Introduction to Electrical Power and Power Electronics

171Power Cable

Solution:

The line current in the motor lines is derived from

Motorinput kVA

3ph

=

×

×

=×

20 000 0 746

097095

31

,.

..

11× I

L

,

which gives I

L

= 850 A.

For 11-kV operating voltage, the next standard insulation class is 15 kV.

Therefore, we use Table 6.13. The heaviest cable in the table is 535 kcmil with

ampacity of 449 A in trays. Therefore, we must use two cables in parallel, each

carrying 1/2 × 850 = 425 A, which is less than the cable ampacity of 449 A.

For a 535-kcmil cable, Table 6.13 gives V

drop

of 0.077 V/A per 1000 ft. For a

200-ft. cable carrying 425 A, V

drop

= 0.077 × 425 × (200 ÷ 1000) = 6.545 V/ph.

The motor line-to-neutral voltage ...