Touch voltage

Question

Hello I have a basic question on touch voltage. I just can't seem to grasp it.. This is taken from guidance note 5 I have attached 3 sketches alongside . 
 1 and 2 are as GN5 describes The third is my confusion. 1 & 2. We calculated touch voltage, in this case 126V without MPB and 94V with MPB Now I understand I think the basic concept the greater the resistance , the higher the voltage current being constant Number 3, is How I imagine it... We have a fault and at the Exposed CP we have a voltage of say 115V The resistance of the CPC to the MET will reduce this voltage again to say half again so 57.5V at the MET Since we have MPB connected to the MET . The MET and Extraneous part will be roughly at 57.5V So our touch voltage would be between the EXP and EXT conductive part so.115V ---- 57.5V A difference of 57.5V I don't understand in the GN5 examples, Why is touch voltage is one amount? 126 volts to what? To Earth? But since the MET will be at a raised potential It can't be to earth? Hope that makes sense.

lyledunn · Accepted Answer

The touch voltage calculated assumes the person subject to it is standing on a floor that is remote from earth, as would be the case inside most normal buildings. The value is the product of the fault current and the resistance of that part of the earth path from the fault to the DNO transformer when no main protective is in place or just the voltage drop along R2 when it is in place. In TN systems with low values of resistance in the earth path back to the transformer, the effect of main protective bonding will be relatively small in reducing touch voltage, but is crucial in TT systems where without MPB, the touch voltage could be close to the full fat system voltage to earth.

geoffsd · Answer

JohnE said: Im I in the right area? 
 More or less, except that - 
 the Zs will be reduced by the SB being fitted, and 
 not quite all of the fault current will flow through the SB; a (small) part of it will flow through R2. 
 So Ut = If (in SB) x 0.05

geoffsd · Answer

Also, with reference to the formula R&le;50/Ia as stated in the regulations: 
 It is not the resistance between exposed and extraneous-c-ps which is the relevant value for touch voltage but, as in the above examples, solely the resistance of the SB. The resistance of the extraneous-c-p between touch point and SB connection is irrelevant.

AJJewsbury · Answer

I used the 126.5V as a constant voltage? 
 
 Probably safer to use 230V (x1.1 if you like = 253V) across the entire loop - 126.5V is based on the assumption that the L and PE sides have the same impedance - which may be true if you have equal sized L and PE or PEN conductors - but having a supplementary bond undermines that assumption (as does using some common cable types, e.g. T&E which have a reduced c.p.c.s). 
 - Andy.

geoffsd · Answer

Nearly - 
 With a fault current of 281.1A flowing through the CPC of 0.5Ω and SB/pipe/MEB of 0.05Ω (say negligible for pipe and MEB) then 90% of the fault current will flow through the SB; i.e. 253A. 
 So Ut = 253 x 0.05 = 12.65V

gkenyon · Answer

JohnE said: Just on the Supplementary though.. 
 geoffsd said: So Ut = If (in SB) x 0.05 
 
 AJJewsbury said: the supplementary bond is in effect parallel with the c.p.c - which reduces the overall earth fault loop impedance, and so increases the fault current 
 No , I disagree with the 'supplementary bond is in effect parallel with the cpc' in all cases. The resistance cpc's of the circuits alone (R 2 of circuit 1 + R 2 of circuit 2) may be of sufficiently low resistance to a common DB earth terminal (or in a smaller installation, all the way back to MET) to meet the requirements of Regulation 415.2, especially if RCDs are used. And, see 544.2.1, protective conductors in sheathed cables, or in conduit or trunking, mean the minimum csa can be less than 4 sq mm; the only issue is if there's a disparity in cpc csa between the circuits ... you'd have to reinforce the smaller one with a conductor of csa not less than the largest of the cpc's. 
 Also, highlights the fact that 415.2 does not necessarily reduce touch-voltage . It 'reduces touch-voltage to 50 V AC or 120 V DC or less for currents that will not operate the relevant overcurrent protective device in 5 s, or alternatively the relevant RCD in either 0.3 s (for non time-delayed devices) or 0.5 s (for S time-delay devices) .' 
 If the fault current is more than I a , the touch-voltage may well be > (or even >>) 50 V AC or 120 V DC, but will be disconnected in a known time. 
 Look closely at the formulae in 415.2.2: 
 415.2.2 The resistance R between simultaneously accessible exposed-conductive-parts and extraneous-conductive-parts shall fulfil the following condition: 
 R &le; 50 V/I a in AC systems 
 R &le; 120 V/I a in DC systems 
 where I a is the operating current in amperes (A) of the protective device or: 
 (i) for RCDs, I &Delta;n 
 (ii) for overcurrent devices, the 5 s operating current . 
 Nothing in BS 7671 limits touch-voltages absolutely in all possible fault conditions, except SELV (and even then, caution if you choose certain modern electronic sources, because they don't all conform to 414.1.1 with respect to voltage limits in certain Sections in Part 7).

mapj1 · Answer

Hmm, well defined this way Ut - the 'touch voltage' is between the case of the faulty kit, and the exposed part - here the vertical pole at the right hand image edge. We seem to neglect any resistance and voltage drop between the point of touch on the pole and the MET and as only a modest current flows in the victim that is OK, They give us a voltage, 57volts, so we know the current in the R2 (half ohm). (have R have V, - get I. ) Now both the substation and the house have electrodes connecting to the soil around them with a non-zero impedance, so if current is flowing in the ground, then the MET will not be at zero volts relative to terra-firma earth potential - an electrode far away if you like- but that is not what is being touched. 
 Mike.

geoffsd · Answer

I think what you are missing is that - with bonding in place - you are using the pipe to touch two points on the same conductor; not a live point and earth.

AJJewsbury · Answer

We have a fault and at the Exposed CP we have a voltage of say 115V 
 
 OK (or using the example, which includes a factor for a possible increase in supply voltage of up to 10%, 126.5V). 
 
 The resistance of the CPC to the MET will reduce this voltage again to say half again so 57.5V at the MET 
 
 Not according to the resistances in the left hand diagram. With the system earth reference at the source, there's Ze (of 0.35&Omega; in this example, half in L and half in the PEN) to take into account - so with an overall loop resistance of 1.35&Omega; (0.35+0.5+0.5) and so 0.675&Omega; (0.175+0.5) between true Earth and the fault, the c.p.c. (between MET and exposed-conductive-part) accounts for 0.5&Omega; of that - so closer to three quarters rather than half. 
 
 But since the MET will be at a raised potential It can't be to earth? 
 
 Correct the MET will not be at true Earth potential (the 0V of the general mass of the earth) - so without bonding, the extraneous-conductive-part may remain at true Earth potential so you get the full 115V (or with Cmax 126V) between the faulty exposed-conductive-part (appliance) and the extraneous-conductive-part (pipe). 
 - Andy.

Touch voltage

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