Ee1352 Monograph Uniti

EE 1352 - POWER SYSTEM ANALYSIS VI semester EEE Branch (2006-2010 Batch) Unit – I Syllabus : Modern Power System – Basic Components of a power system – Per Phase Analysis – Generator model – Transformer model – Transmission line model – The per unit system – Change of base. The Power System – An Overview and Modelling 1.1 Introduction : An electric power system consists of three main divisions namely 1. The generating system 2. The transmission system and 3. The distribution system. Transmission lines are the links between the generating systems and the distribution systems and lead to other power systems over interconnection. A distribution system connects all the individual loads to the transmission lines at substations which perform voltage transformation and switching functions. In the initial stage of development of power systems, they were operated as individual units because they started as isolated systems to meet the requirements of that particular area. Then they spread out gradually to cover the whole country. The demand for large amount of power and increased reliability had lead to the interconnection of the nearby power systems. 1.2 Advantages of interconnection of power systems : 1. Less number of generators are required as a reserve. Hence the reserve capacity of the generating station gets reduced. 2. The spinning reserve of the generating station gets reduced. 3. Most economical operation is possible. 1.2.1 Problems of Interconnection : 1. Short circuit current level increases and so the circuit breakers has to interrupt a large current. 2. The disturbance caused by a short-circuit extends to other neighboring systems also. 3. Synchronism between the alternators of interconnected systems must be maintained. 1.3 Need for Power System Analysis in Planning and Operation of Power System : The successful operation of a power system depends largely on the engineer’s ability to provide reliable, uninterrupted and quality service to loads. The systems being planned are to be optimal with respect to cost, performance and operating efficiency. For this, better planning tools are required. The major power system tools are :

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1. 2. 3. 4.

Load flow analysis Short circuit analysis or fault calculations Stability analysis System protection and relay coordination.

1.3.1 Load flow studies : A load flow study is the determination of the voltage, current, real power and reactive power at various points in an electric network. The main objective of load flow analysis is to identify the potential problems in terms of unacceptable voltage conditions, overloading, decreasing reliability or any failure of the transmission system to meet performance criteria. After this analysis stage, alternative plans are taken to prevent the foreseen problems. The load flow studies are useful in planning the future development of the system 1.3.2 Short circuit analysis : 1. to determine the current interrupting capacity of the circuit breakers. 2. to establish the relay coordination. 3. to calculate the voltages during faulted conditions for insulation coordination.\ 4. to design the grounding systems. 1.3.3 Stability Studies : Stability studies are performed in order to ensure that the system remains stable following a severe fault or disturbance. Power system stability is a condition in which the various synchronous machines of the system remain in synchronism with each other. Stability is classified into a) Steady state stability and b) Transient stability. 1.3 Per Phase Analysis of Symmetrical Three phase System : During normal operations, the three phase generation as well as the three phase transmission and distribution are balanced. Therefore, they can be treated as single phase system for analysis during normal operating conditions. Under faulty conditions, the system becomes a case of unbalanced three phase generated voltage connected to the three phase balanced network. In such situation, it is advantageous to transform the unbalanced phasor quantities into balanced component quantities using symmetrical components transformation method. After transformation, it can be treated as single phase system for analysis purpose as shown in Fig. 1.1 and Fig. 1.2.. Let us consider a simple, balanced three phase network consisting of a three phase generator and a three phase load shown in Fig. 1.1. Since it is a balanced system, the generator and load neutrals are at the same potential, so that the neutral current, In is equal to zero. Thus the neutral impedance Zn does not affect the system behaviour. For the reference phase ‘a’ Ea = (Zs + ZL) Ia The voltages and currents in the other phases have the same magnitude but displaced from each other by 120 electrical degrees. Thus the solution of a single phase network

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completely determines the solution of the three phase network. equivalent network is shown in Fig. 1.2.

The single phase

Fig. 1.1 A balanced three phase network

Fig. 1.2 Single phase equivalent network 1.5 The one-line diagram : A three phase balanced system is always solved as a single phase circuit consisting of one of the three lines and a neutral return. Also, the diagram is further simplified by omitting the completed circuit through the neutral and by indicating the components of the system by standard symbols rather than by their equivalent circuits. Circuit parameters will not be shown in the diagram and a transmission line is represented by a single-line between its two ends. Such a simplified diagram of a power system is called a one-line diagram. It indicates by a single line and standard symbols, the transmission lines and associated apparatus of a power system.

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Fig. 1.3 Standard symbols for power system components. 4

EQUIVALENT CIRCUITS OF COMPONENTS OF POWER SYSTEM Equivalent circuit of generator: The equivalent circuit of a 3-phase generator (Alternator) is shown in fig. 1.2. It consists of a source representing induced emf per phase, a series reactance representing the armature reactance and leakage reactance and a series resistance representing the armature winding resistance.

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1.5 General aspects relating to power flow, short circuit and stability analysis : The purpose of the one line diagram is to provide the significant information about the system in concise form. The details given on the diagram depends on the purpose for which the diagram is intended. For example, for load flow studies, the location of circuit breakers and relays are not needed. Generally the information found on a one-line diagram varies according to the type of study.

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Fig. 1.4 One line diagram of an electric system Generators : MVA, kV, subtransient reactance X” Transformer : Vp / Vs, MVA, X Load : P, Q or S, pf. 1.5.1 Impedance diagram : The one – line diagram is used to draw the single-phase equivalent circuit of the system. In this circuit, the equivalent circuits of all the components of the power system are drawn and they are interconnected. This diagram forms the impedance diagram of the given system. For load flow study, loads A and B are represented by resistance and inductive reactance in series. The impedance diagram for the one line diagram shown in fig. 1.5 is drawn below. The resistance and reactance used for grounding the neutral of the generator is not shown in the impedance diagram under the assumption that no current flows in the ground under balanced conditions and the neutrals of the generators are at the same potential as that of the neutral of the system. Since the magnetizing current of a transformer is negligible compared to the full-load current, the shunt admittance is omitted in the equivalent circuit of the transformer.

Fig. 1.5 Impedance diagram

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1.5.2

Simplified Impedance diagram : While doing fault calculations, resistance values can be omitted, as the resistance is small. Loads which do not involve rotating machinery are omitted, as they have little effect on the total current during a fault. Synchronous motor loads are to be included in fault calculations since their generated emfs contribute to the short-circuit current. Induction motors are taken into account by an emf generated in series with the inductive reactance when the current immediately after the fault is determined. But they are ignored when computing the current a few cycles after the occurrence of the fault, because the current contributed by an induction motor dies out very quickly after the induction motor is short-circuited. 1.5.3

Reactance diagram : After omitting all static loads, all resistances, the magnetizing current of each transformer and the capacitance of the transmission line, the impedance diagram reduces to the reactance diagram as shown below.

Fig. 1.6 Reactance Diagram The impedance and reactance diagrams are also called positive sequence diagrams as they show impedances to balanced currents in a symmetrical three-phase systems. 1.6 Per Unit representation : The per unit value of any quantity is defined as the ratio of the quantity to its base value expressed as a decimal. Actualvalue . BaseValue The base impedance is that impedance which will have a voltage drop across it equal to the base voltage, when the current flowing in the impedance is equal to the base value of the current. Voltage, current, MVA and impedance are so related that selection of base values for any two of them determines the base values of the remaining two. Per unit value =

The advantages of per unit system are : i) The per unit impedance referred to either side of a single phase transformer is the same.

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ii) iii) iv) v) 1.6.1

The per unit impedance referred to either side of three phase transformer is the same regardless of the three phase connections whether they are Δ-Δ, Υ-Υ, ΥΔ. The chance of confusion between line and phase quantities in a three phase balanced system is greatly reduced. The manufacturers usually provide the impedance values in per unit. The computational effort in power system is very much reduced with the use of per unit quantities. Single phase system : The base MVA and base kV ratings given for single phase system are per phase

values. The base current and hence base impedance are calculated as follows : BaseMVA Base current, kA = BasekV If base current and voltage are known, base impedance can be found out. BasekV Base impedance, Ω = BasekA BasekV = BaseMVA / BasekV ( BasekV ) 2 = BaseMVA Base MW = Base MVA actua lim pedance, Ω Per unit impedance of a circuit element = baseimpeance, Ω -------------------------------------------------------------------------------------------------------Solved problem : 1. A single phase transformer is rated 110/440V, 2.5 kVA. Leakage reactance measured from the L.T. side of the transformer is 0.06 Ω. Determine the leakage reactance in p.u. Solution : Actual leakage reactance = 0.06 Ω ( BasekV ) 2 Base impedance = BaseMVA (110 × 10 −3 ) 2 × 1000 = 4.84 Ω 2.5 actua lim peance Per unit impedance of a circuit element = baseimpedance 0.06 = 0.0124 Per unit impedance = 4.84 ---------------------------------------------------------------------------------------------------------Base impedance in the L.T. side =

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1.6.2 Three-phase system : Base voltage / phase = Base kVLL / √3 Base current / phase =

BaseMVA3Φ / 3 3 xBasekVLL

Base impedance, Ω =

=

BaseMVA3Φ 3 × BasekVLL

BasekVLL / 3 Basevoltage = Ba sec urrent BaseMVA3Φ / 3 × BasekVLL

( BasekVLL ) 2 Base impedance = BaseMVA3Φ Note : The formula to calculate base impedance is the same irrespective of single-phase or three phase systems. 1.6.3 Changing the base of per unit quantities : If the values given are already in the p.u. values referred by their own ratings, then to convert them to the selected base value, let us derive a direct formula. Actua lim pedance Per unit impedance = Baseimpeance Z actual xBaseMVAnew Zp.u.given = . . . (1) ( BasekVgiven ) 2 Similarly when referred to new base values Z actual × BaseMVAnew Zp.u.new = ( BasekVnew ) 2 Dividing eqn. (2) by (1), we get, Z p .u .new = Z p.u . given

 BasekVgiven ×  BasekV new 

. . . (2) 2

 BaseMVAnew   × BaseMVA given 

The following points are to be noted while using per unit representation. 1. A base (kV) and base (MVA) are selected in one part of the system. The base values for a 3 phase system are line to line kilovolts and 3 phase MVA. 2. The base MVA will be the same in all parts of the system. 3. For other parts of the system, that is on other sides of transformers, the base kV for each part is determined using the line to line voltage ratios of the transformers. 4. Impedance values for all components given in p.u. will be on the base determined by their own ratings. If p.u. impedance is given on a base other than that determined for the part of the system in which the element is located, must be changed to the proper base by the using the formula Z p .u .new = Z p.u . given

 BasekVgiven ×  BasekV new 

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 BaseMVAnew   × BaseMVA given 

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5. If impedance values are given in ohms, p.u. values are to be calculated from the corresponding base impedances using the formula actua lim pedance, Ω p.u. impedance = baseimpeance, Ω 6. For three single phase transformers connected as a 3 phase unit, the 3 Φ ratings are to be determined from the single phase rating of each individual transformer. MVA rating of 3Φ operation = 3 x MVA rating of 1Φ transformer. 7. Voltage rating of primary and secondary of the three phase unit depends upon the Star or delta connection. Hence the voltage ratings of 3 Φ transformer unit is to be calculated for primary and secondary windings separately. a. When the primary / secondary winding is connected in star, the voltage rating of the corresponding winding of 3Φ transformer is √3 times the voltage rating of 1Φ transformer. b. When the primary / secondary winding is connected in delta, the voltage rating of the corresponding winding of 3Φ transformer is same as that of 1Φ transformer. Note : In delta connection, Vline = V phase. In star connection, Vline = √3 Vphase. ---------------------------------------------------------------------------------------------Solved problems : 1. Three generators are rated as follows : Generator 1 : 100 MVA, 33kV, reactance 10% Generator 2 : 150 MVA, 32kV, reactance 8% Generator 3 : 110 MVA, 30kV, reactance 12% Determine the reactance of the generators corresponding to base values of 200 MVA and 35 kV. Solution : Generator 1 : 100 MVA, 33 kV, X”= 0.1 Z p.u .new

 BasekVgiven = Z p.u . given ×   BasekVnew

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 BaseMVAnew  ×  BaseMVAgiven

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 33  200 X = 0.1 ×   × = 0.1778 p.u.  35  100 '

Generator 2 : 150 MVA, 32 kV, X”=0.08 2

 32  200 X = 0.08 ×   × = 0.0892 p.u.  35  150 '

Generator 3 : 110 MVA, 30 kV, X”=0.12

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 30  200 X ' = 0.12 ×   × = 0.1602 p.u.  35  110 --------------------------------------------------------------------------------------------2. Two generators rated t 10 MVA, 13.2 kV and 15 MVA, 13.2 kV are connected in parallel to a bus bar. They feed supply to two motors of input 8 MVA, and 12 MVA respectively. The operating voltage of motors is 12.5 kV. Assuming base quantities as 50 MVA and 13.8 kV, draw the reactance diagram. The percentage reactance for generators is 15 % and that for motors is 20 %. Solution : Generator 1 : 10 MVA, 13.2 kV, X”=0.15. Z p.u .new

 BasekVgiven = Z p.u . given ×   BasekVnew

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 BaseMVAnew  ×  BaseMVAgiven

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 13.2  50 X ' = 0.15 ×  = 0.6862 p.u.  ×  13.8  10 Generator 2 : 15 MVA, 13.2 kV, X”=0.15. 2

 13.2  50 X = 0.15 ×  = 0.4575 p.u.  ×  13.8  15 '

Motor 1 : 8 MVA, 12.5 kV, X”=0.2. 2

 12.5  50 X = 0.2 ×  = 1.0255 p.u.  × 8  13.8  '

Motor 2 : 12 MVA, 12.5 kV, X”=0.2. 2

 12.5  50 X ' = 0.2 ×  = 0.6837 p.u.  ×  13.8  12

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3. A 100 MVA, 33 kV, 3 phse generator has a subtrnsient reactance of 15 %. The generator is connected to three motors through a transmission line and two transformers. The motors have rated inputs of 30 MVA, 20 MVA and 50 MVA at 30 kV with 20 % subtransient reactance. The 3 phase transformers are rated at 110 MVA, 32 kV/110 kV Y with leakage reactance 8 %. The line has a reactance of 50 ohms. Selecting the generator rating as the base quantities in the generator circuit, determine the base quantities in other parts of the system and evaluate the corresponding p.u. values. Solution : 110 = 113.44kV The base voltage in the transmission line = 33 × 32 32 = 33kV In the motor circuit, the base voltage = 113.44 × 110 ( BasekV ) 2 (113.44) Bus impedance in the transmission line = = = 128.686ohms BaseMVA 100 j 50 = 0.3885 p.u. Per unit impedance = j128.686 Motor 1 : 30 MVA, 30 kV, X”=0.2. 2  BasekVgiven  BaseMVAnew  × Z p.u .new = Z p.u . given ×   BasekVnew  BaseMVAgiven 2

 30  100 ' X 1 = 0.2 ×   × = 0.5509 p.u. 30  33  Motor 2 : 20 MVA, 30 kV, X”=0.2. 2

X2

'

 30  100 = 0.2 ×   x = 0.826 p.u.  33  20

Motor 3 : 30 MVA, 30 kV, X”=0.2. 2

 30  100 ' X 3 = 0.2 ×   × = 0.3305 p.u. 50  33  Transformer : 110 MVA, 32 kV, X”=0.08. 2

 32  100 ' X T = 0.08 ×   × = 0.06838 p.u.  33  110 ---------------------------------------------------------------------------------------------------------

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Bus voltage on the transmission line using the transformation ratio=220kV

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=0.3333p.u.

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Transformer 2 : 15 MVA, 6.9Δ – 115Y kV, X=10 % Transformer 3 : Single phase units, each rated 10 MVA, 6.9/69kV, X=10% Draw an impedance diagram and mark all values in p.u. choosing a base of 30 MVA, 6.6 kV in the generator I circuit.

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Review Questions : (Two mark questions) What is single line diagram? What are the components of power system? What are the components of power system? Draw the symbols used to represent various components in a power system. Define per unit value. Write the equation for converting the p.u. impedance expressed in one base to another? 7. What are the advantages of p.u. components? 8. Draw the equivalent circuit of a 3-phase generator. 9. How the loads are represented in reactance or impedance diagram? 10. Draw the single phase equivalent circuit of a 3 winding transformer. 11. A Υ-∆, 3-phse transformer bank is formed using three numbers of 1-phse transformers each rated at 300 kVA, 127/13.2 kV. What is the kVA and kV rating of the 3-phae bank? 12. If the reactance in ohms is 15 ohms, find the p.u. value for a base of 15 kVA and 10 kV. 13. A generator rated at 30 MVA, 11 kV has a reactance of 20 %. Calculate the p.u. reactances for a base of 50 MVA and 10 kV. 14. A Y-connected generator rated at 300 MVA, 33 kV has a reactance of 1.24 p.u. Find the ohmic value of reactance. 15. The base kV and base MVA of a 3-phse transmission line is 33 kV and 10 MVA respectively. Find the base current and base impedance. 16. How the induction motor is represented in reactance diagram?\ 17. What is impedance and reactance diagram? 18. What are the approximations made in impedance diagram? 19. What are the approximations made in reactance diagram? 20. Give equations for transforming base kV on LV side to HV side of transformer and vice-versa. 21. For the power system shown in fig. below, by taking generator rating as base values, specify the base values of the transmission line and motor circuit. 1. 2. 3. 4. 5. 6.

Motor is 25 MVA 10 kV (16 mark questions) 1. Draw an impedance diagram for the electric power system shown in fig. below. The ratings are given below. All impedances are in p.u. on a 100 MVA base. Choose 20 kV as the voltage base for generator.

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Three-phase power and line-line ratings are given below. G : 90 MVA 20 kV X=9% T1 : 80 MVA 20/200kV X = 16 % T2 : 80 MVA 200/20kV X = 20 % M : 90 MVA 18 kV X=9% Line : 200 kV X = 120 Ω Load : 20 kV S = 49 MW + j 64 Mvar. 2. Draw the reactance diagram for the system shown in fig. below. The specification of the components are given in table.

Generator G

Transformer T1

Transformer Motor M1 Motor M2 Line T2 13.8 kV 13.2/69 kV 69/13.2 kV 13.0 kV 13.0 kV X= 65 Ω 25 MVA 25 MVA 25 MVA 15 MVA 10 MVA X”= 0.15 p.u. X1 = 0.11 p.u. X1 = 0.11 p.u. X”=0.15p.u. X”=0.15p.u. Determine the generator terminal voltage assuming both motors operating at 12kV, 75% full load and unity power factor. 3. The 40 MVA, 25 kV, 3 phase generator has a subtransient reactance of 25%. It is connected through a ∆-Υ transformer to a high voltage transmission line having total series reactance of 50 ohms. At the lod end of the line is Y-Y step down transformer. Both transformer banks are composed of 1 phase transformer connected for 3 phase operation. Each of the 3 transformer composing each bank is rted 16.67 MVA. 13.8 kV/100 kV with a leakage reactance of 20 %. The load represented as impedance is drawing 30 MVA at 24 kV, 0.8 pf lag. Draw the single line diagram of power network. Choose a base of 30 MVA, 24 kV in the load circuit. Determine also the voltage at the terminals of the generator. End of Ist Unit

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K.L.N. COLLEGE OF INFORMATION TECHNOLOGY Pottapalayam – 630 611, Sivagangai District. Year : 2008 -2009, Branch : III B.E. EEE

Semester : 6 Subject : EE 1352 Power System Analysis

Name of the Writer (Collaborator) : R. Ramesh, Professor (EEE), IT Dept. Unit : I

Date of Submission : 14.11.2008

Syllabus : Ist Unit Modern Power System – Basic Components of a power system – Per Phase Analysis – Generator model – Transformer model – Transmission line model – The per unit system – Change of base. --------------------------------------------------------------------------------------

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