Help with Electrical design unit Level 3

I've done most of the stuff but something doesn't add up to me at the adiabatic equation. 

Question 5

Regulation 543.1.3 of BS7671 for Fridge circuit 2

Wired in 70°C single core t/p in pvc conduit

Ambient temperature 35°C 

protective device BSEN60898 16A Type C

Length of run is 9.15m

Quoted fault at pool/cafe 3.7ka

R1+R2 = 14.82x1.06x1.2x9.15÷1000= 0.17Ω

Zdb value = 0.11Ω

 Zdb+ (r1+r2) = 0.11+0.17= 0.28Ω

Zs= 0.28

Ipf = Vo÷0.28 —> 230÷0.28=822A

Adiabatic equation

S=√I2T÷K

S=√8222x0.1÷115= 2.26mm2

The cross-sectional area for the fridge circuit 2 is 2.26mm2 which complies with the regulation 543.1.3 of BS7671.

Hi I am also trainee like you but There are something that I dont understand in your calculations .

you stated R1+R2 = 14.82x1.06x1.2x9.15÷1000= 0.17Ω   could you explain where did you get 1.06 from??? as this circuit is  70  degree so I am assuming its incorportated so I3 would be 1.20  or if not incorportated than it would be 1.04??

Furthermore I found max R1+R2 = Zs max-Ze/ I3*Length multiply 1000  which gave us 115.17 so I went to I1 table Onsite guide and choose 2.5/1  which is 25.51

using this in  Zs=Ze+(R1+R2*Length*1.2/1000) gives us      0.11+0.28  =0.39

IpF=  230/0.39     =  589.7

Adiabetic equation   
S=√I2T÷K

S=√589.7^2*0.1/115

=1.62 mm

so the next available you can choose is 2.5 CPC