The document outlines the methodology for selecting power cables, emphasizing important properties such as flexibility, load capacity, and safety. It details specifications regarding voltage classes, insulation types, installation methods, and design calculations including maximum design current and voltage drop considerations. A supporting example is provided, demonstrating calculations to determine appropriate cable cross-section based on various factors including ambient temperature and installation conditions.

1. POWER CABLE
SELECTION
Presented By: Eng. Wesam Al-
Aslami
CALL: (+967)775497030, EMAIL: W.F.ALASLAMI@GMAIL.COM

2. OUTLINE
•INTRODUCTION
•CABLE PROPERTIES
•POWER CABLES SPECIFICATIONS
•CABLE SELECTION METHODOLOGY
•SUPPORTING EXAMPLE

3. INTRODUCTION
One of the hardest stages of a distribution project is CABLE
SELECTION, in which you wonder What is the proper CROSS
SECTION that you should use?
And this question brings out more and more related QUESTIONS like
What is the ENVIRONMENTAL CONDITIONS you are laying your
cable in?
Hmm.. Did you think that the method you’re LAYING CABLES may
effect cable ampacity?
Did you even noticed when cables are LAID TOGETHER feeding more
than one appliance are disturbing each other which make loses ?
What about “DISTANCE BETWEEN PHASES” does it effect cable
ampacity?

4. PROPERTIES OF SUCH IDEAL CABLE
For such cable it
Must be:
• Flexible.
• Able to Carry load current
without overheating.
• Within the permissible
voltage drop.
Give
• High degree of safety
• The cable must be provided
with suitable mechanical
protection
• The materials used in
manufacturing of cables
should be such that there is
complete chemical and
physical stability throughout.
• Non-inflammable
• Unaffected by acids and

5. POWER CABLES
SPECIFICATIONS
Power Cable Systems are usually characterized
by:
 Voltage classes
 Cable Construction
 Insulation Type
 Methods of Installation
 Environment
• Distribution secondary
systems
(Low Voltage) - 120 V to
4000 V
• Distribution primary system
(Medium Voltage) - 5 kV to
45 kV
• Transmission systems
• Conductor Material
Copper
Aluminum
• Conductor Shape
Single or multicore
Stranded or solid sectorial
conductors
Single or Multicore
For Distribution purpose:
• Polyvinyl chloride (PVC).
• Cross-linked poly-ethylene
(XLPE)
• On Trays
• In Ducts
• In Masonry
• In false ceilings
• In Buried Ducting
• Troughs built into
floor

6.  Methods of Installation
•On Trays
Basic Types:
1.Ladder.
2.Solid bottom
3.Trough
4.Channel
5.Wire Mesh
6.Single Rail

7.  Methods of Installation
•On Trays
•In Ducts
•In Masonry
•In false ceilings
•In Buried Ducting
•Troughs built into floor

8. POWER CABLES
SPECIFICATIONS
Power Cable Systems are usually characterized
by:
 Voltage classes
 Cable Construction
 Insulation Type
 Methods of Installation
 Environment

9. CABLESELECTION
METHODOLOGY

10. CABLESELECTION
METHODOLOGY

11. CABLESELECTION
METHODOLOGY

12. MAXIMUM DESIGN CURRENT
In the case of individual power supply to a device, the
current (IB) will be equal to the rated current of the
device being fed. On the other hand, if the wiring
system feeds several devices, the current (IB) will be
equal to the sum of currents absorbed, taking into
account the installation utilization and coincidence
factors.

13. MAXIMUM DESIGN CURRENT (𝐼 𝐵)
What if there is A COMPENSATION ? How shall we determine the
DESIGN CURRENT ? EXPLAIN.
Well,
If an assumption applied “Compensation in operation”.
CAPACITOR FAILURE WIRING SYSTEM IS PLACED OUT OF
SERVICE.
If an assumption applied “Compensation is out of service”.
CAPACITOR FAILURE  CONDUCTOR CROSS SECTION IS
sufficient to CARRY LOAD CURRENT.
SO…………………………………………………………………………………………………

14. MAXIMUM DESIGN CURRENT (𝐼 𝐵)
FACTORS EFFECTING MAXIMUM DESIGN CURRENT :
• POWER FACTOR AND EFFICIENCY FACTOR (a)
𝒔 =
𝑷
𝜼×𝒑.𝒇
in KVA  𝐀 =
𝟏
𝜼×𝒑.𝒇
• UTILIZATION FACTOR (b)
In an industrial installation, it is assumed that loads will never be
used at their full power level.
B = 0.75 for motors.
B = 1 for lighting and heating.
• COINCIDENCE FACTOR (c)
Use Coincidence factor c
Lighting 1
Lighting and air
conditioning 1
Power outlets 0.1 to 0.2 (for a number > 20)

15. MAXIMUM DESIGN CURRENT (𝐼 𝐵)
FACTORS EFFECTING MAXIMUM DESIGN CURRENT :
• COINCIDENCE FACTOR (c)
Number of circuits having
similar nominal currents
Coincidence factor
2 and 3 0.9
4 and 5 0.8
5 to 9 0.7
10 and more 0.6
• FUTURE EXPANSION FACTOR (d)
In the absence of precise indications, the value of 1.2 is often
used.

16. MAXIMUM DESIGN CURRENT (𝐼 𝑏)
THUS,
𝐼 𝐵 =
𝑃×𝑎×𝑏×𝑐×𝑑
3×𝑉
……for three phase loads.
𝐼 𝐵 =
𝑃×𝑎×𝑏×𝑐×𝑑
𝑉
……for single phase loads.
Subscripts identification
𝐼 𝐵 = Maximum design current Determined by Load.
Calculation
𝐼𝑆𝐶 = Short circuit current of 3 Phase Fault. Calculation
𝐼 𝑍 = Current carrying capacity of Selected Cable. Tables
× 𝒂𝒎𝒑𝒂𝒄𝒊𝒕𝒚. 𝒇
𝐼 𝑛 = Nominal or setting current of Protective Device.
Setting
𝐼𝑆𝐶𝐵 = Breaking capacityProtective Device. Setting

17. PROTECTIVE DEVICES

18. PROTECTIVE DEVICES
•Nominal or setting current
IB ≤ In ≤ IZ, which corresponds to zone a in
the figure.
•Conventional or overcurrent trip
 Case of CB
1. For domestic CB
I2 = 1.45In From standard IEC
898
2. For industrial CB
I2 = 1.3In From standard IEC 947-
2
 Case of Fuses

19. PROTECTIVE DEVICES
•Breaking capacity of circuit breaker.
𝐼𝑆𝐶𝐵 ≥ 𝐼𝑆𝐶3
Thus,
 𝐼 𝐵
 𝐼𝑆𝐶 Etap
 𝐼 𝑛
 𝐼𝑆𝐶𝐵
 𝐼2
𝐼 𝑍 Required to evaluate Cross
section

20. CURRENT CARRYING CAPACITY OF CABLE

21. CURRENT CARRYING CAPACITY OF CABLE
𝐼 𝑍 depends on many factor according to the method of
installations in which many other factors show up.
How to obtain these factors ?
1. Define the installation method. From here
2. Gather the factors belong to such installation method. For
example f1,f5,f6 …etc according to the installation method.
3. Evaluate the overall correction factor f
4. Assume that the current should be carried by cable is
𝑰 𝑩
′
=
𝑰 𝑩
𝒇
5. From tables select the cross section which corresponds to a
current value greater than 𝑰 𝑩
′
. The values in the table show 𝐼 𝑍.
𝐼 𝑍 ≥ 𝑰 𝑩
′
6. Check if the cable able to carry load current by 𝐼′ 𝑍 = 𝐼 𝑍 × 𝑓
𝑖𝑛 𝑤ℎ𝑖𝑐ℎ 𝑰′ 𝒛 ≥ 𝑰 𝑩

22. CORRECTION FACTORS
• Ambient temperature rather than 30 °C (𝑓1).
• ground temperatures other than 20 °C (𝑓2).
• Soil resistivity for buried cables (𝑓3).
• Touching multi-core or groups of single-core cables
(𝒇 𝟒).
• Multi-core cables or groups of single-core cables
arranged in several layers (𝒇 𝟓).
• Number of conduits in air and their arrangement (𝒇 𝟔).
• conduits buried or built into concrete and their
arrangement (𝑓7).
• Non-touching buried conduits run horizontally or
vertically on the basis of one cable or group of 3 single-
core cables per conduit (𝑓8).

23. CROSS-SECTIONAL AREA OF PROTECTIVE
CONDUCTORS (PE)
According to IEC 364
The cross-sectional area of the PE conductor is defined in
relation to the cross-sectional area of the phases.
For 𝑆 𝑃ℎ𝑎𝑠𝑒 ≤ 16 𝑚𝑚2 , 𝑆 𝑃𝐸 = 𝑆 𝑃ℎ𝑎𝑠𝑒
For 16 𝑚𝑚2 < 𝑆 𝑃ℎ𝑎𝑠𝑒 ≤ 35 𝑚𝑚2 , 𝑆 𝑃𝐸 = 16 𝑚𝑚2
For 𝑆 𝑃ℎ𝑎𝑠𝑒 > 35 𝑚𝑚2 , 𝑆 𝑃𝐸 = 𝑆 𝑃ℎ𝑎𝑠𝑒 /2
• In the TT earthing system, the protective conductor
cross-sectional area may be limited to:
25 mm² for copper.
35 mm² for aluminum

24. NEUTRAL WIRE CROSS SECTIONAL AREA
Neutral wire cross section same as phase cross section when
𝑆 𝑝ℎ𝑎𝑠𝑒 ≤ 16 𝑚𝑚2 𝑓𝑜𝑟 𝑐𝑜𝑝𝑝𝑒𝑟 𝑐𝑎𝑏𝑙𝑒𝑠
𝑆 𝑝ℎ𝑎𝑠𝑒 ≤ 25 𝑚𝑚2 𝑓𝑜𝑟 𝐴𝑙𝑢𝑚𝑖𝑛𝑢𝑚 𝑐𝑎𝑏𝑙𝑒𝑠
Have cross section in any other cases for both copper or
aluminum.
𝑰 𝑩
′
=
𝑆 𝑝ℎ𝑎𝑠𝑒
𝟐

26. VOLTAGE DROP CHECK
∆𝑽 = 𝒃 𝒑 𝟏
𝑳
𝒔
𝒄𝒐𝒔 𝝋 + 𝝀𝑳 𝒔𝒊𝒏 𝝋 × 𝑰 𝑩
• Where:
• b : Coefficient (b=1 for three - phase circuit , b= 2 for single -
phase circuit).
• 𝒑 𝟏 : Conductor resistivity during normal service, i.e. 1.25 times that
at 20 °C
• (When: 𝒑 𝟏 = 0.0225 W mm²/m for copper, 𝒑 𝟏 = 0.036 W mm²/m
for aluminum).
• 𝒄𝒐𝒔 𝝋 : Power factor.
• (In the absence of specific indications we can take cos 𝜑 = 0.8 & sin 𝜑
= 0.6)
• 𝑰 𝑩 : Maximum design current, in amps
• 𝝀 : Reactance per unit length of the conductors, in W/m.
• The values of 𝝀 in LV are:
1. for three-core cables: 𝝀 = 0.08 × 103
Ω/m

27. SUPPORTING EXAMPLE
A cable is laid from MDP to another panel board containing
many circuits, depending on the following data calculate
the proper cable cross section.
Given Data:
Voltage level = 380V, Ambient Temperature = 35 °C ,
N°=13
Length (L) = 12 m, Installation method  Perforated
Tray with letter E, installed in parallel with two other
circuits with no layers beside this cable tray.
Insulation type >> XLPE, Allowable voltage drop %  4%,
Panel Load = 224 KVA, assuming unity p.f and efficiency
Utilization factor all attached loads = 0.71
Coincidence factor = 0.95,

28. SOLUTION
1.DESIGN CURRENT
𝐼 𝐵 =
𝑃 × 𝑎 × 𝑏 × 𝑐 × 𝑑
3 × 𝑉
𝑎 = 1, 𝑏 = 0.75, 𝑐 = 0.95, 𝑑 = 1.2, 𝑉 = 0.38, 𝑃 = 224
𝐼 𝐵 =
244 × 1 × 0.75 × 0.95 × 1.2
3 × 0.38
= 316.97𝐴
2.PROTECTIVE DEVICE
Since circuit breakers are manufactured according to
standard values then the closest current to the design current is
320 A.
𝐼 𝑛 = 320 , 𝑆𝑒𝑡 =
316.97
320
= 0.99.
Since setting is close to unity, then the setting won’t change.

29. SOLUTION
3.CURRENT CARRYING CAPACITY OF CABLE .
𝐼′ 𝐵 =
𝐼 𝐵
𝑓
Installation method is “on perforated cable Tray with
letter E”, with ambient temperature of 35 °C
Factors effecting this method are 𝑓0 , 𝑓1 , 𝑓4 , 𝑓5
𝑓0 = 1, 𝑓1 = 0.91, 𝑓4 = 0.88, 𝑓5 = 1
Hence, equivalent of factors is 𝑓 = 1 × 0.91 × 0.88 × 1 =
0.8
𝐼′ 𝐵 =
316.97
0.8
= 396.21𝐴
From tables the cross section that could carry such load
is shown below
SOLUTION
3.CURRENT CARRYING CAPACITY OF CABLE .
𝐼′ 𝐵 =
𝐼 𝐵
𝑓
Installation method is “on perforated cable Tray with
letter E”, with ambient temperature of 35 °C
Factors effecting this method are 𝑓0 , 𝑓1 , 𝑓4 , 𝑓5
𝑓0 = 1, 𝑓1 = 0.91, 𝑓4 = 0.88, 𝑓5 = 1
0.91

31. Hence, the cross section which is able to carry 316.21 is 240
mm².
And corresponding current is 450A which is 𝐼 𝑍
For confirmation 𝐼′ 𝑍 = 450 × 0.8 = 360 A
So, the cable is capable of carrying such load even with such
installation method.

32. • You can continue the solution and you will find that the cable is
able to carry such current with less than 4% voltage drop

33. THANK YOU
CALL: (+967)775497030, EMAIL:
W.F.ALASLAMI@GMAIL.COM