ADAMA SCIENCE AND

HECNOLOGY UNIVERSITY

School Of Electrical Engineering and

Computing
Department Of Electrical &
Computer Engineering

Electrical installation
Electrical regulations and
standards
 The following references contain
provisions which, through reference in
this chapter constitute provisions of the
Ethiopian Building Code
Standards(EBCS-10 1995)on
Electrical Installation of Buildings.
 IEC 364-1:1972 Electrical
Installations of Building, Part 1, Scope,
Object and Definitions.
 b) lEC 50:1982 : International
Electrotechnical Vocabulary.
 c) BS 7671:1992: Requirements for
Electrical Installations, lEE Wiring
Regulations, Sixteenth Edition.
 d) C 22.1:1992: Canadian Electrical
Code, Part I, Safety Standard for
Electrical Installations, Sixteenth 2

Edition.
 (l) The following references contain
provisions which, through reference in this
chapter, constitute provisions of the
Ethiopian Building Code Standards on
Electrical Installation of Buildings.
 (a) BS 8206: 1985 Part 1 - Lighting for
Building.
 (b) DIN 5035:1990 Part 1 - Artificial
Lighting, Terminology and General
Requirements.
 (c) DIN 5035:1990 Part 2 - Artificial
Lighting, Recommended Values for Lighting
Parameters for Indoor and Outdoor
Workplaces.
 (d) DIN 5035:1988 Part 3 - Artificial
Lighting of Interiors, Hospital Lighting.
 (e) DIN 5035: 1983 Part 4- Artificial
Lighting of Interiors, Special
Recommendations for Lighting Educational
establishments
 (f) DIN 5035:1987 Part 5 - Artificial
3

Lighting of Interiors, Emergency Lighting.

 MARKING OF EQUIPMENT
 General
 (1) Each piece of electrical equipment shall bear such of
the following markings as may be necessary to identify the
equipment and ensure that it is suitable for the particular
installation:
 a) The maker's name, trademark, or other recognized
symbol of identification.
 b) Catalogue number or type.
 c) Voltage.
 d) Rated load amperes.
 e) Watts, volt-amperes, or horsepower.
 f) Whether for a.c., d.c., or both.
 g) Number of phases.
 h) Frequency in Hertz.
 i) Rated load speed in revolution per minute.
 J) Designation of terminals.
 k) Whether for continuous or intermittent duty.
 1) Evidence of approval.
 m) Such other marking as may be necessary to ensure safe
and proper operation.
4
 Electrical installation materials:
components and accessories

WIRE AND CABLE

In practice:-
 bare conductors, whether
single or stranded together
are termed as Wire.
Wire
 Conductors covered with
insulation are termed as
cables.
5
The necessary requirements of
a cable are
◦ Conduct efficiently
◦ Cheap
◦ Safe
A Cable consists of three parts
1. The conductor or core:- the
current carrying part of the metal.
2. The insulation or dielectric:- a
covering of the conductor use to
avoid leakage current from the
conductor.
3. The protective covering:- for
protection of insulation from
mechanical damage.
2.2. CONDUCTOR MATERIALS USED IN
CABLES
 The conductor is one of the important items
as most of the capital outlay is invested for i
t. Therefore,
proper choice of material and size of the cond
uctor is of considerable importance.
The conductormaterial used for transmission an
d distribution of electric power should have th
e following properties
 (i) high electrical conductivity.
 (ii) high tensile strength in order to withstan
d mechanical stresses.
 (iii) low cost so that it can be used for long di
stances.
 (iv) low specific gravity so that weight per uni
t volume is small.
 All above requirements are not found in a sin
gle material. Therefore, while selecting a cond
uctor material for a particular case, a compro 8

mise is made between the cost and the requir

Commonly used conductor materi
als. The most commonly used con
ductor materials are copper, alum
inum, steel
cored aluminum, galvanized steel
and cadmium copper.
The choice of a particular materia
l will depend upon the cost, the re
quired electrical and mechanical
properties and the local conditions.
1. Silver: is the best conductor but
due to its higher cost it is hardly
used any where. 9
 Cont….
 2.Copper. Copper is an ideal material for overhead lines o
wing to its high electrical conductivity and greater tensil
e strength.
 It is always used in the hard drawn form as stranded con
ductor.
 It is mechanically strong, hard, extremely tough, durable
and ductile. The electrical resistivity of pure copper at
200c is 1.786 x 10-8 Ωm
 Although hard drawing decreases the electrical conductivi
ty slightly yet it increases the tensile strength
considerably.
 Copper has high current density i.e., the current carrying
capacity of copper per unit of crosssectional area is quite l
arge. This leads to two advantages.
 Firstly, smaller X-sectional area of conductor
is required and
 secondly, the area offered by the conductor to wind load
s is reduced. Moreover, thismetal is quite homogeneous, d
urable and has high scrap value.
 There is hardly any doubt that copper is an ideal materia
l for transmission and distribution ofelectric power.
 However, due to its higher cost and non
availability, it is rarely used for these purposes. 10

 Nowadays the trend is to use aluminum in place of coppe

 3. Aluminium
 Aluminium is cheap and light as compared to copper bu
t it has much smaller
conductivity and tensile strength. The relative compariso
n of the two materials is briefed below :
 (i) The conductivity of aluminium is 60% that of copper.
The smaller conductivity of aluminium
means that for any particular transmission efficiency, th
eX
sectional area of conductor must be larger in aluminum t
han
In copper.
 For the same resistance, the diameter of
aluminium conductor is about
1.26 times the diameter of copper conductor.
 The increased Xsection of aluminum exposes a greater sur
face to wind pressure and, therefore,
supporting towers must be designed for greater transvers
e strength. This often requires the use of
higher towers with consequence of greater sag.
 (ii) The specific gravity of aluminum (2.71 gm/cc) is lowe
r than that of copper (8·9 gm/cc).
 Therefore, an aluminum conductor has almost 1/
2 the weight of equivalent copper conductor. 11

 For this reason, the supporting structures for aluminum

 Cont…
 (iii)Aluminium conductor being light, is liabl
e to greater swings and hence larger cross-
arms are required.
 (iv) Due to lower tensile strength and highe
r co-
efficient of linear expansion of aluminum, th
e sag is greater in aluminium conductors.
 Considering the combined properties of cos
t, conductivity, tensile strength, weight et
c., aluminium has an edge over copper.
 Therefore, it is being widely used as a condu
ctor material.
 It isparticularly profitable to use aluminum
for heavy
current transmission where the conductor si
ze is
large and its cost forms a major proportion o12
f the total
2.3. INSULATING MATERIALS

• The insulation material must

possess the following properties:
 High resistivity
 High flexibility
 Non- inflammability
 High resistivity to moisture, acids


 The various types of insulting materials used
in cables are :-
1. Rubber:- Rubber may be natural or
synthetic.
. Its dielectric strength is above 30KV/mm.
.Though it posses high insulating qualities, it
absorbs moisture readily, softens when
heated to a temperature of 600c to 700c,
swells under the action of mineral oils and
ages when exposed to light.
. Hence pure rubber cannot be used as
insulating materials.
 2. PVC (Polyvinyl chloride)
:- is a man made thermo- plastic which is tough, non-
inflammable and chemically unreactive.

 Its chief drawback is that it softens at a

temperature above 800c.

 It does not deteriorate with age and does not

need to be renewed.

 PVC insulated cables are usually employed for low

and medium voltage domestic and industrial lights
and power installation
Rubber
PVC
 3. Vulcanized India Rubber:- It is
prepared by mixing India rubber
with Minerals such as sulphur,
zinc red lead.
• It absorbs water, which reduces
its insulation properties and
becomes brittle with age.
• The use of VIR cables is limited to
low voltage distribution and
internal wiring as paper-
insulated cables have largely
superseded them.
VIR
4. Impregnated Paper
it is quite cheap, has low capacitance, high
dielectric strength (30KV/mm), and high
insulation resistivity (10Mohm-cm).
The main advantage of paper insulated
cable is that a cable of given size can be
worked out at a higher current density
than a VIR cable.
Impregnated paper insulated cable on its
own would be too fragile to be used
unprotected, and a lead sheath is applied
over the insulation.
 Paper insulated cables are used for
conveying large power in transmission
and distribution and particularly for
distribution at low voltage in congested
Mechanical protection

All the insulating materials used

in the manufacturing of cables
are mechanically weak, so
some form of protection is
required from mechanical injury.
2.4 TYPES OF CABLES USED IN
INTERNAL WIRING

A. Based on number of cores

◦ single core
◦ twin core
◦ twin core with ECC (earth
continuity conductor)
B. According to voltage grading
 250/440 volt
 650/1100-volt cable
 AC cables are designed to be suitable for specific
design voltages, which is called the "Voltage Grade"
(or "Voltage Designation", "Voltage Class" or
"Voltage Rating") of the cable.
 The voltage grade is commonly expressed in the
following form:
 Where is the power frequency voltage between
phase and earth (V rms)
 is the power frequency voltage between two
phase conductors (V rms)
 For example, some standard IEC voltage grades are
0.6/1kV, 1.9/3.3kV, 3.8/6.6kV, 6.35/11kV,
12.7/22kV, 19/33kV, etc.
C. Based on type of insulation

1. Vulcanized Indian Rubber (VIR)

Cables : used for general
conduit wiring.
2.Lead Sheathed Cables : Used for
internal wiring where climatic
condition has moisture.
 PVC
 Colour identification of
bare conductors and cable
cores (EEPCO`s
Regulation)
Function Colour identification of
core of rubber or PVC
insulated cable.
Earthing White
Live of a.c single-phase Green
circuit
Neutral of a.c single-phase Black
or three-phase circuit
Phase R of three-phase Green
a.c. circuit
Phase S of three-phase Yellow
a.c. circuit
Phase T of three-phase a.c. Red
circuit
General specification of
cables
The size of the cable
The type of conductor
Number of cores
Voltage grade
Type of insulation
2.5 CONDUITS
 The commonest method of
installing cables is to draw them
into a conduit.
 The conduit can be steel or
plastic.
 i) Light gauge steel-plain
(unscrewed) conduit
 ii) Heavy gauge steel-
screwed conduit
 iii) Flexible conduit
 Conduit

Conduit Fittings
Conduit Accessories and Fittings
 Conduit Couplers:
 Couplers are used to join two
lengths of conduit.
 screwed conduit are always
threaded at both ends on the outer
side.



Conduit accessories and
fittings
A. Conduit couplers:-
Fig.

 used to join two lengths of
conduit.
a. Tees

 are threaded on both ends.
B. Bends elbows and tees: -are
generally called conduit fittings.
 Bends are usually used for
change
in direction of conduit.
 This should never be sharp.
The minimum allowable radius
b. Elbows

of curvature is 2.5 times the
outside diameter of the conduit.
 Solid elbows and tees should be
used only at the end of the
conduit run.
 Bends, Elbows and Tees:

In general conduit fittings include bends,

elbows and tees.

 Conduit Boxes:

Conduit boxes are used in surface conduit

wiring as well as concealed conduit wiring.
 Conduit boxes are used in surface conduit wiring
as well as concealed conduit wiring.
 Classified as:
Outlet boxes
Inspection boxes
Junction boxes
Lighting accessories and
fittings
 Plugs and socket outlets
 Lamp holders
 Different switches
◦ One-way sw
◦ Two-way sw
◦ Intermediate sw
◦ Gang sw
 Li g
an htin
d
Fi t g A
tin cce
g sso
ri e
s
Fuse and circuit breakers
Designed to interrupt the power
to a circuit due to
◦ The current flow exceeds safe level.
Fuse :
It has narrow strip of metal
which is designed to melt when
current exceeds the rated value.
Two types of fuses
Rewirable Fuses
◦ consists of a porcelain bridge and
base
◦ simple and relatively cheap
◦ Fusing factor about 2
HRC
◦ high breaking capacity
◦ Fusing factor about 1.5
◦ Relatively expensive
Fuse Circuit Breaker
Breakers

It is a mechanical devices, when

the current flow through the
breaker exceeds the rated value,
a bimetallic strip heats up and
bends. By bending it trips the
latch.
Most breakers will carry
◦ 150% of their rated load for 1 min
◦ 200% for about 20 sec
◦ 300% for about 5 sec
Standard Rating
6, 10, 16, 20, 25, 35, 50, 63, 80, 100,
125, 160, 224, 250, 300, and large sizes.
Rating of cables :1.5 2.5 4 6 10 16 25 35
50 70 95 120 150 185 240 300 400
500 630 800
Fuse has higher risk of causing a
fire than circuit breaker due to
◦ loosely screw
◦ contact corroding
◦ wrong size
A circuit breaker has several
advantages over any type of
fuse
◦ All the poles are simultaneously
disconnected
◦ The time can be adjusted
◦ Can be closed again quickly
Distribution board
Is an assemblage of parts for the
distribution of electrical energy
to the final circuits.
It includes
◦ fuses
◦ circuit breakers
◦ main switches
◦ frame
◦ bus-bar
 Electrical Installation design
 wiring design criteria
 Design procedures
 Branch circuit design
 Load tabulation
 Design current
 Cable size design procedures
 Voltage drop calculation
 Service entrance
 Diversity factor
42
 Wiring Design Criteria
 Flexibility:- Able to change to suit new conditions or situations for the
provision for expansion.

 Reliability:- that can be trusted to do something well or with out

interrupting the system

 Safety:- be constantly attentive to an initial safe electrical

installation and such factors as electrical hazards caused by
misuse of equipment or by equipment failure after installation.

 Energy saving and control consideration:- limiting voltage drops,

power factor correction, use of switches for control, etc

 Economic cost: :- initial cost and operating cost.

 Space allocation:- concerned with maintenance ease, ventilation,

expandability, centrality, limitation of access, and noise, in addition to 43
the
basic item of space adequacy.
 Design procedures
 steps involved in the electrical wiring design of any facility.
A. Determine the type and rating of all client furnished equipments.

B. If the designer could not get the exact electrical ratings, determine their ratings
from other consultant.

C. Make an electrical load estimation based on the collected data.

Load Estimation:- estimating total electric loads the building uses after completing the design.
The electrical loads which can be estimated are,
(i) Lighting.
(ii) Miscellaneous power, which includes convenience outlets and small
motors.
(iii) Heating, ventilating, and air conditioning.
(iv) house water pump, air compressors,
(v) elevators, moving stairs,
(vi) Kitchen equipments 44
 Cont…
D. decide the point of service entrance:- type of

service run, service voltage, metering location, and building utilization

voltage.

E. Determine the location and estimate the size of all

required electric equipment spaces including
switchboard rooms, emergency equipment spaces, and
so forth

F. Design the lighting for the facility

G. In your plan Locate all electrical apparatus including

receptacles, switches, motors, and other power
consuming apparatus and signal apparatus such as45
phone outlets, speakers, microphones, TV outlets, fire
 Cont…
H. Make drawing showing all lightings, devices, and power equipments circuit
connection to the appropriate panel board.

I. Prepare the panel schedule (table). This table shows the load distribution over
the three phases and the type of load which is connected on each circuit..

J. From the panel schedule (table) compute panel loads, and make connection
rearrangement so that you will be able to an optimum power balance over the
three phases R, S and T.

K. Prepare the riser diagram. This includes design of distribution panels,

switchboards, a service equipment.

L. Compute feeder sizes and all protective equipment ratings.

M. Cheek the preceding work.

46
 Branch Circuit Design
 Branch circuits are circuits that supplies power from the distribution board(DB)

 Protective devices:- Fuses, MCBs, Main switch,

 Main cables:- carry total current of the installation

 Sub main cables carry current to sections of large installation to SMDB

 Final circuit feeds one type of circuit. It can range from a pair of 1.5 mm2 cables
feeding a light to a very heavy three- core cable feeding a large motor from a CB or
switch at the main DB. Each circuit should have its own protective fuse or CB. the
rating of the protective device must not be less than the designed load current of
the circuit.

The final circuits can be

Lighting circuits 10A, general purpose socket outlets 16A, soket outlets for watter
heater 3kw 16A, s.o for cooker 20/25A, power outlets for feeding motor 16A, bell
circuits 6A etc.
47
 Cont…
 Residential wiring system
Guidelines
a. The NEC requires to supply a load of 3w/sq ft in the building, excluding
unfinished spaces such as porches, garages, and basements.
b. The NEC requires a minimum of two 20-amp appliance branch circuits to feed
all the small appliance outlets in the kitchen, pantry, dining room, family room
etc.

c. The NEC requires that at least one 20-amp circuit supply to be set for laundry
outlets

d. Do not combine receptacles and switches into a single outlet

except where convenience of use dictates high mounting of
receptacles

e. Circuit the lighting and receptacles so that each room has parts of at least two
48

circuits. This includes basements and garages

 Cont…
f. Supply at least one receptacle in the bathroom and one
outside the house

g. Provide switch control for closet lights.

h. In bedrooms supply two duplex outlets at each side of the

bed location.

i. Kitchens should have a duplex appliance outlet every 36 in.

of counter space, but no less than two in addition to the
normal wall outlets

j. A disconnecting means, readily accessible, 49

 Cont…
Non-residential wiring
Guidelines
(a)Schools:-
 LectureHall
 Laboratory
 Shop
 Assembly
 Office
 Gymnasium
 Swimming Pools
 Photographic Labs
50
 Cont…
so it is not possible to generalize on branch circuit design considerations
except for the following
i. To accommodate the opaque and film projectors frequently used in the
classroom, 20-amp outlets wired two receptacles on a circuit are placed
at the front and back of each such room.
ii. Light switching should be provide:

High-low levels for energy conservation and to permit

low-level lighting for film viewing. And With fluorescent
lighting this can be accomplished by alternate ballast
wiring and switching, thus avoiding the high cost of
dimming equipment. And also Separate switching of the
lights on the window side of the room, which is often
51
lighted sufficiently by daylight
 Cont…
iii) Provide appropriate outlets for all special equipment in labs, shops,
cooking rooms, and the like.

iv) Use heavy-duty devices and key operated switches for public area
lighting (corridors, etc.), plastic instead of glass in fixtures, and vandal-
proof equipment wherever possible. All panels must be locked and
should be in locked closets.

v) The NEC requires sufficient branch circuitry to provide a

minimum of 3 w/sq ft for general lighting in schools..

vi) Keep lighting and receptacles completely separate when circuiting.

52
 (b) Office Space

i. In small office spaces (less than 400 sq ft) provide either one outlet for

every 40 sq ft, or one outlet for every 10 linear ft of wall space, In larger
office spaces, provide one outlet every 100 to 125 sq ft.

ii. Corridors should have a 20-amp, 220-v outlet every

50 ft, to supply cleaning and waxing machines.

iii. As with all non-residential buildings, convenience

receptacles are figured at 180 w each.

(c)Stores.
In stores, good practice requires at least one convenience
outlet receptacle for every 300 sq ft in addition to
outlets required for loads such as lamps, show windows,
53

and demonstration appliances

 Load Tabulation
 Arranging facts or figures of loads.
 While circuiting the loads, a panel schedule is
drawn up which lists:
 The circuit numbers
 Load description (the type of the load)
 Wattage (actually in volt-amperes)
 The current ratings
 Number of poles of the circuit-protective device
feeding each circuit and the like
 Spare circuits are included to the extent that the
designer considers them necessary and consonant
with economy, but normally no less than 20% of
54

the number of active circuits.

 Cont…
In calculating panel loads,
(a) Each specific appliance, device, lighting fixture , or other
load is taken at its nameplate rating

(b)Each convenience outlet, is counted as 16 amp

(180 W).

(c)Spare circuits are figured at

approximately the same load as the
average active circuits
55

(d)Free spaces are not added into the load.

 Schedule for lighting
panel

56
 RECOMMENDED ILLUMINATION LEVELS

SERVICE Illumination level (Lux )

Building Areas
- Circulation areas, corridors 100
- Stairways, Escalators 150
- Cloak rooms, Toilets 150
Assembly Shops
- Rough work: Heavy machinery assembly 300
- Medium work: Engine,Vehicle body 500
assembly 750
- Fine work: Electronic and Office machinery
assembly
Offices
- General offices, Typing, Computer rooms - 500
Deep-Plan general offices 750
- Drawing offices 750
Schools
- Class rooms, Lecture theatres 300
- Laboratories, Libraries, Reading rooms and Art rooms 300
Shops, Stores and Exhibition Areas
- Conventional Shops 300
- Self-service Shops 500
- Supermarkets 750
Museums and Art galleries
- Light - Sensitive exhibit 150
- Exhibits insensitive to light 300
Public Buildings
- Cinemas 50
- Auditoriums 150
Theatres and Concert Halls
- Auditorium 100
- Foyer 200
Dwelling Houses
- Bed rooms: - 50
- General 200
- Bed-Head
Nursery 150
 Example
Assume a single floor of an office building 10 m X 20 m.
Calculate the required number of panels, circuits for lighting.

Solution

Office space- illumination=300lux from table (EBCS-10)

Where cu=coefficient of utilization=0.55,

MF=maintenance factor=0.9
 shows luminous flux produced per lamp.
 Let’s select single fluorescent lamp(n=1), wattage =40w which
has efficiency of 60.
 Flux lumen=lumenus efficiency *watt
 =So 60*40 =2400 lumens flux per lamp 59
 Cont…

=51 single
fluorescent lamps

Total wattage=51*40=2040W

Assuming that each lightening branch circuit

2040 is
1300 W rated. 1300

:. Number of branch circuit for lightening

= =1.5~2ckt

60
 Cont…
Receptacles:
 we simply take into account the furniture,
electronic equipments that located in that
room and the functions of rooms. Let ’ s take
total number of receptacles =6(mostly 6
receptacles supplying from one branch ckt ) So,
one branch ckt for socket is required.

Total wattage of
receptacles=6*200=1200W (Each
socket outlet point has wattage of 200 61

W for one gage.)

 Cont…
The total no. of circuits for lighting plus receptacles
is 2 + 1 = 3 ckts

Spares is = 20% of total circuits

3*20% = 3*0.2 = 0.6 ~ 1ckt ~1200W

Total of 4 ckts.

The feeder current is

62
 Riser Diagrams
 This diagram is an electrical version of a
vertical section taken through the building.

 The main switchboard shown in the figure

below constitutes a combination of service
equipment and feeder switchboard. The
service equipment portion of the board
comprises the metering and the 4 main
switches feeding risers, motor control
center (MCC), roof, machine room, and
63
Cont…

64
 Cable Size Design procedure

The correct choice of cable size for any installation is

dependent upon

 Environmental conditions and characteristics of

protection, PVC, weather proof, VIR, LSC

 Current-carrying capacity of the cable and

 Voltage drops of the cable.

65
 Steps for selection of
cable size
1. Determine the design current Ib.
2. Select the rating of the protection In(rating of protective device)
3. Select the relevant correction factors (CFs).
4. Divide In by the relevant CFs to give cable current-carrying
capacity (Iz)
5. Choose a cable size to suit Iz
6. Check the voltage drop
7. Cheek for shock risk constraints
8. Cheek for thermal constraints.

66
 Design current
In many installations, the design current Ib is quoted by the
manufacturer, but there are times when it has been calculated.

 Single phase

 Three phases:

Single phase: Ib = P/V

Three phase: Ib = P/( √3 V)

If an item of equipment has a pf and eff, it will have been taken

into account. Hence:

 Single phase: Ib = (P/(V*PF* eff)

 Three phase: Ib = (P/ (√3*V L*PF* eff )

67
 Cont…
Nominal setting of protection:
 In>Ib. This value may be taken from IEE regulations.
 Standard ratings: both fuse and circuit breakers are available in
standard ratings of 6,10, 16, 20, 25, 35, 50, 63, 80, 100, 125,
160, 224, 250, 300, and large sizes

Correction factors
When a cable carries its full load current, it can become warm.

other influences are

 high ambient temperature:-
 cable grouped together closely
 uncleared over currents and
 contact with thermal insulation.

Because of the high fusing factor of BS 3036 fuses, the rating of the
fuse In, should be less than or equal to 0.725Iz. Hence 0.725 is the
68
correction factor to be used when BS 3036 fuses are used.
Images of BS 3036 fuses
 Cont…
Current carrying capacity of conductor = Iz=In/(relevant
CFs)
Or In
I z 
c g  c a  ci
Ci
 IEE Regulation gives these factors for situations when thermal
insulation touches one side of a cable. However, if a cable is totally
surrounded by thermal insulation for more than 0.5 m, a factor
of 0.5 must be applied to the tabulated clipped direct ratings. For
less than 0.5 m, derating factors should be applied.
o Choice of cable size
Having established Iz of the cable to be used, it now
remains to choose a cable to suit that value.
The IEE regulation also lists all
 Cable Sizes,
 Current Carrying Capacity
 Voltage Drops Of Varies Types Of Cables.
70

(These data is read from table).

 Voltage drop
It is the voltage difference between the voltage measured at DB and the voltage reaching the
load.
R =ρ l/A
The conductor has,
And also, RT= RO (1+ αΔT)
Where, RT = resistance at the required temperature
RO =ρ l/A resistance at room temperature
α -Expansion coefficient
ΔT- Change in temperature
 Vd = voltage drop
 mV = voltage drop in mV per amper per meter obtained from IEE table
 I b = design current
 L = total length of the cable
Then, According to IEE regulations,

Vd = mV/Am* I b* L
71
Cont….
The resistance of a conductor
increases as the length increases
and / or the cross sectional area
decreases.
Associated with an increased
resistance is drop in voltage, which
means that a load at the end of
along thin cable will not have the
full supply voltage available.

72
 The IEE regulation require that the voltage
drop Vc shouldn’t be so excessive that
equipment doesn’t function safely.
 The drop of not more than 4% (IEE)
 Therefore for single phase 220v , the drop
should not exceed 4% of 220v = 8.8V
 For three phase 380V, the voltage drop
should not exceed 4% of 380v = 15.2V
 The drop of not more than 2.5% (EEPCO)
 Single phase It should be less than 2.5% of
220v = 5.5v
 Three phase It should be less than 2.5% of
380v = 9.5v
73
V = V – VL =
 Ri = L/ A
-resistivity of
conductor material
Where L - length from source
to load
A- cross-sectional area
of the conductor
V 
2 xLxI
L

A
R i-resistivity per unit
length
2 xLxI
V  L
cos
A

for pure
resistive
V 
load
2xLx x P
cos 
VA

for 74

reactive load
 For three phase ,

V = Lx  x P
V=380v
AxV
The above formula can be summarized as follows:
Voltage Drop For dc or single phase, For 3-phase ac with
power factor=cos power factor=cos
When the current I is V = 2x xLx IL x V =1.73xLx IL x cos
known cos A
A
When the power P is V = 2x xLxP V = xLxP
known AxV AxV

75
 2 L1 2 L2 2 L3
V  I1( )  I2 ( )  I3 ( )
A A A
2
V  (I1L1  I2 L2  I3L3 )
A
In practice the highest voltage drop is calculated
for the most distant load
 Example
1. A 240 V radial distributor is 70 m long
and has a resistance of 0.0008 per metre
supply and return. Four loads A, B, C and D
rated at 30 A, 45 A, 60 A and 80 A are fed
from the cable at distances of 20 m, 10 m,
15 m and 25m respectively. Calculate the
total current drawn from the supply, the
current in the cable between each of the loads,
and the voltage at load D if all the loads are
connected.
78
 Examples
1. Mr Tadele wants to install an electric
iron of 12kw,220v,50Hz in the kitchen
which is 20m far away from the
distribution board by using PVC
insulated non-armored, copper
conductors multi core cables clipped
directly to the wall at an ambient
temperature of 350c and protection is
by MCB. If only one side of cable is in
contact with thermal insulation.
Determine:
 Design current
 The nominal setting of protection(In)
 The correction factor current(Iz)
 The size of cable and
 Check for the voltage drop 80
Cont…
2. A 16kw, 400V, 50Hz, three-phase
industrial process heater is to be installed in
a factory using a PVC insulated, non-
armored, copper conductors multi core
cables. Length of run is 25m enclosed in a
conduit . Assume a maximum ambient
temperature of 450c, no. of circuits is 4
and protection is by MCB.
 Design current
 The nominal setting of protection(In)
 The correction factor current(Iz)
 The size of cable and
 Check for the voltage drop
2. what length of two core 1.5mm2 copper
cable be used to supply a current to a
combined load of 25 lamps of 80W each on
a 220V supply and at the same time not to
exceeding the allowable voltage drop of IEE 81
regulation i.e. 4% of the nominal voltage.
Cont….
EXAMPLE 4 A circuit is to be
installed using 2.5 mm2, 1.5
mm2 twin and earth 70°C
thermoplastic cables, it is 32
metres long and protected by a
BS 88 fuse. The load to be
supplied is a 4.2kW ,the circuit is
to be installed in trunking
containing one other circuit at an
ambient temperature of 35°C.
Maximum permissible volt drop is
7 V. 82

Calculate the minimum cable that

 SERVICE ENTRANCE
Power is transmitted by means of transmission
lines. And further distributed by means of
distributing lines.

 The consumers are supplied with power by taking

connections (tapping’s) from distributing lines.

The conductors and equipment used for delivering

electric energy from the supply system to the wiring
system of the premises is called the service.
83
 Cont…
Service lines are of two types

1) Overhead service lines :- the service cables are

connected to the line conductors by means of mechanical

connectors called line-taps .

2) Underground Cable Service Lines:- used when the power

to be supplied to the consumer is large (say above 25kw).

84
 Diversity Factor
 The ratio of the maximum demand of a load to the
total connected load.

 it is factor of power utilization

the application of the diversity factor should be

decided by the engineer.

There are some final circuits fed from wiring to

which diversity factor applies:

85
 Cont…
 lighting 0.7- 0.9

 heating:- water heaters (0.2-3) , stoves (0.2) , electric iron(0.3) ,

 cooking appliances which are permanently connected 0.3

 motors (other than lifting motors) 0.7-0.9

 instantaneous-type water heater 0.2

 thermostatically controlled water heater 0.2

 floor-warming installation

 thermal-storage space-heating installation 0.2

 13A fused socket outlets and appliance fed there from and

 Other socket outlets such as 16A sockets. 0.2-0.5

86
 Example
From EBCS-10. Table B.1, select cables of suitable
current-carrying capacity for the following loads and
conditions (p.v.c. cables to BS 6004 into screwed
conduit).
(a) 240 V single-phase sub-mains of lighting load totaling 10.5 kW. Length
of run 10 m. Average ambient temperature 25oC, diversity 66%

(b) 400 V balanced 3-phase power circuit. Load 18.65 kW, efficiency, 80%,

power factor 0.69. Average temperature 30oC. Length of runs 100 m.

87
 Solutions
a)
Given
Power =10.5kw
single phase Voltage =240V
Length of cable run= 10 m=L
Average ambient temperature of 25oC
diversity factor =0.66
Required
 selecting suitable cable size
88
 Cont…
Then,
Current taken by load = Power / Voltage = 10.5kw/ 240v =43.75 A =Ib
maximum current through cables =43.75* 0.66= 28.88
A (using D.F)
From table, If BS 88 32-A circuit breaker is chosen for
protection, 32 A rated circuit breaker can be selected ,
Then, In = 32 Amp
The correction factor for ambient temperature from Table
A.4 for250C is 1.04.

 Therefore the required cable capacity rating:

 Iz = In/Ca = 32/ 1.04 = 30.77 A
From Table B.1, choose a 4 mm2 conductor which carries
32A.
89
 Cont…
Voltage drop = mV/Am * Ib * L
From table B.2 voltage drop for 4mm2 conductor size
= 11mv/Am

 Vd on cable = 11 mv/Am * 28.8 A * 10 m = 3.168

V

Maximum allowable voltage drop( MVd) = 2.5% of 240

MVd =2.5*240/100 = 6V.

Since the actual voltage drop(Vd) < MVd, 4 mm2 can be

90
the correct size selected
 Cont…
If BS 3036 fuse is chosen for protection, this fuse type requires a correction factor of 0.725.

Iz = In/CF

Ca = 1.06, Cf = 0.725

Iz = In/ Ca* Cf =32/1.06*0.725 = 41.64 Amp.

From Table B.1, a 6mm2 conductor carries 41 A. And a 10mm2 conductor carries 57 A.

Take 10mm2 diameter conductor.

Voltage drop = (mV/Am) * Ib * L

From table B.2 voltage drop for 10mm2 conductor size = 4.4mv/Am

Voltage drop on cable = 4.4 mv/Am * 28.8 A * 10 m = 1.27 V

Maximum allowable voltage drop = 2.5% of 240 V = 6V.

Since the actual voltage drop is less than from the allowable maximum voltage drop,
selected size is 10mm2.
91
 Cont…
b) output kw
eff .  
input 3VI cos 
18.65kw
I  48.77 A
3  400  0.8  0.69
From table 9.1, 50 A circuit-breaker of type BS 3871 can use for protection.
then, In = 50 Amp
Load current will be :
→ Iz = In/CF : =Ca = 1
Then,
→ Iz = 50 A
Choose 16 mm2cable which is capable of 92
carrying 52 A.
 Cont…
Testing for Voltage drop:
 Maximum voltage drop = 2.5% of 400 V = 2.5*400/100=10V.
 Voltage drop on the cable = (mV/Am) * Ib * L
= 2.3 * 48.77 * 100
= 11.22 V which is beyond the allowable
voltage drop.
So, choose the next cable size, which is 25mm2.
Voltage drop for 25 mm2 = 1.7 * 48.77 * 100 = 8.29 V
Therefore selected size is 25 mm2.
i.e. voltage drop becomes the main determining Factor
93
 Cont…
when, 50 A BS 3036 fuse can used for protection from table 9.1
→ In = 50 A
→ Correction factor for the fuse is Cf= 0.725
→ Load current Iz = In / CF = In / Cf
→ Iz = 50 A / 0.725 = 68.966 A
From table B.3 select 25mm2 cable which carries 97 A
Testing for Voltage drop:
Voltage drop on the cable = (mV/Am) * Ib * L
Voltage drop for 25 mm2 = 1.7 x 48.77 x 100 = 8.29 V
Therefore selected size is 25 mm2 conductor sizes. 94
Example. Consider the SDB shown
below

95
 No. of light points (lamps) per lighting circuit = 4
 No. of sockets per circuit = 3
 No. of space heater = 1
 No. of water heater = 1
Determine
 a) The estimate of maximum power demand, Pmax
 the rating of the main switch
 the size of the feeder cable
Solution:
 a. Circuit Power with DF Power
without DF
 Lighting 4x220x10 = 8800W 0.7 x
8800 = 6160W
 Socket outlet 3 x 220 x 16 = 10560W 0.2
x10560 = 2112W
 Water heater 220 x 16 = 3520W
0.2x3520 = 704W
 Space heater 220 x 25 = 5500W 0.2
x 5500 = 1100W
 28380 Pmax=10076W
96
 b. P 3 V L I L Cos
P
 IL  ; Cos  1 because most of the loads are resistive
3 V L Cos
10076
  15 .3 A
3 x 380

 Rating of main switch (MCB) is 32 A

 Assuming that the copper cable is enclosed in
conduit, choose cable size of A = 4 x 6mm2.
 Standard Ratings for fuses and CBs:
6,10,16,20,25,35,50,63,80,100,125,160,
224,250,300,etc.
 Maximum load10,676
estimate = 10,676 VA
IL   16.2A (Cos 1)
3 x380x1

In = 32A
 A = 4 x 6mm2 97
The End

98