Assessment of general characteristics
• Before any detailed planning can be
carried out, it is necessary to assess
the characteristics of the proposed
plan.
-New one
-An extension to an existing
-Rewiring an existing building
1. Purpose and Intended use : use of the building
and type of construction will indicate the type of
equipment to be installed.
2.Environmental conditions: Environmental
conditions include: ambient temperature,
altitude, presence of water, dust, corrosion,
lighting and wind hazards. With large
commercial building fire rescues must be dealt.
3.The maximum current demand: It is necessary
to estimate the maximum current demand.
4.Diversity may be taken into account.
• Diversity: The electricity utilized in any
residential building is never hundred
percent of the installed capacity. There
may be some electrical points, which
remain unused even during peak load
periods (when there is high demand of
power).
DF = Utilization capacity
Installed capacity
• Some electrical points in the building are
installed keeping future requirements in
view and some points are used only during
special occasions.
•
However it is applicable for domestic
installation only.
• Some commercial building or business
establishments definitely use 100 % of the
installed capacity during peak load hours
but not for 24 hours .
• The diversity factor is not applicable in the case
of power wring in
➢industries where hundred percent of installed
motors are used for the whole day.
➢For street lighting
➢ Power wiring for agricultural sector
• The application of diversity has advantages:
reducing size of conductors and associated
protective devices
Electrical Points Installed Utilization
Two lamps points(incandescent)
One may not be
used
One fluorescent tube
May be used
Two fan points
One may not be
used
Two 5-ampere socket outlet
One may not be
used
One 15-ampere socket outlet
May or may not be
used
• The DF for various loads are given below:
Type of load
estimate
1.Lighting Circuits
2.Heating loads
2.1. Water Heaters
2.2. Ovens/stoves
2.3. Electric Iron
3.Motor Loads
4.Office equipment
5.General purpose SOs
DF
0.7-0.9
0.2-0.3
0.2
0.3
0.7-0.9
0.3-0.5
0.2-0.5
5.Maintainability: consideration must be
given the frequency and to the quality of
maintenance that the installation can
reasonably be.
• The designer must ensure that periodical
inspection testing and Maintenance can
readily be carried out and the effectiveness
of the protective measures and the reliability
of the equipment are appropriate to the
intended life of installation.
3 Procedures of designing
3.1 sub-division and number of ckt
any installation needs to be divided into
number of ckt
➢Handily and safety
➢For tracing any fault location
3.2 Design current (Ib or IL)
It is the rating of the electrical apparatuses
to be operated it is estimated by the
manufacturer or it has been calculated.
Ib= P/V…………..single phase
Ib=P/√3*V ……….three phase
If there is a capacitive or inductive load
power factor can be produce, and an
equipment has moving parts allowance
must be made for effective (efficiency or
eff).
Ib = P*100/V*pf*eff…………..1Ф
Ib =P*100/√3*VL*pf*eff………3Ф
3.3 Rating of protective device (In)
Having determined Ib we must select the
nominal setting of protection In such that
In>Ib this value may be taken from the IEE
regulation.
Determination of rating of main switch
The current rating of the main switch is
decided as per total current of the circuit (Ib).
3.4 Current carrying capacity
Iz =In/relevant cfs
correction factors(cfs)
Ambiant temperature(ca)……. surrounding
temperature of the equipment and cable
>30 ºc the correction factor needed.
Grouping Factor (cg)…….the same size
cables are grouped together they impart
heat to each other.
Protection by fuse (cf)…….if the fuse type
is BS 3036 it is high fusing factor
In≤0.725Iz hence 0.725 is cf
Thermal Insulation (ci)…….if a cable is
totally surrounded by thermal insulation for
0.5m a factor of 0.5.
3.5 Choice of cable size
Having established the current carrying
capacity Iz to be used then choose a cable
to suit that value.
Voltage drop
The resistance of a conductor increase as the
length increases ,so the cable voltage(Vc)
should not be so excessive no more than 4% of
the nominal voltage
Vc = mv x Ib x L (mv
from the table)
vc =
(220 x 0.04) = 8.8v
vc…….. Voltage drop on the cable is not more
than 8.8v