Equipotential Bonding Explanation

[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.