Over current vs volt drop.

[Monkeyblaine]

I'm having another one of my crazy thought processes that I'd like help clearing up please. This example is completely hyperthetical so please give some consideration for that. If for example a circuit was designed very badly and a length of 1.5mm t+e was used for supplying a 8kw shower and the circuit was protected by a 40A breaker. Obviously this is ridiculous but I'm trying to understand what's going on in the conductors. The cable will be massively overloaded due to the load and in result will become very hot, maybe even melt. Now what I want to understand is, is the over current causing the 230v pressure to struggle with pushing such a large load down a small conductor and in result the voltage drops even more resulting in more current being pulled trying to take its place (ohms law) so is this a vicious cycle as even more voltage is dropped with the increase of current trying to compensate??? Also as low voltages are less of a threat to us in shock terms, how come when there's a massive volt drop and current increase the cable has such incredible thermal effects and melts? When at that voltage probably wouldn't shock us?

 

[D Skelton]

The 1.5mm cable wouldn't struggle at all with 35A, the copper would deal with that just fine, it is the insulation that would eventually break down due to the extra heat generated. As you know from how voltage and current behave when resistors are in series configuration, all the heat in the copper will do is cause the resistance to go up very slightly in the cable making more voltage drop across that so when it reaches the load, there potentially isn't enough to cope. This takes into account the fact that every load is at its most efficient running at the voltage it is designed to be used at.

In theory, if you fed a 8kW shower in bare 1.5mm (let's say a length of the supply cable is 1.5mm bare cable placed out of reach) you wouldn't have a problem at all.

Regarding volt drop and electric shock, if you drop the voltage enough then you won't experience a shock, simples. Let's say for example you wire a 4 core 95mm cable straight in to a 200A supply, if you then run that cable out 20 miles and touch the live ends you aint gonna get a shock. It is the potential difference that allows current to flow, and if that voltage isn't there to 'push' that current through your body so to speak, then you won't get a shock. The reason you never get a shock from touching bare neutral conductors is because the circuit has dropped all of its voltage by the time you touch the neutral. Take the neutral conductor out of its termination however and use your body to bridge the gap, then you essentially become another resistor in that circuit for the voltage to drop across. So long as the voltage dropped across the previous resistor allows for 50V or more to be dropped across you, then the current that circuit is drawing will have enough of a potential difference to allow it to flow through you.

 

[ackbarthestar]

I'll start and then come back to you after my breakfast.....

You can get a shock off of a pp9 volt battery if placed in the right place on your body. Its unlikely to cause heart fibrillation though.
The recognised voltage as the maximum as far as safety goes is 50V ac or 120v DC

 

[telectrix]

Let's say for example you wire a 4 core 95mm cable straight in to a 200A supply, if you then run that cable out 20 miles and touch the live ends you aint gonna get a shock.

yes you will, as until you touch the cable there is no current flow, hence no volt drop.

the 95mm conductor will have a resistance of approx. 2.4kohms for 20 miles. take your body as 1 k ohm and the current flow will be V/R or 230/3400. 68mA. enough to kill.

 

[D Skelton]

A paradox, because the resistance of that cable wouldn't allow for there to be a PD of more than 50V across you.

 

[Monkeyblaine]

Ok getting a little to I depth now. Haha. So voltdrop will only occur when. Current is flowing and it is caused due to insufficiently sized cables with their extra resistance which in turn makes things hotter and more VD?

 

[ackbarthestar]

Let's say for example you wire a 4 core 95mm cable straight in to a 200A supply, if you then run that cable out 20 miles and touch the live ends you aint gonna get a shock.

yes you will, as until you touch the cable there is no current flow, hence no volt drop.

the 95mm conductor will have a resistance of approx. 2.4kohms for 20 miles. take your body as 1 k ohm and the current flow will be V/R or 230/3400. 68mA. enough to kill.

You would be looking at approximately 62V at 68mA with a heat consupmtion in the conductors of 40kJ over an hour across 20 miles of cable. So not much heat per meter then ...

 

[Darkwood]

The 1.5mm cable wouldn't struggle at all with 35A, the copper would deal with that just fine, it is the insulation that would eventually break down due to the extra heat generated. As you know from how voltage and current behave when resistors are in series configuration, all the heat in the copper will do is cause the resistance to go up very slightly in the cable making more voltage drop across that so when it reaches the load, there potentially isn't enough to cope. This takes into account the fact that every load is at its most efficient running at the voltage it is designed to be used at.

In theory, if you fed a 8kW shower in bare 1.5mm (let's say a length of the supply cable is 1.5mm bare cable placed out of reach) you wouldn't have a problem at all.

Regarding volt drop and electric shock, if you drop the voltage enough then you won't experience a shock, simples. Let's say for example you wire a 4 core 95mm cable straight in to a 200A supply, if you then run that cable out 20 miles and touch the live ends you aint gonna get a shock. It is the potential difference that allows current to flow, and if that voltage isn't there to 'push' that current through your body so to speak, then you won't get a shock. The reason you never get a shock from touching bare neutral conductors is because the circuit has dropped all of its voltage by the time you touch the neutral. Take the neutral conductor out of its termination however and use your body to bridge the gap, then you essentially become another resistor in that circuit for the voltage to drop across. So long as the voltage dropped across the previous resistor allows for 50V or more to be dropped across you, then the current that circuit is drawing will have enough of a potential difference to allow it to flow through you.

Think you need to re-educate yourself here.... the reason you dont get a shock from N is because N is sourced from the centre tap of the sub-station transformer as is the earth and either seperated cut-out tncs or the sub station tns .... thus a potential to earth shouldn't exist to N.

 

[D Skelton]

And the reason it has zero potential is because the live circuit it is acting as the return path for has dropped its voltage across it's resistance. Come between the neutral conductor and the load and you will get a belt.

Edit: I don't think I'm explaining it very well, but I am well aware of the fact a neutral has zero potential because of its relationship with earth, what I'm trying to say is a neutral conductor is there for when a load has shed its voltage, seperate and then come between a neutral conductor and one side ceases to be come. Neutral as you become a series resistor in that circuit.

I'm groggy and hung over to buggery! Gimme a break

 

[Darkwood]

And the reason it has zero potential is because the live circuit it is acting as the return path for has dropped its voltage across it's resistance. Come between the neutral conductor and the load and you will get a belt.

I understand what you are trying to explain but you did it badly.. remember N is also a live conductor as well with 230v potential to any phase as well just not to earth the exact reason you get a shock off N if its broken and you either bridge the path back to N or E shows this potential is there. Try to look at neutral as having 230v too but not to earth.
What i was getting at is the reason you dont get a shock off N is not because of the volts drop across the load but because it has no potential difference to earth which is usually the bridge your body makes if you touch N...... it was just wrong and misleading to express it the way you did and yes i know hwat you are trying to say.

 

[D Skelton]

Yes, see my edited post, you must have missed it whilst replying

 

[Outspoken]

Gents, the type of cable is also critical here. Solid conductors have lower CCC that multi-stranded cables because electrical flow is across the surface of a conductor and not through the conductor, thus a 95mm² SWA with say 10 strands has a lower capacity than similar insulated conductor with say 100 strands of finer wire. This is why Tri-rated cable has a higher CCC than a given SWA (although the heat tolerance capacity of the insulation is a minor factor too).

About to leave for work so will address the physics behind this when I am in and have the time.

 

[D Skelton]

Indeed. Welding cable would be a case in point!

 

[Outspoken]

Let's say for example you wire a 4 core 95mm cable straight in to a 200A supply, if you then run that cable out 20 miles and touch the live ends you aint gonna get a shock.

yes you will, as until you touch the cable there is no current flow, hence no volt drop.

the 95mm conductor will have a resistance of approx. 2.4kohms for 20 miles. take your body as 1 k ohm and the current flow will be V/R or 230/3400. 68mA. enough to kill.

Thats strictly correct, the resistance of the cable is wrong, however the rest is spot on.

The volt drop for a 95mm² is 0.41mV/A/m, so assuming the the switch is off and no load connected to the cable the volt drop is 0V;

0.41 x 0 x 32,200 / 1000 = 0V

Where:

0.41 = mV/A/m
0 = current drawn
32,200 = distance in meters

However the average person has a body resistance in the region of 1500Ώ so then we have;

0.41 * 0.153 * 32,200 / 1000 = 2.0199V, thus the volt drop would have fallen only an insignificant amount and the shock would still be potentially fatal depending on the individual and the circumstances of the shock..

However, if you increase the voltage to 400V then you get;

0.41 * 0.2666 * 32200 / 1000 = 3.511V, again there would still be sufficient voltage and energy present to potentially kill

 

[Darkwood]

Where's the local means of isolation?

 

[D Skelton]

Let's say for example you wire a 4 core 95mm cable straight in to a 200A supply, if you then run that cable out 20 miles and touch the live ends you aint gonna get a shock.

Didn't really think that one through did I *facepalm*. Like I said earlier, I woke up this morning groggy and hungover and hadn't really engaged brain before replying. I stand by the rest of what I say and regarding this part, I was purely trying to make a point about voltage drop and distance and made it rather badly it would seem.

Ok let's recalculate, run 1mm T+E for 125 miles, plug one end in to 230V and bridge the two conductors with ya face at t'other end. No shock!

Edit: Assuming bodily resistance of 2kohms

 

[Outspoken]

Where's the local means of isolation?

Ussain Bolt needs a Job after the Olympic you know...