STUDY GUIDE

INSTALLATION RULES
Paper 2
V 5.4

COMPILED BY: G Hendrickse

 Table of Contents
Introduction .............................................................................................................................................................. 3
Syllabus ..................................................................................................................................................................... 3
Example 1: General Knowledge.............................................................................................................................. 4
Example 2: Calculation – Voltage Drop ................................................................................................................. 6
Example 3: Calculation – Voltage Drop ................................................................................................................. 9
Example 4: Calculation – Voltage Drop ............................................................................................................... 12
Example 5: Calculation – Voltage Drop ............................................................................................................... 16
Example 6: Calculation – Voltage Drop ............................................................................................................... 19
Example 7: Calculation – Voltage Drop ............................................................................................................... 22
Example 8: Calculation – General ........................................................................................................................ 24
Example 9: Calculation – General ........................................................................................................................ 25
Example 10: Calculation – Conduit Size ............................................................................................................... 29
Example 11: Loop Impedance & PSCC Test ......................................................................................................... 30
Example 12: Earth Electrode Test ........................................................................................................................ 31
Example 13: Insulation Resistance Test .............................................................................................................. 33
Example 14: Touch Voltage................................................................................................................................... 35
Module 1................................................................................................................................................................. 36
Module 2................................................................................................................................................................. 38
Module 3................................................................................................................................................................. 39
Module 4................................................................................................................................................................. 40
Module 5................................................................................................................................................................. 41
Module 6................................................................................................................................................................. 42
Module 7................................................................................................................................................................. 43
Module 8................................................................................................................................................................. 47
Module 9................................................................................................................................................................. 49
Module 10............................................................................................................................................................... 50
Important Annexure .............................................................................................................................................. 52
Memorandum 1 ..................................................................................................................................................... 53
Memorandum 2 ..................................................................................................................................................... 66
Exam Generator ..................................................................................................................................................... 79
 EXAMPLE 1
GENERAL KNOWLEDGE QUESTIONS

State the Standard Voltage in the RSA.

230 V single phase and 400 V three phase

What is the Standard Voltage for SELV & PELV systems?

The limitations of the voltage in the system to 50 V a.c. rms or 120 V d.c. ripple free

Give the formula to calculate the sustained current carrying capacity of a cable

The sustained current carrying capacity (I) of a cable buried directly in the ground can be calculated
by the following equation:

I = IS x F 1 x F2 x F3 x F4, where:

IS is the standard (tabulated) current rating measured in amperes

F1 is the depth of burial rating factor given in Table 6.16

F2 is the soil temperature at depth of burial rating factor given in Table 6.11

F3 is the thermal resistivity of soil rating factor given in Table 6.12

F4 is the grouping of directly buried cables that are thermally interdependent rating factor given in
Table 6.13

State all the factors that will influence the Current Carrying Capacity of a cable.

Ambient temperature

Grouping and number of cables in a trench

Grouping and number of cables on rack or trays

Grouping and number of cables buried directly in the ground

Neutral imbalance

Harmonics

Direct solar radiation

Power load and Voltage drop

What is the function of the CET?

Consumer’s Earth Terminal (CET), terminal that is effectively and permanently earthed and to which
the earth continuity of an installation is permanently connected. The regulation states that, each
installation shall have a consumer’s earth terminal at or near the point where the supply cables to
the installation enter the building or structure. All conductive parts that are to be earthed shall be
connected to a main earthing terminal, which shall be connected to the consumer’s earth terminal
(for the purposes of carrying earth fault currents). The consumer’s earth terminal shall be earthed
by connecting it to the supply earth terminal or the protective conductor and if installed, the earth
electrode. The effectiveness of the supplier’s protective conductor shall be determined in
accordance with 8.7.5.

What is meant by 06h00 earthing?

Tamper proof (All connections, joints are mechanically sound and shall not be tampered with
especially all bonding and earthing conductors). A main earth terminal or bar must be provided for
each installation to collect and connect together all protective and bonding conductors. It must be
possible to disconnect the earthing conductor from this terminal for test purposes, but only by the
use of a tool. This requirement is intended to prevent unauthorized or unknowing removal of
protection.

What is meant by mixed loading and mixed circuits?

There shall be no mixed loading of circuits, except: A single phase circuit that has over-current
protection rated at not more than 20 A may supply a mixed load of:

(a) Socket-outlets rated at not more than 16 A and fixed luminaries

(b) Socket-outlets and fixed appliances, or Socket-outlets, luminaries and fixed appliances

Note: the number of points need not be limited

Socket-outlets rated at 16 A or more that are connected to circuits with mixed loading shall comply
with the earth leakage requirements of 6.7.5

When will a temporary DB be installed?

Temporary installations for building construction site work, repairs, alterations, extensions and
similar work.

What is the maximum % volt drop for an AC and DC Supply in the RSA?

The voltage drop for a.c and d.c supply shall not exceed 5%

What is the GPS Co-ordinates of your DB at your residence?

GPS co-ordinates of your DB may be obtained at local municipality. You may use the address and
site number
 EXAMPLE 2
VOLTAGE DROP CALCULATIONS

Name FOUR different methods of how the voltage drop of a cable can be determined and give examples
for each method.

ANSWER: Suppose a balance three phase load of 100A is to be supplied by a 25mm2 copper cable over a
distance of 100m. Then calculate the Voltage Drop using FOUR known METHODS.

Given: I = 100A

L = 100m (25mm2 of a copper (Cu) cable)

From Table E1 page 307 R = 0.88 Ω/Km for a 25mm2 Cu

NB: We must express resistance in Ohms, proceed as follows

R (in ohms) = Value of R in Ω/Km x L (in meters) [1Km = 1000m]

R = (Value of R in Ω/Km x L in meters) [Ohms]

1000

Let Vd be the Voltage Drop to be calculated:

METHOD 1

Use this formula from page 308

Vd = (Fv x I x R x L) [Volts]
1000
Where,
Vd is the voltage drop to be calculated in volts

Fv is the multiplication factor determined from Table E3 on page 311

I is the current in amperes

R is the resistance in ohms per kilometer (Ω/Km) from Table E1

L is the length in meters

From Table E3 on page 311 we have Fv = 1 for a balanced 3-phase load

Substituting values to: Vd = (Fv x I x R x L)

1000

Vd = (1 x 100A x 0.88 Ω/Km x 100m) = 8.8 V

1000
METHOD 2

Use the formula: Vd = I x R [Volts]

Where,
R = (Value of R in Ω/Km x L in meters) [Ohms]
1000

R = 0.88 Ω/Km x L = 0.88Ω x L = 0.88 Ω x 100m = 0.088 Ω

1Km 1000m

Vd = I x R = 100 A x 0.088 Ω = 8.8 V

METHOD 3

Use this formula Vd = (mV/A/m) x A x m [Volts]

1000
Where,
mV is the unit of voltage in milli-volts where 1V = 1000mV

A is the unit of current in amps

m is the unit of length in meters

(mV/A/m) = 1.5 for a 25mm2 copper (Cu) determined from Table 6.3(b) on page 105 or Table 6.4(b)
on page 107 of the regulation book – The distinction between the two tables is that Table 6.3(b)
makes provision for a 1 mm2 whereas Table 6.4(b) does not: In the exam you will most likely be
given Table 6.3(b), other than this they are similar in all aspects.

Substituting values to Vd = (mV/A/m) x A x m [Volts]

1000

Vd = (1.5mV/A/m) x 100A x 100m = 15 V

1000

For phase to neutral: Vd = 15/√3 = 8.66 V [NB: See section 6.2.7 page 120]
METHOD 4

Use Vd = (I x ρL) [Volts]

A
Where,
A is area of cable in this case 25mm2
ρ = 0.0223, the resistivity of copper conductors

Vd = (I x ρL) = (100 x 0.0223 x100) = 8.92 V

A 25mm2

ADDITIONAL METHODS

Alternative Method 1:

In the case of a load with impedance, the voltage drop can be calculated from the following

Vd = (Fv x I x (R cosθ + X cosθ) x L)/1000 [Volts]

The phase angle θ of the load is determined by power factor = cos θ and the reactance X may be obtained
from cable manufactures

Alternative Method 2:

Voltage drop = (1.72 x I x R) / √3 [Volts]

Alternative Method 3:

The value of the Voltage Drop may be derived from a monogram.

**If the load is either inductive or capacitive, what methods would you use to determine the actual volt
drop? So if the load is either inductive or capacitive, how do we determine the actual volt drop?

We apply Ohms Law: Vd = I x R [Volts]

Where,

R = XL for an inductive load

R = XC for a capacitive load:

 EXAMPLE 3
VOLTAGE DROP CALCULATIONS

APRIL 2010 EXAM QUESTION

QUESTION 1: SANS 10142 PART 1 OF 2008: ANNEXURE E: VOLT DROP (COMPULSORY)

1.1 With reference to FIGURE 1, on the DIAGRAM SHEET (Diagrams.pdf attached), calculate the following:

1.1.1 The total volt drop of the supply (4)

1.1.2 The total resistance of the supply cable (1)

1.1.3 Test the answer in QUESTION 1.1.1 above by making use of any other method. (3)

1.2 By making use of TABLE E1 (attached) only, calculate the actual length of a 10 mm2 copper ECC
conductor if the total resistance of the conductor is 0,22 ohm. (2)

[10]

APRIL 2010 EXAM QUESTION: ANSWER TO QUESTION 1

Assumption: The installation uses copper cables

We know that the voltage drop shall not exceed 11.5V [5% of 230V] according to the code

Given:

Point of control: Supply voltage of 230V

The resistive load of 50A, 65m away from the point of control

From the Point of control to the joint we have 40m of 16mm2 copper cable

From the joint to the load we have 25m of 10mm2 copper cable

We can get R which is the resistance of the cable in ohms per kilometer (Ω/Km) from Table E1:

For a 16mm2 we have R = 1,4 Ω/Km

For a 10mm2 we have R = 2,2 Ω/Km

1.1.1 The total volt drop of the supply:

Use this formula from page 308

Vd = (Fv x I x R x L) [Volts]
1000
Where,

Vd is the voltage drop, in volts

Fv is the multiplication factor determined from Table E3 on page 311

I is the current in amperes

R is the resistance in ohms per kilometer (Ω/Km) from Table E1

L is the length in meters

From Table E3 on page 311 we have Fv = 2 for a single-phase load

Vd = (Fv x I x R x L)
1000

Vd = (2 x 50A x 1.4 Ω/Km x40m) + (2 x 50A x 2.2 Ω/Km x 20m)

1000 1000

Vd = 5.6V + 5.5V = 11.10 V

1.1.2 The total resistance of the supply cable:

Total Resistance of the supply cable = Resistance of a 16mm2 cable + Resistance of a 10mm2 cable

= 1.4 Ω/1000m x 40m + 2.2Ω /1000m x 25m = 0.111Ω

1.1.3 Test the answer in QUESTION 1.1.1 above by making use of any other method:

Alternative methods:

Method 1:

Use the formula: Vd = I x R [Volts]

Where,
I = 50 A and R = 0.111Ω
Vd = I x R = 50A x 0.111Ω = 5.55V

NB: Remember that we have to account for the voltage drop on neutral.

Vd (Neutral) = I x R = 50A x 0.111Ω = 5.55V

Total Voltage Drop = 5.55V + 5.55V = 11.10V

Method 2:

Use the formula: Vd = (I x ρL) [Volts]

A
Vd = Voltage Drop (16mm2) + Voltage Drop (10mm2)

= (50A x 0.0223 x 40m) + (50A x 0.0223 x 25m)

16mm2 10mm2

= 2.7875V + 2.7875V = 5.575V

NB: Remember that we have to account for the voltage drop on neutral.

Vd (Neutral) = Voltage Drop (16mm2) + Voltage Drop (10mm2)

= (50A x 0.0223 x 40m) + (50A x 0.0223 x 25m)

16mm2 10mm2

= 2.7875V + 2.7875V = 5.575V

Total Voltage Drop = 5.575V + 5.575V = 11.15V
** Method 3:
Use the formula: Vd = (mV/A/m) x A x m [Volts]
1000

Vd = Voltage Drop (16mm2) + Voltage Drop (10mm2)

= 2.8 x 50A x 40m + 4.4 x 50A x 25m

1000 1000

= 5.60V + 5.50V = 11.10V

**Please note that neither Table 6.3(b) nor Table 6.4(b) was given in the exam, so we would not be able
to use this method to authenticate our answer, but it is given here to prove that our answer is indeed
right.

1.2 Calculate the actual length:

Recall that resistance in Ohms can also be expressed in the following form:

R = (Value of R in Ω/Km x L in meters) [Ohms]

1000

If we are given resistance of 0.22 ohms and we have to calculate the actual length of a 10 mm2 (2.2 Ω/Km)
ECC, we can apply the above formula where the only unknown is L in meters:

0.22Ω = 2.2 Ω/Km x L

1000
Therefore,
L = 0.22Ω x 1000m = 100m
2.2 Ω/Km
 EXAMPLE 4
VOLTAGE DROP CALCULATIONS

With reference to FIGURE 1 (See Diagrams.pdf Attached), calculate the following:

1. The minimum cable size

2. Maximum volt drop at load (E).

3. Touch voltage at load (A).

4. The main CB value and kA rating.

1. The minimum cable size

Given:

Red Phase = 160 Amps White Phase = 90 Amps Blue Phase = 110 Amps AND L = 120m

The total voltage drop shall not exceed 5% of 400V = 20V according to the code.

Neutral current (IN) =√ (R2 + W2 + B2) - (RW+RB+WB) ………. See Page 305

= √ (1602+902+1102)-(160*90+160*110+90*110)

= 62.45 A

METHOD 1:

Consider the worst case which is on Red Phase = 160 Amps

Vd = (mV/A/m x A x L) [Volts]
1000

20V = mV/A/m x 160A x 120m

1000

mV/A/m = 20V x 1000

160A x 120m

= 1.042
From Table 6.3(b) or Table 6.4(b) 1.042 is somewhere between a 35mm2 cable and a 50mm2 cable (taking
from the z column – always use the z values).

For a 35mm2 cable mV/A/m is 1.10 For a 50mm2 cable mV/A/m is 0.81
Vd (Neutral) = mV/A/m x A x L Vd (Neutral) = mV/A/m x A x L
1000 1000

= 1.10 x 62.45 x 120 = 0.81 x 62.45 x 120

1000 1000

= 8.2434V = 6.070V

Vd (Red Phase) =Vd at A + Vd at B + Vd at E Vd (Red Phase) =Vd at A + Vd at B + Vd at E

= mV/A/m x A x L + mV/A/m x A x L + mV/A/m x A x L = mV/A/m x A x L + mV/A/m x A x L + mV/A/m x A x L

1000 1000 1000 1000 1000 1000

= 1.10 x 160 x 30 + 1.10 x 100 x 50 + 1.10 x 20 x 70 = 0.81 x 160 x 30 + 0.81 x 100 x 50 + 0.81 x 20 x 70
1000 1000 1000 1000 1000 1000

= 5.28V + 5.5V + 1.54V = 3.888V + 4.05V + 1.134V

= 12.32V = 9.072V

Total Vd = Vd (Red Phase) + Vd (Neutral) Total Vd = Vd (Red Phase) + Vd (Neutral)

Total Vd = 12.32V + 8.2434V = 20.56V Total Vd = 9.072V + 6.070V = 15.14V

20.56 is greater than 20V 15.14V is less than 20V

Therefore 50mm2 is suitable for the installation

METHOD 2:

Consider the worst case which is on Red Phase = 160 Amps

From Table 6.8 on page 114, a 50mm2 cable can handle 169 Amps and thus is selected. To confirm correct
cable size, calculate Voltage Drop (Red Phase and Neutral)

Neutral current (IN) =√ (R2 + W 2 + B2) - ( RW+RB+WB )

= √ (1602+902+1102)-(160*90+160*110+90*110)

= 62.45 A

Use Table 6.3(b) or 6.4(b)

For 50mm2 mV/A/m = 0.81 (taking from the z column – always use the z value)

Vd (Neutral) = mV/A/m x A x L
1000

= 0.81 x 62.45 x 120

1000

= 6.070V

Vd (Red Phase) =Vd at A + Vd at B + Vd at E

= mV/A/m x A x L + mV/A/m x A x L + mV/A/m x A x L

1000 1000 1000

= 0.81 x 160A x 30m + 0.81 x 100A x 50m + 0.81 x 20A x 70m

1000 1000 1000

= 3.888V + 4.05V + 1.134V

= 9.072V

Total Vd = Vd (Red Phase) + Vd (Neutral)

Total Vd = 9.072V + 6.070V = 15.14V

Total Voltage Drop = 15.14V which is less than 5% of 400V = 20V, it is within limits. So main cable size of
50mm2 4 Core copper cable is correct.

The problem with the second method is that Table 6.8 is not normally given in the exam and that we
normally not given the method of installation; therefore it will be difficult to use this method in the exam.
2. Maximum volt drop at load (E).

At Load E L = 70m I=20 Amps

Vd (Neutral) = mV/A/m x A x L
1000

= 0.81 x 62.45 x 70
1000

= 3.5409V

Vd (Red Phase) at E = mV/A/m x A x L

1000

= 0.81 x 20 x 70
1000

= 1.1340V

Total Vd = Vd (Red Phase) + Vd (Neutral)

Voltage Drop at load E = 1.1340V + 3.5409V = 4.6749V

3. Touch voltage at load (A).

From Table E1 page 306 R ECC for 50mm2 is 0.44 Ω/Km L = 30m and ICB = 160 Amps

Touch Voltage = 2 x ICB x RECC

Touch Voltage = 2 x 160A x 0.44 x 30

1000

Touch Voltage at load A = 4.224V

4. The main CB value and kA rating.

Main Circuit Breaker = 160 Amps

For 50mm2 cable, X = 0.076 and R = 0.44 from Table E1

Z = L x √ (0.076)2 + (0.44)2
1000
= (120 x 0.44652)/1000 = 0.05435 Ohms

kA = V/(√3 x 0.05435) = 400/0.09414 = 4.25 kA

 EXAMPLE 5
VOLTAGE DROP CALCULATIONS

With reference to FIGURE 2 (See Diagrams.pdf attached). Calculate the following:

1. The minimum main cable size if Voltage drop L2 not to exceed 2%.

2. The minimum ECC size.

3. The maximum touch voltage at the load

Given:

Load Moment at L2 = 2090.88 A.m

L1 =45m
L2 = 20m
Supply = 250 V [SINGLE PHASE LOAD]

Load Moment = 2 x I x L
Load Moment = 2 x I x L2

2090.88 = 2 x I x 20m

I = 2090.88 / (2 x 20) = 52.272 Amps

52.272 Amps is the maximum connected load according to the code.***

1. The minimum main cable size if Voltage drop L2 not to exceed 2%.

Minimum main cable size (that is L1) if voltage drop L2 is not to exceed 2%

Voltage Drop (B to Load) must be less than or equal to 250V x 2% = 5V

From A to B the length is L1 = 45m

Total voltage drop shall not exceed 5% of 250V = 12.5V according to the code, that is Voltage drop (A to
B) + Voltage drop (B to Load) shall be less than or equal to 12.5V.

Vd (A to B) + 5V =12.5V

mV/A/m x 52.272 x 45 = 12.5V – 5V

1000

mV/A/m x 52.272 x 45 = 7.5V

1000

mV/A/m = 3.188845
3.18845 is between 10mm2 and 16mm2 from Table 6.3(b) (Two-core, single phase a.c. column)

mV/A/m for a 16mm2 is 2.8 mV/A/m for a 10mm2 is 4.4

Voltage Drop = mV/A/m x (A x m) Voltage Drop = mV/A/m x (A x m)

=(2.8/1000) x 52.272 x 45 = (4.4/1000) x 52.272 x 45

= 6.586272 V less than 7.5V = 10.349856V greater than 7.5V

From the above Table 16mm2 minimum main cable size is selected

Please Note: Actual % voltage drop at L2 is 1.17089% of 250V.***

2. The minimum ECC size.

Load Moment =2090.88 A.m

I = 2090.88/L x 2
=2090.88/20 x 2
=52.272 Amps
Say 60 Amps Main Circuit Breaker
ICB =60A
The value of R in Ω/Km for ?? = ??
L = L1 + L2 = 65m

Substituting values we have, T. V. = 2 x ICB x RECC

30V = 2 x 60A x RECC
RECC = 30V/120A = 0.25Ω

Substituting values we have, R = (Value of R in Ω/Km x L in meters) [Ohms]

1000
Value of R in Ω/Km = 0.25Ω x 1000 = 3.846Ω/Km
65m

From Table E1, 3.846Ω/Km is somewhere between a 4mm2 and a 6mm2

For a 6mm2 cable we have 3.6Ω/Km For a 4mm2 cable we have 5.5Ω/Km
I=60A and L=65m I=60A and L=65m
R = (Value of R in Ω/Km x L in meters) [Ohms] R = (Value of R in Ω/Km x L in meters) [Ohms]
1000 1000
R = 3.6Ω/Km x 65m = 0.234Ω R = 5.5Ω/Km x 65m = 0.3575Ω
1000m 1000m
T. V. = 2 x ICB x RECC T. V. = 2 x ICB x RECC
= 2 x 60A x 0.234 Ω = 2 x 60A x 0.3575 Ω
=28.08V =42.90V
28.08V is less than 30V maximum allowed T.V. 42.90V is greater than 30V maximum allowed T.V.
Therefore the minimum area of the ECC is 6mm2
Another way of answering this question is to proceed like this:

Test for the minimum ECC size

First assume minimum ECC size to be 10mm2

Touch Voltage = 2 x ICB x RECC = 2 x 60 x (2.2/1000) x 65 = 17.16V acceptable which is less than 30V

Next assume minimum ECC size to be 6mm2

Touch Voltage = 2 x ICB x RECC = 2 x 60 x (3.6/1000) x 65 = 28.08V acceptable which is less than 30V

Next assume minimum ECC size to be 4mm2

Touch Voltage = 2 x ICB x RECC = 2 x 60 x (5.5/1000) x 65 = 42.9V which is greater than 30V

The minimum size of ECC = 6mm2 is selected.

To confirm that the selected cable size is correct, calculate the actual Touch Voltage

Touch Voltage = 2 x 52.272 x (3.6/1000) x 65 = 24.46V which is less than 30V

3. The maximum touch voltage at the load

Touch Voltage should not exceed 30V according to the code

 EXAMPLE 6
VOLTAGE DROP AND PSCC CALCULATIONS

Calculate with reference to FIGURE 6 (See Diagrams.pdf attached), the following:

1. Volt drop at load (F).

2. The minimum size of the ECC.

3. The three phase PSCC at load (F).

4. The two phase PSCC at load (F).

5. The Load voltage at load (F).

1. Volt drop at load (F):

Given: I = 130 Amps

Length = 100m

mV/A/m = 1.10 for a 35mm2 copper cable

Vd at F = mV/A/m x A x L
1000

= 1.10 x 130A x 100m

1000

= 14.30 Volts

% Voltage Drop at F = 14.30/400 x 100 = 3.575% within 5% tolerance

2. The minimum size of the ECC:

ICB = 160A and L = 100m

The value of R in Ω/Km for ?? = ??

Substituting values we have, T. V. = 2 x ICB x RECC

30V = 2 x 160A x RECC
RECC = 30V/320A = 0.09375Ω

Substituting values we have, R = (Value of R in Ω/Km x L in meters) [Ohms]

1000
Value of R in Ω/Km = 0.09375Ω x 1000 = 0.9375Ω/Km
100m
From Table E1, 0.9375Ω/Km is somewhere between a 16mm2 and a 25mm2

For a 16mm2 cable we have 1.4Ω/Km For a 25mm2 cable we have 0.88Ω/Km
I=160A and L=100m I=160A and L=100m

R = (Value of R in Ω/Km x L in meters) [Ohms] R = (Value of R in Ω/Km x L in meters) [Ohms]

1000 1000

R = 1.4Ω/Km x 100m = 0.14Ω R = 0.88Ω/Km x 100m = 0.088Ω

1000m 1000m

T. V. = 2 x ICB x RECC T. V. = 2 x ICB x RECC

= 2 x 160A x 0.14 Ω = 2 x 160A x 0.088Ω
= 44.80V = 28.16V

44.80V is greater than 30V maximum allowed T.V. 28.160V is less than 30V maximum allowed T.V.

Therefore the minimum area of the ECC is 6mm2

Another way of answering this question is to proceed like this:

Test for the minimum ECC size

First assume minimum ECC size to be 25mm2

Touch Voltage = 2 x ICB x RECC = 2 x 160 x (0.88/1000) x 100 = 28.16V acceptable which is less than 30V

Next assume minimum ECC size to be 16mm2

Touch Voltage = 2 x ICB x RECC = 2 x 160 x (1.4/1000) x 65 = 44.80V acceptable which is less than 30V

Next assume minimum ECC size to be 10mm2

Touch Voltage = 2 x ICB x RECC = 2 x 160 x (2.2/1000) x 100 = 70.4V which is greater than 30V

The minimum size of ECC = 25mm2 is selected.

The easiest method is:

R = 0.096Ω for 160 Amps according to Table 8.1

R = pL/A

A = (0.0223 x 100m)/0.096 Ω = 23.223 say 25mm2 cable size of ECC

3. The three phase PSCC at load (F):

Get the values of R and X from Table E1 on page 307 for a 35mm2 cable

Z = (L x √(R 2 + X2))/1000 = (100m x √(0.632 + 0.0762))/1000 = 0.0634567Ohms

PSCC = V/√3 x Zof cable = 400/√3 x 0.0634567 = 3.6393 kA

4. The two phase PSCC at load (F):

PSCC = VLine
2 x ZLine

= 400 V
2 x 0.0634567

= 3.1518 kA

5. The Load voltage at load (F):

Load Voltage = Supply Voltage – Voltage Drop = 400V – 14.30V = 385.70 V

Touch Voltage at load F = 2 x I x R = 2 x 130 x 0.63/1000 x 100 = 16.38 Volts

 EXAMPLE 7
VOLTAGE DROP CALCULATIONS

Refer to FIGURE 7 (See Diagrams.pdf attached) and calculate the following:

1. Calculate the minimum cable size for the loads.

2. Volt drop at L1, L2 and L3

3. Touch voltage at L3.

4. PSCC at L1.

Given: Single Phase Supply (230V)

Total Current = 140A

Total Length = 96m

1. Calculate the minimum cable size for the loads.

According to the code the total voltage drop at the furthest point of consumption (L2) shall not exceed 5%
of 230V which is in this case 11.5V

Vd = mV/A/m x A x L
1000

11.5V = mV/A/m x A x L
1000

mV/A/m = 11.5V x 1000

140A x 96m

mV/A/m = 0.8556547

0.8556547 is between 50mm2 and 70mm2 from Table 6.3(b) (Two-core, single phase a.c. column)

mV/A/m for 50mm2 is 0.94 mV/A/m for 70mm2 is 0.65

Voltage Drop = mV/A/m x (A x m) Voltage Drop = mV/A/m x (A x m)

=(0.94/1000) x 140A x 96m = (0.65/1000) x 140 x 96

= 12.6336 V = 8.736 V

12.6336V is greater than 11.5 8.736V is less than 11.5

From the above Table 70mm2 minimum cable size for the loads is selected
 2. Volt drop at L1, L2 and L3

Voltage Drop At Load L1 Voltage Drop At Load L2 Voltage Drop At Load L3

L= 18m L= 94m L= 11m
I = 140A I = 60A I = 20A
mV/A/m =0.65 for 70mm2 cable mV/A/m =0.65 for 70mm2 cable mV/A/m =0.65 for 70mm2 cable

Voltage Drop = mV/A/m x (A x m) Voltage Drop = mV/A/m x (A x m) Voltage Drop = mV/A/m x (A x m)

=(0.65/1000) x 140A x 18m =(0.65/1000) x 80A x 94m =(0.65/1000) x 20A x 11m

= 1.638 V = 4.888 V = 0.143 V

3. Touch voltage at L3.

At L3: I = 20A
L = 95m but we use 96m the total length of ECC

From Table E1, R in Ω/Km is 0.31 for a 70mm2 cable

R = (Value of R in Ω/Km x L in meters) [Ohms]

1000

R = 0.31Ω/Km x 96m = 0.02976Ω

1000m

Substituting values we have, Touch Voltage = 2 x ICB x R ECC

T.V. = 2 x I x R = 2 x 20A x 0.02976Ω = 1.1904V

4. PSCC at L1.

R = 0.31 Ohms and X = 0.074 Ohms from Table E1 on page 307

Zof cable = (L x √(R2 + X 2))/1000 = (96m x √(0.312 + 0.0742))/1000 = 0.0305961 Ohms

PSCC = V/Zof cable = 230V/0.0305961 = 7.5172979 kA

 EXAMPLE 8
GENERAL CALCULATIONS

TABLE 8.1 of the Code refers to the maximum resistance of the earth continuity conductor, with reference
to a known protective device. Derive from basic principles the approximate resistance value for the
following protective devices:
1. 100 amps
2. 20 amps
3. 10 amps
4. 1 amp

For questions like this, remember that the maximum allowed touch voltage (TV) according to the code is
30V. Use this formula:
Touch Voltage = 2 x ICB x RECC
Where,
ICB is the value of the protective device (Circuit Breaker or Fuse) = the value given in amps
RECC is the resistance value of the Earth Continuity Conductor =?
Touch Voltage (TV) = 30V

1.

ICB = 100A
Substituting values we have, T. V. = 2 x ICB x RECC
30V = 2 x 100A x RECC
RECC = 30V/200A = 0.15Ω

2.

ICB = 20A
Substituting values we have, T. V. = 2 x ICB x RECC
30V = 2 x 20A x RECC
RECC = 30V/40A = 0.75Ω

3.

ICB = 10A
Substituting values we have, T. V. = 2 x ICB x RECC
30V = 2 x 10A x RECC
RECC = 30V/20A = 1.50Ω

4.

ICB = 1A
Substituting values we have, T. V. = 2 x ICB x RECC
30V = 2 x 1A x RECC
RECC = 30V/2A = 15Ω
 EXAMPLE 9
GENERAL CALCULATIONS

1. Calculate the maximum value of the protection device if the following is given: 100 meters of copper
ECC 4mm2.

2. Calculate from basic principles the maximum value of the protection device, if the following is given:
the length of ECC (earth continuity conductor) is 69 meters of copper with an area of 6mm2.
3. Calculate from first principles, the actual maximum length of the ECC (earth continuity conductor) if
the main protection is 63A using a 6 mm2 copper conductor. What is the actual resistance value for
the above length?
4. Calculate from basic principles the actual length of a 10 mm2 earth continuity conductor, if the main
protection device is 60 amps and the touch voltage not to exceed 30V. What is the actual touch
voltage at the load for this conductor?
5. Calculate from basic principles the aerial area of the ECC conductor if the protection rating is 60A
and the total length is 150 meters.
6. Table 6.28 refers to the minimum size and maximum length of copper earth continuity conductors
with reference to the rated protective device. Calculate, from basic principles the minimum size of
the earth continuity conductor if the main protection is 63A and the total length is 66,66 meters.
Use any other alternative method to calculate the minimum size of the ECC conductor.

For questions like this, remember two facts: the first is that the maximum allowed touch voltage (TV)
according to the code is 30V. Use this formula:
Touch Voltage = 2 x ICB x RECC
Where,
ICB is the value of the protective device (Circuit Breaker or Fuse) = the value given in amps
RECC is the resistance value of the Earth Continuity Conductor =?
Touch Voltage (TV) = 30V

The second fact: if you can recall from the previous discussion that the resistance in Ohms can also be
expressed in the following form:

R = (Value of R in Ω/Km x L in meters) [Ohms]

1000

Please remember to use Table E1 for the value of R in Ω/Km for each given case.
1.

ICB =?
The value of R in Ω/Km for 4mm2= 5.5 Ω/Km [From Table E1]
L = 100m

Substituting values we have, R = (Value of R in Ω/Km x L in meters) [Ohms]

1000
= (5.5 Ω/Km x 100m) = 0.55 Ω
1000

Substituting values we have, T. V. = 2 x ICB x RECC

30V = 2 x ICB x 0.55 Ω
ICB = 30V/1.10Ω = 27.273A
Therefore the value of the protective device is 27A

2.

ICB =?
The value of R in Ω/Km for 6mm2= 3.6 Ω/Km [From Table E1]
L = 69m

Substituting values we have, R = (Value of R in Ω/Km x L in meters) [Ohms]

1000
= (3.6 Ω/Km x 69m) = 0.2484 Ω
1000

Substituting values we have, T. V. = 2 x ICB x RECC

30V = 2 x ICB x 0.2484 Ω
ICB = 30V/0.4968Ω = 60.3865A
Therefore the value of the protective device is 60A

3.

ICB =63A
The value of R in Ω/Km for 6mm2= 3.6 Ω/Km
L=?

Substituting values we have, T. V. = 2 x ICB x RECC

30V = 2 x 63A x RECC
RECC = 30V/126A = 0.2380952Ω

Substituting values we have, R = (Value of R in Ω/Km x L in meters) [Ohms]

1000
L = 0.2380952Ω x 1000 = 66.1376m
3.6
What is the actual resistance value for the above length?
Substituting values we have, R = (Value of R in Ω/Km x L in meters) [Ohms]
1000
= (3.6 Ω/Km x 66.1375m) = 0.2381 Ω
1000