I indicated that the Earthing Systems Design Steps process has (3) main steps:
Earthing Systems Design
Steps
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And I explained the first step: Data Collection in the following Articles:
I explained the second step: Data Analysis in the following Articles:
And I explained What we are going to design for grounding system in any building in the following Articles:
And, in Article " Grounding Design Calculations – Part One ", I indicated the following:
Grounding System Design
Calculations according to type of the building
The procedures for performing the Grounding System Design
Calculations can differ slightly according to the type of the building
as follows:
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First: Domestic,
commercial and industrial premises
We mean by domestic, commercial and
industrial premises, all installations up to 1,000 V ac and 1,500 V dc -
between phases, with some minor exceptions.
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And I started discussion for Methods of Grounding Design Calculations of Domestic, commercial and industrial premises as follows:
Methods of Grounding Design
Calculations
There are many methods can be used for
performing Grounding System Design Calculations But the common methods
are:
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In this Article and Article " Grounding Design Calculations – Part Two ", I explained the first method of grounding design calculations: Equations Method and solved examples.
And I explained the second method of grounding design calculations: Nomographs Method in Article " Grounding Design Calculations – Part Three".
Also, I explained third method of grounding design calculations: Excel Spreadsheets Method in Article " Grounding Design Calculations – Part Four ".
Today, I will explain the Other Methods for Performing of Grounding System Design Calculations.
You can preview the following Articles for more info:
Third Method: Tables Method
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Important
The
Tables Method is used for sizing earthing conductors only, not for
calculating the earth resistance value.
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Methods Of Sizing
Earthing Conductors
There are two methods for sizing
protective conductors including earthing conductors as follows:
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Method#1: By Using
The Adiabatic Equation
This method is used where:
The cross-sectional area of the
earthing conductors, where calculated, shall be not less than the value
determined by the following equation:
Where:
S is the nominal
cross-sectional area of the conductor in mm2.
I is the average fault
current, in A r.m.s.
t is the fault current
duration, in s.
k is the r.m.s. current density, in A/mm2 .
Notes:
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First: Getting K Value
To get values of K, we have (2)
cases to be considered as follows:
Case#1: t= 1 s and initial conductor
temperature of 30 °C
In this case, the value of K is given in below Table-1:
Case#2: t= 1 s and other initial/ final
conductor temperature
In this case, the value of K is obtained from the
following equation:
Where:
T1 is the initial temperature, in °C;
T2 is the final temperature, in °C;
K and β have the values given in
Table-2.
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Second: Getting The Average
Fault Current Value (I)
To get values of average fault
current (I), we have (2) methods:
Method#1: by using Earth fault
loop impedance
The value of the fault current
(I) will be determined from the following equation:
I = Uo / Zs
Where:
Uo the nominal Voltage
Zs the measured value of earth
fault loop impedance corrected to the operating temperature of the conductor.
Getting value of Zs:
Zs = Ze + (R1 +
R2) ohm
Where:
Ze: the impedance of the supply
side of the earth fault loop.( it can be obtained from utility companies but
typical values are 0.35 ohm for TN-C-S supplies and 0.8 ohm for TN-S
supplies),
R1: the resistance of the phase
conductor,
R2: the resistance of the earth
conductor.
Values of (R1 + R2) are given in
the below table (table 9A of the IEE on site guide):
Note:
(R1 + R2) in above table is at 20◦C, for other
temperatures, the value will be increased 2% for each 5◦C rise in
temperature.
Example#1:
The design current for a circuit
is 38A. The current carrying capacity of cable has been calculated and the
circuit is to be wired in 70◦C thermoplastic singles, 6mm2 live conductors and
1.5mm2 circuit protective conductors, Supply is 230V TNS with a Ze of 0.38 ohm, circuit is 28m long. Calculate
Zs
and
fault current.
Solution:
From Table 9A in above, it will be seen
that the (R1 + R2) for 6 mm2/1.5 mm2 copper is 15.2 mΩ per meter at a
temperature of 20◦C.
For cables at their operating
temperature of 70◦C.The
resistance of the copper conductor will increase 2% for each 5◦C rise in
temperature.
50◦C rise in temperature = 2% *
50/5 = 20% increase, so (R1 + R2) need to be multiplied by 1.2
(R1 + R2) = (15.2 mΩ x 1.2 x 28 m)/1000 = 0.51 Ω
Zs = Ze + (R1 + R2) = 0.38 +
0.51 = 0.89 Ω
The value of the fault current
(I) will be determined from the following equation:
I = Uo / Zs = 240 V / 0.89 Ω = 269.4 A
Method#2: from tables and
equations
To get values of average fault
current (I) from tables and equations, we have (3) cases to be
considered as follows:
Case#1: t= 1 s and 3 s and using
a standard size of copper strips, the values of (I) are given in Table-3.
Case#2: t= 1 s and 3 s and using
a standard size of copper strips, the values of (I) are given in Table-4.
Case#3: For other durations, the
values of (I) are calculated from one of the following equations:
Where:
I1 is the fault current for 1 s
duration, in A r.m.s. (given in Table-3 and Table-4).
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Third: Getting (t) Value
The
value of the fault current duration (operating time for the disconnecting
device), can be found from the time / current curve for the disconnecting
device.
For
example, as in below image for 200 A
BS 88 Fuse , with a fault current of 2000 A, the protective device has an
operating time (t) = 0.5 seconds.
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Thermal Constraints For
Earthing Conductors
The initial temperature (T1) may be
increased than the measured value, if there are continuous earth leakage
currents through the earthing conductors and in this case:
Cross sectional area constraints for
earthing conductors:
Thickness constraints for earthing
strips:
Example#2:
An earthing conductor buried in the
ground is protected against corrosion by a sheath, but is not protected
against mechanical damage. The minimum size copper conductor that may be
installed is
Solution:
From the above table, The minimum size
copper earthing conductor is 16mm2.
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Method#2: By Using
Tables
The easier method is to
determine the earthing conductor (protective conductor) size from Tables but
this may produce a larger size than is strictly necessary, since it employs a
simple relationship to the cross-sectional area of the phase conductor(s).
The used tables for determining
of earth conductor size are existing in two standards/codes as follows:
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First: BS 7671 Requirements for
Electrical Installations (IEE Wiring Regulations)
Note:
BS 7671 tables are applied for
(3) cases as follows:
Case#1: The minimum
cross-sectional area of a protective conductor can be determined by selection
from Table-5,
Getting value of K1:
K1 is the value of k for the
line conductor, selected from BS 7671, Table 43.1 in Chapter 43 according to the
materials of both conductor and insulation.
Getting value of K2:
K2 is the value of k for the
protective conductor, selected from BS 7671, Tables 54.2 to 54.6 as applicable.
   
Case#2: Buried earthing
conductor, the minimum cross-sectional area can be determined by selection
from Table-6
Case#3: For a TN-C-S (PME)
supply, the minimum cross-sectional area can be determined by selection from
Table-7.
Example#3:
What
is the CSA of an earthing conductor selected from table-5, where a line conductor
CSA is:
Note
that earth conductor will has same material as the line conductor
Solution:
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Example#4:
Use the adiabatic equation
to determine the minimum cross-sectional area of a circuit protective
conductor suitable for use on a radial circuit protected by a 30 amp fuse to
BS 3036.
The 230 volt circuit is
wired in single-core 70° C PVC insulated cables with copper conductors, which
are installed in steel conduit. The cross sectional area of the phase and
neutral conductors is 4 mm2. The earth fault loop impedance Zs is 1.44 ohms.
Solution:
Fault current, I = Uo / Zs
= 230 V / 1.44 = 160 amps.
The disconnection time, t,
is found by reference to the appropriate time/current characteristic of a 30
amp BS 3036 fuse. This is found in Fig. 3.2A in Appendix 3 of BS 7671.
The disconnection time is
approximately 0.8 seconds.
The value of k is obtained
from Table 54.3 of BS 7671 and is 115.
Now the adiabatic equation
can be used as follows:
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In the next Article, I will explain NEC Article 250 tables for sizing of earth conductors. Please, keep following.
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