Sorry, I've alway's liked the bird on the wire, I just have to change it up a bit.

Bird number 2, weighs 200lbs and lands on a 15kv line on the very top, it's feet are spread about 2 feet apart. What happens to the bird? Is it the same as bird number 1.

The potential is the same all but 2 things (Resistance and Capacitance has changed.)

Just curious,

Regardless of the size of the bird, nothing will happen to it. The capacitance is between the bird and the ground, not between the bird and the wire.

 

The potential and your simple theories can explain the conduction of the super fireball

caught on a cell phone. Transmission lines running behind homes ,with never before seen

fire ball , has anyone explored this fireball theory.....not only would it take out a bird

but the entire community. So - + - + - + has some meaning. Did anyone see the fireball

video,looked like 14,000 volt feeds ,can someone post the fireball since we are thinking

about this subject. A Canada cell phone video ,a first time sighting I think. I googled it,

just google-electric fireball behind homes. By the way ,the fire ball was moving behind

the homes so the POTENTIAL for a disaster was real...I don't think the birds would have a

chance....it was moving real fast and every thing was hot in both ways.

 

...
Besides, nobody transmits DC anymore, not since the early 20th century. I have a feeling the person who wrote that description and said 0.5MVDC had no idea what they were talking about.

Not correct...

For Europe, see: http://en.wikipedia.org/wiki/High-voltage_direct_current

And, the Oregon to LA see: pacific DC intertie
http://en.wikipedia.org/wiki/Pacific_DC_Intertie

 

I highly doubt that's DC. You're not going to be able to have voltage and current that high going through such long, thin cables. Besides, nobody transmits DC anymore, not since the early 20th century. I have a feeling the person who wrote that description and said 0.5MVDC had no idea what they were talking about.

Uhm.... DC monolines are becoming more and more common and are frequently well over 1 million volts. And remember that the whole point of going to as high a voltage as you can is to reduce the current by a corresponding amount. A 1MV DC line only needs to carry 1A of current for every MW of power it transfers.

The reason that DC monolines are in vogue is because of numerous advantages they have. You can go to higher voltages because you don't have the skin effect and EM propagation issues as much, you only need a single line and so the material costs are much less, you don't have to synchronize the phases at all the various tie-in points, and a big one is that you don't have to purchase as wide a right-of-way as you do for three-phase transmission. You do have loses in the rectification and inversion processes at each end, but those have been brought sufficiently under control that DC is quite competitive for long haul power transmission.

 

Thanks everyone who has replied.

I'll let the issue of capacitance pass over my head for now, and just concentrate on voltage.

@ inwo -- you have replied:

1) the bird and the power line DO NOT have a voltage differential between them;
2) the bird IS electrically charged;
3) the bird and the power line ARE at exactly the same state of potential energy
per unit charge; and
4) the reason why is that the bird and the wire are IN CONTACT.

Thank you very much for answering directly on point like that - very helpful.

And does everyone agree with the above, or does anyone dispute any of those points?

But my brain now wants to know:

What exactly is it about being in contact that:

1) causes there to be no voltage differential between the bird and the power line, 2) causes the bird to become electrically charged, and
3) causes the bird's body to have exactly the same state of potential energy per
unit charge as the power line ....

.....and all without ANY current flowing into or out of the bird?

Why doesn't at least a bit of current go into the bird to bring about all of this equalisation? Or since it's AC, I suppose that last question should be: why isn't at least some current flowing in and out of the bird with each cycle?

That question can be generalised like this:

How exactly does the simple fact of being in contact cause precisely the same state of potential energy per unit charge to be communicated from one object to the other? What is the medium of communication - if it is not caused by electrons moving from one object to the other?

In other words:
since contact is just physical touching at the surface of each object, by what mechanism does that cause equalisation of electrical states such that there is no voltage differential between the objects?

Did you read the blog entry I wrote that addresses this, albeit somewhat in passing?

 

since contact is just physical touching at the surface of each object, by what mechanism does that cause equalisation of electrical states such that there is no voltage differential between the objects?

Since contact between your finger and water is just a physical touching at the surface of each object, by what mechanism does that cause equalization of wetness such that there is no wetness differential between the objects?

 

I highly doubt that's DC. You're not going to be able to have voltage and current that high going through such long, thin cables. Besides, nobody transmits DC anymore, .

The province I live in has 12,800 kilometres of transmission lines and are some of the leaders in DC transmission research and development.
For long-distance transmission, power losses are considerably less with DC than with AC.
The cost of a long-distance DC transmission line system uses 2 conductor cables, where an AC system uses 3, making the DC lines approximately 1/3 less expensive. However, DC transmission requires expensive conversion equipment, this cost is offset by the gains made in much lower transmission losses.

A Typical Hydro Generating station 24Kv - 500Kv
Conductor - 7,400 km of 4-cm in diameter cable;
Thyristors - Bipole II 18,432;

 

Sorry, I've alway's liked the bird on the wire, I just have to change it up a bit.

Bird number 2, weighs 200lbs and lands on a 15kv line on the very top, it's feet are spread about 2 feet apart. What happens to the bird? Is it the same as bird number 1.

The potential is the same all but 2 things (Resistance and Capacitance has changed.)

Just curious,

If the transmission line has enough resistance and enough current flowing through it, then you could conceivably develop enough potential between the two feet to cause a current to flow. But this is unlikely at the currents and resistances in a typical power line. Lighting hitting a tree (or some other object) and spreading out through the ground near it is a different matter. Here you have extremely high currents passing through significant resistance and can generate lethal voltages between the feet of humans standing nearby.

I just ran across a document by American Electric Power that gives some useful numbers. For 765kV AC transmission, the maximum power that can be transmitted over a distance of 300 miles is about 2000MW to 2500MW. Now, that's 3-phase power, and I'm assuming that the 765kV is the line-line voltage (could be wrong). If so, then the line-to-neutral voltage would be about 1300kV and, assuming unity power factor, the line currents would be in 600A range. But these are 4 and 6 conductor bundles, so the actual current in a given conductor is only about 100A to 150A. This same document says that their 765kV lines, when loaded to 1000MW (so about 250A of line current), lose about about 3.3MW of power per 100 miles of transmission line distance. So that's 1.1MW per mile of line (since there are three three lines). But, again, because this is a bundled conductor, the loss in a given conductor is only about 200kW to 250kW per 100 miles at a current of about 40A to 50A. In either case, that works out to be 1Ω/mile of conductor, or about 0.2mΩ/ft.

So at 150A, this bird with a two-foot wide stance, would see about 60mV of voltage drop (RMS) between its feet due to resistive drop. At the peak current flow it would see about 85mV.

Another effect would be the fact that the voltage varies with distance along the line. I don't know what the speed of propagation in a power transmission line is, but let's assume it is about 50% of light speed. That means that the wavelength of a 60Hz signal would be about 1500 miles. To two feet represents 90μ° of phase angle. Since the gradient is steepest at the zero crossing, the peak voltage difference due to this effect would be about 5.6mV.

So I don't think you are going to get any appreciable current flowing up through the bird's butt due to having its feet spead out on this wire.

 

Thanks for the calculation WB. I like the extra effort for the 60hz zero crossing voltage.

It is interesting how much is lost per 100 miles.

The producer ends up covering that as a cost of production and then must cover his loses by allocating to the entire customer base by some formula (since those loses dont show up on anybody's power meter). Does anyone know if the more distant customers pay higher prices per kwh to cover those loses? or, do all costomers pay a network-averaged premium to cover those loses? Is there some provision in the Rural Electrification programs that prevent the lucky people that live far from the power plant from paying the fully-loaded cost to serve them?

 

Total losses seem to be in the 7% range, though I have heard numbers as high as 15%, but the 7% seems to be the one I hear most often. These seem to be about evenly split between resistive, corona, and power factor related losses.

I doubt that the marginal increase in line loss associated with getting to distant customers is enough to make it worth doing differentiated billing just on that account (no pun intended).

 

The province I live in has 12,800 kilometres of transmission lines and are some of the leaders in DC transmission research and development.
For long-distance transmission, power losses are considerably less with DC than with AC.
The cost of a long-distance DC transmission line system uses 2 conductor cables, where an AC system uses 3, making the DC lines approximately 1/3 less expensive. However, DC transmission requires expensive conversion equipment, this cost is offset by the gains made in much lower transmission losses.

A Typical Hydro Generating station 24Kv - 500Kv
Conductor - 7,400 km of 4-cm in diameter cable;
Thyristors - Bipole II 18,432;

Uhm.... DC monolines are becoming more and more common and are frequently well over 1 million volts. And remember that the whole point of going to as high a voltage as you can is to reduce the current by a corresponding amount. A 1MV DC line only needs to carry 1A of current for every MW of power it transfers.

The reason that DC monolines are in vogue is because of numerous advantages they have. You can go to higher voltages because you don't have the skin effect and EM propagation issues as much, you only need a single line and so the material costs are much less, you don't have to synchronize the phases at all the various tie-in points, and a big one is that you don't have to purchase as wide a right-of-way as you do for three-phase transmission. You do have loses in the rectification and inversion processes at each end, but those have been brought sufficiently under control that DC is quite competitive for long haul power transmission.

Not correct...

For Europe, see: http://en.wikipedia.org/wiki/High-voltage_direct_current

And, the Oregon to LA see: pacific DC intertie
http://en.wikipedia.org/wiki/Pacific_DC_Intertie

Hmm, that's very interesting guys, thanks for that. I had no idea they still transmitted DC. I feel like that's something I should have known....

 

"Bird on wire meets voltmeter"....

Well if my aim is good when I toss my multimeter at him he will feel it!

 

Hmm, that's very interesting guys, thanks for that. I had no idea they still transmitted DC. I feel like that's something I should have known....

Glad i could contribute.

That's the great part of this site. Put us all together and we can eventually nail down every technical fact and every piece of electronics history.

One member brings up an old topic that I thought I had well understood and then someone else answers from a totally different angle that I had never thought of before. A third person throws in an interesting historical tidbit and interesting conversation ensues.

 

Hello all.

Interesting discussing going here.

Since starting this thread, I have spoken to an electrical engineer about why the bird on the wire doesn't get electrocuted.

Between his answer and the answers on this thread, I find a difference of opinion:

A)
On this thread, the answer seems to be that, not only is there no voltage differential between the points where the bird's feet touch the power line, but there is also no voltage differential between the power line and the inside of the bird - and that is why the bird does not get electrocuted.

B)
But the engineer I spoke to said that the bird's internal voltage state is the same as earth - i.e. there is no voltage differential between, say, the bird's head and the ground - and there does exist a big voltage differential between the bird's internal environment and the power line - however no current flows into the bird because of the combination of 2 factors: 1) the bird's body is like a dead-end - it's not a path on a circuit or a path to ground so it would quickly take in charge until it's voltage state reaches equilibrium with the power line and there is no voltage differential between them, and 2) the resistance of the bird's body is greater than the resistance of the power line. In combination, that prevents current flowing into the bird, even though there is a voltage differential between it's internal environment and the power line.

Now I don't know which of these answers is correct.

Which is it, A or B?

 

I have spoken to an electrical engineer about why the bird on the wire doesn't get electrocuted.

The engineer I spoke to said that the bird's internal voltage state is the same as earth -

it would quickly take in charge until it's voltage state reaches equilibrium with the power line

Which is it, A or B?

I think you should talk the the engineer again and see why you understood him to say, 1) the birds internal voltage is the same as the earth and 2) it reaches equilibrium with the power line. Both of those can't be true.

 

But the engineer I spoke to said that the bird's internal voltage state is the same as earth - i.e. there is no voltage differential between, say, the bird's head and the ground - and there does exist a big voltage differential between the bird's internal environment and the power line - however no current flows into the bird because of the combination of 2 factors: 1) the bird's body is like a dead-end - it's not a path on a circuit or a path to ground so it would quickly take in charge until it's voltage state reaches equilibrium with the power line and there is no voltage differential between them, and 2) the resistance of the bird's body is greater than the resistance of the power line. In combination, that prevents current flowing into the bird, even though there is a voltage differential between it's internal environment and the power line.

This is far from correct.

There is one quote in this thread that sums up everything:

I sense that the OP's question is related to confusion about the meaning of voltage, also called potential. Absolute voltage has no meaning - all voltages are relative to something else. Voltage is potential energy.

If you stuck one probe up the birds butt and the other probe into earth, you would measure 10kV (or whatever voltage is on the power line), because you're measuring relative to earth. If you stuck one probe up its butt and the other probe you attached to the power line, you would measure 0 volts, because you're measuring relative to the power line.

According to Kirchoff's second law, the sum of all voltages in a complete circuit is 0. Since you measure 0V from bird-to-power line, and 10kV from bird-to-earth, you can conclude that the voltage drop over "bird" is equal to 0V, or in other words, the voltage of the bird relative to ground is 10kV.

Some people have already noted this, and I'll re-iterate. The bird does have a capacitance to the surrounding environment, and will conduct very small amounts of currents when standing on an AC wire. These currents are so insignificant, however, they can be disregarded.

 

@TheComet,

Thanks for your reply.

I am not confused about voltage being relative.

I am thinking of 1) the relative voltage between the bird and the earth, and 2) the relative voltage between the bird and the wire.

The mere fact that voltage is a relative measure does not determine which of 1) and 2) above should be zero and which should be 10kV (or whatever).

You've told me that 1) is 10kV and 2) is zero.

Ok, then that gives rise to the next question: Why?

I'd presume that when the bird was standing on the ground eating worms, the voltage between the bird's body and earth was zero. So what is it about landing on a 10kV power line that makes the bird's body now be at 10kV with respect to the earth, when previously it was (I presume) at zero volts with respect to the earth? What happened to cause that change? What mechanism brought that change about?

That I would love to know.

 

I'd presume that when the bird was standing on the ground eating worms, the voltage between the bird's body and earth was zero. So what is it about landing on a 10kV power line that makes the bird's body now be at 10kV with respect to the earth, when previously it was (I presume) at zero volts with respect to the earth? What happened to cause that change? What mechanism brought that change about?

That I would love to know.

I will give you an "ideal" answer, and a "real" answer.

Ideal
In an ideal world where air has infinite resistance and the bird has no capacitance, the answer to this question would be: "There is no change at all". Literally nothing happens when the bird flies up to the wire and sits on it.

But why do you measure 10kV on the bird relative to the ground, you may ask?

Simple: Your voltmeter has a smaller-than-infinite resistance, and by sticking one probe up the birds butt and the other into earth, You are closing the circuit, and the bird acts as a resistor (because its body also has a smaller-than-infinite resistance).

So this scenario:

Is basically the following:

Real
The bird has a capacitance, the air has a resistance, and a small current will flow through the bird in function with the frequency of the wire.

I can imagine a short, higher current flowing into the bird at first contact with the wires, because it would have to charge up to the new potential, but it will be far from lethal.

Also fun fact: You will probably kill the bird if you were to measure it with a voltmeter, because the current produced by closing the circuit would be enough to do so.

 

I can imagine a short, higher current flowing into the bird at first contact with the wires, because it would have to charge up to the new potential, but it will be far from lethal.

Also fun fact: You will probably kill the bird if you were to measure it with a voltmeter, because the current produced by closing the circuit would be enough to do so.

Ok, I'm done with this convesation because of the OPs claim to understand basic concepts of electricity AND still holds the beliefs that...
- the bird would have to "charge up" to the power line potential when he lands on the wire and "current flowing into the bird"

 

the bird would have to "charge up" to the power line potential when he lands on the wire and "current flowing into the bird"

Well technically speaking, if you take the "real" model (last picture) from my previous post, that does actually happen... It's just so insignificant that it's not worth discussing, but I mentioned it anyway so some little know-it-all doesn't pop in and try to correct it.