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# course design-Atif

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course design-Atif

Content

1Main Electrical Connection......................................................................................................... 3

1.1 System and load analysis .................................................................................................... 3

1.2 Main Electrical Connection Scheme Choice ....................................................................... 3

1.2.1 Reliability ................................................................................................................. 3

1.2.2 Flexibility ................................................................................................................. 3

1.2.3 Economy .................................................................................................................. 4

1.3 The choice of main transformer and calculation ................................................................. 6

2The calculation of short-circuit current ....................................................................................... 7

2.1 The purpose of the short-circuit current .............................................................................. 7

2. 2 General provisions of short-circuit current calculation ...................................................... 8

2. 3 Calculation of short-circuit current .................................................................................. 11

2.3.1 per unit calculation of network impedance of each component ............................. 11

2.3.2 Draw the equivalent circuit diagram ...................................................................... 12

2.3.3 Simplified equivalent circuit .................................................................................. 12

2.3.4 Calculation of three-phase short-circuits current and short-circuits impulse current

......................................................................................................................................... 13

3Electrical equipment choice ...................................................................................................... 15

3.1 The general rules and condition of electrical equipment choice ....................................... 15

3. 2 the result of electrical equipment choice .......................................................................... 17 4Conclusion ................................................................................................................................ 18

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Abstract: The course design is based on the design mission book, and is standard on relevant techniques and rules distance ,these notes describes the features of a main electrical connection and other basic things about power system design which illustrate the basi needs to design a electrical connection system and load analysis ,calculations for short circuit current, electrical equipment choice and basic calculations.. it tell us about relevant techniques and rules ,then the concrete work is took into account and the project is confirmed by analyzing entirely attaining the current and advanced technique to the best again economic practical. This design includes four parts: course design (thesis) mission book; course design (thesis) manual; the course design (thesis) computes the book and design the diagram paper. Among them course design manual is principal part, it include the choose of the number, capacity, pattern; the main electric connection; the calculation of the short-circuit electric current and the choose of electric equipment.

Keywords?The main electric connection, the short-circuit electric current

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This manual provides engineering data and criteria for designing electric power plants where the size and characteristics of the electric power load and the economics of the particular facility justify on-site generation.Size of plant considered

in this manual is 110KV.

1Main Electrical Connection

To meet a certain iron core production and life electricity of mining area, build a step down substation of 110KV. 110KV voltage level recent through power is 6 MW, and a forward through power is 10MW. 6KV voltage level recent of the largest load is 6.6 MWthe forward largest load is 22.6MW. Total load calculation of mining area?forward?According to the load material, in forward there are 3 inlet wires and 14 outlet wires. Forward total load?PM=22.6 MW?total apparent power of electrical

?SM=?PM/cosφ=22.6/0.8=28.25MVA

KSn?K?SM=0.85×28.25=24.0125MVAAccording to the data =0.85：?

when choose a main transformerthe main transformer capacity Sn=20MVA?when

choose two main transformersbecause each main transformer capacity should meet 70%Sn=16.8088MVAeach main transformer capacity Sn=20MVA?

''cos0.86x0.183NdGenerator?SN=100MW, ?

Line: the two lines are the same, the length is 50km, reactance is 0.44Ω/km.

Site conditions?elevation of the area is 200mwhich is not area prolific of

earthquake. maximum air temperature is +39?；minimum air temperature is -

18?；The average highest temperature in the hottest month is +30?.

1.2 Main Electrical Connection Scheme Choice

The main electrical connection should meet three basic requirements which are reliability, flexibility and economy.

1.2.1 Reliability

Plant reliability standards will be equivalent to a l-day generation forced outage in 10 years with equipment quality and redundancy selected during plant design to conform to this standard.

1.2.2 Flexibility

ELECTRICAL POWER SCHEME FOR AN ELECTRICAL CONNECTION

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SHOULD BE FLEXIBLE SO THAT IT CAN BE MAINTAINED AS THE USER

DEMAND AND ACCORDING TO THE LOAD,

Power plant arrangement will permit reasonable access for operation and maintenance of equipment. Careful attention will be given to the arrangement of equipment, valves, mechanical specialties, and electrical devices so that rotors, tube bundles, inner valves, top works, strainers, contractors, relays, and like items can be maintained or replaced Power plant arrangement will permit reasonable access for operation and maintenance of equipment. Careful attention will be given to the arrangement of equipment, valves, mechanical specialties, and electrical devices so that rotors, tube bundles, inner valves, top works, strainers, contractors, relays, and like items can be maintained or replaced.

1.2.3 Economy

THE ELECTRICAL CONNECTION SCHEME SHOULD BE

ECONOMICALLY REASONABLE,,IT SHOULD BE UNDER THE BUDGET OF

THE POWER PLANT AND SHOULD MATCH THE RELIABILITY AND

FELXIBILITY NEED,S ALSO ,AND ECONOMOCALLY SHOULD BE

EFFORDABLE.

Scheme selection?

List 1

Scheme Scheme1 Scheme2

Single-bus connection 110kV Double-bus connection with bypass wiring

main 2 sets 2 sets transformer

Single-bus section Single-bus section 6kV connection connection

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Scheme1：?

Scheme2：?

Technical Comparison of the schemes:

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list 2

scheme1 scheme2

1. Choosing single-bus section 1. Reliability of high voltage by

connection and double-bus single-bus with bypass wiring is

worse than double-bus connectionreliability are both

connection. highoperating error is small. reliability 2. There are many devices in 2. Complex looplarge 110KV side, wiring complex. investmentsecondary control

wiring and relay protection

configuration are both complex.

1. Easy to expansion. 1. Operation mode is simple

2. Operation mode diversitydispatch flexibility is well.

2. Two voltage level are suitable dispatch flexibleoperation flexibility for expansionand it not need maintenance convenient.

power outages when expand

equipments.

Relativelyfloor space is largerfloor space is smalleryears

maintenance costs little. investment is largeryears

maintenance costs more. economy

1.3 The choice of main transformer and calculation

The number of main transformer?In order to ensure that the power supply relia

bility, generally substation equipped with two main transformers?When there is onl

y one power supply or substation can supply power to important load from the low voltage side, it can be installed one set?For large hub substationaccording to the e

ngineering specific situation, it can be installed 2 ~ 4 transformers.

The main transformer capacity?The main transformer capacity should be base

d on 5 ~ 10 years of development planning to choose, and consider the transformer normal operation and surge capacity when accident?To installed one transformer s

ubstationthe rated capacity of transformer shall meet the needs of the electrical loadSn?K?SM or Sn?K?PM/cosφ, Snrated capacity of transformerkV

A：?SMPMapparent power and active power of maximum load in substationkVAkW：?cosφpower factor of load?Kload factor, it can gets 0.85. To in

stalled two transformers substation, when one disconnected, the capacity of the other transformer generally guarantees power supply for 60% load. Sn?K?SM/2. The

main transformer capacity choice st-ill should consider the influence of the temperature of the surrounding environment?Sn?0.6SM/KθKθThe ambient temperatu

re correction coefficient. The main transformer type: usually choose three-phase transformer; substation with three kinds of voltage level, such as through main transfor

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mer, the power of each winding can get more than 15% Sn, it can use three-winding transformer.

The choice of main transformer and the calculation verified?

Total load calculation of mining area?forward?according to the load material, 3 in

14 outlet linesthe total load ?PM=22.6 MW? let lines

The choice of main transformer number and capacity?The total apparent power of

The capacity of main transformer?Sn?K?SM=0.85×28.25=24.0125MVA?whe

n choose one main transformerthe capacity Sn=20MVA?when choose two main t

ransformers, because each main transformer capacity should meet 70%Sn=16.8088MVAeach main transformer capacity Sn=20MVA?

Calibration?when choose one main transformerthe capacity Sn=20MVA<24.012

5MVAit not meet the capacity of the choice?when choose two main transformer

0.7K?SMto calisaccording to each main transformer capacity Sn?，0.6(

bration. Because 0.6×24.0125 = 14.4075MVA < 20MVA0.7×24.0125 = 20.0812

5MVA < 20MVAselect main transformer capacity is 2×20MVA. Considering the

influence of the temperature of the surrounding environment?

θp=θmax+θmin/2=10.5?

Kθ=20~θp/100+1=1.095

According Sn?0.6K?SM/Kθ, known 0.6×24.0125/1.095=13.158MVA

Sn=20MVA?13.158MVAit meets requirements. Through calculation and

calibrationfinally choose two main transformerseach capacity is 20MVA.

The main transformer type of choice?

no-loadRated voltagekV Rated no-load load impedance Connection capacityloss loss voltage type group current Low (KW) (%) kVA High voltage (KW) voltage %

SF7-20000 20000 110?2×1.25% 6.3 0.9 23.5 86 10.5 Ynd11 /110

2The calculation of short-circuit current

2.1 The purpose of the short-circuit current

The calculation of short circuit current, essential to the selection of adequately rated protective devices and equipment in industrial and commercial power systems is becoming important to the system designer, today, power systems carry larger blocks of power, are more important to the operation of a plant and building, and have greater safety and reliability requirement, it is necessary to find out the short circuit current because uncontrolled short circuit can cause the service outage downtown, and associate inconvenience, interrupt the essential facilities or vital services, can

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cause personal injury or fatality or can cause possible fire damage, so it is necessary to find out the short circuit current and to isolate it to avoid the above inconvenience.

2. 2 General provisions of short-circuit current calculation

It is interesting to note that nearly all problems in electrical networks can be understood by the applications of its most fundamental law ohm’s law, which states

Current = voltage/resistanceI = v/r

For ac systems

Current = voltage/impedanceI = v/z

This is the ac equivalent of resistance in dc system, and takes into account the additional effects of reactance.

It is calculated by the formula:

Z = R + jX

Where R is resistance and X is reactance,

Impedance

This is the AC equivalent of resistance in a DC system, and takes into account the additional effects of reactance, it is represented by the symbol Z and is the vector sum of resistance and reactance.

It is calculated by the formula:

Z = R + jX

Where R is resistance and X is reactance.

Simple calculation of short-circuit currents

It is to be noted that X is positive for inductive circuits whereas it is negative in capacitive circuits. That means that the Z and X will be the mirror image with R as the base in the above diagram.

Reactance

Reactance is a phenomenon in AC systems brought about by inductance and capacitance effects of a system. Energy is required to overcome these components as they react to the source and effectively reduce the useful power available to a system. The energy, which is spent to overcome these components in a system is thus not available for use by the end user and is termed ‘useless’ energy though it

still has to be generated by the source.

Inductance is represented by the symbol L and is a result of magnetic coupling which induces a back E M F opposing that which is causing it. This ‘back-

pressure’ has to be overcome and the energy expended is thus not available for use by the end user and is termed ‘useless’ energy, as it still has to be generated. L is

normally measured in Henries. The inductive reactance is represented using the

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formula: Inductive reactance= 2Πfl

Capacitance is the electrostatic charge required when energizing the system. It is represented by the symbol C and is measured in farads.

To convert this to ohms, Capacitive reactance = 1 / 2πfc

Where

F = supply frequency,

L= system inductance and

C= system capacitance.

Inductive reactance and capacitive reactance oppose each other by vectors; so to find the net reactance in a system, they must be arithmetically subtracted.

For example, in a system having resistance R, inductance L and capacitance C, its

Impedance,

Z = R+ (j*2 πfL) {j / 2 πfc }

When a voltage is applied to a system, which has an impedance of Z, vectorally the voltage is in phase with Z as per the above impedance diagram and the current is in phase with the resistive component. Accordingly, the current is said to be leading the voltage vector in a capacitive circuit and is said to be lagging the voltage vector in an inductive circuit.

Power and power factor

In a DC system, power dissipated in a system is the product of volts × amps and is

Measured in watts P = V ×I

For AC systems, the power input is measured in volt amperes, due to the effect of reactance and the useful power is measured in watts. For a single-phase AC system, the VA is the direct multiplication of volt and amperes.

Hence VA power for the standard three-phase system is:

VA = ? 3×V×I

Alternatively;

KVA = ? 3 ×V×I

Where V is in Kv , I is in amps, or MVA = ? 3×v×1

Where V is in kV and I is in kA.

Therefore, I= KVA / ? 3 ×KV

Calculation of short-circuit MVA

We have studied various types and effects of faults that can occur on the system in the earlier chapter. It is important that we know how to calculate the level of fault current that will flow under these conditions, so that we can choose equipment to withstand these faults and isolate the faulty locations without major

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damages to the system. In any distribution the power source is a generator and it is a common practice to use transformers to distribute the power at the required voltages. A fault can occur immediately after the generator or after a transformer and depending upon the location of fault, the fault current could vary. In the first case, only the source impedance limits the fault current whereas in the second case the transformer impedance is an important factor that decides the fault current.

Generally, the worst type of fault that can occur is the three-phase fault, where the fault currents are the highest. If we can calculate this current then we can ensure that all equipment can withstand (carry) and in the case of switchgear, interrupt this current. There are simple methods to determine short-circuit MVA taking into account some assumptions.

Consider the following system. Here the source generates a voltage with a phase.

Voltage of E and the fault point is fed through a transformer, which has a reactance Xp.

LET Is = r.m.s short-circuit current

I = normal full load current

P = transformer’s rated power (rated MVA)

Xp = reactance per phase

Ep = system voltage per phase

At the time of fault, the fault current is limited by the reactance of the transformer after neglecting the impedances due to cables up to the fault point. Then from Ohm’s law

Is = Ep / Xp

Now

Short-circuit MVA / rated MVA = ? 3 Ep Is * 10^6 / ? 3Ep I * 10^6

solving the above equation we can find out the value of short circuit MVA.

Though it may look that increasing the impedance can lower the fault MVA, it is not economical to choose higher impedance for a transformer. Typical percent reactance values for transformers are shown in the table below.

Primary Voltage

Reactance % at MVA Rating

MVA Rating Up to 11 kV 22 kV 33 kV 66 kV 132 kV

0.25 3.5 4.0 4.5 5.0 6.5

0.5 4.0 4.5 5.0 5.5 6.5

1.0 5.0 5.5 5.5 6.0 7.0

2.0 5.5 6.0 6.0 6.5 7.5

3.0 6.5 6.5 6.5 7.0 8.0

5.0 7.5 7.5 7.5 8.0 8.5

10.0 & above 10.0 10.0 10.0 10.0 10.0

It may be noted that these are only typical values and it is always possible to design.

Transformer with different impedances, however, for design purposes it is

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