# How do you determine power flow direction in a transmission line?

Below is a picture of a typical transmission line(about 200 kV). Is there a simple physics experiment which can be performed safely near the line, to determine the power flow direction. Or in other words: where is the generator and where a consumer.

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given its AC i think you might struggle as there shouldn't be any difference between the forward voltage and the reverse. –  Nic Sep 28 '11 at 9:38

Actually, you don't need to measure things at two different positions along the wire to determine the power flow. You can do this with measurements taken at a single position. You must measure both current and voltage. The measuring device is built so that it rotates based on the relative phase of the current and voltage; and the direction of rotation corresponds to the direction of power flow. That's what happens in the power meter connected to your house. If you couldn't measure power flow this way it would be very hard for the electric utility to charge you for power.

EDIT: The power meter in your house of course touches the wires, but standing on the ground under a transmission line you can do something similar. You can have two pickups, a coil and a straight antenna, and you could feed them into an xy-oscilloscope. Pure reactive power should show a straight diagonal trace, and real power should show a circular loop. The problem is that you won't be able to see the direction the loop is tracing...clockwise or counterclockwise. You would need to compare the direction of rotation to a known field source, e.g. a short line you've set up purposely for calibration. You've got to make sure you're consistent with the relative physical orientation of the two pickups. In effect you're measuring the Poynting vector.

If you had two guitar amplifiers, you could amplify your pickups and feed them into an ordinary household power meter, which is already set up to convert the relative phases into an actual mechanical rotation that you can see. Of course you've still got to orient your pickups consistently and calibrate the direction of rotation against a known power flow.

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Yes, you are right. I was worried about the fact that there is power returning to the plant, but you get that too. –  Ron Maimon Sep 29 '11 at 15:07
@Marty: The power meter is not a right answer. The question is to determine power flow direction without a direct connection to the line. –  Martin Gales Sep 30 '11 at 6:32
@Martin--- you can determine all these quantities away from the wire by field measurements. This is equivalent to the Poynting vector business you want. –  Ron Maimon Sep 30 '11 at 17:09
@Ron: Not quite. The mentioned power meter measures only the active power. The Poynting vector includes the whole power flow(active + reactive). The active power is transmitted for ex. to your computer. The reactive power can be compared to the foam in a glass of beer : it is not the real stuff, but there is no way to avoid it and the glass must be oversized unless you will have overflow. –  Martin Gales Oct 1 '11 at 7:40
@Martin: The "reactive power" is just the power returning to the plant from the returning line. This is not an objection. –  Ron Maimon Oct 1 '11 at 15:48

The relevant equation is that of complex power in an AC circuit:

$$\vec{S}=\frac{\vec{V} \vec{I}^{*}}{2}=\frac{V I}{2} \angle -\phi$$

This is to say, the complex power flowing through a line is equal to the complex voltage times the conjugate of the complex current divided by two. Those not familiar with AC circuits will have a tough time with the complex variables, but let's turn to the following definitions for illustration. Note that I'm going to establish voltage as the reference angle.

$$S=P+jQ$$ $$\vec{I}=I \angle \phi$$ $$\vec{V}=V \angle 0^{\circ}$$

The real power is $P$ and the imaginary power is $Q$. The interesting thing to note is that $P$ can be either negative or positive. Before applying any of this however, you must have a reference direction. You're looking at the line, you can look either right or left, you establish one of those directions as positive, and then $P$ will follow as either negative or positive and if positive, it flows in the direction you're looking.

Here is an illustration, although I think this differs in convention by the sign of $\phi$.

Now, here is what you need to establish the direction of power flow:

• A reference direction
• The angle between the voltage and current

Exactly how you measure this is certainly the difficult part. I will defer to other answer regarding that. But provided that you do, if $\phi \in (-\pi/2,\pi/2)$ then the power flows in the positive direction and if $\phi \in (\pi/2, 3\pi/2)$ it flows in the negative direction.

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With a high-powered rifle, shoot one of the lines in two at the middle. Observe the two ends. The one that sparks the most is the line going back to the generator.

When these power lines carry DC, I believe it is impossible to easily determine the direction of current flow.

For AC lines, it's fairly easy; one times the 60Hz signal. Use two receivers separated by long enough to get 60Hz out of phase. If the speed of light is 186,000 miles per second, and the speed of wave transmission in the lines is on that order, then a 1% phase difference will need a distance of 0.05 x 186,000 miles/second x (1/60th seconds) = 31 miles.

I learned something from the comments on this question: power companies nowadays adjust the phases of their power lines by putting capacitors and inductors in the line. They do this fairly often and their effect is to modify the relative phase. So if you do measure the field at two points near the wire, make sure that there's not a power station in between your measurements. To learn more about this, try googling "power transmission"+shunt+series

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You're being funny, right? :) You're using the high powered rifle in safe mode by ensuring it's not pointing towards the person using it, but towards the pylon. It snaps the cable in two in the middle, bringing the high voltage cable to the ground, causing criminal damage, a power cut in some local area, as well as your picture in the local newspaper and on police files lol ;) –  Physiks lover Oct 1 '11 at 20:25
@Physics lover: I got to understand the rifle case as it was just a joke. You could remove the down vote please. –  Martin Gales Oct 3 '11 at 6:51