Page 256 - 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 Z a + Z c Z loopc V shock
2
area (mm ) magnitude
(ohms)
16 4.916 + j0.0268 5.9193 227.49
25 5.807 + j0.0250 6.8102 232.91
35 2.221 + j0.0242 3.2244 229.31
50 2.462 + j0.0236 3.4654 232.45
70 2.791 + j0.0224 3.7944 235.03

Z mr = 1.0 + j0.0 ohms
Z c and Z a are given in c)

The resulting Z loopc calculated 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,

2
• Cable cross-sectional area should be at least 50 mm , to comply with volt-drop.
• Fuse rating should be below the rating of the cable since its primary purpose is to protect the
2
cable. Hence the largest fuse should be 125 A for a 50 mm 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 voltage V shock is,

(Z a + Z mr )V ph
V shock =
Z sec + Z c + Z a + Z mr
2
For the 16 mm cable

4.62 + 1.0 415
V shock = √ = 227.49 volts
5.9193 3

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).