Cable selection, calculating earth fault loop impedance

[Stu]

[sorry didn't see page 2]

Hello again, doing this question too.

Step 1

Ref Method C clipped direct

Find Minimum size cable that will comply with shock.

Zs = Ze + ((R1+R2) x mf x L x 0.001

Zs = 0.85 + ((R1+R2) x 1.04 x 15 x 0.001

Zs of fuse must be greater than results from circuit.

Step 2

Ib = P/Uo = 14000W/230V = 60.87A

Using Diversity Table A1 OSG Cooker no Socket

Ib = ((60.87 - 10) x 0.3) + 10
Ib = 25.261A

In = BS 3036 type fuse
In = 30A (from table 41.2 with Zs of 1.09ohms

Therefore corrected cable size resistance value must 1.09ohms - 0.85ohms = 0.24ohms

Zs = 0.85 + ((R1+R2) x 0.156

Using Algerbraic Triangle to find 0.24ohms

0.24 = (R1+R2) x 0.156

0.24/0.156 = (R1+R2) = 15.38ohms

Therefore cables of values 15.38 or lower will comply with shock protection.

Table I1 shows Line conductor of 2.5mm2 with Protective conductor 2.5mm2 has a value of 14.82 and will comply

#####

Hoping this is OK

 

[Richard Burns]

In step one you have calculated the milliohms per metre value of the required cable not the Zs of the circuit and have not corrected the measured Ze to account for temperature (although the question may say this was measured at operating temperature)

In step 2 the 15.38 would therefore be milliohms per metre not ohms.