Elements of Electrical Engr II
Elements of Electrical Engr II ECE 207
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This 125 page Class Notes was uploaded by Dr. Reina Hane on Monday October 19, 2015. The Class Notes belongs to ECE 207 at Rose-Hulman Institute of Technology taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/225088/ece-207-rose-hulman-institute-of-technology in Electrical Engineering & Computer Science at Rose-Hulman Institute of Technology.
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Date Created: 10/19/15
231 Electromagnetic Interference Electromagnetic Interference EMI is a field in which engineers from a variety ofdisciplines work together to solve systemlevel design problems COUPLING PATH SOURCE 4 RECEIVER There must be a source of EM energy a coupling path and a receiver EMI occurs is via electricfield coupling magnetic field coupling or electromagnetic wave coupling Electric eld coupling An electric field is produced by a voltage difference Electric eld coupling typically is via unintentional parasitic capacitance The metal from an interfering circuit forms one plate of the capacitance and and metal from the victim circuit forms the 2nd plate In the above circuit C represents the capacitive coupling between the two circuits How could one predict voltage in the victim circuit present due to electric coupling lecture 23 outline 232 The voltage coupled via the electric eld in the second circuit is directly proportional to the capacitive coupling C The question is how can C be reduced Recall the parallel plate model for a capacitance While not a perfect model for the coupling above the model can be used to point out some important factors in reducing C C eAd are 1St increasing circuit separation increasing d will reduce C The most common way of doing this is to orient the two circuits so that they do not run near and parallel for a distance 2nd decreasing the area ofthe plates reduce A will reduce C In practice this is often not a practical solution Instead the same end is accomplished using an electric shield For example TC circuits are very susceptible to electric eld coupling from other circuits especially of course highvoltage circuits This is due to the Vic being small in the mV range 39 39 conductors for TC electrostatic shield connections lecture 23 outline 233 It is common to route circuits near a conductive plane Placing the circuit conductors next to a conductive plane can diminish electric field coupling by a factor of 10 This is the reason for the common practice of routing cables next to the metal chassis in electronic equipment or next to the metal frame in a car or plane Magnetic field coupling 1 A magnetic field is produced by an electric current Ampere39s law describes this process 2 Atimevarying magnetic field is produced by a timevarying current lecture 23 outline 234 3 These changing magnetic fields can induce voltages in a second circuit Faraday39s law describes this process Vm BoAmcose For a given interfering magnetic field of frequency 0 coupling into the second circuit may be reduced in three ways 1 Reduce the magnitude of B0 by separating the circuits 2 Changee the orientation ofthe victim circuit rule of thumb reduction is limited toa factor of100 3 Reduce the linkage ofthe loop in the victim circuit One way is simply to reduce the area A lecture 23 outline 235 Anothertechnique is to use twistedpairwires rule of thumb reduction limited to a factor of 1000 lecture 23 outline 236 Example The read head ofa disk drive puts out a string of pulses each with a peak amplitude of 1 00 V A pair of wires connects the head to an ampli er over a distance of2 inches Suppose that the disk is located near a television transmitter operating on Channel 2 5525 MHz with an effective power of 100 kW At this standard power level the peak magnetic ux density B A mile away from the antenna is 2gtlt10398 webersm2 i Compute the peak voltage induced in the circuit ifthe wires are spaced at a centertocenter distance of 01 in v What is the peak induced voltage ifthe wires are twisted together at the rate of8 twistsinch Assume the same center tocenter spacing used in i lecture 23 outline Power factor correction voltage regulation and efficiency 0 0 load VS 0 SPjQ P the average power delivered to load is measured by the wattmeter S the apparent power delivered to load is determined by the product ofthe ammeter reading and the voltmeter reading Remember ammeters and voltmeters give amps and volts in rms unless otherwise speci ed The default is rms pf the power factor is the ratio of P to S pf PS Note that the magnitude of the pf does not in itself indicate whether the load is leading or lagging For most industrial loads the power factor will be lagging If this cannot be safely assumed to be the case some means of measuring phase must be in place 8 the complex power delivered to the load 8 846 Q the reactive power delivered to the load Q 8 sin 9 lecture 6 outline 62 Example Suppose in the circuit above that the voltmeter reads 200V the ammeter reads 10A and the wattmeter reads 15kW The load is a resistor and inductor in series Determine i real power ii apparent power iii reactive power iv power factor v oad component values Q for R H for L iff 60 Hz v v lecture 6 outline Power Factor Example A power source has 2 jSQ impedance in its leads lt supplies a load of 1MW at 10kV 60Hz Determine the magnitude of the supply voltage and the average power drawn from the supply when i the load power factor is 05 lag and ii the load power factor is 095 lag 29 13939 r Vs To determine the supply voltage we rst need to nd the current so that we can determine the voltage across the feeder impedance To nd the current we need to determine the load complex power Notice the load power is speci ed in terms of average power P and pf From this information we can determine the loads complex power lecture 6 outline Voltage regulation VR Voltage regulation is a measure of how the load voltage varies with current VR Vnoloadl0 39 Vfullload 100 full load Find the VR at 05 lag and the VR at 095 lag lecture 6 outline Efficiency Ef ciency is the ratio between power in and power out Here Pin is provided by the source Vs Pout is absorbed by the load eff P39 ad 100 P39 ad 100 source load losses Find the ef ciency at 05 lag and 095 lag Power factor correcting capacitance Find the capacitance necessary to correct the pf from 05 lag to 095 lag in the example above lecture 6 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 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 Control Systems Control systems are used to control parameters such as the position ofa cutting tool the temperature ofa chemical bath orthe speed ofa motor Measurement is a critical element in advanced control systems utilizing feedback Openloop control no feedback Consider a system ofa motor together with its load The input Xt is the motor voltage andthe output yt is the shaft speed of the motor Xt motor yt input voltage load shaft speed A control system is a system by which the behavior ofthe output is controlled by controlling the input here the shaft speed ofthe DC motor is controlled by the input motor voltage Consider the motor in steady state with a load such that the input voltageshaft speed relationship is 50 rpmvolt V quotm 50 rpmN Input voltage shaft speed How could the shaft speed be controlled in this case Simple Just divide the desired shaft speed in rpm by 50 to obtain the required input voltage in volts This is an openloop control system It works ne for ho usehold fans etc More sophisticated control systems are often needed The crucial weakness in openloop systems is that they cannot compensate for changes in the motorload system Motor loads vary constantly Motors age The 50 rpmV transfer cha racteristic is only accurate for a particular load at a particular time The design of more sophisticated control systems usually uses feedback that is the output is measured and fed back into the system input Accurate measurement of the output is crucial lecture 26 outline 262 Closedloop control systems with feedback A measurement of the output shaft speed can be used to design a motor speed control system which is relatively insensitive to changes in motor load that is where variations in motor speed are small when compared to changes in motor load Suppose the desired motor speed is 2000 rpm regardless ofthe load to which the motor is attached Take 2V as the input voltage which is to produce 2000 rpm 1 V produces 1000 rpm etc For this case a shaft speed transducer a tachometer is need that produces 2V when measuring 2000 rpm ampli er motor load 0001 Vrpm tachometer The actual speed which corresponds to 2V will be very close to 2000 rpm ifAKm is large n V V 1ooovc forAKmgtgt1000 m 1 0001 1 AK 1000 Two key items in this control scheme are 1accurate measurement ofthe output shaft speed and 2a large forward gain the product ofA and Km lecture 26 outline 263 Example Design a feedback control system for motorload speed control The desired speed is 1500 rpm for a 3V input 1 re Ulred tachometer V q 500 Aom ampli er gain A 5000 gtnm 3V ampli er motor load 0002 Vlrpm tachometer Find the actual motor shaft speed when i Km 50 rpmV ii Km 25 rpmV General Analysis GX HY Y l i X 1GH approximations I i 1H X 1 GH 1 1GH l s l 1 L for iltlt1 L is the error from the quotidealquot of l X H GH GH GH H lecture 26 outline 2674 Example The system shown o tow rs a system tor controng the temperature or a chemrcat reactroh Ah etectrtcaHyrcontroHed yatye mtxes W0 sources of water to produce Water WM a tem eratur Tr wmch then ows through the reactor racket wmch orrhgs the chemrcat Sohmon to the same temperature TH2 rh steaoystate A RTD together wrth a Wheatstohe orroge momtors the temperature T2 and oeyetops a yottage yz whrch rs compared wrth a reterehce yottage yx to produce ah error yottage ywhrch orryes the cohtrouerwhose garh rs 1000 whrch rh turh orryes the y tye The oath or the yatye rs such that my corresponds to do and 5v corresponds to too39c The RTDNVheatstohe comorhatroh has a trahsterruhctroh or 5my C r Draw the dosedr oop otoch oragram that represents the cohtrot systerh Determrhe the peat dosedr oop garh hr Determrhe the exact dosedr oop oath and the percent error rh part u y Determrhe a new yatue or cohtrouer oath that wouto reduce the error h the dosedr oop oath to o 5 mm mm m mum mg a 1mm 25 mm Example cont lecture 26 outline 266 Control of dynamic systems In the previous example T1T2 in steadystate Transient behavior was not considered When transient behavior is of interest system dynamics must be included in the analysis Consider a DC motorload with a constant eld ia Speed nt rpm Energy storage is typically dominated by that stored in mechanical inertia and consequently the system behaves as a firstorder system V th a step function input the resulting speed response is exponential steadyslate speed depends on size of voltage step l nt nF1 e39quot m is the steadystate speed c is the motor time constant lecture 26 outline 267 In the s domain this is modeled as v Km N gt gt input voltage 35m 1 shaft speed Km is the static gain coefficient and rm is the motorload time constant Example The following diagram depicts a motor speed control system in which the preamp has a gain of 2000 with a time constant of 200ps the power amp has a gain of 10 the motor has a gain of 50 rpmV with a time constant of 200ms and the tachometer has a gain of1 mVrpm i Draw the block diagram that represents the control system in the sdomain power amp gtnm motor load tachometer ii Determine an expression for the closedloop gain in polynomial form iii Compare the steadystate value of part ii with the ideal closed loop gain lecture 26 outline Example cont lecture 26 outline Frequency response of dynamic control systems Considerthe motor used in a previous example TF 50 motor 102s Taking the Bode magnitude plot ITleue 40 The plot shows that the rpmV response ofthe motor goes down for frequencies higher than about 5 rs Now consider this motor as part of the closedloop control system power amp 50 V6 gt nm 025 1 motor load 0001 Vrpm tachometer 1 draw the Bode magnitude plot forthe forward path G 2 drawthe Bode magnitude plot forthe feedback path H G i for GgtgtI 1GH G forGltlt lecture 26 outline 120db 1M 1 Dudb 100k 6 Dub 10k 40db 1011 Earth 1k 20db1o 2610 Odb 1 56k 500k 0 1 Notice that for frequencies less than about 5000 rs the system 2 v gain is much as it is at DC Compare this to the response of the motor alone which has a break frequency at 5 rs Using feedback can extend the frequency range of useful operation Put another way using feedback can actually reduce a system s time constant Feedback can make a system quicker Notice that 1H crosses Gs at the 2nd breakpoint which results in a damping ratio 205 If 1H crosses Gs before the 2nd breakpoint then Qgt05 lf 1H crosses Gs after the 2nd breakpoint then Qlt05 which is unacceptable for most control systems lecture 26 outline 261 1 Example For a system with the Bode mag plot shown below determine an expression for the gain without feedback G an expression for the gain with feedback G1GH iii the damping ratio Q 39 an estimate for the gain with feedback when m3000 rs the percent error in the estimate in part iii repeat parts iv and v for 0330 rs IV V V C v 10000 1000 100 ITFI 01 1 10 100 1000 10000 rls lecture 26 outline 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 Temperature Fundamental meaning of temperature Temperature is a measure of the average kinetic energy possessed by a group of particles 201 For example ifthe temperature ofa group of particles is T absolute temperature in K the average energy per degree of freedom is Eav 12 kT where k is Boltzmann39s constant Practical measurement Practical temperature measurement usually involves measuring some physical quantity that varies in a repeatable fashion with temperature Temperature transducers use quantities like volume resistance and voltage to produce quantities that can be used as analogs to the actual temperature RTD thermocouple resistance temperature thermister C sensor detector advantages gtwide range gtmost stable gtsensitive gtmost linear gtinexpensive gtmost accurate gtmost sensitive gtself powered gtmore linear than TC gtinexpensive gtrugged disadvantages gteast stable gtexpensive gtnonlinear gtvery limited range gteast sensitive gtneeds external power gtlimited range gtneeds external power gtnonlinear gtself heating gtfragile gtself heating gtreference required gtneeds external power gtself heating Thermocouples output a voltage analog an RTD produces a resistance analog the thermistor gives a resistance analog and LC sensors are available that output a voltage or current We are going to take a relatively close look at the RTD and the thermocouple The RTD is known for its stability and accuracy the thermocouple for its wide range and its ruggedness The thermocouple is the workhorse of the lot and a sensor that is widely misused lecture 20 outline 202 Resistance temperature detectors RTDs The resrstance 0f meta s rncreaseswrth temperature m an RTD the resrstance 0f reswstor typrcahy a p aunum 100 0 at 0 C resws tor rs used as an ana 09 to emperature P atmum rs the most popmar meta due 0 S hneanty and S stabw My n 1983 the ntemauona E ectrotechmca Commrssron EC adopted the Deutsche nsutute 0f Normung D N standard p atmum T wrth a temperature coemcrent of a 0 00385 ooquotc h U s ndusma Standard rs a 0 003902 omc For T near 0 C 100m 1 on WhereT sm Cetsrus Otherwrse A re auon that remams dose at mgher temperatures s the Cauandaryan Dusen equauon Rm RD1 aT 7 50 01T4001Te30 01T4001T n 1968 a more accurate equauon was deve oped 20 h order Tab es are often used you W m the homework Thetab es m yourtextbook are for the u s ndusma Standard n the 50 We Wm use the more popmarDW standard RTDS The p aunum RTD rs used as the pnmary e ement m aH mghr accuracy resrstance thermometers H s used as the mterpo auon standard from the oomng pomt of oxygen 7182 96 C to the me ung pom 0f armmony 030 74 C 1mm 20 outlme 2073 Signal Conditioning Requirements for the RTD The RTD is orten used in a oneractiverarm Wheatstone bridge which serves to transrorm the changes in resistance to changes in votage taupemac in this arrangemem the resistance or the ieads can cause inaccuracies in the temperature measurement We can eiimiriate this source or error by using three ieads to connect the RTD to the bridge Three iead arrangement in this arrangement Rim appears in R3 and in Far so does not arrect the voitage divider 1mm 20 outline Prove this for yourself Find Vb The lead resistance in series with the voltmeter does not affect Vb too much since the resistance of a voltmeter is high This high resistance will cause the currrent through the voltmeter to be low and therefore the voltage drop across Rlead will be low This arrangement works very well as long as the temperatures being measured do not vary much lfthey do then Rg will not always be close to the value of R and the inherent nonlinear nature ofthe Wheatstone bridge will become a problem Example Estimate the measurement error when using 3wire 1009 RTD at 000385 QQ C when measuring 2 C Assume Rlead 29 Compare this error the error present when measuring 100 C lecture 20 outline AWire Measurement To errmrnatetne proorem ofme rnnerent nonrrnearrty ottne wneatstone brrdge a constant current source can be used to transform tne RTD S cnange rn resrstance to a cnange rn yortage and use a Arwrre yortage measurement to focus armost sorery on tne resrstance of tne R Usrng a constant current source rsr tne cnange rn RE rstransformed to V5 Note Srnce tnere nardry any current owrng rn they rt measures a yortage yery crose to ya Thezwyrre measurement rs a standard precrsron measurement tecnnrque rts adyantage rs accuracy rts Weakness rs cost srnce rt requrres a constant current source Constant current source Tnere are many crrcurts tnat approxrmate a constant current source Eacn rearrzatron nas constrarnts or rrrnrts to the condrtrons underwnrcn rt approxrmates a constant current source Here rs one usrng an amp 1mm 20 mm EM shielding Electromagnetic shielding uses conducting materials that are placed between two circuits to either re ect EM energy away from the receiving circuits or attenuate the energy ofthe EM wave as it travels from source to receiver The effectiveness ofa particular shield is defined in dB as s 20og10quot5 AR sh Vquots interfering voltage without shield Vsh interfering voltage with shield Shielding effectiveness due to absorption A In a conductor the amplitude of the electromagnetic wave decreases exponentially at a rate characterized by the attenuation constant E E e39V The component of shielding effectiveness provided by attenuation is called the absorptive loss A and is measured by comparing the eld with no shield to that with the shield A 20 log10 E 8691 Eoeg 5 6 is quotskin depth It is a function of the material properties and frequency lecture 24 outline Here is the conductivity for several metals Consider the absorptive loss for an aluminum sheet 15 mm thick and an aluminum coating 50 pm at several different frequencies F req ue ncy 5 Asheet Acoating In practice attenuation values in excess of 100 dB are not attainable Shielding effectiveness due to reflections Shielding effectiveness is not due solely to absorption EM energy can also be re ected We can calculate the amount reflected if we know the shield39s impedance For aluminum lecture 24 outline 243 The shielding effectiveness due to re ection can be calculated via the formula below 23779 Z R 20logm 42m where ZW o The total shielding effectiveness can be found by summing the effectiveness due to absorption and reflection SAR At low frequencies the absorptive loss is small because the absorption increases with the number of wavelengths required to transverse the shield Re ection peaks at low frequencies because Zm is small Since the attenuation shielding is small at low frequencies just where the shielding due to reflections is large their combination can result is an effective shield for all frequencies SHIELDING EFFECTIVENESS IN dB 4 10 FREQUENCY lN Hz lecture 24 outline Example Consider an interfering signal at 10 GHz Suppose we wish to provide 100 dB of shielding effectiveness What thickness of copper shielding would be required Recall the impedance of free space ZW is 377 Q lecture 24 outline 245 Shield discontinuities The analysis above is for a continuous shield Real shields are often not continuousthey have gaps intentional or not For a given hole a cutoff frequency can be de ned 15000 K Where the frequency is in MHz and the largest linear dimension of the hole is in cm Above the cutoff frequency little or no shielding is provided 8 can be 0 dB Below the cutoff frequency the shielding effectiveness ofan in nitely thin shield that is no absorption can be approximated as f R 20og10 For holes of nite thickness there is also attenuation For a circular hole A 32 d For a rectangular hole A 2731 K These formulas are for single holes For multiple holes the factor 10 log1oN is subtracted from the single hole shielding effectiveness 8 S S 1Oog10N one aperatu re lecture 24 outline Often in electronic enclosures openings must be present for cooling purposes These opening must be designed so that the shielding integrity is not compromises Many times a wire grid is used Enclosures often also include seams These seams must perform both mechanically and electrically They must allow the current to flow freely across them The seam can be characterized via an impedance CONTACT RESISTANCE CAPACITANCE CONTACT To make an effective seam current must be allowed to freely flow 1 The material should have a low surface resistance Metals such as aluminum that form an oxide layer required a greater contact pressure to perform well The surface should remain free from poor conducting substances such as paint grease or dirt Dissimilar metals can lead to corrosive effects at their junction Care needs to be exercised when testing each seam Be sure to include a variety of environments Adequate contact pressure well past the quotkneequot in the graph below should be maintained to ensure free flow ofsurface currents lecture 24 outline 247 A conductive gasket with sufficient pliancy can help the performance of critical seams CONTACT RESISTANCE APPLIED PRESSURE Example Now suppose we need to cut a1 cm diameter hole in the shield How thick does the shield need to be now to obtain 100 dB of shielding effectiveness at 10 GHZ lecture 24 uutlmz 248 Example A transducer is attached to a machine to investigate its vibration We expect to measure signals as small as 100 V The transducer is located a distance of 20 ft away from the measuring equipment that will analyze the signal The 60Hz magnetic eld produced by other machines in the factory and the wiring for the room lighting has a maximum ux density of 10394 Webersm2 i The accuracy ofthe vibration measurement can be degraded by an interfering signal as small as 110 ofthe smallest signal voltage expected from the transducer Calculate the threshold of interference Let the transducer be connected by two wires that are spaced 1 inch apart Compute the amount of 60Hz voltage induced in the circuit and compare the result to the threshold of interference quot Next place the wires snugly next to each other for a spacing of 01 inches and recompute the induced voltage 2 Finally replace the cable with a twisted pair having 8 twistsinch and compare the resulting induced voltage with the threshold on interference lecture 24 outline 211 Thermocouples adapted from Omega Engineering In 1821 Thomas Seebeck discovered that a current flows when two dissimilar metals joined at Seebeck both ends is heated at one end current metal A metal B Q If the circuit is opened a net voltage the Seebeck voltage is present This voltage is a function of the two metals and the junction temperature metal A VSeEbeck gt g metal B Q The voltagetemperature relationship is not linear for thermocouples but the functional relationships are known and tabulated for common thermocouples coefficients from the NATIONAL BUREAU OF STANDARDS TYPE E TYPE J TYPE K thermocouple TV relation T a0 ajvtc a2 Vic2 Vtc Volts assumes reference junction at 0 C lecture 21 uutlme veterinary probe lecture 21 outlan 213 Measuring the Seebeck voltage When measuring the thermocouple voltage one must be aware that the act of measuring can create additional thermoelectric junctions Consider measuring the voltage VOItmeter J across a CuConstantan type T 3 thermocouple C Q DA Cu J V Cu Constanlan N A Q J2 Consider the equivalent thermoelectric circuit that describes the system above To know the temperature at J1 two things must be known 1 The type ofjunction in this case CuConstantan and 2 The junction voltage V1 In this case V3 0 since the junction is Cu Cu But V2 is CuC and V1 cannot be determined from V the voltmeter reading unless V2 is known V2 can be fixed by placing J2 in a known temperature lecture 21 uutlme 214 We cannot determine the temperature at J1 without knowing the temperature atJ2 This can be accomplished using an ice bath T3 0 v v v2 z 10 3992 a mac27315 T2ac27315 V 0 Trim 712cm 0 Two 0 0 TH v on TH voltmeter V cu Cu V239 c Y1 J1 iceJ2 I Now suppose a different type of thermocouple is bath used suppose Type J which is Fe Constantan If J3 and J4 are at the same voltmeter J3 temperature their junction voltages will cancel and v on TH An isothermal block high thermal isothermal conductIVIty and low electrical voltmeter biock conductivity can help to insure the J3 junction temperatures remain equal e A Cu Fe V Cu Cu fl J1 JA This circuit still requires two thermocouples J1 and J2 It is possible to eliminate the reference thermocouple lecture 21 Butlmz Step 1 Eliminating the ice bath Replace the ice bath with an VOI meter isothermal block at temperature Tref V0LTiiTrer Step 2 Join the isothermal blocks Temperature of isothermal voltmeter block is Tref Cu VDLTM39Tref quotA V Cu H isothermal blocks Step 3 Eliminate J2 Use law of intermediate metal A mm B metals to eliminate J2 voltmeter V0LTii 39Tref isolhermal block TH isothermal J3 block Cu Fe Cu Fe v2 C Vi J1 J 4 J2 metaIC melalA metalC if Cu Fe 3 cu C V1 J1 Tref To measure Tref we can use an RTD or some other convenient temperature sensor lecture 21 uutlme So why use thermocouples Why when another temperature transducer must be used to measure Tref does one even bother with the thermocouple in the first place Why notjust use the other temperature transducer to measure the temperature of interest and be done with it There are two primary reasons temperature range and ruggedness RTDs can be used to measure temperatures that RTDs thermistors and IC sensors simply cannot They are also very rugged thermocouples can be welded to a metal part or they can be clamped under a screw Their wide temperature range and ruggedness makes them the workhorses of temperature measurement in a wide variety of applications be eiai I can 5 be switched so that all use VOKITIeter Fe the same isothermal block C V1 J1 Cu cu This allows one temperature Cu FE sensorto service many N v thermocouples Trev c 1 J1 lecture 21 uutlme 21 7 Type K Chromel lAlumel Type K is the 39general purpose39 thermocouple It is low cost and is available in a wide variety of probes Thermocouples are available in the 200 C to 1200 C range Sensitivity is approx 41 uV C Although Type K is presently most popular with users the improved performance of Type N will serve to supplant Type K as the most popular thermocouple Type E Chromel Constantan Type E has a high output which makes it well suited to low temperature cryogenic use Type J Iron Constantan Limited range 40 to 750 C makes type J less popular than type K The main application is with old equipment that can not accept 39modern39 thermocouples J types should not be used above 760 C as an abrupt magnetic transformation will cause permanent decalibration Type N Nicrosil Nisil High stability and resistance to high temperature oxidation makes type N suitable for high temperature measurements without the cost of platinum BRS types Designed to be an 39improved39 type K it is becoming more popular Thermocouple types B R and S are all 39noble39 metal thermocouples and exhibit similar characteristics They are the most stable ofall thermocouples but due to their low sensitivity approx 10uV C they are usually only used for high temperature measurement gt300 C Type B Platinum Rhodium Suited for high temperature measurements up to 1800 C Type B thermocouples give the same output at 0 C and 42 C This makes them useless below 50 C Type R Platinum Rhodium Suited for high temperature measurements up to 1600 C Low sensitivity 10uV C and high cost makes them unsuitable for general purpose use Type S Platinum Rhodium Suited for high temperature measurements up to 1600 C Low sensitivity 10uVvC and high cost makes them unsuitable for general purpose use Due to its high stability type S is used as the standard of calibration forthe melting point of gold 106443 C Source picotech lecture 21 outline 21 8 When selecting thermocouple types ensure that your measuring equipment does not limit the range of temperatures that can be measured Note that thermocouples with low sensitivity B R and 8 have a correspondingly lower resolution type Range C 01 Cresolution 0025 Cresolution 50 to 1760 330 to 1760 50 to 1760 250 to 1760 B 100 to 1800 1030 to 1800 E 270 to 790 240 to 790 140 to 790 J 210 to 1050 210 to 1050 120 to 1050 K 270 to 1370 220 to 1370 20 to 1150 N 260 to 1300 210 to 1300 340 to 1260 R S Precautions and Considerations for Using Thermocouples Most measurement problems and errors with thermocouples are due to a lack of understanding of how thermocouples work Connection problems Many measurement errors are caused by unintentional junctions Rememberthat any junction of two different metals will cause a junction If you need to increase the length of the leads from yourthermocouple you must use the correct type of thermocouple extension wire for example type K fortype K thermocouples Using any other type ofwire will introduce a junction Any connectors used must be made ofthe correct thermocouple material and correct polarity must be observed Lead resistance To minimize thermal loading and improve response times thermocouples are often made of thin wire For Pt TCs cost is also a consideration This can cause the thermocouple to have a high resistance which can make it sensitive to noise and can also cause errors due to the input impedance of the measuring instrument A typical exposed junction thermocouple with 32AWG wire 025mm diameter will have a resistance of about 15 Qm It is always a good precaution to measure the resistance of your thermocouple before use lecture 21 outline 21 9 Decalibration is the process of unintentionally altering the makeup ofthermocouple wire The usual cause is the diffusion of atmospheric particles into the metal at the extremes ofthe operating temperature Another cause is impurities and chemicals from the insulation diffusing into the thermocouple wire lf operating at high temperatures check the speci cations ofthe probe insulation Noise The output from a thermocouple is a small and is vulnerable to electrical noise Noise mitigation includes integrating that is low pass ltering and shielding lfoperating in an extremely noisy environment such as near a large motor consider using a screened extension cable Source picotech lecture 21 outline 181 Bode Plots Bode plots are widely used as graphical protrayals ofthe sinusoidal steadystate behavior of dynamical systems Bode plots show the relationship in the sinusoidal steadystate between the input and output phasors Recall that in sdomain analysis the input and output are related via the system transfer function To be speci c consider the input and output to both be voltages of course depending on the system they could be temperatures currents or forces Vos TFs Vis The sinusoidal steadystate response is given when s gt ja Notice when s is replaced by joa that V0 TF and Vi all become complex functions Ofoa That is they become phasor quantities Voo TFoVio Think ofthese phasor quantities in polar form V0490 TFAQTF v49 TMTF V 49 490 9i vizei 4 V TF V TF Is a functIon of w QTF 90 9i Qi is a function of w There are two Bode plots 1The Bode magnitude plot gives the magnitude in decibels of TF as a function Ofoa 2The Bode phase plot gives en as a function of o What good are Bode plots What are they used for lecture I 8 outline 1872 Example Suppose a syatem nas the frequency response shown be ow What Sme ouput mne mput 5 vs 10 cos some 15 v0 ts7 Bade Dxagvam Magnnude as Phase deg m2 nequency vansec 1mm 13 outlme 183 Example Suppose another system has a complicated periodic input for whichwe ve found the Fourier series What is the system s output in the sinusoidal steady state vi 9 a m 2 12m W I 139 239 III III 1 anquot 139quot 239 I39ll I39ll I 39 m Irn er mnvlmn mm mm Magnitude 45 pm deg 7 Frequency H11 125mm 13 Wilma 184 Working with Bode plots helps engineers develop a feel for frequency response Bode plots are widely used They will be used this quarter when discussing control systems Forthese reasons engineers need to be able to sketch the frequency response for systems without having to resort to MATLAB Bode plots Begin with the sdomain transfer function The transfer function can be written as a ratio of polynomials in s TF as2 bnsz391 adsp bdsp391 The coef cients are just parameters ofthe system and so are real numbers For polynomials with real coef cients their factors are either real or occur in complex conjugate pairs For this case the transfer function can be manipulated in the form below This form is called Bode form Bode form s s2 2 squot 1 2 z151 TF Kb aEZ1 wn 3221 sm 5 1s 2 s 1 abp1 wnp1 anp1 s is replaced by joa 2 n a a 2 J 2 21Jw1 Kb 21 wn31 wnz1 m Ia S 2 w J 1 2 1w 1 abp1 wnp1 anp1 Notice that the inputoutput relation TF is now complex lecture I 8 outline Bode magnitude plot TFdb 20 Iog10TF ja TFdb 20 Iog10Kb 20nm Iog10jw 20 log wb 1 z1 20Iog J a 1 jw1 20og 2 21 a n21 n21 p1 2 quot1 ja 1 np1 20 log s2 2 wnp1 Bode phase plot 9 4Kb nm4ja LUZ 1 z1 2 4 w 1 AJw1 a a wb n21 n21 p1 2 2 S 4 P1Jw1 wZ np1 np1 Both the magnitude and phase plots are sums of individual terms Let39s consider each ofthese terms in turn Kb db O s db 0 lecture I 8 outline 2 db 1 8 lecture I 8 outline lecture I 8 outline Example THS 2005 s10 lecture I 8 outline Example 105 52 5100s 500000 TFs lecture I 8 outline 21 Nodal analysis Nodal Analysis is a circuit analysis technique which allows circuit analysis with significantly fewer equations than the big gun technique Nodal Analysis is a general analysis technique which can be used to analyze any circuit The rst step in nodal analysis is to choose one node as the reference node For each remaining node we assign a node voltage which has its positive sign at the respective node and the negative sign at the reference node The cleverthing about this choice of variable is that KVL is automatically satis ed This allows analysis to focus on KCL and element relations To repeat nodal analysis is based on KCL and element relations lecture 2 outline Example Use noda ana ysws to nd u check your work yx r A A 2A 1mm 2 mm Example Notice there was no voltage sources in the first example 1009 220 1K9 12vC 1209 check your work V7 5 V V 12 V V3 3207 V lecture 2 umlme V3 Procedure for nodal analysis Example 1 U39PF N Choose a reference node Label node voltages Express control variables in terms of node voltages Circle all voltage sources by supernodes Write KCL equations for all nodes or supernodes not associated with the reference node Write element relations for voltage sources in terms of node voltages The result should be a suf cient set of independent equations That is the number of equations should equal the number of variable node voltages In nodal analysisthe number of equations is equal to the number of node voltages That is the number of nodes minus one There will be one equation for each voltage source the remaining equations come from KCL 5node circuit 2 voltage sources What equations are needed in nodal analysis Four equations in V1 V4 are required They come from KCL and voltage sources Perform nodal analysis on this circuit 6 gt109 2452 50 342 2A lecture 2 outline same circuit labeled as shown L19 249 x 39 t7 10v 350 54 2A Find i and the power delivered by the 10V source check your work V1 10 V2 447 V3 1264V V4 664V i 10V210 0553A Pdemv 10V 553W All these node voltages are positive Is it possible for node voltages to be negative Zero Discuss Compare nodal analysis with the big gun technique for this last circuit How many variables would have been required How many equations ls nodal analysis more ef cient ls nodal analysis worth the trouble of learning lecture 2 outline 131 Exercise Suppose a design engineer is trying improving the efficiency of an internal combustion engine What measurements might be required in experiments on how to improve ef ciency Hints temperature strain flow speed angular velocity force strain stress position timing chemical composition variable to purpose of measurement importance of allowable measurement be measurement error techniques available measured Methods of Measurement Measurement can be split into two types 1direct comparison with a primary or secondary standard and 2measurement via a calibrated system Direct comparison is used when a length is measured by comparing the length to a ruler the ruler being a secondary standard Using a calibrated system is the most common type of measurement voltage with a calibrated voltmeter power with a calibrated power meter and weight with a calibrated scale lecture 13 outline 132 Why Measure Observation is central to the scienti c method and observation requires measurement Likewise engineering design requires experimentation Experimentation requires accurate measurement Measurement is critical for research development production and process control Historical Background In the past measurement was usually based on the length ofa ruler39s arm or foot or hand In fact the word ruler as in a county39s ruler is still used to refer to a standard length In 1790 the French National Assembly directed the French Academy of Sciences to develop a model for weights and measures The Academy39s recommendation the metric system included the stipulation that multiples of all basic units be in factors of 10 In the United States the US Constitution gives congress the authority to maintaining a system of weights and measures In 1821 John Quincy Adams in a report to congress states the necessity for a system ofweights and measures in any society In 1832 the US Treasury Department introduced a system of weights and measures In 1936 these standards were approved by congress In 1875 the US and 16 other countries sign the quotTreaty ofthe Meterquot This established a central bureau of standards in France In 1901 the US Congress passed an act to establish the National Bureau of Standards NBS in 1910 In 1950 the name was changed to the National Institute of Standards and Technology NIST International uniformity is maintained by scheduled meetings every six years by most industrialized counties lecture 13 outline 133 Conceptual Basis for Why Measurement is Important adapted from The Measurement Instrumentation and Sensors Handbook ed John G Webster CRC Press 1999 Quality of product research or organization is based upon Decisions valid ones at least are often based on numerical or quantifiable information VALID DECISIONS are often based upon If our decisions are based upon numerical data it had better be correct CORRECT NUMEthAL DATA are based upon This requires our measurements be accurate ACCURATE MEASUREMENTS are based upon Accurate measurements requires calibrated CALIBRATED NSTRUMENTS instruments sensors signal conditioning FROPERLV UTILIZED display and recording are based upon Accurate measurement is ultimately based on traceable standards TRACEABLE STANDARDS lecture 13 uutlme 134 Overview of Measurement via a Calibrated System Measurement involves using some quantity that has a known relationship to the variable being measured calibration external power external power Ind cator source not always source usually required required recorder I transduced A signal I I measurand gt conditioning computer analogous to input processmg controller figure from Mechanical Measurements 5m ed Beckwith et al AddisonWesley 1993 For example when measuring strain strain is the measurand When measuring strain with a strain gage signal is a change in the strain gage resistance for resistive strain gages Signal conditioning might include transforming the change in resistance to a change in voltage via a Wheatstone bridge and subsequent amplification Since measurement is central to all engineering and scientific work we must take special care to really understand all aspects of measurement from the physical principles of the sensors to proper choices for signal conditioning to knowledge of how to measurement errors interact in a system There are attributes shared by most measurement systems ERROR BIAS ERROR PRECISION ERROR RANGE ACCURACY RESOLUTION STABILITY DRIFT SENSITIVITY HYSTERESIS and LINEARITY lecture 13 uutlme Error 1 35 Error refers to the difference between the measured or observed value and the actual value of a measurand error There are two general types of errors bias error precision error measured value actualvalue average of measurements actual value measurement average of measurements BIAS OR SYSTEMATIC ERROR PRECISION OR RANDOM ERROR Bias errors give the same error in each measurement trial Bias errors can be caused by faulty or outdated calibration ZEROOFFSET errors cause each measurement to deviate a given amount SCALE ERRORS change the sensitivity of the measurement depending upon value the measurement Bias errors can also be introduced by the very act of measurement LOADING ERROR occurs when the act of measurement changes that which is to be measured Placing a thermometer having a different initial temperature on a small piece of metal can alter the temperature ofthe metal Placing a voltmeter in parallel with a high resistance can change the voltage across the resistance Precision errors are errors that uctuate sometimes causing the measurement to be lowerthan the true reading and sometimes causing the measurement to be higher The uctuations can often be characterized by the Normal or Gaussian probability distribution the proverbial bellshaped curve Precision errors can be caused by 1environmental variations 2electrical or magnetic interference 3random variations in the performance of the measurement system 4uncontrolled variations in the measurand lecture 13 outline 136 TRANSFER CHARACTERISTIC Consider some parameters 0 A TRANSDUCER associated with this transducer inputoutput output units characteristic inputoutput curve input units The range of a sensor or measurement system refers to the range of inputs for which the sensor or measurement system functions within specifications Sensitivity refers to change of output per change of input It39s the magnitude of the slope in the above curve The inputoutput characteristic is linear to the extent in which its sensitivity does not change as the input changes Stability refers to the degree of insensitivity that the inputouput relation possesses with respect to environmental changes Stability wrt temperature wrt humidity wrt electromagnetic interference etc Changes in the inputoutput characteristic over time stability wrt time is referred to as drift Drift is especially used to designate changes in the inputoutput characteristic with time Accuracy refers to how closely the measurement corresponds to the actual or true value The lowerthe accuracy the greater the error Resolution is the smallest increment of change in the measured value that can determined from the system39s readout lecture 13 uutlme 137 Hysteresis refers to a dependence of the output units inputoutput relationship depending on whether the input is increasing or decreasing Hysteresis is indicates that the system has 4 memory that the past values of inputs influence present behavior It39s usually associated with some type of energy loss mechanism input units Example Suppose the actual ortrue value of a force is 167 N lnten trials we measure 169 173 170 169 171 168 167 170 172 and 170 N What is the bias error ii What is the range of precision error The material below is a review of ES 205 material We will discuss as time permits lecture 13 uutlme O horder systems x K ft 1S order systems Standard form rx39 X Kft The general solution consists oftwo pieces the forced response and the natural or transient response Let39s look at four cases 1 ft 0 natural response 2 ft Aut step response 3 ft Atut ramp response 4 ft A cos oat e sinusoidal response f t 0 ft Au t t xt KA KAXoe lecture 13 outline 214177710 gz zxmaz vagagvgzgeazazvzzz ELVLZLLBDEDVDZDD 391 01p312eJ s 0 112 edols eu oslv LALU 1 39Lu ed0s euJ Sugpuu pue J SA 12 Buqu pue esuodseJ eledLuooug euJ 1z Bugd0eAep Aq map 3qu ee eu e Bugsn Aq peugelqo eq um 10 samex elemoov uouewusa Jawwemd 1310 ft Atut fgt A cos m t91 lecture 13 outline 131 1 2quotd order systems gtlt Standard form 2 x39 X Kft a a n JNI Just as in the rstorder case the complete solution consists the forced response and the natural or transient response Consider three cases 1 ft O 2 ft Aut 3 ft A cosoat 9 ft 0 Looking at the characteristic equation 22Qquotas a22 O n n solutions for ofthe characteristic equation are S12 Four cases Q O undamped O lt Q lt 1 underdamped Q 1 critically damped Q gt 1 overdamped Q 0 undampedl 0lt lt1 underdam ed lecture 13 outline 1312 Parameter Estimation Logarithmic decrement The natural log of the ratio ofany two successive amplitudes wrt the steadystate value Xt1 Xt1 td 5 For nonsequential max or min 1 N Xt1Ntd Where N is the number ofelapsed cycles C 1 critically damgedl wt xCe n02te gmnt The critically damping case is really never seen It is useful as a limiting case gt 1 overdam ed r t r t 1 2 X C1 e 02 e lecture 13 outline 1313 flt Ault The form of the natural response is not affected by the form or presence ofthe source Sketch of step response picture assumes we have an underdamped system and that X0 O and x390 0 Time Reponse Specifications Steady State value XSS KA Peak Time time to 1St peak nod overshoot of KA 77 2 os 100e V1 39 4 lt2 settling time T8 i if wn lt1 settling time T8 i if an rise time time to cross KA 7239 tan391 n01 4 2 Note A different rise time is utilized in electrical systems where it is de ned as the time required for a signal to go from 10 to 90 of its nal value 12 if lecture 13 outline 1314 ft A cosoat91 Example Given the unit step response below find the differential equation relating the input ft to the output xt Give the differential equation in standard form uulpul vullage 2 A E EiEI1214iEiEZDZZZAZEZESEISZSASESEAEI 5 i Identify the undamped resonant frequency the clamped resonant frequency static gain coefficient the clamping ratio ii Calculate the percent overshoot peak time lt2 settling time and rise time Compare with measurement lecture 13 uutlme 1315 Workspace for example lecture 13 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
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