Page 237 - Handbook of Electrical Engineering
P. 237

CABLES, WIRES AND CABLE INSTALLATION PRACTICES     221

           9.4.3.3.1 Worked example

           A 132 kW induction motor is connected to a 415 V 3-phase 50 Hz supply. The motor has the
           following performance data,
           • Running power factor cos Ø r = 0.88
           • Running efficiency η = 95.2%
           • Starting current = 7.0 × running current
           • Starting power factor cos Ø s = 0.35 lagging

                 Find the smallest cable size and its maximum route length to satisfy a running volt-drop of 5%
           and a starting volt-drop of 15%. Choose an XLPE insulated 600/1000 V 3-core cable to be routed
                                             ◦
           in air. The ambient air temperature is 25 C.

           Solution:
           Running conditions:-
                 The full-load current I fl of the motor is:-
                                                  Rated power
                                I fl = √
                                      3 Line voltage × Power factor × Efficiency
                                            132000
                                  = √                      = 219 amps
                                      3 × 415 × 0.88 × 0.952
                                                                                        2
                 From Table 9.17 the smallest cable size for a running current of 219 amps is 70 mm which
           from Table 9.16 has a specific resistance r of 0.342 ohm/km and a specific reactance x at 50 Hz of
           0.0754 ohm/km.

                 From (9.5) the route length l fl full-load consideration is,
                                                   415 × 5.0
                             l fl =   √
                                  100 3 × 219 × (0.342 × 0.88 + 0.0754 × 0.4750)
                               = 0.1624 km = 162 metres.
           Starting conditions:
                 The starting current I st of the motor is:-

                                    I st = 7.0 × I fl = 7.0 × 219 = 1533 amps

           From (9.5) the route length l st is,
                                                   415 × 15.0
                             l st =  √
                                 100 3 × 1533 × (0.342 × 0.35 + 0.0754 × 0.9368)
                               = 0.1229 km = 123 metres.
                 It can be seen that starting current determines the route length to be no greater than 123 metres.
           If the actual route length is longer than 123 metres then a larger size of cable must be used. If the
           starting volt-drop is allowed to be 20% then the route length l st would be 164 metres, and the running
           current would determine the maximum route length to be 162 metres.
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