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CHAPTER
7
Practical Aspects of
Installation Design
7.1 Design Stages for Electrical Installations
In the previous chapters some basic requirements of Regulations related to
supply and distribution systems were explained. In this chapter practical aspects
of installation design to determine cable sizes and PD ratings for all types
circuits will be outlined. The order of the design follows the steps.
1. Establish details of circuits to determine levels of distribution (power,
main, sub-main, load levels) in the installation (STAGE 1).
2. Determine design current in each levels starting from final circuits
(STAGE 2).
3. If it is required, apply necessary corrections on design currents for
temperature, grouping etc (STAGE 3).
4. Determine current ratings of PDs and cable sizes in each level
according to type of overcurrent protection (STAGE 4)
5. Calculate voltage drop along the longest final circuit from intake
position (STAGE 5)
6. Determine earth fault impedance to check electrical conditions
Fundamentals of Electrical Installation
35
for safe operation against electrical shock (Chapter 5 and 8)
7. Check short circuit conditions for conductor temperature rise (Chapter 8)
7.2 Establish Details of Circuits to Determine Levels of Distribution
(Stage 1)
After deciding on divisions and subdivisions of the building and locations of
distribution boards (DBs) and sub-distribution boards (SBD), it is necessary to
divide installations into a number of circuits independent of its size.
Dividing circuits into several levels have several advantages.
(i) It provides
- safe operation and easy handing of cabling and PDs during
operation
- easy maintanence
(ii) It eliminates requirements for
- voltage drop
- earth fault impedance
(iii) It avoids hazardous situations arising during usage of installation.
Example cases for advantages achieved with leveling installations:
- In small installations, it is advisable to supply lighting circuits from
different PDs, such that in the case of tripping of one PD under fault
conditions eliminates total blackout. Also a separate ring circuit is installed in
kitchen apart from other ring circuits serving other parts of the premise.
- In larger commercial or industrial installations leveling of circuits must be
carefully planned to handle different types of loads such that in case of a fault,
number of loads to be interrupted is minimized. In large installations such as
workshops or factory halls, power and lighting need may be supplied from
different DBs. Such division (Fig. 7.1)
- In all installations, it is economical to distribute power considering load
density. This provides considerable cable and labour cost saving other than the
electrical advantages explained above.
Practical Aspects of Installation Design
(lighting)
SDB
DB
Main DB
Sub-main
supply cable
Final circuits
Intake from
secondary S/S
(power)
Supply cable
(for DB)
Final circuits
36
Main supply or
service cable
DB
DB
Power level
Main-distribution
level
Sub-distribution
level
Load level
Figure 7.1 Leveling an installation.
Leveling of circuits should be based on;
- type of load
- power rating of equipment
- distances of load or loads to the main DB.
A layout of leveling of circuits is shown in Fig. 7.1. Note that a cable
supplying a DB is named with the same as that of the DB, i.e., the cable
supplying a SDB is named as sub-main cable and like.
In a mult-floor office building design, for example, load level includes all
final circuits supplying small fixed and portable loads. These loads are fed from
the DB located in the DB or electrical room at every floor, and this DB
together with the others are supplied from busbar trunking system via tie-off.
The busbar is fed from the main distribution board at main distribution level.
7.3 Determine Design Current (Stage 2)
In order to decide on the type of supply intake it is necessary to estimate the
total power demand of the installation. The procedure starts with calculating
design current at;
Fundamentals of Electrical Installation
37
(i) load level for each final circuit
to determine cable size of DB supply cable,
(ii) sub-main distribution level from total power demand of all DBs to
determine cable sizes of sub-main cables,
(iii) main distribution level from total power demand of all sub-main DBs to
determine cable size of main distribution cables, and
(iv) power level from total power demand of all connected main DBs to
determine cable size of service cable and power rating of secondary
distribution transformer.
Determining design current of is although is a straightforward procedure,
sometimes it needs judgment of the designer. In domestic premises for
example, as the design current of the ring circuit, diversified demand for the
ring socket-outlet circuit, may be taken as half of current rating of PD,
however, in some circumstances, the power ratings of equipment likely to be
installed, may need to be taken as base values for the design.
Power Ratings of Electrical Apparatus
The power rating of an apparatus is commonly expressed as the amount of real
power required to perform a certain task. This rating is given in watts or
kilowatts. Associated with it usually voltage, frequency and efficiency are also
specified by the manufacturers. Ratings of some domestic appliances are given
in Table 7.1. Conductor sizes of cable and ratings of current-carrying
components are usually expressed in table forms in terms of their current
ratings. Cable sizes and ratings of current-carrying components such as PDs,
switches, socket-outlets, plugs etc., must be large enough to carry the maximum
current of the connected apparatus without overheating or being overstressed.
Therefore, current rating of a cable and any component should be selected
relatively larger than design current value in accordance with the design and
protection rules.
Practical Aspects of Installation Design
38
Table 7.1 Power ratings of some domestic appliances.
Appliance
Power rating (watts)
Lamps (filament)
15, 25, 40, 60, 75, 100, 150
Lamps (fluorescent)
15, 20, 30, 40, 50, 65, 75, 85
Water heaters
3,000 - 6,000
Washing machines
2,500 - 3,000
500 - 1,500
Refigerators
5,000 - 7,500
Cookers
100 - 3,000
Vacuum cleaners
1,500 - 2,000
Irons
EXAMPLE 7.1
A three-phase 22.5 kW, 415 V and 50 Hz motor load is supplied through 180 m long 3-core
copper cable. The efficiency of the motor is 87 percent at PF= 0.72 lag. Calculate design
current of the final circuit.
Solution. Input power to the motor is
22500
Pin 
 25860 W
0 .87
The design current becomes
Ib 
25860
 50 A
3 ( 415 )( 0.72 )
EXAMPLE 7.2
A three-phase DB supplies a workshop at 240 m distant. The connected motor loads are 80
kW at 0.7 PF lag and 25 kVA at 0.8 PF lag The voltage at the workshop is to be maintained
at 415 V.
Determine design current in each final circuit and of the main supply cable.
Solution. The design current for each load is;
Load 1:
Load 2:
I b1 
I b2 
P
3 VL  L cos 
25,000
3 (415)

80,000
3 (415)(0.7)
 34.7 A
 159.2 A
Fundamentals of Electrical Installation
39
where, VL-L is the line-to-line voltage at motor terminals. The total design current of the
supply cable to DB becomes
Ib1= Ib1 /-1+ Ib1 /--2=159.2 /-45.5+ 34.7 /-37.8
= 139.3/-134.4o
A
7.4 APPLY NECESSARY CORRECTIONS ON DESIGN CURRENTS
AND DETERMINE CABLE SIZES (Stage 3)
Determining Nominal Current Rating of Protective Devices
The nominal current rating of a protective device is based upon the designed
current for a given load as calculated above. The rating of the protective device
should be chosen close to this value. The Regulation demands that the nominal
current of the PD shall not be less than the designed current value of the
circuit. That is,
In  Ib
(7.1)
The rating of PD should be chosen as close as possible to this value. Tables
41B1, 41B2, 41C and 41D given in IEE Regulation provide data for selection of
standard current ratings for PDs, and are covered in Section .
EXAMPLE 7.3
A distribution board feeds six separate group of lamps used for external lighting lamps. Each
group consists of eight twin 85 W fluorescent lamps. Each circuit is to be protected by Type
1 to BS 3871 MCB. To account for the control gear losses the designed current should be
upgraded by a factor of 1.8. Select a suitable current rating for the MCB to protect the
circuit.
Solution. The total load connected to each lighting circuit for 8 twin 85 W fluorescent lamp
is:
P=8 (2) (85) = 1360 W
Including the control gear losses P becomes
P = 1.8 (1360) = 2448 W
Designed circuit current, Ib = 2448/240=10.2 A
Practical Aspects of Installation Design
40
From IEE Table 41B2, the nearest current rating for the MCB is 15 A.
Regulation requires that the effective current carrying capacity Iz of a cable
conductor shall not be less than the maximum sustained current normally
carried by the conductor, Ib. Thus,
(7.2)
Ib  Iz
Combining with the relation given for the nominal current rating, In, of a PD
(Eq. (7.1), we have
Ib  In  Iz
(7.3)
Factors Affecting Current-Carrying Capacity of Cable Conductor
Regulations provide current-carrying capacities of cables based on
(a) the conductor material
(b) the insulation material
However, the effective current-carrying capacity of a cable conductor is
modified by
(i) Ambient temperature, Ca
(ii) Grouping, Cg
(iii) Thermal insulation, Ci
(iv) Disposition (method of installation)
(v) Class of excess-current protection
The cable rating should be modified for each of these effects as applicable. That
is, when insulation conditions differ in more than one respect from those given
in the appropriate table, a separate factor for each special circumstance must be
applied. The correction factors upgrade the design current value, hence, appears
as divisor.
(i) Ambient temperature. The current ratings given in tables for cables are
based on an ambient temperature of 30oC. Where the ambient air temperature
exceeds this figure, such as in tropical countries or in a boiler house, the
appropriate correction factor Ca should be applied to derate the cable. The
necessary correction factors are given in IEE Tables 4C1 and 4C2.
If the ambient temperature will not be less than 25oC, the cable rating is
increased up to 1.04.
In order to account temperature rise due to direct solar radiation, 20oC
should be added to the ambient temperature and the appropriate correction
Fundamentals of Electrical Installation
41
factor used when determining the
current carrying capacity of the cable.
(ii) Grouping. If a number of cables run together in close proximity to one
another such as in a conduit or trunk, their current-carrying capacity is reduced.
This is because the heat developed during the normal operation is trapped in
the enclosure. The related correction factor Cg is applied from IEE Tables 4B1,
4B2 and 4B3. Correction factors given in dependent upon the type of
installation method used. If the cables are arranged so that there is a space of
one diameter between them, increased factors under the title "spaced" may be
applied, and if spacing of two diameters between cables can be achieved, no
grouping factor need to be used. See footnotes of Tables 4B1 and 4B2 for
corrections according to cable dispositions.
(iii) Thermal Insulation. If a cable run passes through thermal insulating
materials, such as those used in roof spaces and wall cavities, which are fully
covered or surrounded with thermal insulation, ability of the cable to radiate or
shed heat is reduced. Even the cable is in contact on one side with a thermally
conductive surface. As a result its temperature increases hence, it requires usage
of a larger size cable. To account this increase the cable rating must be
corrected by the factor Ci given in IEE Table 52A for a given length of the
insulating material. For example a derating factor of 0.55 applies if a part of the
cable 40 long Iis covered with a thermal insulation material.
h
Thermal insulating
material
Wall cavity
Cable
h
Wall cavity
Non-metallic
sheeth
Figure 7.2 Cases for which correction for thermal insulation apply.
(iv) Method of Installation. Method of installation is one of the factors
determining current carrying capacities of all single core (CU/PVC or
CU/XLPE) and multicore (PVC/PVC/SWA/PVC or XLPE/SWA/PVC)
Practical Aspects of Installation Design
42
armoured or non-armoured cables. Cables can be laid direct in the ground, in
ducts, or fixed to the surface on a cable tray, or fixed to the structure by cleats.
IEE Table 4A summarizes the existing methods applied in lying cables. The
installation methods are basically grouped according to cable enclosures and
surroundings. They are given in Table 7.2.
As noticed in Tables 4D1A, 4D2A and 4D4A, the choice of installation
method has a marked effect on current-carrying capacity of cables. For
example, from Table 4D2A it can be estimated that current-carrying capacity of
2
2.5 mm three-core non-armoured cable installed in conduit (Method 3) is 83%
of that for the same cable mounted directly on a wall (Method 1).
Table 7.2 Summary of cable installation methods (IEE Table 4A)
Reference
Installation Procedure
Method(s)
Open or direct clipped
1
Embedded direct in building materials
1
Enclosed in conduit or in
trunking
a wall or ceiling
3
Enclosed in conduit or in trunking in
a wall or ceiling but thermally insulated
4
On trays
11 or 13
Trenches
3, or 4
12 or 13
18 or 19 or 20
In air, on cleats, brackets or a ladder
12
Sometimes along the route of a cable one method of installation for part of
a cable route and another method for the reminder is applied. For example,
if a part is installed underground and the other part in air, the currentcarrying capacity shall be determined from the part producing higher cable
cross section.
(v) Class of excess-current protection. Certain thermoplastic materials (e.g.
PVC) deteriorate if subjected to sustained high temperatures. Therefore, the
current rating of cables insulated with PVC or synthetic rubbers are determined
not only by the maximum conductor temperature allowed for continuos rating
Fundamentals of Electrical Installation
43
but also by the probable duration of
excess current.
If a cable is protected by a semi-enclosed (rewirable) fuse to BS3036, the
current ratings stated in tables must be derated by applying the factor 0.725.
This is due to the longer operating time of semi-enclosed fuses.
Overall Correction for Current-carrying Capacity of a Cable Conductor
As can be seen from the ongoing notes on derating affects on current-carrying
capacity of a cable, a number of correction factors are applied. As one factor
may be applied for one condition, more of them may also be applied to the
same cable. In this case, if the overloading is not allowed, the designed current
value of the circuit is divided by each of these factors as
(I b )c  I b .
1 1 1
1
.
. .
C a Cg Ci 0.725
(7.4)
where Ca a is correction factor for ambient temperature
Cg is correction factor for grouping
Ci is correction factor for thermal insulation
In this case to select the appropriate cable size, the tabulated current-carrying
capacity It, should be compared with (Ib)c, that is,
It  (Ib)c
(7.3)
EXAMPLE 7.4
A three-phase 25 kW, 415 V furnace load is to be protected by a semi-enclosed fuse. The
connecting cables are multi-core and non-armoured and are clipped onto a perforated metal
cable tray and touching with two other single-phase circuit cables on the tray. Determine fuse
rating and cable size.
Solution. The designed current of the load is
Ib =25,000/[3 (415)] =34.7 A
(a) From Table 41B1, the nominal rating of the fuse is 45 A
Practical Aspects of Installation Design
44
(b) From Table 4B1, for three circuits a grouping factor of 0.81 must be applied, so the cable
must be able to carry
(I b ) c  34.7 / 0.81
Since protection is to be provided by a semi-enclosed fuse, applying the necessary correction,
the cable must be able to carry the current
I b c 
34.7
 59.22 A
0.81( 0.725 )
The appropriate cable size, It can be selected from Table 4D2A for the reference Method 11
using Eq. (7.5), i.e.
It  59.2 A
yields the tabulated current value of 60 A, which corresponds to conductor size of 10 mm2.
EXAMPLE 7.5
A 5 kW, 230 V water heater is to be wired with PVC-insulated single-core cables through
PVC conduit. The conduit is thermally insulated from a hot water pipe over 15 cm of its
length. If the proposed isolation is for a boiler house where temperature is normally 40oC,
determine the size of cable.
Solution. The designed current of the circuit is, Ib =5,000/230 = 21.7 A.
Correction for thermal insulation on Ib is performed taking h=15 cm and applying Ci= 0.68
(Table 52A). The ambient temperature is accounted by applying the correction factor Ca=
0.87 (Table 4C2). Therefore, the corrected designe current Ib becomes
I b c
21.7
 36.6 A
0.680.87 
The cable size may be selected from Table 4D1A according to reference installation Method

3, hence
It 36.6 A
and the appropriate cable size is 6 mm2.
7. 5 DIVERSITY FACTOR (IEE Reg. 311)
It is a factor usually applied to industrial loads supplied from a DB to determine
size of cable conductor of supplying DB. It is also applied for large scale
residential and commercial commercial loads.
The size of a cable or accessory is not necessarily determined by the total
power rating of all the current consuming devices connected to it. It depends
Fundamentals of Electrical Installation
45
on what percentage of the connected
load is likely to be operating at any one time, or the "diversity" of the loading.
That is, the ratio of real power of the load used at any time to the value of the
real power of the connected load. For
any load it may be expressed as
(7.9)
P
DF  u
Pc
where Pu is average power used by the load at one time and
Pc is the value of connected load
For example, if n different loads are connected to a submain cable, each rated
with (Pc)1 ...,( Pc)n and power used by each load at any one time is (Pu)1 ...,( Pu)n
respectively, diversity factors applied to individual loads are
DF1 
( Pu )1
(P )
,..., DFn  u n
( Pc )1
( Pc ) n
(7.10)
and to the submain is
n
 ( Pu ) i
DF  i 1
n
 ( Pc ) i
(7.11)
i 1
where (Pu)i is average power used by i-th load at one time and
(Pc)i is the value of each connected i-th load
In assessing the maximum current demand of circuits, allowances for
economical installation design diversity is permitted, but should also be used
with care for any particular installation. Its usage needs a degree of experience
and knowledge about the
type of installation, and should be decided by the engineer responsible for the
design.
Table 1B of IEE On-Site Guide gives approximate maximum current
values that are expected to flow in an installation and may be increased or
decreased for any given installation.
- In calculating maximum current demand of radial or ring circuits
Practical Aspects of Installation Design
46
supplying many socket outlets half of the nominal rating of PD or 400 W per
socket can be taken as diversified load for the circuit. For example, if 8 socket
outlets are connected to a ring circuit, diversified demand may be taken as
30/2= 15 A or 8 x 400 W /240 = 14 A.
- For an appliance manufactures data or 80-90% of power rating of the load
may be taken as diversified current demand of the load.
EXAMPLE 7.6
For a flat comprising the loads indicated in the following table, calculate the maximum
expected demand of the submain cable supplying the loads.
Connected load
Lighting
Cooking appliance
Total
Demand (A)
Diversity
limit
15
-
25
10
Applied diversity
factor
9.9
66 %
+30% of fl in excess of
10+7.5
10 A
Heating
18
10
+50% of any current
excess of
10+4
+50% of any current
10+5
demand in
A/C (same as heating
and power)
Installed ring circuit
feeding 8 socket outlets
10
10 A
demand in
30 (20)
-
Maximum expected
demand (A)
excess of
10 A
15(10) or
13.3
Half of the maximum
demand or 400 W
Total demand (A)
108(98)
Diversified demand (A)
68.4 (63.4) or 66.7
When calculating voltage drop in a circuit or feeder cable to a group of circuits,
the current demand can be used after diversity has been taken into account.
7.6 DETERMINATION OF THE SIZE OF CABLE ACCORDING TO
TYPE OF PROTECTION (STAGE 4)
Having established the design current (Ib) of the circuit, determination of the
conductor size is related to cable overloading and its protection.
(i) Overload protection is not required
Fundamentals of Electrical Installation
47
The cable under consideration is
not required to be protected against overload. In this case the design current of
the circuit (Ib) is to be divided by any appropriate correction factors, the size of
the cable to be wired is to be
such that its tabulated current-carrying capacity (It) for the installation method
concerned is
not less than the value of Ib as given as
It  Ib.
1 1 1
.
.
C a C g Ci
(7.6)
(ii) Overload protection is required
For single circuits this condition is met by dividing the nominal current of the
protective device (In) by any applicable correction factor such that the size of
cable to be used is to be such that its tabulated current-carrying capacity (It) is
not less than the value of the nominal current of the PD adjusted as in Eq.
(7.6):
1
1 1
It  In.
.
.
Ca C g Ci
(7.7)
If the circuit to be protected by a semi-enclosed fuse to BS 3036, the right hand
side of the inequality given in Eq. (7.7) should be further divided by 0.725 to
correct cable size for longer trip-off time of the fuse.
If grouped cables are not simultaneously overloaded, grouping factor Cg
may be expressed by the formula
It 
In
2
2
2 1 Cg )
 0 . 48 I b (
Cg2
(7.8)
EXAMPLE 7.7
A three-phase 17 kW, 415 V at 0.866 lag motor load is to be wired with a four-core PVC
insulated non-armoured cable. The cable is directly clipped on a metal tray where there are
Practical Aspects of Installation Design
48
two other three-phase circuits wired with similar cables. All cables are touching to each other.
Determine the cable size and rating of MCB (BS3871 type 1) for cases;
(i) overloading is not required and
(ii) overloading is required.
Solution. The design current of the circuit is
17,000
Ib 
 27.31 A
3 (415)(0.866)
From Table 41B2 rating of the MCB can be selected as 30 A.
Also from Table 4B1, the correction factor for grouping of multicore cables installed as
single-layer on a metal tray and touching to each other is Cg=0.81.
(i) Since overloading is not required, from Eq. (7.6),
It 
27.31
 33.7 A
0.81
Referring to Table 4D2A with reference to installation Method 11, the current-carrying
capacity of the cable is 34 A and its size is 4 mm2.
(ii) Similarly if overloading is to be afforded, from Eq. (7.7) we have
It 
30.0
 37.03 A
0.81
From Table 4D2A, this results in the cable size of 6 mm2 which corresponds to currentcarrying capacity of 43 A.
7.7 VOLTAGE DROP ALONG FINAL CIRCUITS (STAGE 5) (IEE
Reg. 525.01)
The resistance of a conductor increases as the length increases and/or as the
csa decreases. Associated with an increase in length is a reduction in voltage at
the receiving end of the cable run. This reduction is dependent on the
impedance of the cable, values of which are given in mV/A/m.
The Regulation requires that the voltage drop should not be so high that
equipment does not function safely. The allowed voltage drop should be no
more than 4% of the nominal voltage at the origin of the circuit. This means
that:
Fundamentals of Electrical Installation
49
1. For single-phase 240 V, the
voltage drop should not exceed
4% of 240 V = 9.6 V.
2. For three-phase 415 V, the voltage drop should not exceed
4% of 415 V= 16.6 V.
EXAMPLE 7.8
A 4 kW, 240 V at 0.9 lag washing machine is supplied by a 40-m long single-core cables
installed through conduit. If no correction is applied to the design current, calculate the
voltage drop for permissible limits.
It 
4,000
Solution. The design current is
 18.5 A
240(0.9)
This requires 20 A MCB (type 1). Since the cable is enclosed in conduit and if overloading is
not required, from Table 4D1A, the size of the cable becomes 2.5 mm2. Reading from the
voltage drop Table 4D1B, the voltage drop of the cable 15 mV/A/m. The voltage drop
along the circuit may be calculated as
Vc  (mV  I b  L)10 3
 (5  18.5  40)10  3  11.1
V
This voltage drop exceeds 4% (9.6 V) and so is not satisfactory. The next largest cable size
will be used instead, which is 4 mm2 with a voltage drop of 9.5 mV/A/m. Hence, repeating
the calculation the voltage drop along the cable now becomes
Vc = 9.5(18.5)(40) 10-3 = 7.03 V
This result is within the voltage limit required.
If the load has a power factor less than unity, for cables up to 16 mm2, the voltage drop
figures in mV/A/m given in the regulation tables are directly used. However, for cables
having sizes larger than 16 mm2, the voltage voltgae drop is approximately calculated using
resistive r and reactive, x components of the cable impedance z in the formula:
V  (r cos   x sin )  L  I 10 3
volts
where (r cos +x sin) is in mV/A/m and L is the length of the cable in meter.
50
Practical Aspects of Installation Design
EXAMPLE 7.9
A 415-V wye-connected 105 A fan motor at PF=0.8 lag is to be fed along a 65 m long
Cu/PVC/SWA/PVC armoured multi-core cable installed on cable tray. What should be the
size of the cable in order to keep the voltage drop along the cable less than the limiting
value?
Solution. From Table 4D4A and column 5, 25 mm2 cable is sufficient to carry the load
current 105 A. From the voltage drop table Table 4D4B, the resistive and reactive
components of the cable impedance are r=1.5 mV/A/m and x=0.145 mV/A/m
respectively. Finally, the voltage drop along the cable becomes
V=[0.8(1.5)+0.6(0.145)] (65) (105) 10-3 = 8.783 V
which is less than the acceptable voltage limit.