Assignment Answers Pdf Of Designing Of Earthing System:ELEC5204

Assignment Answers Pdf Of Designing Of Earthing System:ELEC5204

Power system 1
POWER SYSTEM AND PROTECTON
Student Name
Affiliated Institution
Instructor
Course
Date
EARTH DESIGN SYSTEM
ABSTRACT
This work pres …

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Power system 1
POWER SYSTEM AND PROTECTON
Student Name
Affiliated Institution
Instructor
Course
Date
EARTH DESIGN SYSTEM
ABSTRACT
This work presents of the designing of earthing system of a22kv substation for calculations of
the expected parameters. In most substation, Earthing is more important not only to give safety
to the people around or passing gin process of earthed structures and equipment against the
electric system. Therefore, this paper is to provide information concerning the safety of earth
activities in asubstation, commonly in a22KV transformers.
Introduction
This paper focuses on the earthing practices and design of asubstation system for the earthing
system of 1500KVA. The main purpose of this report is to check the earthing system on the
single phase calculations, which is proportional to the ground, look at the earth grid of the
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substation also seek to check and elaborate briefly on the touch and step potentials of the system
(the system in figure 1in the assignment). The paper also elaborates briefly the field checks
needed to asses ’soil resistivity, procedures and formulas and calculations, phase to ground
calculations, asses and simulate grid conductor size that helps in designing the earth system.
Body
Analysis part
The following formulas are used in the calculations, that is when calculating the phase to ground
our system, the earth grid for the substation, touch and step voltages.
=1973.52 ×√ ( × × × 〖10 〗^4 )/( /  {1+(( − )/(
+ ))} )
Where I=rms value in KA
Amm^2 =conductor sectional size in mm^2
= ℃
= ℃
= ℎ 0℃
= â„Ž
= 1
0

0−
TCAP =thermal capacity factor
For ground resistance, the formula used is
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= 1
+ 1
20 1+ 1
1+ â„Ž 20
Where =
Lt =total length of grid conductor
A =total area enclosed of the earth grid
h=depth of the earth grid conductor
for calculating the grid current, the equation below is used
= × 30
Where =maximum grid current
3 =symmetrical fault current of the substation for conductor sizing in A
=current diversity factor
When calculating the step voltage, the formula below is used. [1,3]
=
+1,15
Where =step voltage point
=spacing factor of the step voltage
=current factor for grid geometry
Most of the factors being used in these formulas are provided in the manual IEEE 80 reference.
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.
= OIc=Ic=0
= =
=fault current
=earth fault current
=normal phase voltage at faint location
Z1 =positive phase voltage sequence network impedance to the fault.
Z2 =negative phase sequence network impedance to the fault
Zo =zero phase sequence network I
Z= 2+ 2
1= 1500
2 = 1500
=0.415
Select 1500KVA to be the study base
− = 15000 /100 × 1000
1000MVA into KVA
100 ×1000 =100000KVA
= 15000 /100000 − 0.015
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= (0.415) 15000 /1500 = 0.415
= +
= 0.015 + 0.415
= 0.43
= 0.433 × 1500
1500
( 22
0.433 ×10)× 1000=0.0508 Ω
= 0.0508( 1500
1500 ) = 0.0508
Z= 2+ 2
Z= 0.43 2+ 0.0508 2
= 0.4329
=
3 = 1500
3(0.4529)(0.022)
=90,932.75 â„Ž
Q2.TWO-LAYER SOIL MODEL CALCULATION ON A GIVEN SOIL RESISTIVITY
i. Earth grid resistance ,given that
=max fault current of 1500 KA for 1second at 22KV
=1second
Resistivity =1372.9
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h= 0.3 meter is the thickness of the surface layer mean soil resistivity =3 Ω
Number of ground rod =60, length of ms rod
h= 0.3m is the depth of the underground grid
area occupied by ground grid after design =60105.5 2
=[1
+ 1
√20(1+ 1
1+ â„Ž 20/)]
thus =1372.9[ 1
16.5 + 1
√20×60105.5 (1+ 1
1+ 0.3 20 60105.5
)
=0.10 Ω
ii. Ground potential rise
GPR=1G =1g =
Sf= fault current division factor
= 1.033
= 0.10 Ω,Ig is the value earth electrodes=28.21 when considered= 0.09
GPR= 1×0.7 ×40000 ×0.09= 252OV
3. Earth conductor sizing
Conductor size ignoring the station resistance, the symmetrical ground fault current = 3 is
computed i.e.
=
3 +1+ 2+ +1+ 2+ for the 115KV bus fault.
3 =
3 115000
3
3 0 + 4.0+4.0+10.0 + (10.0+10.0+40.0)
Thus 3 =3180 hence X/R ratio =3.33
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It should be noted that due to the delta wye connection of the transformer, only the positive
sequence 155KV fault impedance is transferred thus
1= 13
115
2
4.0+ 10.0 +0.034+0.034+ 1.014=0.085+ 1;142
=0.034+ 1.014
3 =
3 13000
3
3 0 + 0.085+0.085+0.034 + (1.142+1.142+1.014)
=6814 ,â„Žâ„Žâ„Ž â„Ž
For the case of copper-clad steel wire and the diameter d=0.3m. [2,3]
= = 197.4

× ln( +
+
= 197.4
3.85
0.5 0.00378 5.862 ln 245+700245+40
=65
Question 4Simulation of earth grid
Calculating ground grid resistance, the following formula maybe used
= [1
+ 1
20 (1+1/(1+ â„Ž 20
)
â„Ž =
= â„Ž
= â„Ž
= â„Ž â„Ž
But = …
where
Power system 8
, ,
â„Ž
â„Ž
3555 = 1372.9 ×0.416 ×3.21 ×3870

=60
Also,
(II) space between the loads
â„Ž = …
where lm is the effective space between the rods
= . . .
â„Ž
= 1372.9 × 0.416 × 3.212 × 3870
732.6
=16.5
(III) ℎ ℎ .. = −
â„Ž â„Ž â„Ž
â„Ž â„Ž + â„Ž = 16.5
â„Ž â„Ž â„Ž = 14.2
ℎ , ℎ ℎ = 16.5 − 14.2 = 2.3
Q5. CALCULATE STEP VOLTAGE
i. E step is the step voltage safe limit in volts.
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The difference in surface potential experienced by aperson bridging adistance of 1m with the
feet without contacting any other grounded object.
thus Estep =(1000+6 × )× 0.116
√
where is the function of and â„Žis the factor for computing effective foot impedance in a
finite thickness material
=the resistivity of the surface layer Ω
=the duration shock current in seconds
At =1372.9
=IG/Ls
=substation soil resistivity
K and L are based on physical grid parameters
Where, Ki= 3.21, Lm =2646.0, Ls= 1995.8, =3000 Ωm, Ks= 0.416, Cs= 0.77
Estep= (1000+6 ×0.77 ×3000) 0.116
√1=2430.13V
ii. E touch voltage is the touch voltage safe limit in volts.
This refers to the shock hazard present on an exposed metal work or protective conductor.
The protection different between the ground level lies by ground potential rise and the
surface potential at the point where aperson is standing at the same time having ahard in
contact with agrounded level.
E touch= (1000+1.5 ×0.754 × 1372.9) 0.116
√1=732.6
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iii. Tolerable voltage refers to the minimum abnormal interface voltage that acircuit can
sustain without being damaged.
Relating to E touch and E step
=
Where
IG is symmetrical grid ground current in amps
Ig is symmetrical grid ground current in amps
is the symmetrical factor that depends upon X/R ratio of the fault
= 1+ /(1 − [− 2
] where Ta=X/Wr
w=frequency
But √=human tolerance constant.
Determine the factor at o.5 seconds
= 1+12.6/377 × 0.5[1 − −2× 377 × 0.5
12.6 ]
=1.033 thus Ig=7500
Tolerable voltage=(1.033)(7500)(0.707)=5478V
Q6. BONDING PROTECTIVE
Introduction
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The purpose of main protective bonding is to come up with an earthed equipotential zone. All the
parts that are left naked or exposed within this zone are circuited to the Main Earth Terminal by
ways of the circuit protective conductors of the main protective bonding. Thus bonding is
effective to bond all the steel structures to ensure that all the above stated reason in this
introduction is achieved.
This is because grounding alone is not that much effective to have adequate planning hence
reducing grounding integrity. Thus two important applications are needed for steel grounding i.e.
grounding indoors and outdoors structures, reinforcing bars in concrete construction, fence and
gateposts, steel enclosures and power; signal and power bands to rail and bonds on crane rails.
For us to have effective bonding of steel structures, the following requirements should be
considered-;
Surface separation -regardless of the connection type being used, all the insulating materials
must be eliminated i.e. grease, paint etc. on the steel and any type of rust so as we have a
favorable connection. All the layers must be removed by galvanizing. For electro-galvanized
surface, we need only to brush to asparkling end.
Furthermore, exothermically welded connections do away with the conductor-to-steel
bimetallic interface. A circuit that is welded has no mechanical interface in the electrical path
between two dissimilar metals. Hence, no bimetallic corrosion can occur in the current path.
Connections to rebar present aspecial situation, due to deformations on the bars. Hence on
present aspecial situation, due to deformations on the bars. Hence an exothermically welded
connection doesn ’trequire you to grind off the deformation.
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Moreover, another bonding steel requirement is the mechanical connections .Mechanical lugs
have two bimetallic interfaces. One is between the conductor and the lug and the steel surface.
When two steel surfaces contact. They touch only at the ends of the peaks. This results to an
actual area of current flow being much smaller than the apparent area of InTouch. [2] This
results in contact resistance between the mating surfaces of mechanical connections. Since
grading conducts are usually copper, and grounding clamps are normally alloy or steel, you come
up with abimetallic couple consisting of copper and steel using clamp. Whenever traces of
moisture are observed, corrosion can occur in which steel corrodes to protect the copper, hence
resistance connection is increased.
There are two styles of mechanical connections i.e.
a). the gadget holds the conductor within the connector and holds the surface of the connector
against the structure.
b). the device holds the conductor against the structures. [2,3] In addition, within these styles
they are many shapes, for example clamps to around surface i.e. afence post i.e. normally uses a
U-bolt around the post.
Conclusion on bonding
in conclusion, bonding to steel is one of the hardest types of grounding connections. Therefore,
several requirements are very important in doing so, so that we have an effective steel bonding
structures. And one of these requirements is the use of mechanical connections where the device
can hold on the conductor or itcaptures the conductor within the connector and holds to the
surface of the connector against the structure.
Discussion on the voltage and step voltage
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