Page 227 - 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

2
V s = OD = (OA + AB + BC) + (DF − BE) 2
Unless the cable is exceptionally long the bracketed terms can be compared as:-

2
(OA + AB + BC) (DF − BE) 2 (9.2)
Therefore the right-hand bracket can be ignored, and:-

V s OA + AB + BC

= V r + 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:-
√
3I(Rcos Ø + Xsin Ø)100
V % (9.3)
V
which is the equation often quoted in cable data publications.

Note R = rl and X = xl

Where, r is the specific resistance and x is specific reactance in ohm/km or m ohm/m and l is
the route length in km.

9.4.3.1.1 Worked example
2
A 120 mm 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 resistance r and reactance
◦
x for 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:-

0.197 × 150
R = rl = = 0.0296 ohms/phase
1000
0.072 × 150
X = xl = = 0.0108 ohms/phase
1000
2
Note that for low voltage cables R is greater than X until the size is in the order of 300 mm .