Handbook of Electrical Engineering

240 HANDBOOK OF ELECTRICAL ENGINEERING

Table 9.34. Earth loop impedance results for the worked example with

braided armouring

Nominal conductor

area (mm^2 )

Za+Zc Zloopc

magnitude

(ohms)

Vshock

16 4. 916 +j 0. 0268 5.9193 227.49

25 5. 807 +j 0. 0250 6.8102 232.91

35 2. 221 +j 0. 0242 3.2244 229.31

50 2. 462 +j 0. 0236 3.4654 232.45

70 2. 791 +j 0. 0224 3.7944 235.03

Zmr= 1. 0 +j 0 .0 ohms

ZcandZaare given in c)

The resultingZloopccalculated from the circuit for each cable is given in Table 9.34.

Comparing the tabulated results above with those for the fuses in d) shows that all the cables

have an earth loop impedance much greater than that permitted by the fuse, by a ratio of approx-

imately 10:1.

Hence an earth leakage circuit breaker should be used in the MCC to protect the circuit against

electric shock hazard.

The most appropriate choice of cable cross-sectional area and fuse rating are,

• Cable cross-sectional area should be at least 50 mm^2 , to comply with volt-drop.


• Fuse rating should be below the rating of the cable since its primary purpose is to protect the
  cable. Hence the largest fuse should be 125 A for a 50 mm^2 cable. (If a larger fuse is needed
  the cable size would need to be increased.)

h) Calculate the electric shock voltage

From Figure 9.9 the shock voltageVshockis,

Vshock=

(Za+Zmr)Vph

Zsec+Zc+Za+Zmr

For the 16 mm^2 cable

Vshock=

4. 62 + 1. 0

5. 9193

(

415

√

3

)

= 227 .49 volts

i) Replace the braided armour with round steel wires.

Assume the resistances of the armour wires to be 0.72, 0.50, 0.46, 0.40 and 0.36 ohms for the

200 m route length. Repeat the calculations of g).