Equipotential Bonding Explanation

[KeenPensioner]

Thank you all again - maybe I'm not asking the right question so I've attempted to draw what I mean, bearing in mind all the advice that has been given regarding equipotential bonding, ADS activation times etc. The image tries to show a fault in a kettle where the casing becomes live and develops a 230V potential as does the tap due to the bonding (am I correct in that?). The fundamental question for me is if stick man touches both kettle and tap at the same time as the fault develops and before the MCB/RCD trips, will he get a shock?

 

[Devonchris]

No he won't get a shock. Yes the tap will become live.

The effect of bonding is too make all the exposed metal work one big conductor, with little resistance, flowing to earth, the path of least resistance. So any two parts of it will be at the same potential.

Its the same as putting your test leads on two seperate points of the same live conductor. The meter will not register a voltage as it won't register a difference.

 

[sinewove]

Am I missing something,or was the requirement to bond sinks and steel tables dropped in an earlier edition for safety reasons in cases of contact with a live appliance and the sink?

 

[Pete E]

No he won't get a shock. Yes the tap will become live.

In that diagram, wouldn't the voltage from the kettle case effectively fall to zero if it has a direct low resistance path to earth?

 

[Richard Burns]

Just to confuse further I have done another diagram to show touch voltages with and without bonding.
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.

 

[static zap]

Main Earth Conductor ... now MET - Main earth terminal !

 

[ELECNEWT]

Just to confuse further I have done another diagram to show touch voltages with and without bonding.
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

Richard, what a superb and comprehensive illustration!!

If you can spare the time, could you calculate the current distribution for your third diagram to demonstrate why the resistance of the main bonding conductor, whether it is 0.05 ohms or 5 ohms, has no effect on the touch voltage?

 

[Midwest]

Richard, what a superb and comprehensive illustration!!

Agree, just need somebody to explain it to me now

 

[KeenPensioner]

Just to confuse further I have done another diagram to show touch voltages with and without bonding.
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

Richard, thanks very much, I understand the maths behind this now. I've spent ages on this and eventually the penny has dropped thanks to all the responses here but in particular your excellent diagram. One point (a typo or me?)...for a resistance of 1000 ohms and a touch voltage of 57.5 volts is the current not 57.5 mA?

 

[Richard Burns]

Quick sketch (which i never worked out before!) to show that there is minimal volt drop over the bonding conductor because the current through the bonding conductor is practically zero compared to the fault current on the cpc.
The extraneous part is not really involved in the current path at all.
(Better check the maths though!)

Richard, thanks very much, I understand the maths behind this now. I've spent ages on this and eventually the penny has dropped thanks to all the responses here but in particular your excellent diagram. One point (a typo or me?)...for a resistance of 1000 ohms and a touch voltage of 57.5 volts is the current not 57.5 mA?

"Do not do calculations late at night without thinking", it would be 57.5mA so an "ouch that hurt" shock and muscle contraction after 0.4s. Below 50V touch voltage the shock should be perceptible without adverse effects for the duration of the fault (0.4s)
However these are just random resistance numbers used for convenience, real life may be very different.

 

[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.