Equipotential Bonding Explanation

[ELECNEWT]

Thanks again Richard.
I had not taken account of the 500 ohm resistance connection to earth.
What is the calculation to derive proportion of fault current to 0.45 amps ?

 

[Pete E]

In the diagram with the example values the touch voltage would be 57.5V and for the resistance of a human body at 1000Ω the current through the body would be 5mA, only just perceptible, therefore someone would not feel a significant shock, perhaps tingling.
View attachment 40726

Thanks for that..I think I need to o back to TEC and refresh my Electrical Theory as I am struggling with the calculations in the second and third diagram ! lol I need to sit down with a pencil, paper and calculator and work through the maths myself....

 

[KeenPensioner]

Richard...thanks again for your patience but I don't get the maths this time. I understand parallel conductors but am I correct in this....where you draw the fault current path in the small dotted line (from just under the MET to earth) one route is 0.1 ohms for the "earth path" and the other (parallel?) path is 0.05 ohms ("bonding conductor") and 500 ohms "connection to earth. If the voltage drop is 57.5 V across this parallel circuit and you say there is a voltage drop of 0.02V over the "bonding conductor" does this not suggests a current of 0.4 amps flowing in that leg of the parallel circuit which would mean 574.6 amps flowing through the "earth path". If you can be bothered humouring me, I'd appreciate it.

 

[Richard Burns]

The calculation is not that precise, rough figures only.
The earth fault loop impedance is (0.1+0.1+0.1+0.1 Ω) = 0.4Ω, the 500Ω from any additional earth paths is ignored.
Ipf = V/R = V/Zs = 230/0.4 = 575A
If you were to calculate the parallel resistance of the extraneous part and the earth cable back to the transformer this would be ((500*0.1)/(500+0.1)=0.09998Ω which is effectively 0.1Ω to 1 dp.
Therefore in a fault 575A should flow through the circuit.
Because the extraneous Conductive part is by definition connected to earth with a resistance of less than 22,000Ω (assumed to be 500Ω here) there is a potential split of current at the MET. For a 230V supply this would then be 230/500=0.46A, because the actual voltage to earth at that point would be only 57.5V the current would actually be 0.11A but as a worst case look at 0.46A. The bonding conductor at 0.05Ω with a current of 0.46A would drop 0.02V, an insignificant amount compared to 57.5V.
The fall in current on the earth path would be 0.46A and so the earth path would take 574.54A, which is insignificantly different from 575A.

The diagram is there to show that any volt drop on the bonding conductor can be ignored for the purpose of calculating touch voltage as it would err on the side of caution. The comparative resistances are orders of magnitude different and so the worst case values are used ignoring minor variances from elsewhere. Voltages and resistances can change over time and so this is a text book calculation using specific figures.

Thanks for that..I think I need to o back to TEC and refresh my Electrical Theory as I am struggling with the calculations in the second and third diagram ! lol I need to sit down with a pencil, paper and calculator and work through the maths myself....

I have just done a quick calculation on the fly, I am not doing a full mathematical treatment so please check the maths and take account of any errors made, however the background information should allow you to understand the theory enough to make any needed corrections.

 

[ELECNEWT]

For a 230V supply this would then be 230/500=0.46A, because the actual voltage to earth at that point would be only 57.5V the current would actually be 0.11A but as a worst case look at 0.46A.

Thank you Professor Burns. I calculated 0.11A and I did not realise that you had instated 0.46A as the worst case scenario.
You must be a very patient individual to respond to such amateurs as I am.

 

[KeenPensioner]

Richard......I can't thank you enough for your comprehensive explanation. I now understand what bonding is all about and can back it up with the maths behind it. I echo ELECNEWT's sentiments above. I'm now moving on to supplementary bonding! Thanks again.

 

[johnduffell]

Maybe I'm missing something but I feel like people are mixing up ADS and EEB.
ADs is where the cases of class 1 equipment and other electric casings eg swa armour and steel conduit are all connected to the neutral at the substation (via the cpcs in the installation and on via the dnos CNE conductor, or separate CPC in tn-s or even the installation earth rod and the substation earthing.)
The purpose of ads is to remove the supply withing 0.4s or 5s of a fault, thus reducing the danger to occupants.

eeb is earthed equipotential bonding, and refers to the connection together of the main earth and any extraneous conductive parts. Those are anything likely to introduce a potential (usually thought of as earth potential, but could be other than earth in some situations with tn-c-s) from Outside the building.
So as you can see the fault inside the installation is handled by ads, and faults outside are handled by equipotential bonding.

Now I'd agree that the bonding will also reduce voltages between the water pipe and the kettle during those 0.4 seconds, but that is because the water pipe is introducing earth potential. But even so, it will be cleared within 0.4s so that is classed as sufficient for safety.

I hope that makes sense as I'm on my phone so it's basically impossible to edit. And thanks for the interesting discussion.

 

[KeenPensioner]

Richard.....I can't thank you enough for your comprehensive explanations. I "get it" after many hours (as an enthusiastic amateur) trying to figure it out. I now move on to supplementary bonding.....wish me luck. Thanks again.

 

[KeenPensioner]

Richard, thanks very much for comprehensive explanations.

 

[Richard Burns]

Thank you all for the acknowledgements, it does encourage me to respond again I am pleased that it has helped and hopefully it may help others to follow the reasoning.
I am glad that the explanation makes sense, but do always remember to check that I have it correct, I am just posting quickly on a forum and not writing a textbook.

Supplementary bonding should be straightforward now as it is just shortening the length of the cpc over which voltage is dropped so that the local area has even lower touch voltages.

 

[ELECNEWT]

I can understand that when supplementary equipotential bonding is applied, you can establish a bonding connection from very near to origin of the fault and so the voltage at the extraneous part (say a radiator) will be very close to the voltage at the fault.

Taking the example of a circuit with a 20 amp breaker the current to operate the breaker within 5 seconds is 60 amps. Textbooks state that if the resistance between exposed or extraneous conductive parts is 0.83 ohms or less then supplementary bonding is not required.

I understand that 50volts divided by 60 amps gives the resistance value of 0.83 ohms.

I have tried drawing a realistic circuit diagram to prove that 0.83 ohms is safe, but I just tie myself in knots.

Would any member be willing to find the time to draw such a diagram (similar to Richard’s hint hint) showing the position of the 0.83 ohms and an alternative diagram using the value of say 5 ohms?

 

[ELECNEWT]

I can understand that when supplementary equipotential bonding is applied, you can establish a bonding connection from very near to origin of the fault and so the voltage at the extraneous part (say a radiator pipe) will be very close to the voltage at the fault.

Taking the example of a circuit with a 20 amp breaker the current to operate the breaker within 5 seconds is 60 amps. Textbooks state that if the resistance between exposed or extraneous conductive parts is 0.83 ohms or less then supplementary bonding is not required.

I understand that 50 volts divided by 60 amps gives the resistance value of 0.83 ohms.

I have tried drawing a realistic circuit diagram to prove that 0.83 ohms is safe, but I just tie myself in knots.

Would any member be willing to find the time to draw such a diagram showing the position of the 0.83 ohms and an alternative diagram using the value of say 5 ohms?

 

[Richard Burns]

From the previous information you should be able to piece this together.

They are saying that if 60A flows in the circuit and it is all flowing through the supplementary bonding (not very likely if there is a cpc) then the volt drop over that supplementary bonding would be 50V and so safe. This is erring on the side of caution (very very much so) but makes things easy to check.
For the calculation V = IR = 60*0.83 = 50V
Above 3.83Ω, in this case, the voltage drop would be the supply voltage and the current would be reduced below the 5s disconnection time by the resistance of the supplementary bonding. However current will flow through the cpc as well so there would be 2 resistors in parallel, but this is discounted for a safer approach.
V = 60* 3.83 = 230 V

The 60A is for a 20A rewireable fuse, a 20A type B breaker would be 100A.

 

[ELECNEWT]

Richard, thank you for that response.
I should have realised 60 amps related to a fuse.

Referring to what you said:
“They are saying that if 60A flows in the circuit and it is all flowing through the supplementary bonding (not very likely if there is a cpc) then the volt drop over that supplementary bonding would be 50V and so safe.”

Have I misunderstood the situation?
I thought that the stipulation was “if the resistance between exposed or extraneous conductive parts is 0.83 ohms or less then supplementary bonding is not required”.

Thanks for your patience.

 

[Richard Burns]

50V touch voltage is the level considered not dangerous and this is the level to which most of the calculations regarding bonding are aiming for.
If the resistance between an exposed conductive part and an extraneous conductive part is less than 0.83Ω (with this protective device) then the touch voltage will be less than 50V and so safe because it is as if supplementary bonding is in place.
If the resistance is greater than 0.83Ω then the touch voltage will be above 50V and so not safe therefore supplementary bonding is required to reduce the resistance between the two points.
I have worded my post in another way because I am thinking of supplementary bonding and its effectiveness; this is the situation described in guidance note 3 using the formula R<=50V*Ia to confirm the suitability of supplementary bonding.

 

[ELECNEWT]

Thank you Richard I understand now.
I have been trying to teach myself.
But as the saying goes "when you teach yourself you have a fool for a teacher".

 

[Richard Burns]

The problem with the values used for safety calculations by BS7671 is that they always aim to provide greater safety by ignoring minor (or sometimes not so minor) items that would affect the results.
Therefore doing a text book calculation may not yield the expected result.
I attach a diagram illustrating the supplementary bonding issue.

 

[ELECNEWT]

Thank you Richard.Another brilliant diagram.
It has helped me to understand fully.

If I can add another maybe naive observation:
even at 50V touch voltage, a current of 0.05A through the body (if you could not let go for five seconds) would be uncomfortable to say the least!

 

[Richard Burns]

It would be uncomfortable and possibly fatal but it is possible that at 50V the skin resistance is strong enough that it will not break down and cause a significant reduction in resistance and the consequent increase in current that may be experienced at the higher voltages that can jump gaps more easily, however at this point I am onto supposition and not definite information.
50V is considered a safe voltage but someone else would have to provide an accurate rationale for the base premise.

 

[ELECNEWT]

Thank you again Richard.
You have been an excellent teacher on this topic and a model of patience.