Can Transmission line be dangerous during rain ?

[I_am_learning]

Can Transmission line be dangerous during rain ?!

Suppose you are standing beneath 132KV transmission line. A pretty big water drop drops from the transmission line into your bald head. Is there any chance you get shock?

My understanding of physics tells me that, if the drop left the transmission line at the exact instant the AC voltage there was at peak, then it would be carrying a charge.
The charge it carries can be found out from its capacitance. Assuming the droplet radius of just 0.1 inch, http://deepfriedneon.com/tesla_f_calcsphere.html, gives capacitance of 0.285pf.
So, now Q = CV = 37.62 nC
Can this provide any significant shock? How do I find it out?
I am interested in the physics behind this
Thanks for reading.

 

[I_am_learning]

Ok, I thought about it again and recalled one empirical relation that gives the threshold level of Current for human death risk depending upon time. It says
I = 0.165 / sqrt(t) . where t is the duration of current.
So, for our case, I = Q / t
so, threshold of Danger, Q(threshold) = 0.165 * sqrt(t)
So, it appears that, if the charge on the droplet moves to human very very fast (t very very small), then Q(threshold) will be small, meaning even a tiny amount of charge could be lethal.
But I don't really know, if that's even distantly likely ?

 

[jim hardy]

I googled on "electric current threshold of perception" and got only abstracts.

Here's what i can report -

in high school days i did some experimenting with small capacitors and a 90 volt radio "B" battery
I couldn't detect by feel less than about 90 nanocoulombs.

So i don't believe your 37 nanocoulombs is likely to hurt you.

CAVEAT regarding experiment - Don't do that ! You could hurt yourself .
900 nanocoulombs was very noticeable, 9000 was quite painful.
You can feel a milliamp and twenty can kill you.

We've all experienced static electricity on winter days, can you find anything about the amount of charge involved there?

 

[sophiecentaur]

Ok, I thought about it again and recalled one empirical relation that gives the threshold level of Current for human death risk depending upon time. It says
I = 0.165 / sqrt(t) . where t is the duration of current.
So, for our case, I = Q / t
so, threshold of Danger, Q(threshold) = 0.165 * sqrt(t)
So, it appears that, if the charge on the droplet moves to human very very fast (t very very small), then Q(threshold) will be small, meaning even a tiny amount of charge could be lethal.
But I don't really know, if that's even distantly likely ?

Not sure that your use of "t" is correct. It doesn't matter how long the drop took to reach you. What counts is the duration of the Current. A bucket full of charged drops, one after the other could constitute a current, I suppose but the polarity would keep changing so no net charge. I think that threshold value that you are using refers to a shock with a current for time t, which doesn't apply for a single small dollop of charge.

If a large enough object could become detached instantly then it could carry a lethal charge but it would need a pretty high capacity to be lethal.
You could consider what happens the other way round. When they do live-line working, they actually do need to wear a Faraday suit to eliminate a shock when they actually connect, due to an initial charging current. But, of course, they are dealing with AC so there will be a constant flow of AC between the worker's body and the line, once an arc strikes, until they actually connect, producing possible burns rather than a shock. So this approach would, I think be very pessimistic and possible totally irrelevant. It certainly doesn't bother birds who land on power lines.

The Capacitance of the human body is around 100pF (which would apply to anything of 'human size') so this falling object would carry some charge if it approached human dimensions (as well as squashing you flat haha) Capacitance is approximately proportional to the linear dimension so the capacitance of a raindrop would be about 1/1000 of the capacity of a body. That's 0.5pF. Q=CV, so the charge would be about 10^-7C if charged at 400kV. For a full sized human body, the charge would only be 100 times that. Still not significant, I think. The actual energy involved (which is really what counts) is in the order of mJ (from 1/2 CV²).

 

[I_am_learning]

Not sure that your use of "t" is correct. It doesn't matter how long the drop took to reach you. What counts is the duration of the Current. A bucket full of charged drops, one after the other could constitute a current, I suppose but the polarity would keep changing so no net charge. I think that threshold value that you are using refers to a shock with a current for time t, which doesn't apply for a single small dollop of charge.
.

Sorry for not making this clear. I was taking 't' to mean the time it takes 'the charge in the droplet' to completely discharge to Earth through human body after landing on the 'head'.
Blue: Yes, I believe the same.

If a large enough object could become detached instantly then it could carry a lethal charge but it would need a pretty high capacity to be lethal.
You could consider what happens the other way round. When they do live-line working, they actually do need to wear a Faraday suit to eliminate a shock when they actually connect, due to an initial charging current. But, of course, they are dealing with AC so there will be a constant flow of AC between the worker's body and the line, once an arc strikes, until they actually connect, producing possible burns rather than a shock. So this approach would, I think be very pessimistic and possible totally irrelevant. It certainly doesn't bother birds who land on power lines.

The Capacitance of the human body is around 100pF (which would apply to anything of 'human size') so this falling object would carry some charge if it approached human dimensions (as well as squashing you flat haha) Capacitance is approximately proportional to the linear dimension so the capacitance of a raindrop would be about 1/1000 of the capacity of a body. That's 0.5pF. Q=CV, so the charge would be about 10^-7C if charged at 400kV. For a full sized human body, the charge would only be 100 times that. Still not significant, I think. The actual energy involved (which is really what counts) is in the order of mJ (from 1/2 CV²).

I couldn't grasp the Bold parts. I understood other parts.
And for the red part,
I_lethal = 0.165 / sqrt(t)
Energy_lethal ~= I ^2 * t = 0.027225 = Constant.
Oh! I agree.
the Empirical formula used such a straight forward relation. Thanks for bringing this to light for me. :)
So, it appears that, it don't really matter whether the charge in the droplet discharges into human body in 0.1 sec or 0.01 sec, what matters is the net energy transfer.
For our case of 0.1 inch R drop, Energy in the droplet = 1/2*C*V^2 = 0.00248 (below the threshold level, safe)
After some iterative maths, I found out that, if the droplet had radius of 1.2 inch (massive drop), it would have capacitance of 3.42pf and that could carry lethal amount of charge.
Please do check, if I am very much wrong.
Thanks.

 

[sophiecentaur]

The "bold parts".
What I meant is that, with AC, there will be a current flowing all the time that there is an AC arc this will mean Power is dissipated all the time an arc exists. The current will be charging / discharging your body every cycle during the connection process. This arc will burn you. I have seen a movie of the operation in which a man holds an 'earthing' wand for some time with lots of sparking, until he actually dares to touch the conductors.
For a one-off discharge (raindrop landing or charged man), the current flows just once so not much energy transfer. For live working on a DC power line, I think the situation would be more like the raindrop situation.

The red part.
When you get down to it, it's always energy that counts. Whatever the rate of transfer of energy (i.e. power), there is always a limit to the amount of damage you can do with 1J of energy. To achieve anything at all, there is usually some lower limit to the actual power needed in order to make this change. That's a general principle.

I seem to remember the tables for electrical risk have more or less the same 'current times time' value over quite a range. Below a certain current you can go on for ever but, at higher currents, there's another factor and your body is more susceptible.

I have just one reservation about the way you use the formula. That formula was not, I suspect, produced for Capacitor Discharge safety but for exposure to Mains Supply hazard. Do you have a better reference than "as far as I can recall" haha? Let's see if we can thrash this out.
btw 'as far as I can recall', the Leyden Jars on a Whimshurst machine (such as we were allowed to use at School in the 60s - but no longer) could kill you on a bad (/good) day. That could give us a figure to go on.

 

[I_am_learning]

The "bold parts".
What I meant is that, with AC, there will be a current flowing all the time that there is an AC arc this will mean Power is dissipated all the time an arc exists. The current will be charging / discharging your body every cycle during the connection process. This arc will burn you. I have seen a movie of the operation in which a man holds an 'earthing' wand for some time with lots of sparking, until he actually dares to touch the conductors.
For a one-off discharge (raindrop landing or charged man), the current flows just once so not much energy transfer. For live working on a DC power line, I think the situation would be more like the raindrop situation.

The red part.
When you get down to it, it's always energy that counts. Whatever the rate of transfer of energy (i.e. power), there is always a limit to the amount of damage you can do with 1J of energy. To achieve anything at all, there is usually some lower limit to the actual power needed in order to make this change. That's a general principle.

I seem to remember the tables for electrical risk have more or less the same 'current times time' value over quite a range. Below a certain current you can go on for ever but, at higher currents, there's another factor and your body is more susceptible.

I have just one reservation about the way you use the formula. That formula was not, I suspect, produced for Capacitor Discharge safety but for exposure to Mains Supply hazard. Do you have a better reference than "as far as I can recall" haha? Let's see if we can thrash this out.
btw 'as far as I can recall', the Leyden Jars on a Whimshurst machine (such as we were allowed to use at School in the 60s - but no longer) could kill you on a bad (/good) day. That could give us a figure to go on.

Thanks for clarifications and bunch of information there.
Bold: I agree. In-fact I had that in mind all along, but kept on using it due to lack of any better way. Either way, I was mostly doing order-of-magnitude estimates, so I thought it could work.
Green: :) . I tried searching around, but all I came was graphs of Time-Currents separating the safe-side from dangerous sides, but not that exact relation.

Thanks again.

So, I hope we can conclude that, 'No, you should practically never get concerned about getting electric shock from rain drops from the lines, but be aware that a falling bird (or anything larger) might be dangerous, both mechanically and electrically. :) "

 

[sophiecentaur]

I think that's a pretty safe conclusion. I think the 'authorities' agree with us or they'd have fenced-off areas below all power lines!
It is satisfying to do the sums, however, even if only approximate.

 

[0xDEADBEEF]

Just a quick calculation:

The threshold for feeling a current seems to be around 1 mA. The current flowing through the body for DC current is:

$$I=\frac{C_{\text{droplet}}\cdot R \cdot A \cdot U}{V}$$

Where $$C_{\text{droplet}}$$ is the capacity of a rain drop. R is the rate of the rainfall in m/s. A is the exposed area in $$m^2$$, U the Voltage and V the volume of a drop.

If we use drops of 1mm (it gets more dangerous the smaller the drops are because the number of drops goes up fast). Heavy rainfall of 0.1m/h. U = 135kV. An area of $$1m^2$$ I get a current of $$I=1.5\cdot 10^{-2}mA$$. Alas you won't feel a thing. And this is assuming that you really manage to get the whole rain that hits you charged, in reality the powerline is much slimmer than your body and most of the rain will not be charged. The current is usually AC, so drops tend to cancel each others charge. Oh and if you calculate something where you get a deadly current of something like .1A. Imagine this goes for every meter of wire. A short powerline of 10km would have losses of 133 Megawatt! That is a fifth of a small powerplants output. I think they would turn off the line if that ever happened.

 

[dlgoff]

Not in the rain, but some interesting information on Approach Distance from the Occupational Safety and Health Administration.

 

[Pablo Verdugo]

A pretty big water drop drops from the transmission line

Is this possible?
I would think that at that voltage, it's not possible to get even near the line because of the electric field.