3.

Electrical Installation Design

3.1. The electrical work standards and regulations
3.2. deign procedure
3.3. Electrical motor installation and control
3.4. Electrical drawing
 3.1. The electrical work standards and regulations
Utility companies and building code standards describe
service requirements. For example the following is
extracted from EBCS-10 pp144.
Clearance from Building opening
i. Service conductor installed as open conductor or
multi-conductor cable without any overall other jacket
shall have a clearance of not less than 925 mm from
windows, doors, porches, fir escape, or similar locations.
ii. Service conductors shall not be installed beneath
openings through which materials may be moved, such
as openings in farm and commercial buildings.
iii. Overhead wire shall not be run such that they
obstruct entrance to these buildings openings.
3.2 Deign procedure
1. Sub-division and number of ckt: any installation
needs to be divided into number of ckt
▪ Inside the consumer’s premises there are basically
two ways of distributing power
A. Only from MDB to SDB
B. from MDB to SDB and from SDB to SDB
▪ Example: Let there be three buildings B1, B2 and B3
and they consume powers of 30kW, 20kW and 15kW
respectively.
▪ The arrangements of the DB’s using In a) failure of
SDB 1 does not affect SDB 2 while in b) it does affect.
In b) SDB 1 should have a capacity of 35 kW for the
20kW branch and for SDB 2. In a) while the over all
cables length is relative to(b) larger, its sizes are small.
 (A) Procedures of designing cont..

(B)

Electrical installation design in general is the

engineering process of selection and decision on the
required materials, accessories and protective devices.
 Procedures of designing cont..
▪ In electrical installation terminology, a branch
circuit or a final circuit is the last portion of the
installation in which one or more parallel
connected loads are supplied with electric power
as well as use one common isolating and protective
devices.
▪ However, each load in a branch circuit is controlled
separately with its own control such as a switch or a
starter/contactor.
▪ Thus, the following constitute a branch circuit:
a high power socket outlet (such as for cooking
range, stove or boiler),
a motor supply circuit,
a group of light points,
a group of general purpose socket outlets,
2. Cable selection
▪ The selection or knowledge of voltage level of
service-entrance from utility company is the first
step for the design of the service-entrance
equipment.
▪ Cable from the utilities operating company line to
the energy meter at the consumer’s intake point is
called the service cable or line.
▪ The service line voltage level is determined by the
power requirement of the consumer premises.
▪ In Ethiopia power is supplied to industrial premises
at 15 kV which is the distribution voltage while it is
at 380/220 Volts for low power consuming
commercial and residential consumers.
 Cable selection cont..
The size of a cable to be used for an installation depends
upon:
▪ the current rating of the cable under defined
installation conditions and
▪ the maximum permitted voltage drop
The factors which influence the current rating are:
1. Design current: cable must carry the full load current.
2. Type of cable: copper conductors or aluminum
3. Installed conditions: clipped to a surface or installed
with other cables in a trunking.
4. Surrounding temperature: cable resistance increases
as temperature increases and insulation may melt if
the temperature is too high.
5. Type of protection : for how long will the cable have
to carry a fault current?
Step to select cable Procedures of designing cont..
2.1 Determine design current (Ib): It is the rating of the
apparatuses to be operated. It may be given by
manufacturers or if not it should be calculated as
follows:
P
For single phase: Ib= p-power consumed by the
V
apparatuses V= single phase supply (220 v
𝑃
For three phase Ib=
3∗Vl
Vl = voltage between two lines ( 380 v)
▪ But ac circuits may be capacitive or inductive can
produce poor power factor.
▪ In addition to this, allowance must be made for
efficiency (eff) and this will have to be allowed for.
So that: Procedures of designing cont..
P∗100
Ib = for single phase
V∗pf∗eff
P∗100
Ib = for three phase
3𝑉1 ∗pf∗eff
2.2. Rating of protective Device (In)
▪ After having established the design current of the
circuit, the next stage is to determine current rating
or setting protective device.
▪ Allowance should be made for possible additional
future load, the rating of the protective device.
▪ It should not be less than design current of the circuit
(In>=Ib)
▪ The reason for this is the protective device must be
capable of passing sufficient current for the circuit
function at full load without the device operating
 Procedures of designing cont..
2.3. Current carrying capacity
In
Iz =
relevant cfs
correction factors(cfs):
✓ Cables can be overheated when over currents are
carried due to the following main reasons:
Overloading
Short circuit occurred
✓ The other reason is increase in temperature when
number of cables are bunched together, for instance
in conduit and trunking, which is a situation in which
each cable contributes its heat to that of others.
✓ So to choose cable size or to determine current
carrying capacity of conductor the following factors
should be considered.
 Procedures of designing cont..
A. Ambiant Temperature (ca)
▪ It is the temperature of the immediate surroundings
of the equipment and cables before the temperature
of the equipment or cables contribute to the
temperature rise.
▪ The cable ratings in the IEE regulations are of the
time based upon 300c, and hence it is only above this
temperature that adverse correction is needed.
▪ table below relates to temperature correction factor
(Ca) for ambient temperature.
 Procedures of designing cont..
B. Grouping Factor (cg)
▪ When cables are grouped together they impart heat
to each other.
▪ So if cables are grouped they have to be derated, the
amount they have to be derated is dependant upon
how they are installed.
▪ Tables should be referred for correction factors Cg to
account for grouping.

Thus, if the current rating of one of the cables in

non-grouped condition is Iz, then the new current
capacity when grouped with other cables will reduce to
Iz = Cg x Iz
 Procedures of designing cont..
▪ Correction factors need to be applied in the
calculation of cable sizes for lighting system in a
large factory since many cables of the same size and
loading could be expected to be fully loaded at the
same time.
C. Thermal Insulation (ci)
▪ When cables are in contact with thermal insulation,
thermal insulation reduces rate of flow of heat from
the conductors, thus rising the conductor’s
temperature.
▪ This means that the current carrying capacity of the
conductor has to be reduced to compensate for the
reduction in heat loss.
▪ The size of conductor increased in size is dependant
upon the manner in which it is in contact with the
 Procedures of designing cont..
D. protective device
▪ 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 correction factor to be used.
Application of Correction Factors
▪ Some or all of the conditions outlined may affect a
cable along its whole length or part of it, but not all
may affect it at the same time. So consider the
following:
a) If the cable ran for the whole length, grouped with
others of the same size in a high ambient temperature
and was totally surrounded with thermal insulation, it
would seem logical to apply all cfs, as they all affect the
In
whole cable run. Iz =
Cg ∗Cf ∗Ci ∗∗Ca
 Procedures of designing cont..
▪ But if the ambient temperature will have no effect on
the cable, if the thermal insulation prevent heat
reaching to the cable. Only Cg, Cf, and Ci will be
used.

In
Iz =
Cg ∗Cf ∗Ci
 Procedures of designing cont..
b) If the cable first runs grouped with, then leave the
group and runs in high ambient temperature, and finally
enclosed in thermal insulation, three conditions, each
affecting the cable in different areas.
▪ The Bs 3036 fuse affects the whole cable run and
therefore Cf must be used, but there is no need to
apply all of the remaining factors as the worse one
will automatically compensate for others.

In
Iz =
Ci ∗Cf
 Procedures of designing cont..
c) In the figure below: combination of cases a and b is
considered.
▪ The effect of grouping and ambient temperature is
0.7 x 0.97 = 0.69. The factor for thermal insulation is
still worse than this combination, and therefore Ci is
the only one to be used.

Having chosen the relevant correction factors, we now

apply them to the nominal rating of the protection
current In.
In
Iz=
 2.4. Choice of cable size
▪ Having established the tabulated current carrying
capacity Iz of the cable to be used, it is now remains to
choose cable to suit this value
2.5 Voltage Drop Calculation
▪ The resistance of a conductor increases as the length
increases and / or the cross sectional area decreases.
▪ Associated with an increased resistance a load at the
end of along thin cable will not have the full supply
voltage available.
▪ The voltage drop caused by the resistance of the cable
of the required length must not exceed a limit given
as a standard. Usually drops of up to 2.5% of the
supplied voltage is tolerable(EEPCO).
▪ Single phase, it should be less than 2.5% of 220v= 5.5
▪ Three phase, it should be less than 2.5% of 380v= 9.5v
 Voltage Drop Calculation cont..
DC circuits
a) Using line constants :
Consider the two wire circuit (single phase) shown next

The voltage drop on the pair of cable must be

∆V = V1 – V2 <=2IR
Where R = L/𝜎A ; R in ohom, L in meters, A in mm2 and
𝜎 in m/W- mm2
For copper 𝜎 =57 m/W-mm2 for aluminum 𝜎=33m/W-
mm2
▪ Thus cross sectional area of the wire can be related to
the loading must satisfy
∆V< =2IR =2IL/ 𝜎A or A (mm2 ) >= 2IL(m)/ 𝜎∆V
▪ If ∆V is in percentage of the rated value
[i.e (∆V/V)100% = C or ∆V = C V/100], then
A (mm2 ) >=200 x I [A] x L[m]/ 𝜎CV
▪ If power is given instead of current
A (mm2 ) >= 200 x P[W] x L[m]/ 𝜎CV
Ex.
A 1 kW electric appliance is to be supplied from a
220V from a single phase source 200m away.
Assuming an allowed 2.5% voltage drop, what cross
sectional area of copper wire must be used.
 Voltage Drop Calculation cont..
General ac load (reactive + resistive)
▪ The triple wire power cable of length L is assumed
to have only resistance R in each line but negligible
inductance. Lost power on cable
∆P = Ps - Pr = 1.73 I x Vs cos 𝜑𝑠 - 1.73 I x Vr cos𝜑
= 1.73 I (Vscos 𝜑𝑠 - Vr cos𝜑)
= 3 I2 R

With proper phasor diagram of the receiving end as

well as the sending end phase, line and resistance
drop voltages it can be shown that the line voltage
difference b/n Vs and Vr becomes: ∆V = 1.73IRcos𝜑
 Voltage Drop Calculation cont..
The above formula can be summarized as follows:
Voltage Drop For dc or single For three phase ac with
phase, pf= cos𝜑 pf=cos𝜑
When I (current) V =
2𝜎LPIL cos𝜑
V=
2𝜎LPIL cos𝜑
A A
is known

When P (power) V =
2𝜎LP
V=
𝜎LP
AV AV
is known

Table reference method

A simpler approach in electrical installation design is
to use pre calculated values of voltage drops of the type
prepared by the Institute of Electrical Engineers (IEE).
See table for instance Voltage Drop Calculation cont..
EX
1. A 7kW water heater is to be supplied at 230 V by a
PVC sheathed and insulated copper cable, 10 meters
in length. The cable run in a roof space which is
thermally insulated with glass fiber. The roof space
temperature is expected to rise to 45°C in summer,
and where it leaves the consumer unit and passes
through a 50mm thermal insulation material, the
cable is grouped with three other cables of similar
size and loading. The protection is by means of a BS
3036 fuse and the voltage drop is limited to 2.5% of
the supply voltage. (Ci =0.89, Cf =0.725 )
a. Calculate the minimum cable size required.
b. Calculate the minimum cable size required if the
cable is not grouped with other cables.
2. A 2 kW 230 V 50 Hz single-phase fan has an efficiency
of 75% and works at a power factor of 0.8. It is connected
to its starter by single-core copper conductor; the length
of run is 15m and for about half this run the cable is
clipped to wall surfaces. For the remainder it shares a
cable tray, touching two similar cables across the top of a
boiler room where the ambient temperature is 50°C. The
voltage drop in the cables must not exceed 5V. Calculate
the minimum cable size required. (Ci =0.89)
3. A 400 V 50 Hz three-phase pump motor has a rating of
15 kW at 0.8p.f. lagging and is supplied from a BS 88
distribution board 25m distant. The cables run with
three similar circuits. The ambient temperature is 35◦C
and that the voltage drop in the cables is limited to 2.5%
of the line voltage. Calculate the minimum cable size
required.
4. A p.v.c.-insulated (copper conductors) cable runs
between a 230 V distribution board at the origin of an
installation and a 10kW heater. The cable passes through
the following environmental conditions:
(a) on its own in a switch room with an ambient
temperature of 35◦C;
(b) on its own in an outdoor area with an ambient
temperature of 25◦C;
(c) bunched with three other cables on a wall surface in
an area with an ambient temperature of 40◦C;
(d) finally on its own passing through a thermally
insulated wall section for a distance of 2m, in an ambient
temperature of 30◦C. Protection is by BS 3036 fuses,
length of run is 30m and the voltage drop is limited to
5.5V. Calculate the minimum cable rating and select
suitable cable for voltage drop limitation.(Ci =0.5)
Cable size & protective device capacity selection for

Variable load
➢ All connected loads to a distribution center may
not function simultaneously.
➢ For example, a residential building service cable
may have to supply the lighting system, socket
outlets and air conditioners and we don’t switch on
all lighting in our residential building at the same
time.
➢ In order to select cable size there are rules for
different applications which are available on
standards.
 Cable size selection for Variable or Distribution Centers cont..
i) Demand factor and diversity factor method
➢ Demand factor is the ratio of the maximum demand
of a system, or part of a system, to the total connected
load on the system, or part of the system under
consideration.
➢ “Demand factor” is a percentage by which the total
connected load on a service or feeder is multiplied to
determine the greatest probable load that the feeder
will be called upon to carry. It is always less than one.
➢ Diversity factor is the ratio of the sum of the
individual maximum demands of the various
subdivisions of a system, or part of a system, to the
maximum demand of the whole system, or part of the
system, under consideration.it is usually more than
one.
 Cable size selection for Variable or Distribution Centers cont..
➢ The sum of the connected loads supplied by a feeder-
circuit can be multiplied by the demand factor to
determine the load used to size the components of the
system.
➢ The sum of the maximum demand loads for two or
more feeders is divided by the diversity factor for the
feeders to derive the maximum demand load.
➢ Demand factor allows to reduce the demand on
various equipment for feeders and services due to
intermittent, periodic, short-time, or varying duty
operation.
➢With the application of demand factors, smaller
components can be utilized in the electrical system
and greater savings can be passed on to the consumer.
➢The maximum demand can be determined by the
total connected load in the consumer premises and
knowledge of their operating schedule or demand
factor.
➢ 2011 NEC Articles 210.19(A)(1) and 215.2(A)(1)
require the branch circuit and feeder conductors to
have an ampacity not less than the sum of 100% of the
non continuous load plus 125% of the continuous
load. Continuous load is defined in the 2011 NEC 100
as "A load where the maximum current is expected to
continue for three hours or more.
 Cable size selection for Variable or Distribution Centers cont..
ii) Method of operation schedule knowledge of
connected loads
➢ In this method, estimation of maximum load is
determined based on the knowledge of operation
schedule of the connected loads which are all
assumed known.
iii) Premises total area method
➢ In this method, the design rating for a particular
application in a premises depends on the total area
of the particular area of the premises.
➢ According to EBCS-10, recommended power per
unit area of the different parts of a premises such as
schools, hospitals, stores, etc. is estimated and
provided in table. Refer to EBCS-10 Section 6 pp
150 to 156
EX:
1. Calculate the size of transformer having following
details:

Calculate the size of transformer if the diversity factor

is 1.3, 1.7 and 1.65 for the load connected with beaker 1,
breaker 2 and breaker 3 respectively.
2. consider that a feeder supplies five users with the
following load conditions: On Monday, user one
reaches a maximum demand of 100 amps; on Tuesday,
two reaches 95 amps; on Wednesday, three reaches 85
amps; on Thursday, four reaches 75 amps; on Friday,
five reaches 65 amps. The feeder’s maximum demand
is 250 amps. Determine the diversity factor of the
system .
3. Calculate the size of a main feeder from substation
switchgear that is supplying five feeders with
connected loads of 400, 350, 300, 250 and 200 kilovolt-
amperes (kVA) with demand factors of 95, 90, 85, 80
and 75 percent respectively. Use a diversity factor of
1.5.
 3.3. Electrical motor installation and control
The following six steps can be used for the installation of
electrical motors.
1. Determining the motor full load current.
▪ The maximum demand for the circuit depends on the
duty of motor.
▪ Continuous load is defined in the 2011 NEC 100 as "A
load where the maximum current is expected to
continue for three hours or more."
▪ For the continuous time duty 125% of the motor
rating current is taken as the maximum demand
current.
▪ For intermittent and short time duties standard
tables have to be used.
 3.3. Electrical motor installation and control cont….
2. Determine the size of the motor branch circuit
conductor
3. Determine the fuse size (dual element) to be used as
motor branch-circuit short circuit and ground fault
protection.
▪ Maximum setting or rating of short circuit and
ground fault protection devices is given by NEC.
4. Determine the rating requirement for the motor
disconnecting switch
▪ Motor disconnecting means must be sized in
accordance with the requirements of Part H of NEC
Article 430. The disconnecting means for motor
circuits rated 600V, nominal, or less, must have an
ampere rating of at least 115% of the full-load current
rating of the motor.
 3.3. Electrical motor installation and control cont….
5. Determine the motor and branch circuit overload
protection requirement
▪ Current rating of over current relay to respond to
motor overloading is also of the above rating (125%
full load).
6. Determine the requirement for the motor control
circuit over current protection
▪ The control circuit of the motor may require
protection because it may extend to remote
start/stop button.
▪ If the control cable is of small size the impedance
associated is normally high.
▪ Therefore; the branch circuit protection device may
not clear any fault on the control circuit.
 3.3. Electrical motor installation and control cont….
NEC Code 430.22 (Size of Cable for Single Motor)
▪ Size of Cable for Branch circuit which has Single
Motor connection is 125% of Motor Full Load Current
Capacity.
Example: What is the minimum rating in amperes for
Cables supplying 1 No of 5 hp, 415-volt, 3-phase motor at
0.8 Power Factor. Full-load currents for 5 hp = 7Amp
Ans=8.75 Amp.
NEC Code 430.6 (A) (Size of Cable for Group of Motors
or Elect.Load)
▪ Cables or Feeder which is supplying more than one
motors other load(s), shall have an ampacity not less
than 125 % of the full-load current rating of the
highest rated motor plus the sum of the full-load
current ratings of all the other motors in the group.
 3.3. Electrical motor installation and control cont….
▪ For Calculating minimum Ampere Capacity of Main
feeder and Cable is 125% of Highest Full Load
Current + Sum of Full Load Current of remaining
Motors
Example: What is the minimum rating in amperes for
Cables supplying 1 No of 5 hp, 415-volt, 3-phase motor
at 0.8 Power Factor , 1 No of 10 hp, 415-volt, 3-phase
motor at 0.8 Power Factor, 1 No of 15 hp, 415-volt, 3-
phase motor at 0.8 Power Factor and 1 No of 5hp, 230-
volt, single-phase motor at 0.8 Power Factor?
Full-load currents for 5 hp = 7Amp
Full-load currents for 10 hp = 13Amp
Full-load currents for 15 hp = 19Amp
Full-load currents for 10 hp (1 Ph) = 21Amp.
 3.3. Electrical motor installation and control cont….
NEC Code 430.24 (Size of Cable for Group of Motors
or Electrical Load)
▪ It may not be necessary to include all the motors into
the calculation. It is permissible to balance the motors
as evenly as possible between phases before
performing motor-load calculations.
Example
What is the minimum rating in amperes for conductors
supplying 1No of 10 hp, 415-volt, 3-phase motor at 0.8
P.F and 3 No of 3 hp, 230-volt, single-phase motors at 0.8
P.F. The full-load current for a 10 hp, 415-volt, 3-phase
motor is 13 amperes.
The Full-load current for single-phase 3 hp motors is 12
amperes.
 3.3. Electrical motor installation and control cont….
NEC 430/32 Size of Overload Protection for Motor
▪ To Find the motor running overload protection size
that is required, multiply the F.L.C. (full load
current) with the minimum or the maximum
percentage ratings as follows;
Maximum Overload
Maximum overload = F.L.C. (full load current of a
motor) X allowable % of the maximum setting of an
overload, 130% for motors, found in NEC Article 430/34.
Increase of 5% allowed if the marked temperature rise is
not over 40 degrees or the marked service factor is not
less than 1.15.
Minimum Overload3.3. Electrical motor installation and control cont….
▪ Minimum Overload = F.L.C. (full load current of a
motor) X allowable % of the minimum setting of an
overload, 115% for motors found in NEC Article
430/32/B/1. Increase of 10% allowed to 125% if the
marked temperature rise is not over 40 degrees or the
marked service factor is not less than 1.15.
▪ When calculating motor loads, convert a motor's
current rating (given in amps) to a VA rating.
▪ To do this, multiply the motor's nameplate amperage
by the supply voltage.
▪ If there no nameplate data on the motor, then
multiply the supply voltage by the current rating
shown in the appropriate table in 430,
 3.3. Electrical motor installation and control cont….
Noncoincident loads
▪ Heaters and air conditioning don't run
simultaneously on the same feeder. In most cases,
either heat or A/C. run Therefore, heat and A/C are
normally examples of noncoincident loads.
▪ This kind of load is calculated by choosing the larger
of the VA ratings and disregard the other.
Example: Which is the larger load: 240kW heating unit
or a 97A A/C unit? The supply is a 480V 3-phase system.
Step 1: Select larger load per Sec. 220-21. Heating load:
240kW x 1000 x 100% = 240,000VA A/C load: 480V x 97A
x 1.732 = 80,642VA
Solution: The 240,000VA-heating unit is the larger load.
Thus, this will used for overall load calculations and
the other load is not used.
Exercise
1. Determine the size of the feeder conductors and
thermal magnetic circuit breaker feeding a motor
control center that has a total connected motor load
of 215A, with the 60-hp,77amp motor being the
largest motor. In addition, there are 45A of
continuous load and 65A of non continuous load.
Conductors should be copper with insulation
installed in an area where the ambient temperature
is less than 30°C.
2. It is desired to design and installs a squirrel cage
induction electric motor fed from a 380V bus-bar. The
electric motor nameplate values are pn = 2.2 kW,
nn =2850 rpm, Vn=380V, pf=0.8. The motor is for
continuous duty operation.
Table I: Single core PVC insulated copper conductor
with sheath, current carrying capacity and voltage
drop.

Table: Maximum rating or setting of motor Branch

circuit short-circuit and ground fault protective devices