Answer Posted / dpl

A cable said to adequately sized when all the following
conditions aer satisfied:
1. Current carrying for steady state operation
2. Short circuit withstand capacity
3. Voltage Drop (steady state as well as during motor
starting)

As cable data provided by manufacturers/standards (ERA
report 69.30 etc) is for certain test conditions, which may
be very different from actual site conditions where the
cable is being installed.

Hence, to obtain the actual current carrying capacity of
the cable during steady state operation of the instalation
coditions of the cable shall be considered and suitable
derating factors (derating factor for group of cables,
Ambient temperature correction factor, derating factor for
thermal resistivity of soil, derating for depth of
installation) shall be multiplied to the manufacturer
supplied data. tehse values could be supplied by the
manufacturer or u can refer standards such as ERA 69.30 or
IEC.

The short circuit withstand level of the cable must be
greater than the energy let through value of the protective
device.
i.e. k2S2 >I2t where,
k = cable core insulation constant (143 for
XLPE)
S = Cable cross sectional area in mm2
I = Prospective fault current, rms in A
t = Disconnection time of protective device (in
seconds)

The voltage drop in A.C. cables shall not be more than 5%
based on the continuous maximum current loading and rated
voltage. During motor starting or re-acceleration the
transient voltage dip at the terminals of any motor will
not exceed 18-20% of the rated equipment voltage.

A cable of length L (km) having a resistance of RC (ohm per
km) and a reactance XC (ohm per km) operating in a three
phase ac system carrying a balanced current I (Amps) with a
power factor of Cos Φ will develop a voltage drop ΔV
(Volts):
1. For three phase ac systems
ΔV = √3 x I x L x (RC.Cos Φ + XC.Sin Φ)
2. During starting of 3 phase motors ( with a starting
current of IST at a power factor of Cos Φ) the voltage drop
will be:
ΔV = √3 x IST x L x (RC.Cos ΦST + XC.Sin
ΦST)

3. For single phase ac systems:
ΔV = 2 x I x L x (RC.Cos Φ + XC.Sin Φ)
4. For DC systems:
ΔV = 2 x I x L x RC
The percentage volt drop relative to the rated voltage is
given by:
Vd = (ΔV x VL) /100 %

Also, one may need to size the cable for earth loop
impedance considerations.

WTA used cb?

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