Power Systems I
Power Systems I ECE 470
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l N Complex Math Operations on the T 183 Calculator Keith Hoover December 8 2005 So our T183 calculators all behave the same way in this class please set the mode of your calculator as follows Note that all keys besides numeric keys on your calculator are bracketed by lt gt in this document Note that some of the indicated keys require two keys to be pressed For example the ltigt key requires the yellow 2nd key leftmost key in top row to be pressed followed by the key bottom row 3rd from the right Likewise any alphabetic key such as A requires the green ALPHA key to be pressed before the alphabetic key is pressed 4 gt7 Start by pressing the ltONgt button if the calculator is not on and then pressing the ltMODEgt key Use the cursor arrows and the ltENTERgt key to select the following options Normal Sci Eng Float012 456789 Radian Degree Func Par m Seq Connected Dot Seg uential Simul Real abi re Oi m Horiz G T Press the ltCLEARgt key to eXit the setup mode and return to a blank calculator screen These defaults should not go away when you turn off your calculator so theoretically you will only have to set them once at the beginning of the quarter Find the phasor representation of vat lOcos20t 30 Volts Solution By inspection thus the phasor representation of the sinusoid vat is Va 10eXpj30 Instead of boldface please use an overbar to indicate phasor you are writing Qhasors by hand To enter this into your T183 enter when 10 ltequotgt ltigt 30 lt717gt lt gt 180 ltgt ltEntergt Note that we must enter j sqrtl as i and that there is a special calculator key for this The multiplication sign is optional when multiplying two dissimilar quantities such as i times the phase angle so I have left it out When using the eXponentiating function key ex with an imaginary phase angle as we are here the phase angle must always be expressed in radians thus we must convert the given 30 degree angle into radians using the standard technique of multiplying it by 71 180 You should now see the following two lines on your calculator display 10e i307r180 10000eA524i E 4 2 You may convert this complex number from polar to rectangular form by entering ltMATHgt ltCPXgt lt6 Rectgt ltENTERgt ltENTERgt The displayed result is Ans gt Rect 86605000i To store this compleX number in our calculator as variable A we may assign up to 27 real or compleX variables using oneletter variable names from A to Z so that it can later be retrieved without having to manually reenter it enter ltSTOgt ltAgt ltENTERgt The displayed result is Ans gt A 20000e 524i To recall this number at a later date you may type ltRCLgt ltAgt ltENTERgt Find the phasor representation of vbt 5sin20t 30 Volts Solution Since vbt is expressed as a sine function rather than as a cosine function we must rst use the trig identity sinX cosX 90 Please memorize this trig identitv as it will be needed often in this course Thus vbt 5cos20t 300900 5cos20t 60 and Vb 5eXpj60 Thus we must enter 5 ltexgt ltigt ltgt 60 ltngt lt gt 180 ltgt ltEntergt Note that when entering negative numbers such as here we enter 60 you must use the negation key lt gt which is di erent from the subtraction key lt gt To store it as variable B enter ltSTOgt ltBgt ltENTERgt Find vat vbt using phasor addition A and Vb as Variable B simply enter Because we have previously stored Va as Variable ltAgt ltgt ltBgt ltENTERgt The result should be displayed on the calculator as AB 11180e 0060i Thus taking this result back to the time domain yields vat vbt 11180cos20t0060 radians Of course we can multiply the phase angle 0060 radians by 1807 to get 343 80 if we desire to express the phase in degrees but if you do this please be sure to include the degrees symbol on the phase If there is no symbol after the phase value it will be assumed to be in radians 5 Given that vst 10sin20000t volts R 250 ohms L 10 mH and C 01 uF nd vct vLt and vRt using phasor analysis in the following circuit VR t R1 250 ohms L1 Vs 10mH lOCOS ZOOOOt V Solution Store inductive impedance in variable L ltigt 20000 ltxgt 10 ltEEgt lt gt 3 ltENTERgt ltSTOgt ltLgt ltENTERgt Store capacitive impedance in variable C 1 ltgt ltgt ltigt 20000 ltxgt 01 ltEEgt lt gt 6 ltgt ltENTERgt ltSTOgt ltCgt ltENTERgt Calculate current I and store this in variable I 10 lt gt ltgt 250 ltgt ltLgt ltgt ltCgt ltgt ltENTERgt ltSTOgt ltALPHAgt ltIgt ltENTERgt Calculate VR 250 ltxgt ltIgt ltENTERgt Answer 6402e00876i gt VRt 640200s20000t 0876 Volts Calculate VL ltLgt ltxgt ltIgt ltENTERgt Answer 5121e02447i gt VLt 512100s20000t 2447 Volts Calculate Vc ltCgt ltxgt ltIgt ltENTERgt Answer 12804e00695i gt vca 12804cos20000t 7 0695 Volts Threephase power AC power is typically generated and delivered via 34 systems There are two principal advantages to using 34 power 1 The power delivered to a balanced 34 load does not vary with time as do 14 loads This results in less vibration and lower stresses in 34 motors and in systems driven by them I A 34 system can deliverthe same amount of power at lower current levels than can 14 systems This allows the use of smaller conductors A threephase source is made from three independent sources Each ofthese sources is a sinusoidal source ofa certain frequency all three sources have the same 0 and differing phases thus the term threephase Threephase source This 34 source is connected in the Y con guration 3 Here the quotYquot can be viewed sideways Another 3 39 39 39 I 4 connection the A configuration Is also 0 Vpccosmt 9c possible For sources the YcoanguratIon Is most often used Vpacos0t 98 The source is quotbalancedquot if Vpa1 Vpb Vpc Vp and if ea 9b and Go differ by 120 6 Vphcosmt 9b lfany ofthese conditions are not met then the 34 source is unbalanced b To be a balanced 34 source the source amplitudes must be equal and their phases must differ from each other by 120 lecture 7 outline ThreePhase Loads Y configuration The Y connected 3 load is said to be quotbalancedquot if all impedances are equal That is if 28 Zb 20 Zv If these impedances are not equal the load is unbalanced A configuration Likewise the A connected 3 load is said to be quotbalancedquot only if all impedances are equal That is if Zeb ZbC Zca ZA If these impedances are not equal the load is unbalanced A balanced 3 Aconnected load can be converted to an equivalent Y connected load by dividing ZA by 3 Athree phase system the source and load is said to be balanced only if both the source and load are balanced lfeither is unbalanced the system is also unbalanced For balanced 3 systems we can simplify ourtask quite a bit if we are familiar with the characteristics of balanced 3 system On the other hand the way to treat an unbalanced 3 system is like any other phasor circuit lecture 7 0141le YA system Phase Quantities Phase currents Lb and phase voltages V are associated with the individual loads and sources For the Y connection the phase currents are Ia lb IQ and the phase voltages are Van Vbn Von For the A connection the phase currents are lab Ibo lea and the phase voltages are Van Vbc Vca Line Quantities Line currents IL and line voltages VL are the currents in and the voltages between the lines between load and source They are the same for Y or A systems the line voltages are Vab Vbc Vca while the line currents are Ia lb Io In 34 systems voltages and currents are understood to be line quantities unless otherwise speci ed Powers are understood to be 34 powers lecture 7 outline 74 KVL and KCL give the relations between phase quantities and line quantities a the line voltage s phase leads the phase of phase voltages by 30 KVL gives Vb Vab Vbn 39Van 0 Rearranging gives Vbn A VL Vi IL x3 ll 0 the line currents phase lags the phase of phase voltages by 30 KCL Ia39lablca0 ab Rearranging gives Ia lab 39 Ica 39ca lecture 7 outline Example A 208V 34 supply feeds a balanced Aconnected load with z120 35 o Determine i the line and phase voltages and currents ii the real reactive and apparent power supplied to the 34 load lecture 7 outline YY system Forms to be a ba anced system w 3o system 23 zs 29 z The source vo tages have the same magntude and dwffer120 m b age For a ba anced YVY systern KCL shows that In 0 Tms 5 mponantsmce the vo tage between h and n rnust be Zero The connecuon between h and n has no current through I and no vo tage across 1 Nothing would be changed irtne connection were removed n pamcu an the vo tage between n a d n wou d remam Zero even meoutme connecuon present 1mm 7 mm Balanced YY System with neutral connection removed Analyzing balanced 34 systems 14 equivalent from 34 If loads are given as impedances convert all As to Ys by dividing ZA by 3 If loads are given as powers the powers will be 34 powers Divide 34 power by 3 to obtain 14 power Divide all line voltages by v3 to produce Y phase voltages Analyze the 14 equivalent system Multiply resulting phase voltages by v3 to produce line voltage results Multiply resulting powers by 3 to give 34 power 14 equivalent circuit lecture 7 outline maman ltgt 683 0 9 V ltuk No lt35 03 ltuko a lt35 3 m 0 C Fa ltukLNg lt23 496 U 3 magmas 08 303 n QSNEw Example Two 34 loads are connected in parallel and supplied at 480 V Load 1 is 15 kW 07 lag pf Y connected Load 2 is 15 kVA 08 lag pf Aconnected The feeder from the source has an impedance of 025 j05 Ophase Determine i VR ii P Q and S drawn from the source iii pf seen bythe source lecture 7 outline lecture 7 outline There are usually two main factors in the cost ofelectricity i Demand charge which is based on the maximum kVA demanded in the month ii Energy charge which is based on the total kWh consumed in the month Electric bills are greatly reduced ifthe pf is corrected to near unity It is worth noting that power factors are not corrected to exactly unity because this would lead to the pf being overcorrection when the load is small and produce a leading pf Example The following diagram shows two loads supplied from a common 34 feeder Load 1 is 10 MW 065 lag while load 2 is 15 MVA 075 lag Both loads are supplied at 345 kV 29 Q 2 Q 2 Notice that the meter is placed before the feeder This means that the feeder losses will be part ofthe electricity bill Determine i Current in the feeder v Voltage regulation iii Cost ofelectricity yr for constant loads energy costs 25 kWh and demand charge is 700KVAmth iv The capacitance uFph of a wyeconnected capacitor bank needed to improve the combined load pfto 098 lag v V v Repeat parts i iii with the new pf lecture 7 outline lecture 7 outline 713 General formulas nomenclature units and conventions in power Time domain vt Vm cos oat 6V Vp cos oat 6V it it Im cos oat 9i p cos oat 9 Phasors with magnitude in peak value vt Vp W 9V Ip Ipz 9i Phasors in rms V V49V V N2zev I Izei IplDZMGi If a phasor current or voltage is given without other quali cation it s assumed to be in rms Power 3 34eVei2 84953 PaV jQ 8 complex power in VA 8 apparent power in VA PaV average power in W Q reactive power in VAR es 3 ev el 92 If poweris referred to without any other quali cation it s assumed to be average power 8 V V49V Iz e VI4eVe V cosmvei j V sin9Vei S W PaV V coswvei V cos as Q V sinIPBVei V sin as power factor cosIIBVei lagging for evgte egt0 QgtO inductive loads leading for evlte elt0 QltO capacitive loads lecture 7 outline 714 34 power Unless otherwise speci ed voltages and currents are assumed to be line quantities Power unless otherwise speci ed is assumed to be average 3 power Any power unless otherwise speci ed is assumed to be 53 A 34 system is balanced if 1 Phase voltages are equal in magnitude and displaced 120 in phase from each other 2 Phase load impedances are equal for balanced Y connection VL V3Vp IL lp for balanced Aconnection VL Vp IL V3lp 95 92 9V 9 s 8495 P jQ V3 VI V3 Vl cos as jV3 Vl sin es lecture 7 outline 111 34 AC induction motors Motors transform electromagnetic energy into mechanical energy Before developing equations and models let s first discuss their principles of operation The basic mechanism of almost all rotating motors is very similar no matter what particular type they happen to be Rotating machines can be viewed as two magnets One magnet is hooked to the motor39s shaft and the second magnet surrounds the rst magnet An induction machines works by causing the outside magnet to rotate which in turn pullsthe magnet hooked to the shaftalong with it This induced rotation is what is seen when the motor is turned on the shaft begins to rotate How are these two magnets are produced in the AC induction machine 1St magnet the stator Stator operation is based on two ideas i A magnetic field can be produced by a current loop ii A magnetic eld produced from a threephase source can produce a rotating magnetic eld i Suppose the current I is sinusoidal It would grow large and positive achieve its maximum value grow smaller become zero then begin to grow large again this time negative ach39eve its maximum value grow smaller become zero again grow large again positive etc lecture 11 outline 112 This would in turn cause the magnetic ux 4 to grow large pointing upward achieve its maximum value grow smaller become zero then become larger this time pointing downward achieve its maximum value grow smaller become zero grow large again upward etc This magnetic eld is not rotating A 34 source can produce a rotating magnetic eld The three currents differ in phase by 120 and are also spatially displaced from one another by 120 When la in and iC are positive the resulting magnetic fields are as shown below lecture 11 outline 113 Let s see how the resultant eld acts as a function oftime iat l cos oatA ibt l cos oat 120 A ict l cos oat 120 A Let39s look at the resultant magnetic elds for oat 0 oat 60 oat 120 oat 180 oat 240 oat 300 oat 360 We39ll look at the first four to get the idea m 113 c la it l b b ml 130 lib d39c l c l 1 0b a lecture 11 outline 2nd 114 This is how the rotating magnetic eld is produced in a 34 induction machine The stator windings are spatially diplaced by 120 from one another More precisely this is the way ifthe machine is to have a synchronous speed of 3600 rpm For slower speed motors we adjust a bit more on this later magnet the rotor The 2nd magnet is mounted on the shaft and can rotate It s called the rotor magnet The rotor is wound by wrapping wire around magnetic material How does this wire and magnetic material ever cause the current that is necessary for a 2nd magnetic eld to exist Recall Faraday39s law ofelectromagnetic induction the same one giving rise to transformer action ijdl loop f dt V loop The rotor does not rotate at synchronous speed the speed of the stator s rotating magnetic eld It rotates a bit more slowly How much more slowly Well that depends on the torque required by the load The more slowly it rotates the greater d dt and the greater its current and magnetic field strength The rotor comes to a speed that just balances the load and output torque Slip is a measure ofthe difference between the rotor speed and the synchronous speed lecture 11 outline m geherah the synchronous Speed HS and shp 5 0f ah mouctroh motorrs gwen by theformu55 oerow H m rpm r SumereqencymHZ p number of pores quots Shp 5 Working with AC induction motors mouctroh machmes are the most common of 5H erecthcar machmes They are srmpre m cohstructroh rugged rehaore nexpenswe and compact They come rh smge e and three age cohstructrohs We Wm omy be concerned wrth threephase mducuon motors A Zpo e motor HS 3000 rpm for r 00Hz compared wrth a 47p0e motor rp 1800 rpm for r 00Hz twopore mouctroh motor founpo e mouctroh motor For 60 HZ Systemsr H can be 3600 1800 1200 900rpm etc 1mm 11 mm 116 Example For a 60 Hz 8 pole motor nd the slip at i standstill nm 0 ii 855 rpm Example A 60 Hz induction motor runs at 1164 RPM and develops 003 slip Determine the number of poles The power rating given for a motor is the mechanical power available from the motor usually given in hp A motor rated 10 hp provides 10 hp ofoutput mechanical power at full load Power in Power out V3 VL IL pf T03 Efficiency The quantity of mechanical power available from the motor is less than the electrical energy supplied to it Losses are due to heating and friction The efficiency of large machines is usually well above 90 but can drop signi cantly when the motor is operated far from its rated load n 100 In Electrical power is input to the stator Losses associated with the stator are winding 2R losses and hysteresis losses in the core lecture 11 outline 1177 The remarmng power rs transferred across the arrgap of the macmne and becomes the rnput power to the rotoricaHed the arr gap power Some ofme arrgap power rs orssrpateo by the rotor FR tosses and the remamder rs converted rnto mecbamcat power Some of the mecbamcat power rs orssrpateo by the rotatronat tossesimcuon wscous damprng The remarnoer rs the mecbamcat output power Eteomcet Power Input m i 3 t P e Vllcose ta L s S or 0552 Power out of Stator Powe Into Rotor Pslnlcv PCave Pwmng P P t m Te Rotor Losses Mechamcal Powar Devetoped Pm sPu Pa 1sPuw To Mechamcal Losses Mechanlca Powe Output rm PM mNTW P r F m pm P e rm 2mm 1mm 11 mm 118 Example A threephase 10 hp 480 V 60 Hz induction motor has a rated speed of 1 152 rpm Determine i of poles and rated slip output torque and ef ciency if it draws a line current of 133 A 075 lag iii airgap power ifthe stator losses are 90 W iv mechanical power developed and rotor loss v developed electromagnetic torque v39 rotational losses v 5quot lecture 11 outline 4 1 Power in AC steadystate power in phasor circuits For a circuit with sinusoidal sources all voltages and currents in steadystate have the same form All are cosines with various amplitudes and phases it it l cos oat 9 vt V cos oat 9 The power absorbed by the above element at any instant of time is as always the product of the voltage and current when the voltage and current are labeled with the passive sign convention PSC pt vt it V cos oat 9 l cos oat 9 Consider the voltage and current associated with a 539 resistance The amplitude of the voltage is 5V the amplitude of the current is 3A f 50Hz m314 rads Instantaneous Power Voltage amp Current in a Resistance iner Mn Vnilzue v a Currenl 1A Tlme lmsl The voltage and current are in phase so their product is always positive As a consequence the power dissipated is never negative power is always absorbed by the resistance resistance never supplies power The average power dissipated is half the maximum value 252mg 4 ouzlmz 42 Consider nowan inductor The current lags the voltage by 90 Taking the voltage as phase reference vt V cos oat it Icos oat 90 pt vt it V cos oat l cos oat 90 V th vt and it having the same amplitudes and frequency as previously the power absorbed by the inductance will vary with time as shown below Instantaneous Power Voltage amp Current in an Inductance PowerW Voltage V a Current A n 39 E Time ms Notice that the average power dissipated is zero This is because the inductor stores energy takes it from the source in the 1St half cycle shown and then returns all ofthe stored energy to the source in the 2nd half cycle As in the resistor case the power oscillates at twice the line frequency The maximum amplitude now is ileml2 lecture 4 outline 43 How about a capacitance Time ms lecture 4 outline 44 Consider now a more general impedance one that has a phase 9 between zero and 190 Taking voltage as reference in sinusoidal steadystate the current and votlage are vt V cos oat it l cos oat e pt vt it V cos oat l cos oat 9 Taking V and l to have the same amplitudes and frequency as previously and e 600 Z 460 Q then power will vary with time as shown below Instantaneous Power Voltage amp Current for an Impedance 0 0 Power W Voltage V a Current A o 0 Time ms Power oscillates at twice the line frequency and passes through zero ifeither voltage or current is zero and in general can be off set from the time axis The average power dissipated over one cycle is the mean ofthe maximum 1125 W and the minimum 375 W which is 375 W in the example lecture 4 outline 45 In cases where pt varies many times each second the quantity of interest is often not the value of pt at each instant oftime but rather the average value of power Pav Since in this case pt is periodic the average can be found by nding the average over a period 1 PW ij cos wt Q I cos wt 91dt where T is the period Using the identity cosa cos coso 00so withozwt6vand atQ PaV 12 V l cos 6V 6i can also be expressed in terms of phasors without integration vt V cos oat 6V gt V V49v phasorin peak not RMS it l cos oat 6 gt I Izel phasorin peak not RMS PaV Re 2 VI Re 12V l coseV 9i j 12V l sineV 9i PaV 12 V l coseV 6 Note Idenotes the complex conjugate of I This is an important result since it allows average powers to be readily calculated using phasor analysis lecture 4 outline 476 The reat part ot the comptex ouanttty Vetr phasors not tn PMS ts the ayerage power dtsstpated tn the toad Dtsstpated power ts etectrornagnettc energy that ts changed to another formimermat energy or rnechantcat energy tor examp e The trnagtnary part ts atso rneantngmt The trnagtnary part ot the quartmy Vetr agam wtth the facmr 0f phasors are not tn PMS ts retated to the ttow ot energy that ts not oetng dtsstptated tt ts retated to energy that ts swapped between the tn the toad and the source on as we shatt see stored at dttterent potnts tn the power cycte by dtfferertt eternents at the toad ptut ut stnusutdat yuttage A physical look at energy storage Constder the yottage wayetorrn hown above and supposethatthe yottage ts across a capacttance From prevtous Work the energy Mass stored by a capacttance ts Et 12 c v20 Start at to Here ytt o and so Et o Fort gt 0 ytt oegtnsto grow urtm ytt reaches a maxtmum at sornethtng tess that 2 seconds M2 to be exact Durtrtg thts ttrne tnteryat energy ts oetng dettyered to the capacttance trorn the rest ot the ctrcuttithe capacttance ts outtdtng up tts etectrtcttetd Then tor ttrne greater than 1r 2 seconds ytt grows srnatter urtm tt reaches zero tort 1r Forthts ttrne the energy stored by the capacttance ts agatn zero so that for the ttrne tntetyat 1r 2 lt t lt 1r the capacttance gaye backme energy that ts had stored durtngn lt t lt 1r 2 1mm 4 mm The same process repeats itself again and again Forthe time interval 11 lt t lt 3 11 2 energy is delivered to the capacitance which then gives it back in the time interval 3 11 2 lt t lt 2 11 This process then storing energy and giving it back occurs twice in each period ofthe voltage waveform Thus in a 60 Hz power system this process occurs 120 times each second Doing the same thing for an inductance which stores energy in the form ofa magnetic eld would result in Et 12 L i2t Phasor Analysis in Power Now getting back to the main story The imaginary part of 12 Vi is related to this ow ofenergy to and from energy storage devices in circuits We can even tell what type of device dominates energy storage by the sign ofthe imaginary part inductance dominates if it is positive and capacitance dominates if negative Since both parts of12 VI real and imaginary turn out to be useful the parts of this quantity have been given names and so here we must de ne several terms The quantity 12 Vi is called the complex powers 8 12 VI Sis complex power with units ofvoltampere VA The complex power can be split into real and imaginary parts 8 PEV j Q Pav is average power with units ofwatts W Q is reactive power with units of voltampere reactive VAR lecture 4 outline 48 The complex power can also be express in polar form 3 12vf 12VI4eVei 8495 S the magnitude of S is called the apparentpower with units of voltampere VA For cases where Pav is positive for loads that do dissipate average power ratherthan supplying average power 8 occupies the 1St and 4th quadrants in the complex plane so that 90 lt es lt 90 In this case cos as coseV 9 is called the power factor pf The pf is further characterized as to whether as is positive or not If as is negative then 6 gt 6 so that the current leads the voltage in phase lfes lt 0 we say the power factor cos as is leading Note for a leading pf Q is negative If as is positive then 9 lt 6V so that the current lags the voltage in phase lfes gt 0 we say the power factor cos as is lagging Note for a lagging pf Q is positive lecture 4 outline 4 9 Let39s explore these relationships by looking at specific examples Let39s start with the complex power absorbed by a resistance Resistance I S zVI 2IRIRlzeilziei 12RI24e e12RI240 V R VR which is areal number The complex power absorbed by a resistance is all real which makes sense if we considerthat the imaginary part ofS is associated with energy storage that the resistance does not do For a resistance s1ZVI1ZVI 12 IZR 12v2R wherel and V are the amplitude of it and vt lecture A umlme Capacitance s 12 VI 12 joaC 1 12 lmC l 49 90 Iz e v v c I 12 l2oaC 490 12 j IZImC WC 1 which is an imaginary number The complex power absorbed by a capacitance is purely imaginary which makes sense if we considerthat the real part of S is associated with energy dissipation and that the imaginary part of S is associated with energy storage Since the capacitance does not dissipate energy but only store it the complex power associated with the capacitance must be purely imaginary ln power engineering the imaginary component of complex power is called reactive since it is associated with the reactive elements L and C For a capacitance s 12 VI 12j VI 12 j l2oaC 12 ijVZ wherel and V are the amplitude of it and vt Note The reactive power absorbed by the capacitance is negative and the currents phase that is the phase of it leads that of vt Be careful here Absorbed reactive power is describes stored energy being swapped between the element and other parts of the circuit Reactive power does not referto average power being absorbed Work is not done with reactive power lecture A umlme 411 Inductance S V2 VI 12joaLI I V2 oaLl 49 90 Ize 12 03L I24 90 12j 03L I2 v ij v ij I which is an imaginary number The complex power absorbed by an inductance is purely imaginary which makes sense if we consider that the real part of S is associated with energy dissipation and that the imaginary part of S is associated with energy storage Since the inductance does not dissipate energy but only store it the complex power associated with the inductance is purely imaginary purely reactive For an inductance S zVI 2jV 12ij l2 2jV2oaL wherel and V are the amplitude of it and vt Note The reactive power quotabsorbedquot by the inductance is positive and the currents phase that is the phase of it lags that of vt In particular notice how the reactive power absorbed by the inductance is 180 outofphase with respect to the reactive power absorbed by the capacitance This merely means that when the inductance is ready to store energy the capacitance is ready to give up stored energy And when the capacitance is ready to store energy the inductance is ready to give up stored energy lecture A umlme 412 General Impedance S12VI12IZI12Zl4eezlze v z 12Zl2492 which is a complex number In general the complex power absorbed by an impedance has nonzero realand nonzero imaginary parts That is in general both energy dissipation and energy storage are involved If capacitance dominates energy storage eZ and Q the reactive power will be negative current will lead the voltage in phase This is why the term leading or lead is used for the powerfactor cos 91 If inductance dominates energy storage eZ and Q the reactive power will be positive current will lag the voltage in phase This is why the term lagging or lag is used forthe power factor cos 91 For a general impedance S zVI12VZez 1ZIZZzeZ 12v2ZzeZ where l and V are the amplitude of it and vt and Z is the magnitude of the impedance Note It is assumed here that the real part of Z is positive This causes the real part ofS aborbed complex power to be positive That is the load Z is assumed not to be a source for average power lecture A 0141le Using RMS values of current and voltage When performing sinusoidal steadystate analysis in circuits especially in power the convention is to use RMS values for voltages and currents For sinusoids the rootmeansquare RMS value is the amplitude divided by the square root oftwo In the study of power systems RMS values for the current and voltages are usually used rather that peak values When RMS are used the 12 in the power relation is eliminated Is 1 v P 2 2 J5 rms rms lt Ohm39s Law is unchanged just divide both sides by V2 When performing circuit analysis with sources given as RMS values the currentvoltage relationships are unchanged so that circuit analysis proceeds as usual The voltages and currents obtained will be RMS values so that when calculating powers the 12 should be left out ofthe power relationship When moving into the time domain remember to replace the V2 That is suppose we nd a phasor current in RMS I Izei Ams The corresponding it will then be mwlmummA Since the amplitude is V2 times the rms value lecture 4 outline 414 General formulas nomenclature units and conventions in power Time domain vt Vm cos oat 6V Vp cos oat 6V it it Im cos oat 9i p cos oat 9 Phasors with magnitude in peak value vt Vp W 9V Ip Ipz 9i Phasors in rms V V49V VpaZMGV I I49 lpa2zei qualification it39s assumed to be in rms If a phasor current or voltage is given without other Power 8 849692 8462 PaV jQ 8 complex power in VA 8 apparent power in VA PaV average power in W Q reactive power in VAR 9 eVei If power is referred to without any other quali cation it39s assumed to be average power s VI we Iz ei Vlzev ei VI cosGBVei j VI sin9Vei S VI PaV V cost vei V cos 9 Q V sianei V sin 9 power factor costevei lagging for evgte egt0 QgtO inductive loads leading for evlte elt0 QltO capacitive loads lecture 4 outline 415 Example 1 1 Find the phasor circuit forthe circuit shown below use rms value for source Use phasor analysis to nd 2 land it 3 88 P and Q delivered to RL load 4 load power factor 5 compare P with 2R it 39 39 39 39 39 39 39 39 39 39 39 39 5 I2 cos2n1000t v 220 9 lecture 4 outline 416 Example 2 1 Find the phasor circuit forthe circuit shown belowuse the rms value for source Use phasor analysis to nd 2 land it 3 S S P and Q delivered to RC load 4 load power factor 5 compare P with V2R 39 5 12 cos21t10001 v 047 HF 2 130 o E 39 I lecture 4 outline 81 Magnetic circuits The source of magnetic flux is current Not just current but the product of current and the number of times the current is wrapped around a core NI The units for NI are ampereturns W This quantity magnetomotive force MMF is analogous to voltage in electric circuits 5N Magnetic ux 4 plays the role of current By convention the pole from which magnetic ux leaves is called the north pole and the one to which it enters is called the south pole The earth is a notable exception In magnetic circuits eluctance l2 plays the role that resistance does in electric circuits Like resistance reluctance is determined by material property and geometry The inductance of a coil is determined by the core s reluctance and the number of turns L N2l2 Example iii Determine the inductance ofthe coil i Find the current needed to produce a ux of5 me in the air gap ii What is the direction of the ux in the core based on the shown direction of the current 83 Relays and solenoids These are devices that produce mechanical movement from an electrical signal Its basic components are coil which when connected to a supply creates the MMF which is the source of magnetic flux in the core core which carries the flux from the coil to the airgap sometimes more than one airgap is used The core is usually fixed and has a low reluctance armature which carries the flux between the airgaps It is movable and like the core has a low reluctance The armature is usually connected to the mechanical device that has to be actuated eg shutoff valve spring which holds the armature in the nonactuated position Armature J L l moveable V l Spring WW UUUUK Core fixed In the ideal case the reluctance of the steel in the core and armature is negligible this means that all MMF is dropped across the two airgaps which are in series 222g The magnetic circuit is The energy stored by an inductance is ELF Using this to find the energy stored in each air gap Now taking the derivative with respect to the gap usually denoted as X or g the force per air gap is obtained 85 Example For the relay shown previously the spring exerts a force of 02 N The gap length is 5mm when fully open and 2mm when fully closed The coil has 5000 turns and is wound on a core of 1cm square cross section Predict the pickup and dropout currents Example A lifting magnet is shown on the left The core has a square crosssection of6 X 6 cm2 The coil has 300 turns and a resistance of6Q Determine the lifting force when the airgap is 5mm and a 120V DC supply is used Neglect the reluctance ofthe core and the mass being lifted V030 Mass to be lifted magnetic material Example Repeat with a 120 V 60 Hz AC supply instead of DC The reactance of the AC coil reduces the current which greatly reduces the lifting force Lifting magnets are nearly always DC 101 Ideal Transformers Transformers are enormoust important V thout them electrical power would not be as available and widespread as it is today They allow changes in voltage levels In particular voltage can be increased to allow power to be transported at lower current levels Transformers have two principal components A core of high permeability read ferromagnetic material which is able to con ne a magnetic field V ndings which are wound about the core An externally supplied AC current on one winding call the primary winding produces a magnetic eld in the core by Ampere s law AC voltages will then be induced by Faraday s law on other coils secondary winding that are wound on the core The ideal transformer model is based on two assumptions The core of an ideal transformer has in nite permeability The result is that all flux is con ned to the core There is no power loss in the ideal transformer The result is that power in must equal the power out 50 E Equotquot39 LL 2 39 12 N N 3 V2 V7 I D 1 2C I s gt c D lecture 10 outline 102 Transformer action The sinusoidal currents i1t and i2t flow in the primary coil and secondary coils the coils having N1 and N2turns respectively The two currents working together produce a sinusoidal magnetic ux t in the core Since the core is assumed to have infinite permeability all the flux is confined to the core no flux leaks out The result is that the same ux links both coils Sinusoidal voltages are induced in the coils via Faraday s law Since they are caused by the same ux they must be in phase Their magnitudes are related by the turn s ratio V1t N1 dt V20 N2 dt v1t V1 cos oat 6V v2t V2 cos oat 6V V1t N1 V1 a a is called the turns ratio V2t N2 V2 For an ideal transformer the permeability of the core is in nite and therefore has no reluctance The magnetic KVL gives N11 N2I2 For an ideal transformer Pin Pout no losses V11COS9V9i1 V2I2 COS9V9i2 Therefore em 92 i1t l1 cos oat 9i i2t l2 cos oat 9i lecture 10 outline 103 Phasor relations for the ideal transformer N2 390 394 ll Z INZ Nlt Alt The ideal transformer application impedance reflection lecture 10 outline 104 Example Determine the impedance seen by the source and then determine the primary and secondary currents and voltages an l1 j459 110 I2 1 new V1 lt U and U 007 Notice that when the load was referred to the primary side its value reduced This is because the load was taken from the high voltage side across to the low voltage side Impedance will go up when it is moved to a higher voltage and will go down when it is moved to a lower voltage lecture 10 outline 105 EXAMPLE This example demonstrates how transformers reduce the cost of supplying electricity A singlephase load of 1OMW with pf1 is to be supplied at 24kV via a distribution feeder with 015 jO1Q impedance Electric energy costs 3 kWh Calculate the yearly cost of supplying the line losses ifthe load is constant and continuous if a no transformers are used and b two transformers each with a 10 are used original system 015 j01Q 1 Vs 24mn kV 1 1quot 1 V system with stepup and stepdown transformers 110 015 IF 11119 101 439 10 MW V a man voltagel8 low current 24 W pf 1 um in this loop lecture 10 outline Energy Costs VR lecture 10 outline 107 Practical Transformer Model A model that accounts for energy storage and energy dissipation is shown below 1 R1 39 a1 R2 1 jX2 The model includes an ideal transformer with components added to account for nonideal behavior of real transformers 0 Real transformers are not lossless R1 and R2 account for energy dissipation in the primary and secondary windings o In real transformers the ux is not entirely con ned to the core some ux leaks out X1 and X2 account for the energy stored in this leakage ux 0 The core is not lossless Some energy is lost magnetizing and demagnetizing the core Rc represents the real power that is dissipated in the core of the transformer due to hysteresis and eddy currents o Xm accounts forthe magnetic energy stored in the core 0 Vp and V2 are the primary and secondary voltages that appear at the terminals ofthe transformer lecture 10 outline 108 The circuit above is known as the exact equivalent circuit A simpler model that still gives accurate results is shown below Ip R iX I1 a 1 2 L 39 V2 2 VI v2 d The parallel elements RC and M are moved in front of R1 and X1 and placed across the supply This does not introduce much error since RC and Xm are much larger impedances than R1 and X1 and so appear to be almost open circuits R2 and X2 are referred to the primary side by impedance reflection They are added to R1 and X1 since they are in series with them R R1 a2R2 and x x1 a2X2 This circuit is called the approximate equivalent circuit It s the one we ll work with Efficiency and Regulation Efficiency and regulation are gures of merit for a transformer they measure well a transformer performs how close to ideal it is For VR use Vp and V1 For n use Pout ReV1I1 and Pin ReVpIp lecture 10 outline General Analysis Ploss hastwo components Poo and Pcore Poo varies quadratically with current goes up with load Pcore is almost constant since the voltage Vp doesn t vary much Example A 300 kVA singlephase transformer is rated to transform 24 kV to 600 V and has the following equivalent circuit parameters referred to the high voltage side RO75 2 X15 2 Rc500 2 Xm6OQ Calculate VR and n when the transformer supplies rated load at 07 lag pf lecture 10 outline 1010 Example cont lecture 10 outline 1011 Example Rework with the load pf corrected to unity P remains the same The results are VR 288 n 921 Poo 796 kW Pcore 122 kW lecture 10 outline 1012 Experimental determination of equivalent circuit The equivalent circuit parameters can determined from the open circuit 00 and shortcircuit SC tests Required measurements are Vp lp and Pin 00 test The test is performed at rated voltage and no load the load is an opencircuit I1 O R and X can be neglected Only RC and Xm load the supply Vac VIEth 0 Voc be and PDC are measured once the supply has been adjusted to Vrated The procedure is 1 determine 6 C from pfo VPoc then 72060054pr c 000 2 determine IOC from IOC locze0c C jm The negative signs associated with 6 C and in are due to he lagging current How does one determine it s lagging Calculate Rc Y and xm c m lecture 10 outline SC Test 1013 This test is performed at rated current with the secondary terminals short circuited V1 0 The supply voltage is turned down to zero and gently increased until rated current flows RC and Xm are usually ignored in this test since they are very large compared to R and X That is ISO 39 I2 vac Ve N Inw Vsc LC and PSC are measured once the lSC has been adjusted to lrated The procedure is 1 Determine ago from 2 Determine I1 from 3 Determine 2 from pf P50 then 500 1 sc Cos pfsc I1 Iscl39 980 vsczo R X Z 4 J SC 39 SC lecture 10 outline 1014 Example Find the approximate equivalent circuit for the following transformer ratings 100 WA 2500 V 125 V 00 test V0C 2500 V 0C 25 A P0C 2750 W sc test VSC 178 V SC 40 A PSC 2250 W lecture 10 outline