Power system 1

POWER SYSTEM AND PROTECTON

Student Name

Affiliated Institution

Instructor

Course

Date

EARTH DESIGN SYSTEM

ABSTRACT

This work pres …

### Preview text

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

Power system 2

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 â€™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 Ã—âˆš ( Ã— Ã— Ã— ã€–10 ã€—^4 )/( / âŸ{1+(( âˆ’ )/(

+ ))} )

Where I=rms value in KA

Amm^2 =conductor sectional size in mm^2

= â„ƒ

= â„ƒ

= â„Ž 0â„ƒ

= â„Ž

= 1

0

0âˆ’

TCAP =thermal capacity factor

For ground resistance, the formula used is

Power system 3

= 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

= Ã— 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.

Power system 4

.

= 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

âˆ’ = 15000 /100 Ã— 1000

1000MVA into KVA

100 Ã—1000 =100000KVA

= 15000 /100000 âˆ’ 0.015

Power system 5

= (0.415) 15000 /1500 = 0.415

= +

= 0.015 + 0.415

= 0.43

= 0.433 Ã— 1500

1500

( 22

0.433 Ã—10)Ã— 1000=0.0508 â„¦

= 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

Power system 6

h= 0.3 meter is the thickness of the surface layer mean soil resistivity =3 â„¦

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

âˆš20(1+ 1

1+ â„Ž 20/)]

thus =1372.9[ 1

16.5 + 1

âˆš20Ã—60105.5 (1+ 1

1+ 0.3 20 60105.5

)

=0.10 â„¦

ii. Ground potential rise

GPR=1G =1g =

Sf= fault current division factor

= 1.033

= 0.10 â„¦,Ig is the value earth electrodes=28.21 when considered= 0.09

GPR= 1Ã—0.7 Ã—40000 Ã—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

Power system 7

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

Ã— 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 Ã—0.416 Ã—3.21 Ã—3870

=60

Also,

(II) space between the loads

â„Ž = …

where lm is the effective space between the rods

= . . .

â„Ž

= 1372.9 Ã— 0.416 Ã— 3.212 Ã— 3870

732.6

=16.5

(III) â„Ž â„Ž .. = âˆ’

â„Ž â„Ž â„Ž

â„Ž â„Ž + â„Ž = 16.5

â„Ž â„Ž â„Ž = 14.2

â„Ž , â„Ž â„Ž = 16.5 âˆ’ 14.2 = 2.3

Q5. CALCULATE STEP VOLTAGE

i. E step is the step voltage safe limit in volts.

Power system 9

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 Ã— )Ã— 0.116

âˆš

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 â„¦

=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 â„¦m, Ks= 0.416, Cs= 0.77

Estep= (1000+6 Ã—0.77 Ã—3000) 0.116

âˆš1=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 Ã—0.754 Ã— 1372.9) 0.116

âˆš1=732.6

Power system 10

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 âˆ’ [âˆ’ 2

] where Ta=X/Wr

w=frequency

But âˆš=human tolerance constant.

Determine the factor at o.5 seconds

= 1+12.6/377 Ã— 0.5[1 âˆ’ âˆ’2Ã— 377 Ã— 0.5

12.6 ]

=1.033 thus Ig=7500

Tolerable voltage=(1.033)(7500)(0.707)=5478V

Q6. BONDING PROTECTIVE

Introduction

Power system 11

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 â€™trequire you to grind off the deformation.

Power system 12

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

Power system 13

- Assignment status: Already Solved By Our Experts
*(USA, AUS, UK & CA PhD. Writers)***CLICK HERE TO GET A PROFESSIONAL WRITER TO WORK ON THIS PAPER AND OTHER SIMILAR PAPERS, GET A NON PLAGIARIZED PAPER FROM OUR EXPERTS**

**NO PLAGIARISM**– CUSTOM PAPER

## Recent Comments