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

[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 resistance 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

 

[John3509]

Thanks for all the help everyone.