something to take up at your leisure.

 

[John3509]

Thank you, that answers that question. Now to wrap up the original question how does voltage propagate along a wire in AC and does it negatively effect the voltage drop? What did mfb mean by "voltage synchronization problem in a long wire?

 

[mfb]

AC needs to be synchronized: You can't have positive voltage on a cable at the same time your neighbor has negative voltage on that cable, that just doesn't work as the cables are connected to each other. While that is no problem within a smaller region: Changes in the voltage cannot propagate faster than the speed of light. If your power plant is 1000 km away from you, then whatever it does arrives the earliest 0.003 seconds later at your place, in practice more like 0.005 seconds. Sounds small - but with 50 Hz the voltage changes its sign 100 times per second (50 times up, 50 times down), or every 0.010 seconds. Your grid stops being synchronized. Among other issues this increases the losses in the grid.

 

[sophiecentaur]

"voltage synchronization problem in a long wire?

Imagine taking an AC generator which is going +-+-+-+- and connecting it to a generator that's going -+-+-+-+. These things can deliver thousands of amps. At the peak of the cycle, it would be the equivalent of taking two car batteries and connecting them +to- and - to + terminals; each one would see a short circuit and you would blow the sides of the batteries ( sometimes literally!).

Any slight phase difference between two sinusoidal AC generators will involve something of the above and current will flow from one into the other (like when a fully charged battery will discharge into a partly discharged battery until the volts are the same). The effect will be that the generator that's slightly advanced in time will end up actually driving the other one. Whenever two are connected, this happens to a tiny extent and the speed control kicks in and brings them together. A long wire between two generators will introduce significant phase delay.

Scary scenario:
There is a strange associated problem with very long lines; Assume both generators are synchronised to a reference time source, half way in between them. They are 'in phase' with the waveform from that sync source. However, each generator gets a delayed waveform from the other. What happens then is that power ends up sloshing up and down the line, forming a standing wave with scary high voltage peaks appearing along the line. To deal with this, one of the generators is allowed to be 'slave' to the other and is allowed to be driven by the volts down the line and not using any coal in its boiler.

But, in real life, there are massive loads all over the network (that's why the generators are there in the first place) and small voltage drops exist all over the lines (series resistance) and this helps the system to stabilise and both generators end up in appropriate phase and supplying appropriate power.

There have been some spectacular disasters on grid systems due to this effect and the Grid Controllers spend their time making sure they foresee the possible problems and make suitable adjustments.
This is not a problem with a DC system because there is no 'phase' delay,

 

[vanhees71]

For compact circuits you neglect retardation all together and the voltage along the wires is just synchronous. For really long wires, i.e., when their length is not entirely negligible compared to the wave length of the electromagnetic waves at the frequency of your AC, $\lambda \simeq 2 \pi/k =2 \pi c/\omega=c/\nu$. For usual house-hold current ($\nu=50 \; \text{Hz}$ in Europe that's $\lambda \simeq 6000\;\text{km}$) you can use the telegrapher's equation, which follows from the quasistationary theory by assuming the resistivity, capacitance, and inductance to be distributed along the wires, which should be close together, so that you can apply the quasistationary field equations for integration regions along an infinitesimal part along the wires:

https://en.wikipedia.org/wiki/Telegrapher's_equations