Handbook of Electrical Engineering

CABLES, WIRES AND CABLE INSTALLATION PRACTICES 211

EF=IX sin Ø

DF=IX cos Ø

AC=AB+BC=AB+EF=IR cos Ø+IX sin Ø (9.1)

DC=DF−CF=DF−BE=IX cos Ø−IR sin Ø

Vs=OD=

√

(OA+AB+BC)^2 +(DF−BE)^2

Unless the cable is exceptionally long the bracketed terms can be compared as:-

(OA+AB+BC)^2 (DF−BE)^2 ( 9. 2 )

Therefore the right-hand bracket can be ignored, and:-

VsOA+AB+BC

=Vr+IR cos Ø+IX sin Ø volts/phase

The ‘volt-drop’V is normally considered as a per-unit or percentage quantity with respect
to the sending end line-to-line voltage V, therefore:-

V

√

3 I(RcosØ+XsinØ) 100

V

% ( 9. 3 )

which is the equation often quoted in cable data publications.

NoteR=rlandX=xl

Where,ris the specific resistance andxis specific reactance in ohm/km or m ohm/m andlis
the route length in km.

9.4.3.1.1 Worked example

A 120 mm^2 3-core XLPE insulated cable 150 m in length feeds a 110 kW induction motor that has
a starting current of 6.5 times the full-load current of 180 amps. The starting power factor is 0.35
lagging. The sending end line-to-line voltage is 400 volts. The specific resistancerand reactance
xfor the cable are 0.197 and 0.072 ohm/km respectively at 90◦C and 50 Hz. Find the percentage
volt-drop on starting the motor.

The cable.

The series impedance is:-

R=rl=

0. 197 × 150

1000

= 0 .0296 ohms/phase

X=xl=

0. 072 × 150

1000

= 0 .0108 ohms/phase

Note that for low voltage cablesRis greater thanXuntil the size is in the order of 300 mm^2.