Voltage drop

[HacksawParticipant @hacksaw]

Kinda daft question .. I read somewhere years ago ,and i don't think it was just theory , if you had a long length of 2.5 twin&earth type cable with a plug on it and plugged in , at some point, somewhat less that a mile long , there would be no juice at all at the other end ? ? So how did the Transatlantic cable work then ?

[HacksawParticipant @hacksaw]

[Nick_GParticipant @nick_g]

.

Voltage drop is linked to how much load (amps) there is being drawn.

If you put 240 volts into a length of 2.5 a mile long I for one would not be grabbing hold of the other end.

If you had the same cable and put a load of just a few amps onto it the voltage would collapse. – There is a formula for working out voltage drop per amp for size of cable vs distance.

No idea how the transatlantic cable worked as there is losses in the conductors themselves.

Nick

[JAParticipant @ja]

Transatlantic cables did not work until the great Lord Kelvin invented the mirror galvanometer which responds to microamps.

Bed time!

JA

Edited By JA on 23/02/2017 23:26:51

[John RuddParticipant @johnrudd16576]

Generally speaking,

The cable conductors have resistance, ergo if the cable has a resistance of 1 ohm for every metre length and carries a current of 1 amp, the volt drop is 1 volt…..

Cable manufacturers supply this sort of data….the resistance is also a function of cross sectional area, the greater the area the lower the resistance….

Transmission losses are minimised by using high voltages….I'm referring to power transmission where electricity is distributed over a network….

Edited By John Rudd on 24/02/2017 01:00:55

[not done it yetParticipant @notdoneityet]

Transmission cables have power (energy) losses according to he standard laws of physics. The power loss is calculated as I^2R, where I represents current in Amperes and R represents the resistance in Ohms.

The energy loss is as heat, measured in Joules.

Likewise, the maximm current is calculated according to Ohm's Law where V=IR, where V is measured in Volts.

From the second calculation the current carrying ability of any conducyor is reduced to one half for each doubling in length. Halving the current means reducing the energy generated (lost) by a factor of four. That is for a short circuit situation, although the same voltage would be available anywhere along the line if the current were zero.

There is always some 'juice' at the end of any cable – it may just be so small as to be economically useless.

The simple analogy for electricity flow is a water pipe, where voltage, and current are equivalent to pressure and water flow respectively. Flow at a open tap (without pipe attached – so zero length) is far greater than at the flow though a very long length of pipe.

For any power transmission line (electricity or water) high pressure and low resistance to flow (a large diameter conductor) and low flow rate are required to minimise the losses within the length of the item. For transatlantic cables, the current is infinitesimally small, hence the requirement of a galvonometer to measure same. But voltage changes are more easily detected and measured, fed to an amplifier and converted to a useful signal – hence the first transmissions were of voltage and zero voltage (on or off signal) like morse code signals.

Separately, for high power transmission such as the international interconnectors for grid power, to other parts of Europe per eg, large diameter (so low resistance) conductors and very high voltages are required. They are made more efficient by transmitting direct current at high voltage. This is because with direct current the voltage is constant (100 times every second the voltage would be zero if transmitting alternating current). Furthermore, the peak voltage for any sinewave alternating current powerline would need to be 1.41 times that of a direct current line of the same dimensions and material.

[Michael GilliganParticipant @michaelgilligan61133]

As hinted by JA …

Insert the word 'Telegraph' into the name 'Transatlantic Cable'

It was a remarkable achievement; but intended for signals, not power transmission.

MichaelG.

.

https://en.m.wikipedia.org/wiki/Transatlantic_telegraph_cable

Edited By Michael Gilligan on 24/02/2017 08:04:53

[John HaineParticipant @johnhaine32865]

Posted by not done it yet on 24/02/2017 07:41:23:

Transmission cables have power (energy) losses according to he standard laws of physics. The power loss is calculated as I^2R, where I represents current in Amperes and R represents the resistance in Ohms.

The energy loss is as heat, measured in Joules.

Actually I^2 x R has dimensions of Watts, not Joules.

…………

Separately, for high power transmission such as the international interconnectors for grid power, to other parts of Europe per eg, large diameter (so low resistance) conductors and very high voltages are required. They are made more efficient by transmitting direct current at high voltage. This is because with direct current the voltage is constant (100 times every second the voltage would be zero if transmitting alternating current). Furthermore, the peak voltage for any sinewave alternating current powerline would need to be 1.41 times that of a direct current line of the same dimensions and material.

From what I recall reading at the time the channel power link was built, they used DC because it meant they didn't have to synchronise the AC grids on either side of the channel. After all, the cross-channel distance is comparable with a typical AC grid spans; and in most of Europe the length across the borders is zero. Using DC imposes extra conversion losses but those are probably less than or comparable with the AC/AC conversion technology at the time. Whether that's still true with static conversion I don't know.

Sorry, my inner pedant at work… It's interesting to note that they have to take care that the magnetic field produced doesn't affect ships' compasses by using a bifilar cable.

[KWILParticipant @kwil]

Going back to the OP.

If you have a ring main, there is a length to a distant double socket at which you reach the maximum allowable voltage drop with 2x13A being drawn at that point. I would have to think hard (or look it up). I have one such ring in my house, because of the physical length of the building.

[MuzzerParticipant @muzzer]

An engineer would reach over from his armchair and do some quick sums.

For a 1 mile (1600m) length of 2.5mm2 copper, the resistance would be about 11 Ohms per wire. So about 22 Ohms for the return path. With 240V, you'd be seeing around 10A if you shorted the ends together. Total dissipation around 2.4kW over the length. I hope I've got the sums right…

The maximum power you could dissipate would be when the load resistance was the same as the cable resistance ie current of 5.5A and voltage of 120V >> 600W. Not something you'd put your tongue across, then – or only once if you did. You could run US 120V equipment on the face of it but the load regulation would be appalling and the no-load voltage would still be 240V.

If it were immersed in the sea, the current would be slightly higher due to the lower temperature and resistance of the copper.

Murray

[Clive IndiaParticipant @cliveindia]

[Edit – Sorry Muzzer, we were posting at the same time, but with a slightly different approach….]

Hacksaw, as the guys are saying, it depends on the current you take. You need to go back to your school days and Ohms Law.
A 2.5mm area copper cable which is a mile long has a resistance of about 11 ohms. This gives 22 ohms cable loss because the current flows there and back.
The normal rating 2.5mm cable is 10Amps so if we passed 10A down the twin cable 1 mile long, the voltage drop across the cable would be 10 x 22 = 220V.
So if we put 240v into the cable at 10A, at the other end we would have 20V at 10A = 200W to use, although at the source we had 240 x 10 = 2400W.

If, like the communication cable, you take a very small current, there will be less voltage drop across the cable per mile and there will still be a useable voltage at the other end.
All of the above is simplified, since a cable is a complex impedance, not just a simple resistance but not a bad approximation – and your initial statement is near enough correct.

Edited By Clive India on 24/02/2017 10:22:39

[Ian PParticipant @ianp]

Presumably there would be slightly different results (Murray's and Clive's) depending on whether the voltage was AC or DC.

If it was AC the capacitance between the two wires (or to surrounding water if submerged) would act as a load so theoretically over an extreme length the cable resistance combined with the capacitance might reduce the voltage at the far end to zero.

Ian P

[TobyParticipant @toby]

and if you want to know the ratings of different cables a quick google search found this:

**LINK**

The voltage drop in those tables is per Amps per meter but that is basically the same as Ohms per meter if you want to use ohms law and relate it to Clive and Muzzer's posts.

[MuzzerParticipant @muzzer]

Ian – unlikely unless you have a VERY long cable. To get any appreciable AC effect you'd need to get somewhere close to the wavelength at 50Hz. Even a 1/4 wavelength at 50Hz would be about 12,000 km. The inductance will be pretty low, as the conductors are close together. Not quite interwound but close all the same.

Murray

[MuzzerParticipant @muzzer]

Double post. Finger trouble!

Edited By Muzzer on 24/02/2017 13:10:04

[Ian PParticipant @ianp]

Murray, my conjecture was really in the same vein as Hacksaw's original question. Although to be fair to him his inquiry related to 2.5mm cable and mains voltages whereas I was rather more vague. In theory though, with a very long (but not infinite) cable length there would be nothing at the far end.

Another 'old wives tale' type of thing is that I remember being told by a teacher as school, that it was possible to cut glass with scissors underwater. I thought then, and still do now, that its not possible. In the back of my mind though I wonder whether maybe at great depth, with extreme water pressure, the physics change and glass can be cut?

Ian P

[Martin KyteParticipant @martinkyte99762]

Posted by Muzzer on 24/02/2017 13:09:40:

Ian – unlikely unless you have a VERY long cable. To get any appreciable AC effect you'd need to get somewhere close to the wavelength at 50Hz. Even a 1/4 wavelength at 50Hz would be about 12,000 km. The inductance will be pretty low, as the conductors are close together. Not quite interwound but close all the same.

Murray

There are two effects relating to AC apart from the IR losses.

Firstly the cables radiate and lose energy. We have all heard the mains hum on radio 4.

Secondly for very long cable you can get reflections from the load(s) which set up standing waves. Site your tap off point at a node and you see zero voltage. The cable length does have to be extremely long as in many thousands of kilometers and at does depend on the termination. (End load).

I suspect that the original poster is remembering the schoolboy stories of power lines in Russia. This scenario is unlikely as you would no doubt build your power stations a lot closer to where you need the power.

[HacksawParticipant @hacksaw]

Ah, the 1 mile of 2.5 cable , zero output , must have been fake news then. If it was 1 mile of 1mm cable , could i put my tongue on that ?

[Martin KyteParticipant @martinkyte99762]

Now if if was copper pipe a few years back there would have been so many pinholes in the mile that you really would not have got any water out of the end.

Martin

[KWILParticipant @kwil]

Martin,

I remember that problem, imported from Greece if I remember rightly.

[Brian OldfordParticipant @brianoldford70365]

Posted by Martin Kyte on 24/02/2017 14:33:20:. . . . . .

Secondly for very long cable you can get reflections from the load(s) which set up standing waves. Site your tap off point at a node and you see zero voltage. The cable length does have to be extremely long as in many thousands of kilometers and at does depend on the termination. (End load).

. . . . . ..

The wavelength of 50Hz (in free space) is 6000kms

[Martin KyteParticipant @martinkyte99762]

mmm. but the propagation velocity depends on the dielectric constant so it's slower for 2.5mm twin and earth.

Martin

[Russell EberhardtParticipant @russelleberhardt48058]

Posted by Martin Kyte on 24/02/2017 15:09:37:

mmm. but the propagation velocity depends on the dielectric constant so it's slower for 2.5mm twin and earth.

Martin

Yes, the propagation velocity will be about 60% of that in free space so the wavelength at 50 Hz will be about 10,000 km.

Russell

[Brian OldfordParticipant @brianoldford70365]

Posted by Russell Eberhardt on 24/02/2017 16:38:51:

Yes, the propagation velocity will be about 60% of that in free space so the wavelength at 50 Hz will be about 10,000 km.

Russell

I'm afraid that is incorrect. The cable length will be shorter by the velocity factor i.e.approximately 3600km.

Regards

Brian (Retired Telecommunications Professional)

Edited By Brian Oldford on 24/02/2017 17:45:53

[Geoff TheasbyParticipant @geofftheasby]

Haha! Why am I thinking of Cerenkov radiation? Anyway, coaxial stubs, filters, Hi-Q breaks, on 144 & 432 MHz

Geoff

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