eletric systems

7/24/2019 Eletric systems

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Introduction

A power system can be subdivided into three major parts before supplying electricalenergy to consumers :

Generation

Transmission and Subtransmission Distribution

A generation of electric power is done by using a generators where one of theessential components of power systems is the three phase ac generator known assynchronous generator or alternator ac generators can generate high power at high voltage. typically 30 kV the size ofgenerators can vary from 50 MW to 1500 MW.In a power station several generators are operated in parallel in the power grid to

provide the total power needed. They are connected at a common point called a bus.

After the electric power is generated by a generators a step-up transformers are usedfor transmission of power to reduce the power losses . These transformers connectingto the transmission network is used to transfer electric energy from generating unitsat various locations to the distribution system which ultimately supplies the load.Transmission voltage lines operating at more than 60kV are standardized at 69 kV,115 kV, 138 kV, 161 kV, 230 kV , 345 kV, 500 kV, and 765 kV line-to-line.Transmission voltages above 230 kV are usually referred to as extra-high voltage(EHV).

At the receiving end of the transmission lines step-down transformers are used to

reduce the voltage to suitable values for distribution or utilization.The distribution system is that part which connects the distribution substations to theconsumers ' service entrance equipment.Loads of power systems are divided into industrial, commercial, and residential.In addition to generators, transformers, and transmission lines, other devices arerequired for the satisfactory operation and protection of a power system. Some of the

protective devices directly connected to the circuits are called switchgear. Theyinclude instrument transformers, circuit breakers, disconnect switches, fuses andlightning arresters. These devices are necessary to deenergize either for normal

operation or on the occurrence of faults. The associated control equipment and protective relays are placed on switchboard in control houses.


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For a power system to be practical it must be safe, reliable, and economical. Thusmany analyses must be performed to design and operate an electrical system.However, before going into system analysis we have to model all components ofelectrical power systems

Power system representation

The complete circuit diagram for a three phase system is seldom necessary to conveyeven the most detailed information about a system. Therefore the interconnectionamong the components of the power system may be shown in a simplified diagram iscalled a single line diagram.

Single line diagram

A single-line diagram of a power system shows the main connections andarrangements of components . Power system networks are represented by single-linediagrams using suitable symbols as shown below :


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The impedance diagram on single-phase basis for use under balanced operatingconditions can be easily drawn from the single-line diagram such as shown

The impedance diagram can be abbreviated to reactance diagram when theresistances and capacitors are neglected.


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The per unit quantities (pu)

Because of the large amount of power transmitted , kilowatts or megawattsand kilovolt-ampere or megavolt-ampere are the common terms. However,these quantities as well as amperes and ohms are often expressed as apercent or per unit of a base or reference value specified for each.

In the per-unit system, all quantities are represented as a fraction of the basevalue:

actual valueQuantity in per unit

base value of quantity

For example

V(pu) =actual value of voltage (v)base value of voltage (v)

I(pu) =actual value of current (A)base value of current (A)

Z(pu) =actual value of impedance()base value of impedance()

S(pu) = actual value of appearant power(VA)base value of appearant power(VA)

If any two of the four base quantities are specified, the other base values canbe calculated. Usually, base apparent power and base voltage are specifiedat a point in the circuit, and the other values are calculated from them. Thebase voltage varies by the voltage ratio of each transformer in the circuit but

the base apparent power stays the same through the circuit.Real power systems are convenient to analyze using their per-phase (sincethe system is three-phase) per-unit (since there are many transformers)equivalent circuits. The per-phase base voltage, current, apparent power, andimpedance are

1 ,

,

basebase

LN base

S I

V


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2,,1 ,

LN base LN basebase

base base

V V Z

I S

Where V LN,base is the line-to-neutral base voltage in the three-phase circuitS 1 ,base is the base apparent power of a single phase in the circuit.

The base current and impedance in a per-unit system can also be expressedin terms of the three-phase apparent power (which is 3 times the apparentpower of a single phase) and line-to-line voltages (which is 3 times the line-to-neutral voltage):

3 ,

,3base

base

LL base

S I V

2,,3 ,3

LL base LL basebase

basebase

V V Z

S I

Change base impedanceThe per-unit impedance may be transformed from one base to another as:

2

old newnew old

new old

V S Per unit Z per unit Z

V S


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Example : a power system consists of one synchronous generator and onesynchronous motor connected by two transformers and a transmission line.Create a per-phase, per-unit equivalent circuit of this power system using abase apparent power of 100 MVA and a base line voltage of the generator G 1 of 13.8 kV. Given that:

G 1 ratings: 100 MVA, 13.8 kV, R = 0.1 pu, X s = 0.9 pu; T 1 ratings: 100 MVA, 13.8/110 kV, R = 0.01 pu, X s = 0.05 pu;

T 2 ratings: 50 MVA, 120/14.4 kV, R = 0.01 pu, X s = 0.05 pu;

M ratings: 50 MVA, 13.8 kV, R = 0.1 pu, X s = 1.1 pu;

L1 impedance: R = 15 , X = 75 .

Solution: To create a per-phase, per-unit equivalent circuit, we need first tocalculate the impedances of each component in the power system in per-unitto the system base. The system base apparent power is S base = 100 MVAeverywhere in the power system. The base voltage in the three regions willvary as the voltage ratios of the transformers that delineate the regions.These base voltages are:


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,1

,2 ,1

,3 ,2

13.8 Region 1

110110 Region 2

13.8

14.4 13.2 Region 2120

base

base base

base base

V kV

V V kV

V V kV

The corresponding base impedances in each region are:

22,

,13 ,

22,

,23 ,

22,

,33 ,

13.81.904 Region 1

100

110121 Region 1

100

13.21.743

100

LL basebase

base

LL basebase

base

LL basebase

base

V kV Z

S MVA

V kV Z

S MVA

V kV Z

S MVA

Region1

The impedances of G 1 and T 1 are specified in per-unit on a base of 13.8 kVand 100 MVA, which is the same as the system base in Region 1. Therefore,the per-unit resistances and reactances of these components on the systembase are unchanged:

R G1,pu = 0.1 per unit

X G1,pu = 0.9 per unit

R T1,pu = 0.01 per unit

X T1,pu = 0.05 per unit

There is a transmission line in Region 2 of the power system. The impedanceof the line is specified in ohms, and the base impedance in that region is 121

. Therefore, the per-unit resistance and reactance of the transmission lineare:


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,

,

150.124

12175

0.620

121

line system

line system

R per unit

X per unit

The impedance of T 2 is specified in per-unit on a base of 14.4 kV and 50MVA in Region 3. Therefore, the per-unit resistances and reactances of thiscomponent on the system base are:

2

new given given new new given per unit Z per unit Z V V S S

2

2,

22,

0.01 14.4 13.2 100 50 0.2380.05 14.4 13.2 100 50 0.119

T pu

T pu

R per unit X per unit

The impedance of M 2 is specified in per-unit on a base of 13.8 kV and 50MVA in Region 3. Therefore, the per-unit resistances and reactances of thiscomponent on the system base are:

2

new given given new new given per unit Z per unit Z V V S S

22,

22,

0.1 14.8 13.2 100 50 0.219

1.1 14.8 13.2 100 50 2.405

M pu

M pu

R per unit

X per unit

Therefore, the per-phase, per-unit equivalent circuit of this power system isshown:


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node equation

When the per-unit equivalent circuit of a power system is created, it can be used tofind the voltages, currents, and powers present at various points in a power system.The most common technique used to solve such circuits is nodal analysis.

In nodal analysis, we use Kirchhoffs current law equations to determine the voltagesat each node (each bus) in the power system, and then use the resulting voltages tocalculate the currents and power flows at various points in the system.

For example a simple three-phase power system bellow

The per-unit equivalent circuit of this power system:


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Note that the per-unit series impedances of the transformers and the transmissionlines between each pair of busses have been added up, and the resultingimpedances were expressed as admittances (Y=1/Z) to simplify nodal analysis.

The voltages between each bus and neutral are represented by single subscripts (V 1,V2) in the equivalent circuit, while the voltages between any two busses are indicated

by double subscripts (V 12).The generators and loads are represented by current sources injecting currents intothe specific nodes. Conventionally, current sources always flow into a node meaningthat the power flow of generators will be positive, while the power flow for motorswill be negative.

According to Kirchhoffs current flow law (KCL), the sum of all currents enteringany node equals to the sum of all currents leaving the node. KCL can be used toestablish and solve a system of simultaneous equations with the unknown nodevoltages.

Assuming that the current from the current sources are entering each node, and thatall other currents are leaving the node, applying the KCL to the node (1) yields:

1 2 1 3 1 1a b d V V Y V V Y VY I

Similarly, for the nodes (2) and (3):

2 1 2 3 2 2

3 1 3 2 3 3

a c e

b c f

V V Y V V Y V Y I

V V Y V V Y V Y I

Rearranging these equations, we arrive at:

1 2 3 1

1 2 3 2

1 2 3 3

a b d a b

a a c e c

b c b c f

Y Y Y V Y V Y V I Y V Y Y Y V Y V I

Y V Y V Y Y Y V I

In matrix form:


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1 1

2 2

3 3

a b d a b

a a c e c

b c b c f

Y Y Y Y Y V I

Y Y Y Y Y V I

Y Y Y Y Y V I

Which is an equation of the form:

busY V I where Y bus is the bus admittance matrix of a system, which has the form:

11 12 13

21 22 23

31 32 33

bus

Y Y Y

Y Y Y Y

Y Y Y

Y bus has a regular form that is easy to calculate:

1) The diagonal elements Y ii equal the sum of all admittances connected to node i.

2) Other elements Y ij equal to the negative admittances connected to nodes I and j.

3) The diagonal elements of Y bus are called the self-admittance or driving-pointadmittances of the nodes; the off-diagonal elements are called the mutualadmittances or transfer admittances of the nodes.

4) Inverting the bus admittance matrix Y bus yields the bus impedance matrix:

1bus bus Z Y

Simple technique for constructing Y bus is only applicable for components that are notmutually coupled. The technique applicable to mutually coupled components can befound elsewhere.

Once Y bus is calculated, the solution to (10.15.1) is


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1

bus

bus

V Y I

V I

A number of techniques can be used to solve systems of simultaneous linearequations, such as substitution, Gaussian elimination, LU factorization, etc.

A system of n linear equations in n unknowns

Ax b where A is an n x n matrix and b is and n -element column vector; the solution will be

1 x A b where A-1 is the n x n matrix inverse of A.

Example 10.3: a power system consists of four busses interconnected by fivetransmission lines. It includes one generator attached to bus 1 and one synchronousmotor connected to bus 3.


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The per-phase, per-unit equivalent circuit is shown.

We observe that all impedances are considered as pure reactances to simplify the casesince reactance is much larger than resistance in typical transformers, synchronousmachines, and overhead transmission lines.

Find the per-unit voltage at each bus in the power system and the per-unit currentflow in line 1.


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The first step in solvingfor bus voltages is to convert the voltage sources into the equivalent current sourcesby using the Nortons theorem. Next, we need to convert all of the impedancevalues into admittances and form the admittance matrix Y bus then use it to solve forthe bus voltages, and finally use voltages on buses 1 and 2 to find the current in line1.

First, we need to find the Norton equivalent circuits for the combination of G1 and T 1.The Thevenin impedance of this combination is Z TH = j1.1, and the short-circuitcurrent is

1.1 101.0 80

1.1oc

scTH

V I

Z j


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The combination of M 3 and T 2 is shown.

The Thevelin impedance of this combination is Z TH = j1.6, and the short-circuitcurrent is

0.9 220.563 112

1.6

oc sc

TH

V I

Z j


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0.9 220.563 112

1.6oc

scTH

V I

Z j

The Nortons equivalent circuit.

The per-phase, per-unit circuit with the current sources included


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The resultingadmittance matrix is:

12.576 5.0 0 6.667

5.0 12.5 5.0 2.5

0 5.0 10.625 5.0

6.667 2.5 5.0 14.167

bus

j j j

j j j jY

j j

j j j j


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The current vector for this circuit is:

1.0 80

0

0.563 1120

I

The solution to the system of equations will be

1

0.989 0.60

0.981 1.580.974 2.62

0.982 1.48

busV Y I V

The current in line 1 can be calculated from the equation:

1 1 2 1 0.989 0.60 0.981 1.58 5.00.092 25.16

line I V V Y j

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