This may not be an issue in newer bathrooms that comply with current editions of BS7671 as supplementary bonding in bathrooms may be omitted when regulations 701.415.4.5 (iv), (v) and (vi) are met.
However, supplementary bonding the radiator has always been a grey area especially in older bathrooms because the radiator may or may not be an extraneous-conductive part. This is because in certain conditions it may not introduce an earth potential into the environment (for example with plastic heating pipes).
How can we check if a radiator (or anything) is an extraneous-conductive part?
A radiator may be insulated from earth due to the building/construction materials used. To check the effectiveness of this insulation the following formula can be used…
Rcp > (Uo/Ib) – Zt
Where…
Rcp = Resistance Ω of the radiator to the Main Earth Terminal (MET)
Uo = Nominal Voltage to Earth (230V)
Ib = Max value of current through human body, 30mA (lower values of 0.5mA and 10mA can be used)
Zt = Impedance of human body usually 1000Ω
So the minimum resistance between the radiator and the MET of the building is
(230/0.03) – 1000 = 6666Ω or 6.67kΩ
Anything less than this and the radiator can be considered to be an extraneous-conductive part and as such if it also doesn’t comply with 701.415.4.5 (iv), (v) and (vi) then supplementary bonding will be required.
How to check if existing supplementary bonding is effective
To check the effectiveness of the connection of extraneous-conductive parts to earth in a bathroom have a look in Part 7 at the NOTE at the bottom of regulation 701.415.2
This sends you to regulation 415.2.2
R ≤ 50/Ia
If there are RCDs protecting the bathroom then Ia would be 0.03 (30mA) so R ≤ 50/0.03 = 1667 ohms
If it’s an older bathroom and there are no RCDs then R ≤ 50 divided by the current needed to disconnect the protective device (fuse or circuit breaker) within 5 seconds. See the current / time disconnection graphs in Appendix 3.
So if its a bathroom with two circuits protected by Type B, BSEN 60898 circuit breakers, Lights 6A and Shower 40A then the amount of current needed to disconnect the devices is 5 x their rating (see page 525) so Ia would = 30A or 200A ‘you always use the highest’. So in this case R ≤ 50/200 = 0.25 ohms.
Thank you to BS7671 and Guidance Note 8 and hours of fun.