Selection of Cables/Conductors

SELECTION OF CABLES/ CONDUCTORS

Technically, there are at least 12 factors to be considered:

a) Purpose b) Ambient condition and environment

c) Degree of protection (weather, chemical, mechanical, particle,

liquid, etc.)

d) Insulating material (Cable Type)

e) Conductor material

An Old 2-core Cable. Note new colour code is Brown / Blue.

f) Method of installation

g) Thermal insulation

h) Type of protective device

i) Short-circuit capacity

j) Current-carrying capacity

k) Voltage drop

l) Minimum size requirement

All factors are covered by I.E.E. Wiring Regulations (BS7671) and C.O.P. a) to c) would establish the types of cables. d) to k) shall be discussed in class, and each of them would affect sizing of cables. d) to g) plus i) are factors affecting heat dissipation and temperature tolerance, and hence affecting factor j) current carrying capacities of the cables.

HKUEEE Electrical Installations p 149

Busbars



Polyvinyl-Chloride (PVC) is a thermo-plastic synthetic resin

and is being widely used as an insulation on electrical

equipment. It has high electrical resistance, good dielectric

strength and mechanical toughness over the common range of

room temperatures. While mechanically tough, bending is quite

flexible. PVC is also the general insulation material of small

cables. An additional layer of PVC is called a sheath and is

acceptable for general mechanical protection. And where

higher risks exist, metal enclosure accommodating the PVC

cables is considered necessary.

For larger cables, Cross-linked polyethylene, XLPE, is

considered to be a better insulating material than PVC. A

XLPE cable can also withstand higher temperature, and

therefore has a higher current rating than a PVC cable of the

same size. In the past, XLPE is significantly more expensive.

Now the cost is moving down. Thus more and more installations

use XLPE cables, especially for larger cables in the communal

installation of a building. XLPE are quite inflexible and hence

may not be suitable for cable routes with many bends.

Mechanical protection requirement is the same as that for PVC

cables.


Metal sheath or metal armour provide much tougher protection

by themselves on any surface, or even at trenches and

underground.

Except for cables’ x-sect. not exceeding 2.5 sq.mm., conductors are usually, but not definitely, stranded; i.e. each cable conducting material is in a bunch of cores.

The general numbers of strands are 3, 7, 19, 37, 61,………

Cables that carry an protective conductor is also available.


Recall: SHORT-CIRCUIT CAPACITY

Recall: The constraint of short-circuit capacity of a cable is bounded by

Let through energy I2 t ≦ K2 S2

constant cross-sectional for a given area

cable type

short-cct time All of them are + ve current quantities

Knowing I – fault current, t – operating time, and K – thermal constant of the specified conductor, then work for S. You may either treat S as an unknown, or substitute a S value into the condition to test its feasibility.

Values of K and reference fault levels are available in BS7671 and CoP respectively.

Operating times are observed by I-t curves.

Energy let through is obtained from I2 t tables.

Operating Temperature is the temperature at which the conductor is utilizing its full current carrying capacity (full-load). At this temperature, the conductor can perform without deterioration. This is also the initial temperature we shall assume for fault and Energy-let-through calculation.

The final limiting temperature is the temperature which the conductor can withstand for a short time. Beyond this temperature the conductor shall deform instantly.


PVC, max. continuous operating temp. 70º C

XLPE, max. continuous operating temp. 90º C

K = 115 K = 143

Busbars in Switchboard

K = 159


From IEE Wiring Regulations/ BS7671. Similar Tables availble at CoP Table

11(2).



Current-carrying Capacity

Let IL current of total load

Ib design current (i.e. IL × diversity)

In nominal rating of protective device

Iz effective current carrying-capacity of cable

Ita feasible (or tabulated) current carrying-capacity of cable

Their relationships are : Ib = IL × diversity factor of load (see table 7(1) of COP)

Ib ≦ In ≦ Iz To determine Iz :

Including type of protective device, there are 8 factors which must be taken into consideration. They are :

i) type of protective device (Cf = 1 or 0.725)

ii) method of installation (In general, we group similar methods in the same category, and assume they are equivalent. In particular cases, we use Cm as correction factor)

iii) type of cable

iv) no. of phases

v) a.c. or d.c.

vi) grouping of cable ( Cg )

vii) ambient temperature ( Ca )

viii) thermal insulation ( Ci )


CORRECTION FACTORS

Let Ca, Cg, Ci, 0.725 be the correction factors for the current-carrying capacity of cable, then

I n I z ≧

Cf Ca Cg Ci

Apply other correction factors if applicable.

Note that although the correction factors are for the cable, the solution is worked from the nominal rating (or current setting) of the protective device. Then from the appropriate table and column in IEE Regulations or

COP, or otherwise, find Ita ≧ I z , The corresponding conductor size is the lower bound in respect of this requirement.

Since Ita ≧ I z we use Ita instead of Iz in the equations from now on. Please also read ii) to note an additional correction factor for the current. i) Type of protective device

No correction is required for MCB, MCCB, ACB, IDMTL relay and HRC fuse as we assume that the fusing faster of each is less than 1.45. (i.e. Cf = 1)


However, fusing factor of semi-enclosed fuse is 2.

∴ I2 ≦ 1.45 Iz

2 In ≦ 1.45 Iz

In ≦ 0.725 Iz i.e. Cf = 0.725

For semi-enclosed fuse only

I n or I ta ≧

0.725 Ca Cg Ci

ii) Method of installation

20 typical methods of installation are specified.

In general, we group similar methods in the same category, and assume they are equivalent. Different columns of a table for each type of cable are used to differentiate the change in current carrying capacity. For installation methods 1 to 17, no correction factor is required.

For installation methods 18 to 20, current carrying capacities are obtained by treating them as methods 12 or 13 as appropriate, but with suitable correction factors applied. These correction factors Cm are additional to the Ca, Cg, Ci, Cf, Ct factors, and are obtainable from TABLE A5(6).

iii) Type of cable

Different tables are provided for different types of cables, no correction factor is required.


iv) Number of phases

Refer to different columns in tables

v) a.c. or d.c.

Refer to different columns in tables vi) Grouping [ Table A5(3) ]

When Cables are grouped or bunched in the same route. They are very close to each other, and thus shall affect heat dissipation. A correction factor Cg is required.

vii) Ambient temperature [ Table A5(1) ]

More energy is required to raise a cable to its maximum permissible temperature when the ambient temperature is lowered. The opposite is also true.

Therefore, an ambient temperature factor Ca is required.

Values of the factors are different for different installation methods.

viii) Thermal insulation

Where a cable is to be run for a significant length in a space to which thermal insulation is likely to be applied, the cable shall wherever practicable be fixed in a position such that it will not be covered by the thermal insulation. Where fixing in such a position is impracticable, the current-carrying capacity of the cable shall be appropriately reduced, i.e. a thermal insulation factor Ci is required. For full insulation ≧ 0.4 m, Ci = 0.55. Smaller values for shorter insulations. Refer to C.O.P. Appendix 5(3) and Table A5(4)


Correction for current carrying capacities (single & group)

(1) Where overload protection is by fuse to BS 88 or BS 1361 or a MCCB or MCB:

a) for single circuits

I n I ta ≧ Ca Ci

b) for groups

i) circuits may be simultaneously overloaded

I n I ta ≧ Ca × Cg × Ci

ii) circuits which are not liable to simultaneous overload

1 – Cg2

I b I n

2 + 0.48 Ib2 (

Cg2 )

I ta ≧ larger Ca Cg Ci , Ca Ci


(2) Where overload protection is by semi-enclosed fuse to BS 3036:

a) for single circuits

I n I ta ≧

0.725 Ca Ci

b) for groups

i) circuits liable to simultaneous overload

I n I ta ≧

0.725 Ca Cg Ci

ii) circuits not liable to simultaneous overload

1 – Cg2

I b 1.9I n

2 + 0.48 Ib2 (

Cg2 )

I ta ≧ larger Ca Cg Ci , Ca Ci

(3) Where overload protection is omitted:

i.e. where IEE Regulation 473-01-04 applies

I b I ta ≧

Ca Cg Ci

Variation of installation conditions along a cable route Where various factors apply to different parts of the route, each part shall be treated separately, or alternatively only the factor or combination of factors appropriate to the most onerous conditions encountered along the route shall be applied to the whole of the route.


Voltage Drop

Maximum permissible voltage drop at receiving end is 4% of the nominal voltage. i.e. If nominal voltage of 3-phase 4-wire installation is 380 V / 220V

Max permissible voltage drop is 8.8V for phase-voltage, and is 15.2V for line voltage.

Voltage drop is a cable = current in the cable × impedance of the cable

( V drop = I b × Z cable )

Z cable = f (temp, material, x-sect area, length) Voltage drops per ampere per metre are given in tables A6 of COP. Note resistive parts are affected by temperatures of conductors. When O/C protective device ≠ BS 3036 fuse ambient temp ≧ 30

Correction for operating temperature (for voltage drop)

t p – (Ca

2 Cg2 –

Ib2

Ita2 ) (t p – 30) t1

=

where t p = max permitted normal operating temp


Resistivity at t1 230 + t1 Design mV/A/m

Resistivity at tp =

230 + t p =

Tabulated mV/A/m

= Reduction Factor = Ct

Adjust Resistive component by

Ct =

230 + t p – (Ca2 Cg

2 – Ib

2

Ita2

) (t p – 30)

230 + t p

When conductor ≦ 16 mm2 Then with correction for volt drop, (mV/A/m) actual = C t (mV/A/m) tabulated When conductor > 16 mm2 then (mV/A/m)r actual = C t (mV/A/m)r tabulated

mV/A/m x is unaffected by temp

When conductor is very large

x such that

r ≧ 3 ,

no correction is required.


Correction for load power factor ( for voltage drop )

When conductor ≦ 16 mm2 Apply Cp = cos Φ

[ i.e. (mV/A/m) actual = cos Φ (mV/A/m) tabulated ] When conductor > 16 mm2 (mV/A/m) actual = cos Φ(mV/A/m)r tabulated + sin Φ (mV/A/m)x tabulated not for

cables in flat formation x-area > 240 mm2

& p.f. > 0.8

Combined correction for both operating temperature and load power factor

When conductor ≦ 16 mm2 combined correction = C t cos Φ When conductor > 16 mm2 (mV/A/m) actual = C t cosΦ(mV/A/m)r tabulated + sinΦ(mV/A/m)x tabulated


For calculation of the impedance, Assume:

For conductors 16 mm2 , assume resistive only. Source impedance independent of temperature when

specific parameters not known. Using tables for cable ratings/sizes, we may obtain per length impedance from taking a half of tabulated mV/A/m values with temperature adjustments. The reason for taking a half is because the length in table refers to circuit length, but for resistance/impedance, we talk about conductor length. That is: Resistance in mV at tx

oC is given by

Tabulated mV/A/m 230 + tx 2

x ( 230 + tp ) x length in m

Conversely, knowing other parameters may permit us to calculate size of protective conductor. Conductor resistances at 20C in milliohms/metre (from Electrical Installation calculations) X-sectional area mm2 Copper Aluminium

1 18.1 1.5 12.1 2.5 7.41 4 4.61 6 3.08

10 1.83 16 1.15 1.91 25 0.727 1.2 35 0.524 0.868

Multipliers to be applied for protective conductor: 70C pvc 1.24 + 0.002ta 85C rubber 1.36 + 0.002ta XLPE 1.42 + 0.002ta


From: Guidelines on Energy Efficiency of Electrical Installations, 2007 page 16 of 35

TABLE 4.2B

Single-core PVC/XLPE Non-armoured Cables, with or without sheath (Copper Conductor)

Conductor Resistance at 50 Hz Single-phase or Three-phase a.c.

(Based on BS7671, Requirements for Electrical Installations, Table 4D1B & 4E1B)

Conductor

cross- Conductor resistance for PVC and XLPE cable

in milliohm per metre (mΩ/m) sectional

area (mm2) PVC cable at max. conductor

operating temperature of

70°C

XLPE cable at max. conductor

operating temperature of 90°C

Enclosed in

conduit/trunking

Clipped

direct or

on tray,

touching

Enclosed in

conduit/trunking

Clipped

direct or

on tray,

touching

1.5 14.5 14.5 15.5 15.5

2.5 9 9 9.5 9.5

4 5.5 5.5 6 6

6 3.65 3.65 3.95 3.95

10 2.2 2.2 2.35 2.35

16 1.4 1.4 1.45 1.45

25 0.9 0.875 0.925 0.925

35 0.65 0.625 0.675 0.675

50 0.475 0.465 0.5 0.495

70 0.325 0.315 0.35 0.34

95 0.245 0.235 0.255 0.245

120 0.195 0.185 0.205 0.195

150 0.155 0.15 0.165 0.16

185 0.125 0.12 0.135 0.13

240 0.0975 0.0925 0.105 0.1

300 0.08 0.075 0.0875 0.08

400 0.065 0.06 0.07 0.065

500 0.055 0.049 0.06 0.0525

630 0.047 0.0405 0.05 0.043

800 - 0.034 - 0.036

1000 - 0.0295 - 0.0315


Steps for Determination of Cable Size

1. Determine Ib and its electrical properties

2. Determine In

3. Choose protection device

4. Determine cable route and circuit length

5. Determine method of installation

6. Determine type of cable

7. Work out ambient temperature

8. Check thermal insulation

9. Determine Iz

10. Check short-circuit capacity, select cable size -- S1

11. Select cable size ( by looking at Ita in tables ) -- S2

12. Check voltage drop, select cable size -- S3

13. Check if there is a minimum size requirement -- S4

14. S ≧ max [ S1, S2, S3, S4 ]

If S1 = size of cable calculated from short-circuit capacity S2 = size of cable calculated from continuous loading consideration

S3 = size of cable calculated from volt-drop consideration

S4 = minimum according to mechanical, ambient, purpose consideration

then S = the size of cable which should be selected is given by: S ≧ max S1, S2, S3, S4 Alternatively, find S1 first, and then test it against volt drop and I 2 t requirement.


NEUTRAL CONDUCTOR Make & break:

Makes before the phase conductors

Breaks after the phase conductors

or simultaneously make and break

No switch (unless inherently linked) nor fuse shall be connected in neutral conductors, including those of control circuits.

Size of neutral depends on

a) neutral current, how balancing the phase currents is;

b) fault level of phase to neutral fault;


Overcurrent protection of neutral conductor is

a) by the protective devices in phase conductors if size of neutral conductor is not less than that of a phase conductor;

b) by the protective devices in phase conductors if load is shared evenly by the 3 phases and the neutral conductor can meet the let through energy requirement;

c) by its own protective device if a) & b) are not met. But the protective device should also disconnect phase conductors.


Circuit Arrangement of Neutral Conductors

a) Neutral conductor of a single phase circuit should not be shared with any other circuit.

b) Neutral conductor of a three phase circuit should only be shared with its related phases in a three phase four wire system.

c) For a polyphase circuit, the neutral conductor should have a suitable current carrying capacity to cater for any imbalance or harmonic currents which may occur in normal services.

(Note that Triplen harmonics cannot be cancelled by each other in 3-phase 4-wire systems. Nowadays much harmonic is generated from electronic non-linear loads)

d) Where an autotransformer is connected to a circuit having a neutral conductor, the common terminal of the winding should be connected to the neutral conductor.


Capacity of conduit & trunking

Main purpose of conduit and trunking is for mechanical protection. Conduits are for small cables only, and trunking can be for both small and medium cables. Methods to determine their sizes are fully discussed in Code 14 of C.O.P. 1) by comparing sum of cable factors with conduit & trunking

factors 2) space factor ≦ 45% For conduits, two sets of tables are available, one is for short straight run, the other is for long run (straight or with bends) Space Factor is: The ratio (expressed as a percentage) of the sum of the overall cross-sectional areas of cables (including insulation and any sheath) to the internal cross-sectional area of the conduit or other cable enclosure in which they are installed. The effective overall cross-sectional area of a non-circular cable is taken as that of a circle of diameter equal to the major axis of the cable.

Space factor

=


Example 1 of Cable Size for discussion

Three identical sets of three-phase circuits have cables installed in the same trunking. Each circuit

has a balanced full load of 83 A at unity power factor. The circuits are subjected to simultaneous

overload.

The end of each circuit is connected to an electrical load that has an exposed conductive part. Each

circuit is controlled and protected by a MCCB at its origin, with its earth terminal installed close to

the MCCB. The type of MCCB will operate within 0.02sec by a magnetic trip when the current

flowing through it is not less than 10 times its current rating.

Other information of each of the circuits is as follow:

Earthing system: TN-S

Supply voltage: 380/220 volts

Supply frequency: 50 Hz

Ambient temperature: 10C to 45C

Length of circuit: 70 metres

MCCB instantaneous operation time: 0.02 s.

MCCB magnetic sensitivity 10 times current rating

Highest fault level in the circuit: 8 MVA

Earth fault loop impedance measured

at the earth terminal at the MCCB: 0.11 ohm

Voltage drop between source & MCCB 5 volts

Cables for the circuit: single core XLPE copper cable to BS 7889

Wiring method: Cables and protective conductors in trunking.

Estimate, with justification, assumptions and comments clearly stated:

a) the desirable current rating and breaking capacity of the MCCB;

b) the maximum earth fault loop impedance based on your suggestion in a);

c) the correction factors used for assessing current carrying capacities of cables;

d) the desirable size of the live conductors;

e) the desirable size of the protective conductors;

f) the touch voltage when an earth fault occurs at the exposed conductive part of the load.


ANSWER:

The answer shall involve both quantitative and qualitative analysis.

Assume:

Operating times of earth fault devices (MCCB) OK, and is 0.02 s

Resistivity is given by volt drop per amp per m

Take ambient temp as 45C for tabulated values.

Length of cpc equals circuit length

a) Given, max current in a phase = 83 /0 A

Therefore, MCCB In = 100 A

Breaking capacity of it should be 12.12 kA which is larger than system symmetrical

fault current IF = 12.12 kA (8 MVA).

b) Earth fault Loop Impedance Consideration:

The minimum E/F current to effect tripping of MCCB = 100A x 10 = 1000 A

Max permissible E/F loop impedance = 220 / 1000 = 0.22 ohm

c) Technically, to determine size of live conductors, consider 5 factors: i. current carrying capacity ii. purpose iii. ambient temperature, condition & environment iv. voltage drop v. short-cct capacity

Factors for current carrying capacity:

Ca = 0.87 for 45C, Cg = 0.7 for 3 sets, Cf = 1 for MCCB,

As not given, assume Ci = 1,

And there’s no Cm for this method of installation.


d) Consider current carrying capacity,

Recall Ib = 83 A

Have choosen MCCB rating In = 100 A

Iz = In / [Ca x Cg ] = 164.20 A

From table, choose 50 mm2 Cu cable, with 175 A (Ref Method 3, 3-phase)

Consider voltage drop,

Should meet the max 4% requirement.

Please note that the 4% volt drop is measured from nominal. Thus volt drop in the up-stream must also be taken into account.

Also note that the designed load is balanced 3-phase, hence voltage drop is a 3-phase voltage drop.

In balanced 3-phase load, and assume no triplen harmonics, the neutral has zero voltage drop as the current in it is zero. Thus the total voltage drop in the 3-phase circuit (phase + neutral) is smaller than a single phase circuit with the same load across the phase.

Also note that the 4% volt drop allowance applies to phase-phase, hence larger tolerance too (15.2 V against 8.8 V).

Max volt drop from source to circuit load =380 x 4% = 15.2 V

Max volt drop in this circuit = 15.2 V – 5 V = 10.2 V

Hence Max volt drop per A per m in circuit = 10.2 V/ (83 x 70) V = 1.756 mV

Also start with assumption and approximation: voltage correction is not necessary for power factor

and temperature. (Indeed p.f. given to be 1 in this example)

From table, 25 mm2 Cu cable with volt drop 1.6 mV (method 3) which is less than above.

(Once OK, no need to check with adjustment by temp correction. If not, correction is reqd.))

Consider short-circuit capacity,

With IF = 12.12 kA (from M ). Assume IF for the whole of t = 0.02 s, 8 VA

From table K = 143,

S (IF2 t)_

K

S = 11.99 mm2 Cu cable, Hence Choose 16 mm2 Cu cable

Compare the three lower-bound sizes in the 3 considerations (current carrying capacity, voltage drop,

short-cct capacity), take the maximum value or higher, Hence Choose size 50 mm2 Cu cable.


e) Resistivity of live conductor (approx from table) = 1.05/2 = 0.525 m-ohm/m

Hence max impedance of protective conductor = 0.22 – 0.11 – 0.525 x 70 / 1000

= 0.07325 ohm

Thus max resistivity of c.p.c. = 0.07325 x 1000/70 = 1.0464 ohm

Check cable size for 2 x 1.0464 = 2.0928 mV and use it to check cable size from table.

It gives 25 mm2 Cu cable or larger as protective conductor.

Try protective conductor size 25 mm2 Cu cable to match the live 50 mm2 Cu cable

Resistivity of live conductor (approx from table) = 1.05/2 = 0.525 m-ohm/m

Resistivity of c.p.c. conductor (approx from table) =1.9/2 = 0.95 m-ohm/m

Earth fault loop impedance = 0.11 + (0.525+0.95) x 70 / 1000 = 0.2133 ohm

Acceptable to match with the permissible value.

Consider short-circuit capacity of protective conductor,

Max IEF = 220/ 0.11 = 2000 A

With IEF = 2000 A

Using let thru energy condition, assume c.p.c. is XLPE cable

S = 1.978 mm2 Cu cable

Choose 25 mm2 Cu cable (larger than 1.98 mm2) as protective conductor is acceptable,

(or any size not less than 1.98 mm2 Cu cable as long as Zloop not less < 0.22 ohm.

Of course, mechanical and chemical protection requirement must be observed.

f) Touch Voltage Consideration:

Earth fault current at exposed conductive part of load = 220/0.22 = 1000 A

(This earth fault loop impedance may produce current to operate magnetic trip of MCCB.)

Assume installation earth and supply earth have the same impedance.

Largest possible touch voltage is 1000A x [0.11/2 + 0.95x70/1000] = 121.5 V

Largest possible touch voltage between MCCB and earth terminal is 1000A x [0.11/2] = 55 V

These values are higher than 50 V, hence the followings are functional:

Equipotential bonding within premises;

Partition between MCCB and main earth terminal to avoid simultaneous access;

Operation times are safe according to touch curves.


Example 2 of Cable Size for discussion A three-phase circuit has a 3-phase full load of 20 A, (36 + j15) A and (20 – j15) A. The end of the

circuit is connected to a distribution board that distributes electricity to its loads. The distribution

board has an exposed conductive part. The circuit is controlled and protected by a moulded case

circuit breaker (MCCB) at its origin, with its main earth terminal installed adjacent to this switch.

Other information of the circuit is as follows:

Earthing system: T-T Supply voltage: 380/220 volts Supply frequency: 50 Hz Ambient temperature: 20C to 35 C Length of circuit: 50 metres MCCB instantaneous operation time: 0.02 s. Min. magnetic trip current for MCCB: 10 times its setting (10 In) Highest fault level in the circuit: 6.6 MVA Voltage drop measured at MCCB: 1.2 volt Earth loop impedance measured at

earth terminal beside the MCCB: 0.1 ohm Cables for the circuit: single core XLPE cable to BS 5467 Wiring method: in trunking

Determine:

a) the desirable current rating and breaking capacity of the MCCB;

b) the maximum permissible earth fault loop impedance;

c) the correction factors for current carrying capacity of the cables;

g) whether 10 sq,mm. is a desirable size of the live conductors;

h) whether 6 sq.mm. is a desirable size of the protective conductors;

i) the earth fault loop impedance after selection of conductors; j) the fault current that flows in an earth fault;

State all your assumptions.


ANSWER:

a) Given, max current in a phase (phase L2) = 36 + j15 A or 39 /22.6 A

Therefore, In = 50 A (or a current not less than 39A)

Given fault level = 6.6 MVA, and system voltage = 380/220V

Fault level in terms of current = (6600/3/220) = 10kA

Breaking capacity must be larger than fault current, which is 10 kA.

(Note: In real design, we should practically balance the loads.)

b) Max. permissible earth fault loop impedance is the impedance that shall produce an earth fault

current just enough to break the circuit within the max. tolerable time (in this case 5 sec.) by the

selected protective device.

Assume the instantaneous trip of MCCB occurs at I ≥ 10 In .

Max permissible earth fault loop impedance = 220 V / 10 x 50 A = 0.44 Ω

c) Given max amb. temp = 35° C. Assuming no grouping and no thermal insulation,

Factors: Ca = 0.96, Cg = 1, Ci = 1,

Correction factors for volt drop are to be discussed as part of d).

d) There are no rules saying which consideration goes first, let us try a sequence different from ex.1

* Consider short cct capacity,

With I = 10 kA, K = 143, t = 0.02 s,

(You should refer to the “t” provided by manufacturer. We take this as 0.02 s as MCCBs are

current limiting devices nowadays.)

Substituting values into boundary condition I2 t = K2 S2

Smallest acceptable S = 9.89 mm2 Cu 3-ph cable

Choosing ≥10 mm2 Cu cable is acceptable.

* Consider current carrying capacity,

Ca = 0.96, Cg = 1, Ci = 1,

In = 50 A

min Ita = In / (0.96 x 1 x 1) = 52.1 A

Assume cables are in trunking

From Table, cable 10 mm2 Cu is acceptable.


Therefore 10 mm2 Cu cable is OK.

* Consider volt drop,

Note that the designed load is unbalanced 3-phase, hence voltage drop is different among the

3-phases.

The single-phase voltage drop consideration is more appropriate.

In this situation, we have two options.

Choose the highest single phase volt drop, and treat it as the reference.

Calculate the neutral current, and then determine the highest single phase volt drop

(largest volt drop in phase + volt drop in neutral). Of course, when you want more exact

value, vector calculation should be used.

A lot of engineering problems involve many parameters. Hence usually there are more than one

approach or one direction for a solution. We have realized it in the overcurrent setting

determination. Let us use this answer as another example.

1st method: assume 10 mm2 Cu cable (taken from above). We assume a size first because

this data is required for calculating the temp correction factor.

tp = 90oC, Ib = 36 + j15 A (39A), p.f. = 0.923

Take max volt drop = 4% x 220 V = 8.8 V

From table,

Volt drop (3-phase) per A per m = 4.0 mV (Note for 1-phase, it would be 4.7 mV). None

of them is perfect, as the loads are unbalanced three-phase.

If no exact calculation is preferred, then play safe and use 1-phase value.

Without temp and p.f. correction, and no voltage vector adjustment,

Volt drop from source = (39 x x 4.7 x 50 mV) + (1.2 V)

= 9.17V + 1.2V = 10.37 V.

It’s larger than 8.8 V!!

Note 1: Please note that the 4% volt drop is measured from nominal. Thus volt drop in the

up-stream must also be taken into account.

Note 2: There are two voltage drop correction factors: temperature and power factor. Since both

voltage drop correction factors are not larger than 1 (unless the ambient temp is low), thus when

the calculation does not take the voltage drop correction factors into account, the calculated volt

drop value without correction may be larger than actual voltage drop value which has taken


correction factors into account. Hence applying voltage drop correction factors may give us an

opportunity to select a smaller cable size, or similarly, not to select a larger cable size.

If the total voltage drop without correction factors (and also the unbalanced condition in Note 4)

does not exceed 8.8 volt for the selected size, then we can stop calculation, and consider voltage

drop is acceptable.

Note 3: When the total voltage drop exceeds permissible voltage drop, don’t immediately assume

that the cable size is unacceptable. Apply correction factors to check whether the corrected volt

drop is within the limit. When the drop still exceeds the limit, engineers should change the cable

to a larger size.

Note 4: Moreover, when load currents in 3-phases are unbalanced, the largest voltage drop

occurs at the phase having the largest current. Its exact volt drop is obtained by the volt drop

along its line conductor, plus the volt drop along the neutral conductor. The volt drop along the

neutral conductor is calculated by the combined neutral current. But usually for simplicity, we

use 1phase voltage drop calculation, and assume the neutral current same as line current. For

exact calculation, we must firstly determine the neutral current, and the neutral voltage drop.

Since volt drop > 8.8 V, then we have to rethink the aforesaid volt-drop by one of the following:

i) The easiest way is to try larger size: 16 sq.mm. Students may try it and show that

mathematically it will be OK.

ii) Calculate exact volt drop by considering phase 2 current and neutral current

separately. Then work out their respective volt drops. The total volt drop is their sum

plus 1.2 volt (VECTOR sum).

Also, Work out temp correction, and p.f. correction,

Combining the above and see whether the corrections may lower the above calculated volt

drop to smaller than 8.8 volts.

Correction factors of the phases:

Ct (phase L1) = [230 + 90 – [(0.962) – 202/632)(90 – 30)]/ (230 + 90) = 0.846

Ct (phase L2) = [230 + 90 – [(0.962) – 392/632)(90 – 30)]/ (230 + 90) = 0.899

Ct (phase L3) = [230 + 90 – [(0.962) – 252/632)(90 – 30)]/ (230 + 90) = 0.856

(It is obvious that the higher the current, the larger the correction factor)

Apply the largest factor, i.e. the factor of L2-phase, (not need to calculate phase by phase).

Using, Ct (L2) = [230 + 90 – [(0.962) – 392/632)(90 – 30)]/ (230 + 90) = 0.899

And p.f. 0.923

Then with both temp and pf corrections,


Volt drop = 39 x (0.899 x 0.923) x 4.7 x 50 mV + 1.2 V = 8.8 V

which is just acceptable.

(Note that in 3-phase, by considering volt drop due to phase current and neutral current

separately will show a value even smaller than 7.6 V because neutral current is often

smaller (Note 5).)

Since the voltage drop in its up-stream does not exceed 1.2 V. (assume same p.f. at

upstream and downstream), Choosing 10 mm2 Cu cable is acceptable.

Note 5: For using volt drop values in tables, engineers must clearly understand the difference

between 3-phase volt-drop values and those for 1-phase. The Volt-Drop Considerations and

Tolerances are different. 1-phase V-drop is 4% of 220 V and 3-phase is 4% of 380V. (See Note 4

above).

2nd method: Alternatively, assume the size of cable is an unknown,

To start with, as earlier alternative, use 1-phase value,

Permissible volt drop per A per m = (8.8 – 1.2) V / 39A/ 50m

= 3.90 mV

Thus any cable size giving a volt drop per A per m less than 3.90mV is acceptable.

Therefore from table, ≥16 mm2 Cu cable is OK.

Again, for marginal consideration, the correction factors may help our consideration to save

cable size. And indeed 10 mm2 Cu cable can be used.

Compare the cables sizes by the three considerations (current carrying capacity; volt drop; &

short-cct capacity), take maximum value (10 mm2, 10 mm2, 10 mm2, also compare min. size

for mech protection)

Choose 10 mm2 Cu cable. Existing cable size is acceptable.

e) Consider size of protective conductor is 6 mm2 Cu cable,

For simplicity of calculation, we take no temperature adjustment (But temp adjustment

gives more accurate and appropriate result, unless you allow tolerance consideration in

choosing the size),

Resistivity of c.p.c.(approx from table) = 7.9/2 = 3.95 m-ohm/m

Resistivity of live conductor.(approx from table) = 4.7/2 = 2.35 m-ohm/m


Earth fault loop impedance = 0.1 + (3.95 + 2.35) x 50 / 1000 = 0.415 ohm

For Max permissible earth fault loop impedance consideration:

The max. value is 0.44 ohm,

thus 0.415 ohm is acceptable.

Min earth fault current = 220/0.415 = 530 A

Max earth fault current = 220/0.1 = 2200 A

For Energy-Let-Through consideration:

Check energy-let-thru condition by the max E/F current (Assume using XLPE cable, K = 143)

S = 2.18 mm2 Cu cable

Choose 6 mm2 Cu cable as protective conductor is acceptable, (or any size not less than

2.18 mm2 Cu cable)

For Touch Voltage consideration:

IEF to effect 5 sec operation = 10 times In = 10 x 50 A = 500 A

Max Z loop = 220 V/ 500 A = 0.44

Max Z protective = (50 / 220) x Max Z loop = 0.1

Compare with Z protective of 6 mm2 Cu cable = (3.95) x 50 / 1000 = 0.1975

Which is larger than 0.1 , hence not acceptable.

A larger protective conductor is required.

Choose 16 mm2 Cu cable as protective conductor.

Check Z protective of 16 mm2 Cu cable = (2.9/2) x 50 / 1000 = 0.0725

This is now acceptable.

Thus, take all things into consideration, choose 16 mm2 Cu cable as protective conductor.

f) Using 16 mm2 Cu cable as protective conductor, and consider the circuit end,

Earth fault loop impedance = 0.1 + (2.9/2 + 4.7/2) x 50 / 1000 = 0.29 ohm

which is less than max permissible Z-loop 0.44 ohm.


g) Earth fault current is between

Min. current = 220/0.29 = 758.6 A

Max. current = 220/0.1 = 2200 A

Use min. current to determine actual touch voltage.

Use max. current to verify size.

Note: Read also discussion in example 1.

Inside the equipotential zone,

Touch voltage at exposed conductive part between the limb and the distribution board

= 759 A x c.p.c. resistance = 759 x 0.0725

= 55 V slightly above 50 V.

(Even with the high touch voltage calculated as above exceeds 50V, it is still safe. Check the

touch voltage curves, a touch voltage of 55 V should not harm persons under protection by

MCCB operating at 0.02 sec.)

(Nonetheless, strictly speaking under CoP, we should use a larger size of c.p.c. With this

change, then of course, we have to recalculate everything from e) and onwards).

Note that persons outside the fault zone may also receive a touch voltage.

Assume earth resistance of installation is half of the up-stream earth fault loop impedance.

Touch voltage = 759 x earth resistance

= 759 x (0.1/2) = 37.9 V

This is less than 50 V, thus acceptable for T-T system with MCCB.

0.1/2 ohm is based on assumption that the upstream protective conductor plus an earth

resistance contributed half of the earth fault loop impedance at the origin.)

As touch voltage is acceptable, thus the size of protective conductor is acceptable.



Where the question also asks for accommodation of cables, then either:

1) Use the Cable Factor Method

Adding up the cable factors of all the cables in the enclosure;

Compare to check which enclosure has a enclosure factor not less than this sum of cable factors;

Then this enclosure is the minimum size of enclosure to accommodate the cables.

2) Use the Space Factor Method

Calculate the cross-sectional area of each cable including its insulation/ sheath;

Add up all the cross-sectional areas;

Divide the sum by 0.45 to get the minimum cross-sectional area of the enclosure.

Documents

Selection of Cables/Conductors