Albert, here we go, get your thinking hat on fella and put your brain in gear..bit unfair on a Friday night, but you asked the question
You have a circuit wired in 2.5mm2 T&E, it is 20m in length and protected by a BS88-2.1 16A HRC Fuse. The Regs state this protective device must operate in 0.4s in order to comply.
From Table 9A (OSG-P158) we see that 2.5/1.5 has a R1+R2/meter value of 19.51mOhms per meter.
So, 19.51 x 20 / 1000 = 0.39 Ohms.
In your original post you state that Zs = 0.32 Ohms, I am now questioning where this assumption came from, as you can see R1+R2 is greater than that.
Now I will assume that what you actually meant was Ze, which would mean that Ze = 0.32 Ohms.
That being the case Zs = Ze+(R1+R2) = 0.32 = 0.39 = 0.71 Ohms
Therefore the Earth Fault Loop Impedance of this circuit is 0.71 Ohms.
Now from Table 41B1 (BS7671 -P45) section (a) we see that in order to comply with Regulation 413-02-09 and 413-02-10 the maximum permissable Earth Fault Loop Impedance for the Overload protective device in this circuit (16A BS88-2.1) must not exceed 2.82 Ohms.
Therefore as 0.71 is LESS that 2.82, this circuit will comply with the Regulations.
If you wish to then discover the fault current Ipz = Uo/Zs = 230 / 0.71 = 323.94A.
Then from graph Fig 3.3b (BS7671-P198) we see that this fault current will operate the protective device in less that 0.01 seconds, well within the 0.4 seconds stated in the Regs. In fact the graph indicates that the protective device will only require a Fault current of 85A in order to comply with the 0.4 second rule.
Please don't tell you don't understand that!!