Adding an earth rod IN ADDITION to a TN system

[HappyHippyDad]

I have just watched one of John Ward's interesting videos -

It all makes sense apart from one of the final points involving adding an earth rod to an existing TN system. Not making a TT, but using the rod in addition to the TN system. This part is at 20 minutes and 8 seconds if you want to skip to it.

In relation to a broken PEN conductor he mentions that a suitably low Ra would be required for this rod to be of use. He mentions 10Ω's or less. He then says it may need to be 4Ω's or less if you have a substantial load on the system (i.e EVCP). I realise this is so a dangerous voltage is not reached but I cannot see how that relates to his figures of 10Ω and 4Ω.

For example. Lets say we have 20A (not even a substantial load) flowing throw the system (with a PEN fault) and we have a rod with an Ra of 10Ω

Using V = IR

V = 20 x 10 = 200V

I realise I must be using ohms law in the wrong context here as if the load was 40A the equation gives 400V which is impossible on a single phase system. Confusing!

 

[davesparks]

For example. Lets say we have 20A (not even a substantial load) flowing throw the system (with a PEN fault) and we have a rod with an Ra of 10Ω

Using V = IR

V = 20 x 10 = 200V

I realise I must be using ohms law in the wrong context here as if the load was 40A the equation gives 400V which is impossible on a single phase system. Confusing!

That 20A load isn't constant, it is itself the product of the resistance and voltage.

This is a big over-simplification but will hopefully help make some sense of it.

If 20A is flowing under normal conditions then think about that as being an 11.5 ohm load (assuming perfect 230V supply and 0 ohms supply impedance etc etc)

If you then introduce the PEN fault you are effectively connecting that 11.5 ohm load in series with the resistance Ra giving you a simple circuit of two resistances in series.

For simple maths we can call Ra 11.5 ohms so we have two equal resistances in series.
This would give us a combined resistance of 23ohms, so the current that flows will be 10A. Then the voltage drop across the load resistance will be 115 volts and the voltage drop across Ra will be 115 volts. This gives us a voltage of 115V at the MET.

As I said this is a very simplified explanation and in reality things will be somewhat different.
A number of loads, appliances etc, will probably just shut down as a result of the sudden reduction in voltage, some SMPSUs will attempt to compensate for the reduced voltage by drawing more current. So ode expect the voltage at the MET to vary quite a bit as everything tries to carry on working.

I'd also expect the sustained flow of current through the earth rod to potentially cause the Ra to vary thus upsetting the whole thing further.

 

[HappyHippyDad]

That 20A load isn't constant, it is itself the product of the resistance and voltage.

This is a big over-simplification but will hopefully help make some sense of it.

If 20A is flowing under normal conditions then think about that as being an 11.5 ohm load (assuming perfect 230V supply and 0 ohms supply impedance etc etc)

If you then introduce the PEN fault you are effectively connecting that 11.5 ohm load in series with the resistance Ra giving you a simple circuit of two resistances in series.

For simple maths we can call Ra 11.5 ohms so we have two equal resistances in series.
This would give us a combined resistance of 23ohms, so the current that flows will be 10A. Then the voltage drop across the load resistance will be 115 volts and the voltage drop across Ra will be 115 volts. This gives us a voltage of 115V at the MET.

As I said this is a very simplified explanation and in reality things will be somewhat different.
A number of loads, appliances etc, will probably just shut down as a result of the sudden reduction in voltage, some SMPSUs will attempt to compensate for the reduced voltage by drawing more current. So ode expect the voltage at the MET to vary quite a bit as everything tries to carry on working.

I'd also expect the sustained flow of current through the earth rod to potentially cause the Ra to vary thus upsetting the whole thing further.

Thankyou for the example Dave.

I understand where the figures come from until you mention the VD across the load resistance and VD across Ra. Where did the 115v come from? Is it 115V for both because they are both at an equal resistance, i.e 11.5ohms?

Could you do the same for an example where the resistances are different, then I can see where the maths come from....

Lets say we have 40A flowing (normal conditions), giving us an 5.75ohm load.

Lets say we have an Ra of 10ohms.

combined resistance = 10 + 5.75 = 15.75ohms

Current that flows = 230V / 15.75 = 14.6A (to 1 dp)

How do we use this figure (i.e 14,6A) to get to the VD figures?

Perhaps this figure isn't used and it is just the ratio between the 2 resistances? If so.....

We had 5.75Ω and10Ω. Giving us 83.98V and 146.02V (equalling 230V), but this is really just guesswork and I don't know which of those voltages applies to load resistance, Ra or the MET. I would assume the MET must be the 146.02V, but I really am clutching at straws here.

 

[Lucien Nunes]

How do we use this figure (i.e 14,6A) to get to the VD figures?

Multiply it by the resistances involved to get the voltages across each.

Voltage from MET to true earth = I x Ra = 14.6 x 10 -=146A
Voltage across the load = I x 5.75 = 84V.

Your observation that the voltage divides in the same proportions as the resistances is correct. If Ra is twice the load resistance, then the touch voltage will be twice the voltage across the load. This is the nature of a voltage divider.

However, as noted by Davesparks, many loads do not behave as resistances, and to know the actual division of voltage one needs to understand their behaviour, which may be complex and non-linear. Especially complex are electronic and electonically controlled loads that might start up, apply some load that pulls the L-N voltage down, shut down in response to the falling voltage, and then reboot, creating a series of load spikes. Connect ten of those all competing with each other for the limited supply and the voltage to earth at the MET will be chaotic. Some of the largest domestic loads are heating elements, which are true resistances and will behave according to the rules of simple voltage dividers whilst they are energised, although increasingly their electronic controls will react in their own unpredictable ways to the unstable L-N voltage.

Your original premise is also valid i.e. the resistance of typical electrodes is at least an order of magnitude too high to offer worthwhile protection against dangerous touch voltages in the event of an open PEN. I am not sure why JW picked those specific figures, which would produce dangerous touch voltages at typical loads on electrical systems. I think he was trying to make a point conceptually and the numbers didn't carry any particular significance. The value of distributed earth rods is safety in numbers, not being able to protect each individual system from disconnection from the PEN.

 

[davesparks]

Thankyou for the example Dave.

I understand where the figures come from until you mention the VD across the load resistance and VD across Ra. Where did the 115v come from? Is it 115V for both because they are both at an equal resistance, i.e 11.5ohms?

Could you do the same for an example where the resistances are different, then I can see where the maths come from....

Lets say we have 40A flowing (normal conditions), giving us an 5.75ohm load.

Lets say we have an Ra of 10ohms.

combined resistance = 10 + 5.75 = 15.75ohms

Current that flows = 230V / 15.75 = 14.6A (to 1 dp)

How do we use this figure (i.e 14,6A) to get to the VD figures?

Perhaps this figure isn't used and it is just the ratio between the 2 resistances? If so.....

We had 5.75Ω and10Ω. Giving us 83.98V and 146.02V (equalling 230V), but this is really just guesswork and I don't know which of those voltages applies to load resistance, Ra or the MET. I would assume the MET must be the 146.02V, but I really am clutching at straws here.

Sorry, I didn't explain that too well, I used the same number to make things easy.

I'm simplifying this down to treating it as a simple circuit of two resistors in series connected to a perfect power supply.

With resistances in series the total restance is the sum of the resistances, if we call the resistance of the load Rl then that would be R=Rl+Ra, where R is the total resistance.
With resistances in series the current is simply I=V/R.
The voltage drops across each resistance proportionally, so the voltage drop across the load is Vl=I*Rl and the voltage drop across Ra is Va=I*Ra.

Hopefully this makes sense?

The voltage can also be worked out based on the ratio of the resistances yes, both methods should yield the same result.

So for you example of a 40A load yes it would be 5.75 ohms, total resistance 15.75 ohms and current of 14.6A

So voltage drop across the load will be 14.6*5.75=84V (rounded) and the voltage drop across Ra will be 14.6*10 = 146V.

Thus the MET will be at 148V above earth (assuming earth is the 0V reference of the supply etc etc)

 

[HappyHippyDad]

Understood. Thankyou both. Also, thanks for emphasising that the voltage will not be behaving as lovely and neatly as the maths presumes (what a shame it's not that easy).

 

[davesparks]

Understood. Thankyou both. Also, thanks for emphasising that the voltage will not be behaving as lovely and neatly as the maths presumes (what a shame it's not that easy).

It gets even more fun if youve got a three phase supply and you loose the neutral.

In that case you get a floating star point which will sit at some voltage between the three phases relative to the loads connected.
If the loads are perfectly balanced the star point will float at exactly the zero point between the phases, hence 3 phase motors or heaters don't necessarily need a neutral connection.

If the loads aren't balanced, for example a lot of single phase circuits connected to a DB which loses the neutral from the submain, it gets fun. At least one of the phases will end up rising well above 230V until some of the connected load explodes/catches, then the star point shifts further so more loads die etc etc until there's a nice big insurance claim!

I know of one electrician who forgot to replace the neutral test link in a fused switch feeding a TPN DB and wrote off a brand new IT suite in a university.

 

[HappyHippyDad]

I suppose really we need to assume maximum demand. lets say 60A (unlikely I know, but I guess you have to use it when the outcome may be death).

Also we want to limit the touch voltage to 50V.

Given these 2 known figures we can say...

We have a 3.83Ω load (230V / 60A).
We have to a have a VD of 180V across load in order to have 50V on MET, therefore we can see the current that flows must be 47A (47A x 3.83Ω = 180V)

Therefore...

230V / (Ra + 3.83) = 47A

rearrange the equation

Ra = (230 / 47) - 3.83 = 1.1Ω

Blimey, that's never going to happen. Still it's been a fun Sunday morning

Edit.. Of course, most of the time you will have a load <60A, meaning a rod with a higher Ra will still lower the touch voltage compared to no rod at all, so perhaps it is worthwhile, as long as you do get a fairly low Ra.

So, after all that, the highly mathematical and precise answer is to get a rod with a 'fairly' low Ra

 

[timhoward]

You might find an old thread where I pretty much exactly followed the same learning curve interesting:

Calculating broken PEN touch voltage with backup earth electrode

My brain has been exercised a bit by this. Am I on the right lines with this extremely simplified example: If we take: -TNCS installation with a Ze of 0.2 ohms. -Plastic services, nothing bonded, keeping it simple -Line conductor resistance from supply transformer should be similar to PEFC and...

www.electriciansforums.net

This was long before A2, and I arrived at conclusion that an additional earth electrode on a PME system would be pretty useless most of the time in open-PEN conditions. And now it's in A2....go figure! (I was told DNO pressure rather than electrical engineering advice led to it)

 

[davesparks]

I suppose really we need to assume maximum demand. lets say 60A (unlikely I know, but I guess you have to use it when the outcome may be death).

Also we want to limit the touch voltage to 50V.

Given these 2 known figures we can say...

We have a 3.83Ω load (230V / 60A).
We have to a have a VD of 180V across load in order to have 50V on MET, therefore we can see the current that flows must be 47A (47A x 3.83Ω = 180V)

Therefore...

230V / (Ra + 3.83) = 47A

rearrange the equation

Ra = (230 / 47) - 3.83 = 1.1Ω

Blimey, that's never going to happen. Still it's been a fun Sunday morning

Edit.. Of course, most of the time you will have a load <60A, meaning a rod with a higher Ra will still lower the touch voltage compared to no rod at all, so perhaps it is worthwhile, as long as you do get a fairly low Ra.

So, after all that, the highly mathematical and precise answer is to get a rod with a 'fairly' low Ra

Ready for the next layer of complexity? So far we have assumed we have a single installation with no bonding.

Add in some parallel paths from bonding of any extraneous parts, these all effectively act as earth rods.

Then consider that there may be more than one installation on the downstream side of the fault, so there you have more than one earth electrode in parallel.

And the DNO side of things will have various earth rods along the length of the distributing main.

 

[davesparks]

This was long before A2, and I arrived at conclusion that an additional earth electrode on a PME system would be pretty useless most of the time in open-PEN conditions. And now it's in A2....go figure! (I was told DNO pressure rather than electrical engineering advice led to it)

I think most other countries that use TNCS supplies have had this requirement for a long time now. Its my opinion that we are behind the times on this and should have been doing it from day 1.

By now it could have easily become normal practice to make a connection to the rebar in the foundations of every new build and large extension.

Yes a single additional earth rod may be quite useless, but if every installation connected to a PME distributing main has an additional earth electrode it could make a difference.

 

[Baddegg]

I love expanding my knowledge with these types of threads….but you havnt half given me an headache on a Sunday HHD!

 

[davesparks]

I love expanding my knowledge with these types of threads….but you havnt half given me an headache on a Sunday HHD!

It's given me a little relief from a very boring job

 

[HappyHippyDad]

You might find an old thread where I pretty much exactly followed the same learning curve interesting:

Calculating broken PEN touch voltage with backup earth electrode

My brain has been exercised a bit by this. Am I on the right lines with this extremely simplified example: If we take: -TNCS installation with a Ze of 0.2 ohms. -Plastic services, nothing bonded, keeping it simple -Line conductor resistance from supply transformer should be similar to PEFC and...

www.electriciansforums.net

This was long before A2, and I arrived at conclusion that an additional earth electrode on a PME system would be pretty useless most of the time in open-PEN conditions. And now it's in A2....go figure! (I was told DNO pressure rather than electrical engineering advice led to it)

Haha... we do go around in circles on this forum don't we. Oh well, a little something new comes up with each similar thread.

 

[HappyHippyDad]

I love expanding my knowledge with these types of threads….but you havnt half given me an headache on a Sunday HHD!

I love the maths. I just wish my physics was even half as good.

I did get a bit carried away! I've got yet another cracked rib from battling with the youngsters (trying jiu jitsu now), so am at a loose end.

It's @davesparks fault for edging me forward with more and more complicated scenarios

 

[HappyHippyDad]

Ready for the next layer of complexity? So far we have assumed we have a single installation with no bonding.

Add in some parallel paths from bonding of any extraneous parts, these all effectively act as earth rods.

Then consider that there may be more than one installation on the downstream side of the fault, so there you have more than one earth electrode in parallel.

And the DNO side of things will have various earth rods along the length of the distributing main.

Gosh Dave. Too many variables there for me to plug into a neat equation

 

[Baddegg]

I love the maths. I just wish my physics was even half as good.

I did get a bit carried away! I've got yet another cracked rib from battling with the youngsters (trying jiu jitsu now), so am at a loose end.

It's @davesparks fault for edging me forward with more and more complicated scenarios

Yeah I’m the same I just believe in the total immersion theory….I just keep reading and trying even though I’m out of my depth and slowly but surely things make sense …..

 

[HappyHippyDad]

It's given me a little relief from a very boring job

Well I'm glad you're back on the forum Dave, and a big thankyou for taking the time to answer all my questions.

 

[timhoward]

Haha... we do go around in circles on this forum don't we. Oh well, a little something new comes up with each similar thread.

I just love that there are clever people around who are willing to help.

Add in some parallel paths from bonding of any extraneous parts, these all effectively act as earth rods.

A TT house I was looking at yesterday was giving a Zs of 0.43 ohms at a socket.
It turned out to be via copper gas pipes and the PME supply next door (there was external meter box, they were out, so I had a sneaky look!)

Yes a single additional earth rod may be quite useless, but if every installation connected to a PME distributing main has an additional earth electrode it could make a difference.

I agree, but I'd like to see it rolled out in a more coordinated and considered way. Depending on where a supplier N breaks, and where the DNO's own electrodes are you could end up with several properties diverted neutral current going via the one house where an enlightened sparks fitted the extra rod.

 

[davesparks]

I agree, but I'd like to see it rolled out in a more coordinated and considered way. Depending on where a supplier N breaks, and where the DNO's own electrodes are you could end up with several properties diverted neutral current going via the one house where an enlightened sparks fitted the extra rod.

Is there a more coordinated way to roll it out?

Unfortunately it has to start somewhere and yes such situations will occur but in the fullness of time they will sort themselves sout