Volume 8 – Mechanical and Electrical Services - Malaysia Geoportal

Chapter 7 ELECTRICAL SERVICES
The regulation stipulates that there must be no more than a 4 % voltage drop of the nominal voltage
from the Main Switch Board to any point in the installation. Thus for a 240V system, the maximum
voltage drop permissible is 9.6 volts while 16.6 volts is permitted on a 415V system.
Table 4D, 4E & 4F etc of the IEE Regulations for the Electrical Equipment of Buildings give current
ratings and voltage drop characteristics of various conductors. The voltage drop can be calculated
from the formula:-
Voltage = Ic x Vd x l (7.5)
1000
Where Ic = current normally carried by the conductor
Vd = mV drop per A per M
L = length of cable in M
It is pertinent to point out that the cable rating must always be higher than that of the circuit
breaker that is supposed to protect that part of the installation.
It is standard practice to use PVC/PVC wires for concealed (not in conduit) wiring. For wiring in
conduit PVC wires are used. The widely used underground cables are:
PVC/SWA/PVC cable (Polyvinyl chloride/ steel wire armoured / Polyviny chloride cable). This is used
with cable gland terminations and is available both in the 3 phase 4 core variety as well as in the
single phase 2 core variety. It is generally used when the single phase current demand is less than
60A.
An immersion heater rated at 240V, 3kW is to be installed using twin-with –earth PVC insulated and
sheathed cable. The feed will be from an existing 15A spare way Circuit breaker and will run for
much of its 14m length in a roof space which is thermally insulated with glass fibre. Ambient
temperature is expected to be 35 C. when leaving the consumer unit the cable will be bunched with
seven other twin-with–earth cables.
Solution:
Determine whether the 15A circuit breaker will be adequate.
I=P/V=3000/240=12.5A
The protection rating (15A), not the circuit rating, must be divided by each of
a) The group correction factor – from table B1 eight multi-core cable on a wall, the factor = 0.52
b) The ambient temp. factor - from Table 4C1 of this chapter the ambient temp. factor 0.94
c) The thermal insulation factor- since the cable can be cooled on one side this the thermal
insulation factor is 0.96.
The calculation then becomes:
Required Rating = 15 = 31.97A
0.52 x 0.94 x 0.96
The next step is to find from table 4D1A (Refer of IEE 16th edition). Since the cable concerned will
be clipped direct to surface, it can be seen that 6mm cable must be selected with a current rating of
43A fore one twin cable with protective conductor. For other of cable or conduct or the relevant
tables can be referred from the IEE 16 th Edition.
March 2009 7-7