The "symbols" do not align with BS 7671 and associated standards.

The first expression for fault current for a fault to cpc or exposed-conductive-part is incorrect, as AJJewsbury  says it normally accounts for the external source impedance. The expression is usually written in the UK as follows:

I_f = CU₀/(Z_e+R₁+R₂)

where:

I_f is the fault current

C is a correction factor to take into account the maximum or minimum voltage. We choose either C=C_max or C=C_min depending on what we are trying to calculate, to obtain worst-case conditions. For public LV supplies according to the ESQCR in the UK, C_max = 1.1 and C_min = 0.95. As an example, C_max would be used to obtain maximum prospective fault current, along with assessing touch-voltage (A722.3 in BS 7671), whereas C_min would be used when calculating acceptable earth fault loop impedance for protective devices used for automatic disconnection of supply (Reg 411.4.4 of BS 7671).

U₀ is the nominal voltage Line to to Earth.

Z_e is the external earth fault loop impedance (source impedance) outside the installation.

R₁ is the resistance of line conductor between the origin and the point of fault.

R₂ is the resistance of cpc between the origin and the point of fault.

Why is C_min not 0.94?

geoffsd said:

Why is C_min not 0.94?

It's not because it says so in BS 7671 (but it does) ... the initial place voltage factor was introduced is BS EN 60909, and in particular  BS EN 60909-0:2016 (Clause 5.3.1 and Table 1). The use of C_max of 1.05 for supplies with permissible upper variation of +6 % and C_min of 0.95 for supplies with  a permissible lower variation of - 6 % is given in the paragraph just before Table 1 in the standard:

If there are no national standards, it seems adequate to choose a voltage factor c according to Table 1, considering that the highest voltage in a normal (undisturbed) system does not differ, on average, by more than approximately +5 % (some LV systems) or +10 % (some HV systems) from the nominal system voltage U_n.

So, it's a "rule of thumb" based on a "near enough statistically" explanation.

The "rule of thumb" has been confirmed for use in PD IEC 50480:2011 and also for use in BS 7671 since BS 7671:2008+A3:2015.

You could use 0.94 in the UK for public supplies, I suppose, but this would be erring even more on the side of caution.

Thank you.

gkenyon said:

You could use 0.94 in the UK for public supplies, I suppose, but this would be erring even more on the side of caution.

Well, yes but it would at least be consistent with the pretend permitted voltage range of +10%/-6%.

It just raises the question of why then is  C_max not 1.09?

Well, yes but it would at least be consistent with the pretend permitted voltage range of +10%/-6%.

But they are different things in a way - for fault loops we're using the impedance of the entire loop - so the voltage is really that of the source (Uoc in old money) whereas the permitted voltage range is at the consumer's supply terminals, so has to include voltage drop in the supplier's lines. To use Ze and the voltage at the consumer's end would be to double-count the impedance of the supplier's lines (if under different conditions).

   - Andy.

Well, yes but it would at least be consistent with the pretend permitted voltage range of +10%/-6%.

But they are different things in a way - for fault loops we're using the impedance of the entire loop - so the voltage is really that of the source (Uoc in old money) whereas the permitted voltage range is at the consumer's supply terminals, so has to include voltage drop in the supplier's lines. To use Ze and the voltage at the consumer's end would be to double-count the impedance of the supplier's lines (if under different conditions).

   - Andy.