Power Systems - Basic Concepts and Applications - Part I

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1 PDHonline Course E104 (1 PDH) Power ystems Basic Concepts and Applications Part I Instructor: hihmin Hsu PhD PE 01 PDH Online PDH Center 57 Meadow Estates Drive Fairfax A Phone & Fax: wwwpdhonlineorg wwwpdhcentercom An Approved Continuing Education Provider

2 Power ystems Basic Concepts and Applications Part I By hihmin Hsu PhD PE Power ystems Basic Concepts and Applications Part I Module 1 Page 1

3 MODULE 1: Introduction to Power ystems Overview This module provides an introduction to power systems It discusses a basic structure of power systems the fundamentals of AC circuits mathematical notations balanced threephase systems and per unit values Basic tructure of Power ystems A power system is an interconnected network with components converting nonelectrical energy continuously into the electrical form and transporting the electrical energy from generating sources to the loads/users A power system serves one important function and that is to supply customers with electricity as economically and as reliably as possible It can be divided into three subsystems: Generation Transmission Distribution Generating and/or sources of electrical energy Transporting electrical energy from its sources to load centers with high voltages (115 k and above) to reduce losses Distributing electrical energy from substations (44 k ~ 1 k) to end users/customers This basic structure of a power system is shown in Figure 11 The generator converts nonelectrical energy to electrical energy The devices connecting generators to transmission Generation Transmission Distribution G 18:115 k 115:1 k Fig 11 A basic structure of a simplified power system system and from transmission system to distribution system are transformers Their main functions are stepping up the lower generation voltage to the higher transmission voltage and stepping down the higher transmission voltage to the lower distribution voltage The main advantage of having higher voltage in transmission system is to reduce the losses in the grid ince transformers operate at constant power when the voltage is higher then the current has a lower value Therefore the losses a function of the current square will be lower at a higher voltage Power ystems Basic Concepts and Applications Part I Module 1 Page

4 DC vs AC The current of a direct current (DC) circuit as shown in Figure 1 consisting of a battery and a pure resistive load can be calculated as DC I DC = R and the power provided by the voltage source equals where and P DC = DC = DCIDC = IDCR R I DC = DC current in Amperes (A) DC = DC voltage in olts () P DC = DC power in Watts (W) R = the load resistance in Ohms (Ω) The voltage and current waveforms are shown in Figure 1 I DC DC R Fig 1 A simple DC circuit I DC I DC 0 Fig 1 oltage and current waveforms of the simple DC circuit t Power ystems Basic Concepts and Applications Part I Module 1 Page

5 Example 11: A DC circuit as shown in Figure 1 has a DC voltage of 1 volts and a resistor of Ω What are the DC current in the circuit and the power consumed by the resistor? olution: 1 I DC = = 6 (A) P DC = 1 6 = 7 (W) It is worth mentioning that these DC quantities are real numbers not complex numbers There is another category of circuits the alternating current (AC) circuits ince in power systems the sinusoidal voltages are generated and consequently most likely sinusoidal currents are flowed in the generation transmission and distribution systems sinusoidal quantities are assumed throughout this material unless otherwise specified In general a set of typical steadystate voltage and current waveforms of an AC circuit can be drawn as shown in Figure 14 and their mathematical expressions can be written as follows: and where and v( t) = m cos( ωt) i ( t) = Im cos( ω t θ ) m I m = the peak or the maximum values of the voltage and current waveforms ω = angular frequency in radians/second θ = phase angle with respect to the reference in degrees or in radians v i v I m 0 m i ωt θ Period Fig 14 Typical voltage and current waveforms The period in Figure 14 can be 60 or π In some cases the period can be in time for instance second for 60 Hz Power ystems Basic Concepts and Applications Part I Module 1 Page 4

6 There is an important quantity called root mean square value or rms and is defined as 1 T rms = v ( t) dt T 0 For a sinusoidal voltage its rms value equals π 1 T m π 1 cos ωt m ωt sin ωt rms = [ cos( )] ( ) T 0 m ωt dt = = = T d ωt 0 ω π 4 0 For example a typical household voltage of 10 volts is rms value and its peak value is 10 = 170 volts ince the frequency in U is 60 Hertz (Hz) the angular frequency ω = π 60 = 77 uch a voltage can be expressed as v ( t) = 10 cos(77t) = 170cos(77t) volts If the current lags the voltage by 0 and its magnitude is a half of the voltage then it can be written as follows i ( t) = 85cos(77t 0 ) amps m Phasor Representations It may not be convenient to express the voltages and currents in instantaneous forms all the time As utilized in AC circuits phasor representations are used in power systems because of convenience Recall Euler s identity jθ e = cosθ jsinθ Then the current can be rewritten as i( t) = Re[I m Im jθ jωt cos( ω t θ ) jim sin( ωt θ )] = Re( e e ) = Re(Ie where I is defined as the phasor (or polar) representation of i and is a complex number in general It has two parts the magnitude and phase angle namely Im I = I I = θ = I θ where I is the rms value of the current and the subscript rms is commonly neglected Figure 15 shows the graphical representation of I The direction of the phase angle is defined as counterclockwise for a positive value and clockwise for a negative value jωt ) Power ystems Basic Concepts and Applications Part I Module 1 Page 5

7 Imaginary Axis I = I θ Isinθ θ Icosθ Real Axis Fig 15 A graphical representation of I From Figure 15 it is obvious that a phasor can be expressed not only in a polar form but also in a rectangular form namely I = I θ = Icosθ jisinθ One of the advantages of using phasor representations instead of instantaneous forms is that one can add sinusoidal functions of the same frequency by expressing them as phasors and then adding the phasors by the rules of vector algebra More information on the phasor (or polar) representation can be found in the Appendix 1A Example 1: What are the phasor representations of the following instantaneous quantities? v ( t) = 170cos(77t) volts and i ( t) = 85cos(77t 0 ) amps olution: 170 = 0 = 10 0 volts 85 I = 0 = amps Unlike in DC circuits the loads in AC circuits as shown in Figure 16 can be expressed as its impedance consisting of resistance R and reactance X as follows where 1 Z = = R (jωl ) = R j(x L XC ) = R jx = Z θ Ω I jωc and Z = R X X θ = tan 1 ( ) R Power ystems Basic Concepts and Applications Part I Module 1 Page 6

8 I R Z L C Fig 16 RLC circuit Example 1: A 60 Hz 10 volts AC voltage source is connected to a 10Ω resistor a 18 mh inductor and 166μF capacitor as shown in Figure 16 Find (1) The total impedance Z () The current I in polar form () The voltage and current in instantaneous forms olution: (1) ince the frequency is 60 Hz the inductive and capacitive reactances can be obtained as X = L ω L = = 1Ω X C = 1 ωc 1 = = Ω The total impedance seen by the voltage source Z = R j(x L X C ) = 10 j(1 ) = 10 j10 = 1 10 ( ) tan = () To calculate the current the angle of the voltage is set to be the reference namely 0 Then 10 0 I = = = amps Z () To convert the phasors to the instantaneous forms v ( t) = 10 cos(77t) = 170cos(77t) volts i ( t) = 8485 cos(77t 45 ) = 1cos(77t 45 ) amps Ω When the imaginary part of the impedance is positive the load is called an inductive load and the current lags the voltage On the other hand if the imaginary part of the impedance is negative the load is called a capacitive load and the current leads the voltage Power ystems Basic Concepts and Applications Part I Module 1 Page 7

9 An admittance is the inverse of its impedance and can be expressed as 1 I Y = = = G jb Z where G is the conductance and B is the susceptance both in siemen Power in inusoidal teadytate Conditions If the voltage and the current of an AC circuit as shown in Figure 17 are expressed in their instantaneous forms as follows and v t) = cos( ωt θ ) ( v i t) = I cos( ωt θ ) ( i i A v B Fig 17 Power transfer between two systems Then the instantaneous power from A to B is the product of v and i as follows p t) = v( t) i( t) = I[cos(ωt θ θ ) cos( θ θ )] ( v i v i The real power or the active power is defined as the average of the instantaneous power in one period 1 T P = p( t) dt = Icosφ (W) T 0 where the angle φ = θ v θ and is defined as the phase angle of the current lags the voltage i The complex power from A to B can be calculated as where and * * = I = ( θ v )(I θ i ) = I φ = Icosφ jisinφ = P jq (A) P = Re{} = I cosφ is the active power in watts (W) Power ystems Basic Concepts and Applications Part I Module 1 Page 8

10 Q = Im{} = Isinφ is the reactive power in voltamperes reactive (var) Furthermore can be expressed in Polar form as follows = 1 Q ( P Q ) tan = φ P The magnitude of the complex power is defined as the apparent power = = I = P Q (A) A graphical presentation of complex power active power and reactive power is called a power triangle (or PQ triangle) as shown in Figure 18 It is worth mentioning that these power quantities are directional For instance when P is positive the power flow is as defined from A to B If the P is negative the power flow is actually from B to A = φ jq φ The power factor pf is defined as P Icosφ pf = = = cosφ I Fig 18 Power (PQ) triangle Therefore the phase angle φ is sometimes called the power factor angle A lagging power factor indicates an inductive impedance and therefore a positive value for φ imilarly a leading power factor implies a capacitive impedance and therefore a negative value for φ P inglephase and Balanced ThreePhase ystems The abovementioned concepts are intended for singlephase systems However electric power is generated transported and distributed mainly in a symmetrical (balanced) threephase structure In particular the power is generated by threephase synchronous generators It is then transmitted and distributed in the form of threephase power except for the lowest voltage levels of distribution system where singlephase systems are used ymmetry is one of the most important characteristic features of the threephase systems This is because threephase symmetrical systems are more effective than other ones in particular singlephase systems with respect to the capability of power transmission Power ystems Basic Concepts and Applications Part I Module 1 Page 9

11 In general the phase voltages of a balanced threephase voltage source with positive sequence can be expressed as v an ( t) = cos( ωt) T vbn ( t) = van ( t ) = cos( ω t 10 ) v cn T ( t) = van ( t ) = cos( ω t 40 ) = cos( ωt 10 ) Their waveforms are shown in Figure 19 Their phasor representations are obtained as discussed for singlephase systems an = 0 = 10 = 10 bn an = 10 = 10 cn an v v an v bn v cn m 0 ωt Period Fig 19 Waveforms of phase voltages of balanced threephase systems The above notation is obtained with the assumption that the symmetrical voltage source has a positive (or abc) sequence If a negative (or acb) sequence is assumed then leads an by 10 and cn lags an by 10 Threephase voltage sources are generally connected in two different configurations namely a wye connection and a delta connection An ideal threephase voltage source with a wye connection is shown in Figure 110 and its linetoline voltage for instance ab can be obtained by Kirchhoff s voltage law (KL the sum of all phasor voltage drops around any path in a circuit equals zero): bn Power ystems Basic Concepts and Applications Part I Module 1 Page 10

12 ab = an bn = 0 10 = [1 ( 05 j0866)] = 0 = an 0 imilarly the linetoline voltages bc and bc = 90 = bn 0 ca = 150 = cn 0 ca can be obtained Therefore for a wye connected balanced threephase voltage source the linetoline voltages or line voltages are times the phase voltages in magnitude and the line voltages are 0 ahead of their coresponding phase voltages Figure 111 shows a graphical presentation of these relationships v an i a a n v bn i b v ab b v ca v cn i c v bc c Fig 110 Threephase voltage source with wye configuration ca cn ab = 0 an 0 an bn Fig 111 The relationships of phase voltages and linetoline voltages in a threephase system with wye configuration bc Power ystems Basic Concepts and Applications Part I Module 1 Page 11

13 As one can see from the Figure 110 the phase currents are the same as the line currents for a wye connected voltage source The other configuration of the threephase voltage source is to connect them in a delta configuration as shown in Figure 11 As one can see the phase voltages are the same as their corresponding line voltages in such a delta configuration However the line currents and phase currents are different and their relationships can be derived similar to the voltage relationships in a wye configuration namely at node a by Kichhoff s current law (KCL the sum of all phasor currents into any nodes in a circuit equals zero) Ia = Iab Ica = I 0 I 10 = I[1 ( 05 j0866)] = I 0 = Iab 0 imilarly at nodes b and c the other two line currents can be obtained as follows Ib = I 150 = Ibc 0 Ic = I 90 = Ica 0 Therefore for a delta connected balance threephase voltage source the line currents are times the phase currents in magnitude and the line currents are 0 behind of their coresponding phase currents Figure 11 shows a graphical presentation of these relationships i a a v ca v ab i ca v bc i ab i bc i b i c Fig 11 Threephase voltage source with delta configuration b c I c I ca 0 I ab I b I bc I a = I 0 Fig 11 The relationships of phase currents and line currents in a threephase system with delta configuration ab Power ystems Basic Concepts and Applications Part I Module 1 Page 1

14 A balanced threephase system has their phase/line voltages mutually shifted by 10 and their phase/line currents has the same property ince it is balanced the power for each phase is the same for instance the per phase complex power or = = = an ab bn bc cn ca 1φ = = = 1φ Therefore the threephase complex power equals three times of its per phase value namely φ = 1 φ = φ Iφ φ where and I are the phase voltage and the phase current respectively As discussed earlier φ φ for a wye connected voltage source the magnitude of the line voltage is times of the phase voltage while the line and phase current are the same Therefore the threephase apparent power is equal to = L φ = φ Iφ = IL L IL where L and IL are the rms values of the line voltage and line current respectively This equation can be applied to a delta connected source as well Then the threephase active power reactive power and power factor can be obtained as follows P = I cosφ I cosφ φ φ φ = L L and Q pf = I sinφ I sinφ φ φ φ = Pφ P1 φ = = cosφ φ = φ 1φ L L Example 14: A balanced threephase load of 50 ka pf = 085 lagging is supplied from a balanced threephase wye connected voltage source of positive sequence such that L = 4157 volts Calculate: (1) IL φ and Iφ () φ Pφ and Q φ olution: L 4157 (1) φ = = = 400 volts Power ystems Basic Concepts and Applications Part I Module 1 Page 1

15 φ Iφ = IL = = = 694 amps 4157 L 1 () φ = 50 cos (085) = ka = 50 φ ka P φ = = 45 kw 1 = Q φ = 50sin[cos (085)] = 50sin kvar Per Unit alues In power system calculations it is very often to normalize actual values such as voltages and currents to per unit values It is very convenient if many transformers and voltage levels are involves The per unit value is defined as the ratio of the actual value to the selected base value namely X Actual X pu = X Base where X can be the power voltage current and/or impedance However usually the base voltage and base power (A) are given quantities while the base current and base impedance are to be determined accordingly Example 15: A motor is rated 416 k and can be operated / 10% of its rated voltage (11~09 pu) What is the range of the operational voltage in k? olution: The high limit voltage is 11 pu and its actual voltage can be calculated as Hi = = 4576 k imilarly the low limit voltage is Lo = = 744 k The motor can be operated in the range of 4576~744 k For singlephase systems: (1) elect = 1φ and Base 1φ = φ () Then I Base 1φ = = 1φ φ and Z = = I Power ystems Basic Concepts and Applications Part I Module 1 Page 14

16 Actual φ PActual QActual () Therefore pu = = j = Ppu jqpu Actual pu = IActual I pu = and I ZActual Z pu = Z Once the per unit values are calculated the actual values can be obtained by multiplying the per unit values with their corresponding base values Example 16: A singlephase system as shown in Figure 16 is given in actual values Find: (1) I Base and Z Base with selecting Base = 7 ka and Base = 10 volts () pu I pu Z pu R pu and X pu () The current in amperes olution: 700 (1) I Base = = 60 amps ZBase = = Ω 700 () et the voltage as the reference 0 Then 10 0 pu = = 1 0 pu j10 Z pu = = 5 j5 = R pu jx pu pu R pu = 5 pu X pu = 5 pu 1 0 I pu = = pu 5 45 () I = I I = 60 ( ) = Base pu amps which is the same as in the Example 1 imilarly for threephase systems: (1) elect = and Baseφ = φ 1φ Baseφ BaseL = L = φ = Base 1φ = Power ystems Basic Concepts and Applications Part I Module 1 Page 15

17 () Then 1 Baseφ Baseφ I Baseφ = = = and 1 Baseφ Baseφ 1 φ Base Baseφ Z Baseφ = = = 1 Baseφ Baseφ () Therefore pu Actual φ = Baseφ Actual pu = Baseφ IActual I pu = and I Baseφ ZActual Z pu = Z Baseφ From time to time it is necessary to change the per unit value of the impedance from one base to another The following equation can be used for this conversion: Base(old) Base(new) Z pu(new) = Zpu(old) Base(new) Base(old) Proof: (Hint: when changing the impedance per unit value from one base to another its actual value does not change) Z pu(new) = Z Z Actual Base(new) = Z pu(old) Base(old) Base(new) Base(new) Base(old) = Z pu(old) Base(old) Base(new) Base(new) Base(old) Example 17: An impedance of Z pu = 005 j05 pu on 00MA and Base L = pu Base φ = 18k Calculate the new Z if new base values for and are given as (a) 100MA 18 k (b) 00MA 1 k Baseφ Base L Power ystems Basic Concepts and Applications Part I Module 1 Page 16

18 (c) 100MA 1 k respectively olution: 100 (a) Z pu = (005 j05) = 005 j0 5 pu (b) Zpu = (005 j05) = j pu (c) Zpu = (005 j05) = 007 j0 7 pu 1 00 Example 18: A threephase equipment is rated at 800 ka 1 k and has an impedance of 0005j01 pu referred to its ratings Calculate the impedance in ohms olution: ZActual Z pu = pu Z Baseφ Therefore Z Actual 1 10 = Zpu ZBaseφ = j ( ) ( ) 0005 j01 = ( 0005 j01) 180 = 09 Ω Power ystems Basic Concepts and Applications Part I Module 1 Page 17

19 Appendix 1A Complex Numbers and Polar Coordinates A complex number can be expressed in terms of its real component and its imaginary component in the complex plane For instance j 4 can be placed on the complex plane as shown in Fig A1 For the same complex number it can be expressed in terms of its modulus (magnitude) and angle namely j4 = 1 4 ( 4 ) tan = 5 51 The magnitude is the distance of the complex number to origin and the angle (or argument) is the counterclockwise angle from the realaxis (0 ) of the complex number Imaginary Axis 5 4 j 4 1 RealAxis Fig A1 j 4 in complex plane The same complex number is shown in polar representation in Fig A Imaginary Axis RealAxis Fig A Polar representation of j 4 in complex plane Example A1: Convert 115 j 09 into its Polar coordinate olution: j09 = tan = ( ) Power ystems Basic Concepts and Applications Part I Module 1 Page 18

20 Example A: Convert 10 0 into its trigonometric form olution: 10 0 = 10 cos0 jsin0 = j05 = 866 ( ) ( ) j5 Example A: Adding 10 0 and 115 j 09 and express the sum in trigonometric form and polar form olution: The sum in trigonometric form: j09 = 866 j5 115 j09 = 981 j59 The sum in Polar form: j09 = 981 j59 = 1 59 ( ) tan = Power ystems Basic Concepts and Applications Part I Module 1 Page 19

Power Systems - Basic Concepts and Applications - Part I

Power Systems - Basic Concepts and Applications - Part I PDHonline Course E104A (1 PDH) Power Systems - Basic Concepts and Applications - Part I Instructor: Shih-Min Hsu, Ph.D., P.E. 01 PDH Online PDH Center 57 Meadow Estates Drive Fairfax, VA 030-6658 Phone

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