Chapter 3 - Waveforms, Power and Measurement

Transcription

1 Chapter 3 - Waveforms, Power and Measurement Recommended problems to study: Problem Page Concentrates 3: Low-pass filter/fourier series < : Two wattmeter 3-f power measurement Timed 3: DC ammeter 4< 5: Power factor correction 6 8: AC meter calibration 8< Concentrates (from two versions of textbook) : DC voltmeter : AC voltmeter < 4: AC meter measurement 4 5: Power factor correction 5 6: AC impedances/single phase power 6< --

2 Reference formula Part - periodic signals average values: f(t) t f avg t + T f t dt positive area -negative area T Ú () t rms values: f rms t + T T Ú t f () t dt Fourier series: f() t a + Âan n n Ê pn T t ˆ b n T t n Ë + Ê p cos ˆ Â sin Ë --

3 Instantaneous power: pt () vtit ()() Vmax cos( wt+ q) Imax cos( wt+ f) Using the trig identity cos( a) cos( b) cos( a -b)+ cos a + b V I pt () max max [ cos ( q -f)+ cos ( wt + q + f) ] Average power t + T P pt T Ú t VmaxImax cos( q f) () - For rms quantities P VrmsIrms cos( q -f) where cos q - f ( ) is the power factor Complex power: * P Re{ VI } Re{ Vrms qirms -f} VrmsIrms cos( q -f) Note the - sign for f ( ) this becomes +jq S q P apparent power * S VI VAs (Volt-amperes) real power * P Re{ VI } Watts reactive power Q Im VI * VARs (Volt-amperes reactive) P PF.. cosq Q { } Part - Measurement of DC and Periodic Signals Part 3 - Power Trigonometric relationships Complex numbers -3-

4 The ability to convert complex numbers from a rectangular format (a+jb) to a polar form Ae jq is essential to describing the behavior of ac electrical networks. This conversion is provided by Euler's identity which states that a + jb a + b Tan - b a and can be readily understood by the diagram shown below. where a+jb a + jb a + b Tan - b a and b tanq b a q a The rectangular form is a+jb; the polar (or phasor) form is c q where c a + b Phasors () ( + f ) sinusoidal voltage vt V cos w max t jq Euler s formula e cosq + jsinq j t j t j () { ( w q ) w q max } { max } + vt Re V e Re V e e For 6Hz power systems all voltages are at the same frequency so we ignore the e jwt term q () { max } vt Re V e j, or vˆ Vmax q For power systems ˆv q where V V rms rms Vrms Complex power, power factor, power factor correction For sinusoidal signals the power factor is defined by pfcosq where q is the phase angle of the voltage or current relative to some reference. For power circuits, the generator (or line) voltage is usually taken as the reference since the loads are usually connected in parallel. For phasors we can define complex (also called the reactive) power as: CP v i* real power ± j reactive power P ± jq where real power or P v i cosq -4-

5 and reactive power or Q v i sinq The relationship between P and Q determines the power factor as shown below. inductive load capacitive load CP P +jq LEADING P LAGGING CP -jq Complex power relationships for inductive and capacitive loads I V V I lagging power factor leading power factor (inductive) (capacitive) Relationship between voltage and current in complex loads imaginary CP in units of kva q real power CPcosq imaginary power CPsinq in units of kvar in units of kw Relationship between real and imaginary power in a complex load. Note that the diagram is drawn for an inductive circuit. It would be reversed for a capacitive load. real -5-

6 Delta/ wye conversions Delta circuits can be transformed into wye circuits and vice versa to simplify circuit analysis. A A a b 3 c B C B C With reference to the circuits shown above (Note that that A corresponds to A, B to B and C to C in the two circuits.) the conversion formula are: Delta to Wye R * R a + R b + R c R R ar c R * R R ar b R * R 3 R br c R * Wye to Delta R * R R + R R 3 + R R 3 R a R* R 3 R b R* R R c R* R Three phase circuits Three phase power is very complex for non-electrical engineers and usually accounts for several questions in the morning and afternoon sections of the exam. Do not attempt to understand how these expressions are derived simply use them and you should do well on that part of the exam. A three-phase load always has three terminals which I have labeled A, B and C in the drawings below. The connections to the voltage sources are called lines and the voltages between the terminals (lines) are called line-to-line voltages. The current passing through each terminal is called the line current. Other voltages and currents can be defined internal to the different loads possible, i.e. wye and delta. The lineto-line voltages are out of phase relative to each other as shown in the diagram below. V CA V AB V BC -6-

7 I A A V AB V CA I C C I CA Z Z I AB V BC Z I BC B I Balanced Delta-connected Load B The delta load connects to the three-phase voltage source through the terminals A, B and C as described above. The delta load is balanced when all three load impedances are identical. The current through each load is called the phase current. line-to-line voltages: phase currents: line currents: V AB V AB V BC V BC - V CA V CA + where q is defined by I AB I AB -q I BC I BC -q- I CA I CA -q+ I A I A -q-3 I B I B -q-5 I C I C -q+9 Z Z q The magnitudes of the phase and line current are related to each other for a balanced system. I I phase line 3 The total power consumed by a balanced load is P total 3 V line-to-line I line cos q I A A V CA Z V AB C I C Z O V BC I B Z Balanced Wye-connected Load B The wye load connects to the three-phase voltage source through terminals A, B and C as shown above. The wye load is called balanced when all three load impedances are identical as shown above. A unique characteristic of the wye load is that the three load impedances are connected together at a common node labeled O. The voltage across each load impedance, i.e. the voltage betwwen the terminal A, B or C and the common node O is -7-

8 called a phase-to-phase voltage. The phase relationships between the voltages and currents using the line-to-line voltage V AB as the reference is then given by: line-to-line voltages: V AB V AB V BC V BC - V CA V CA + phase-to-phase voltages: V AO V AO -3 V BO V BO -5 V CO V CO +9 In a balanced system the three load impedances are identical in magnitude and phase resulting in the magnitudes of the different line-to-line voltages being equal. V AB V BC V CA The same result is true for the phase-to-phase voltages. V AO V BO V CO The magnitude of the phase-to-phase and line-to-line voltages are related to each other for a balanced system. V V phase-to-phase line-to-line 3 The total power consumed by a balanced load is P total 3 V line-to-line I line cos q where q is defined by Z Z q Note: for a wye load the line current is equal to the phase current. Complex power, power factor, power factor correction For sinusoidal signals the power factor is defined by pfcosq where q is the phase angle of the voltage or current relative to some reference. For power circuits, the generator (or line) voltage is usually taken as the reference since the loads are usually connected in parallel. For phasors we can define complex (also called the reactive) power as: CP v i* real power ± j reactive power P ± jq where real power or P v i cosq and reactive power or Q v i sinq The relationship between P and Q determines the power factor as shown below. -8-

9 inductive load capacitive load CP P +jq LEADING P LAGGING CP -jq Complex power relationships for inductive and capacitive loads I V V I lagging power factor leading power factor (inductive) (capacitive) Relationship between voltage and current in complex loads imaginary CP in units of kva q real power CPcosq imaginary power CPsinq in units of kvar in units of kw Relationship between real and imaginary power in a complex load. Note that the diagram is drawn for an inductive circuit. It would be reversed for a capacitive load. real -9-

10 Chapter 3 Concentrates 3. A train of rectangular pulses is applied to an ideal low-pass filter circuit. The pulse height is 5 volts, and the duration of each pulse is ms. The repetition period is ms. The low-pass filter has a cut-off frequency of 5 Hz. What percentage of the signal power is available at the output of the filter? The ideal filter fo5hz f +5V T t(msec) As drawn this pulse train is an even function. That was my option since I prefer even function series. Ê pn T t ˆ b n T t n Ë + Ê p cos ˆ Â sin Ë f() t ao + Âan n n The last term of this expression for f t function and sine is odd. vt ao () + Â an n Ê pn cos Ë T t ˆ () can be ignored. All b n since f t () is an even a a o n favg [( 5V)( msec) ] V msec t + T n f t Ê p T Ë T t ˆ Ú cos dt () t Do the integral from t-.5 to t+.5 seconds. For that case, the integral reduces to --

11 a n... Ú Ê pn t ˆ 4 dt 5Ê. ˆ Ê p 5cos sin nt ˆ Ë.. Ë pn Ë. Ê p sin n ˆ pn Ë 5 w n pn p n T. n f n n n T. Hz The ideal filter will pass a o, a, a, a 3, a 4 and a 5 Power is rms voltage squared. Vrms È ao a a a a a Pout Ê ˆ R R ÍË Î The first term is different since the rms value of the dc term is the dc term a o. For all other Vpeak terms the rms voltage is given by Vrms. Computing the terms we get a 3 a 3 Ê p sin ˆ 3p Ë 5. a Ê p sin ˆ 87 p Ë 5 4. a 4 Ê p sin ˆ 47 4p Ë 5. a Ê p sin ˆ 5 p Ë 5 5. a 5 Ê p sin ˆ 5p Ë 5 The output power is then Vrms Pout Ê ˆ R R Ë + ( ) + ( ) + ( ) È + ( ) Í Î R Using the definition of rms to determine the power for the input pulse waveform P in Vrms È Í R R T ÎÍ +. Ú -. dt 5 R 5( ) 5.. R The percentage power passed by the filter is then 45. Pout R 9% P 5 in 5 R --

12 . The measurement system shown below is used for a balanced load of kw with a lagging power factor of.8. Determine the wattmeter readings. See Section 3-6 for an explanation of the two-wattmeter method of three phase power measurement. Ia Pa Van - Ib + Vbn- neutral Pb + Vbc - - Vcn + Ic The neutral is an artificial point used to make the two-wattmeter analysis easier. Since it is not specified assume an ABC phase sequence. For this problem PF cos q 8. lagging so that q Vc Va Vb The voltages and currents are then specified by V an V I I I a - q V bn V I I I b -q V cn V I I I c - q Recall the power triangle and write the expressions for the voltages and currents as seen by the wattmeters. --

13 imaginary S q P Q real For wattmeter A: Vac Van - Vcn V - V + V( ) I a I * * Sac VacIa V( ) ( I ) VI( ) ( ) VI( ) Or, in rectangular form S ac VI ( 7. + j7. ) For wattmeter B: Vbc Vbn - Vcn V - - V + V( ) I b I * * Sbc VbcIb V( ) ( I ) VI( ) ( ) VI( ) Or, in rectangular form Sbc VI( j59. ) Vline-line For a balanced load P 3Vline-lineIline cosq and Vphase 3 In this problem V an, V bn, and V cn are phase voltages. V ac and V bc are line-line voltages. Then P 3 3VphaseIline cosq 3VphaseIline cosq Using the numbers for this problem watts 3VphaseIline(. 8) or VI46.67 watts As a check on our calculations Pa Re { Sac} 7. VI. 7( ) watts Pb Re { Sbc} 68. VI (. 67) These powers add up to exactly watts so the answer looks good. -3-

14 3. An ammeter is being designed to measure currents over the five ranges indicated in the accompanying illustration. The indicating meter is a. milliampere movement with an internal resistance of 5 ohms. The total resistance R + R + R3 + R4 + R5 ohms. Specify the resistances R a and R through R 5. ( ) is to be.5ma 5mA 5mA R R Ra R3 5A.5A R4 5W R 5 For.5mA.5mA.mA There is.5 milliampere through the kw resistor. Since the voltages must be equal: Ra (. 5mA) ma( R a + 5W) or W R a 45W 5W For 5mA.mA ma( R + 45W+ 5W) ( 5 -) ma - R ( ) 5mA W-R R 5W 45W R + 5W 4-4R 5R R 9-4-

15 For 5mA ma( R + 9W+ 45W+ 5W) ( 5 -) ma - 9W+ R.mA R + 4W 49 - R ( ( )) ( ) 9W 45W 5R mA R R 9 W-(9W+R) 5W For.5A ma( R + 9W+ 9W+ 45W+ 5W) ( 5 -) ma - 9W+ 9W+ R ( ( )) 3 3.mA.5A 9W 9W R3 45W R ( ) + 49W R 3 3 5R 3 35 R 3 9W W-(9W+9W+R3) 5W For 5A ma( R + 9W+ 9W+ 9W+ 45W+ 5W) ( 5 -) ma - 9W+ 9W+ 9W+ R ( ( )) 4 4.mA 9W 9W 9W R + 499W R 4 4 5R 4 35 R 4 9. W ( ) 5A R4 W-(9W+9W+9W+R4) 5W And finally R 5 W-( 9W+ 9W+ 9W+. 9W). W -5-

16 A load is connected to a voltage of 3 volts at 6Hz. The load dissipates kw with a.867 lagging power factor. Specify the capacitance needed to correct the power factor to: (a).895 lagging (b).95 leading (Give the voltage and volt-ampere-reactive ratings at 6Hz for the capacitors.) 3V, 6Hz kw, PF.867 lagging Determine the initial reactive power before correction. imaginary S q kw Q real PF. 867lagging cosq therefore q From the power triangle Q kwtanq tan (. 5747)+ j57. 47kVAR This reactive power must be corrected as per the problem specification. The desired power factor is PF. 895lagging cos q '. Therefore, the new angle must be q' The new reactive power for this angle comes from the new power triangle. imaginary S q kw Q real Q' kwtan q' tan (. 4984)+ j49. 84kVAR The difference in reactive powers must be supplied by the correction capacitor. Q+ Qcapacitor Q' Q Q '- Q + capacitor j j j7. 63kVAR -6-

17 P capacitor VI * - ( ) Ê Ë Á ˆ j * X C Solving for the capacitive reactance: ( ) X * 3 C 3 - j763. jp 6 C - ( ) Solving for the required capacitance C p 6 ( ) ( ) m f We already know the power rating to be 7.6kVAR and the voltage rating to be 3 volts. -7-

18 8. The signal shown below is measured with the following voltmeters: (a) DC voltmeter, (b) an RMS reading AC voltmeter using a d Arsonval meter in a full-wave bridge in the feedback circuit of an opamp, (c) an RMS reading AC voltmeter using a d Arsonval meter in series with a diode in the feedback circuit of an opamp, (d) a true RMS voltmeter such as an electrodynanometer, (e) an RMS reading AC voltmeter using a peak detector, and (f) an RMS reading AC voltmeter using a peak-to-peak detector. Determine the reading on each meter. +5V V t(msec) The trick in this problem is to know that old style meters were always calibrated with sine waves. The meter always actually read the average value, but the scale was generated with the appropriate correction factor. Thus the procedure is calculate the scale for a sine wave as compared to the average value for a sine wave, and the calculate the average value of the waveform. The product of these two quantities will be the meter reading. (a) a DC voltmeter reads the average value of a waveform For this case the meter reading is T msec 5 T f t dt V Ú tdt msec Ú m sec 5 t () ( ) () (b) a d Arsonval meter reads the average value volts The full wave bridge in the feedback loop of an opamp is a fancy way of telling you that it is a precision rectifier and you can neglect the voltage drop across the diode. The waveform that the meter will see for a sine wave is 3 4 The average value for this waveform is cos pt V Ú sin pt dt Úsin ptdt p V avg [- cosp + cos] [-(-)+ ] p p p avg ( ) ( ) - ( ) -8-

19 The rms value of this waveform (which is the units in which the meter is calibrated) is V rms tdt tdt m Ú ( ) sin sec p Úsin ( p ) ( ) - ( ) V t rms Ú - ( t ) dt Ú dt Ú dt cos p cos p The calibration constant for the meter is then Vrms p V avg p The meter reading is given by multiplying the average value by the (sine wave) scaling factor: Vrms meter reading ( ) V V p.. Volts avg, waveform 5 39 avg (c) This is essentially the same as part (b) except that we have an ideal half wave rectifier and the sine wave calibration constant changes. Note that the ramp waveform is always positive and will give the same reading through either rectifier circuit. 3 4 The average value for this wave is half that of a full wave rectifier, so Vavg Vavg full wave Ê ˆ Ë, - p p Computing the rms voltage for this waveform we get V ( rms t ) dt Ú Ú - ( t ) dt Údt Ú t dt sin p cos p cos p 4 4 The calibration constant for the meter is then 4 ( ) - ( ) Vrms p V avg p The meter reading is given by multiplying the average value by the (sine wave) scaling factor: -9-

20 Vrms meter reading ( ) V V p.. Volts avg, waveform 5 96 avg (d) the true RMS is pretty easy. This is computed directly from the definition with no scaling factors. 3 t Vtrue- rms Ê 5 t ˆ Ú dt tdt Volts Ë Ú (e) RMS using a peak detector is also pretty easy. This is just a different scaling factor based upon the peak value of the waveform. The meter reads the peak value for the waveform of 5 Volts. Vpeak The calibration for sinusoids is Vrms. 5 meter reading 354. Volts (f) This is essentially the same as (e) except using peak-peak values. Vpeak - peak Vpeak peak The calibration for sinusoids is Vrms -. Note that the meter still reads 5 Volts as the peak-peak value. 5 meter reading 77. Volts --

21 Concentrates. A voltmeter is being designed to measure voltages in the full-scale ranges of 3,, 3 and volts DC. The meter movement to be used has an internal resistance of 5 ohms and a full-scale current of ma. Using a four-pole, single-throw switch, design the voltmeter. The meter circuit is easily designed using the equivalent circuit of the volts Rext V Rcoil5W ma FS R V volts I ma 5 W R ext 5 W- 5 R V volts I 3 ma 3 W and R ext 3-5 R V volts I ma W and R ext - 5 R V volts I 3 ma 3 W and R ext W You can design several different types of meter circuits using this data. 9995W 995W 3V V 95W 7W W 7W 995W 95W V 3V ma FS 3V V 3V ma FS V --

22 . An AC voltmeter consists of a d Arsonval meter with a full-scale current of.5ma and a series resistance of 6 W. A full-wave bridge of silicon diodes (Vthr.6volts) is used to rectify the AC voltage. The full-scale needle deflection is 5 degrees. Give the scale increments in degrees from to volts in volt increments for the RMS vaue of a pure sinusoid. This is somewhat of a tedious problem. The crucial items to note are: full-wave bridge non-ideal diodes with a threshold RMS meter circuit Otherwise the problem is fairly similar to the previous meter calibration problem. If you follow the current path through the meter circuit you see that the current flows through two diodes and we have two diode drops to include in our calculations. 6W.5mA FS RMS meter Note that there can be no output when the AC input voltage is less than two diodes drops, i.e., *.6. volts. Vpeak.V p p 3p 4p A d Arsonval meter will read the average value of the voltage waveform shown below. Note that I want to write the peak value in terms of the RMS value for convenience. --

23 sqrt()vrms.v wt wt wtp/ wt This drawing shows what is happening. I have to subtract. volts from the sine wave. As a result I will only get a contribution to the average from the shaded region of the waveform. Only a quarter cycle is shown because of symmetry. The actual average will then be given by T wt V T f t dt V t d t avg Ú () Ú[ rms cos w -. p ] ( w ) wt is the point in time when the input voltage drops below the threshold of the diode bridge and there is no output. This can be solved for as V cos w t - rms. or -Ê. ˆ wt cos Á Ë V rms Using this result the average voltage becomes wt wt Vavg [ Vrms sin wt-. ] Vrms sin wt -. wt p wt p [ ] The full scale deflection of the resistor-meter combination is volts Assuming the meter deflection is linear we have an angular deflection sensitivity of 5 5 and a meter deflection of q 5volts 5volts V avg This is not something I wanted to calculate by hand so I used a spreadsheet to compute it. Vrms wt sqrt()*vrms*sin(w t).*w t difference angular deflection (degrees)

24 4. A full wave rectifier type VTVM (vacuum tube voltmeter) is set to an RMS AC scale with a range of 5 volts. The meter is connected to a symmetrical (zero average) triangular waveform of volts peak-to-peak. What does the meter read? The input waveform is input to an AC VTVM with a full wave bridge We we assume an ideal rectifier circuit. +5V -5V The output of the full wave rectifier will look superficially the same but with different voltage levels. +5V -5V T Compute the input to the meter. T V T vtdt avg Ú () ( 5) T 5Volts T The meter is designed and calibrated for sinusoidal waveforms. V rms V rms Computing the calibration relationship for the meter. p Vrms t Vavg Vrms tdt - cos Ú sin V p p p p The meter will read an average (actual) of 5 volts and display it as the appropriate RMS value, i.e., 5V or p V 5p rms 7. 77volts V rms rms -4-

25 5. An impedance receives a line current which lags the voltage by 3. When the voltage across the impedance is volts (RMS), the impedance dissipates watts. Specify the reactance of a capacitance to be places in parallel with the impedance which would make the line current be in phase with the voltage. The given circuit is V RMS, dissipates watts The current lags the voltage 3 current voltage From the power specification P V or R V ( ) 5. This gives G.mhos. R P At this point we have I YV ( G+ jb) V (. + jb) V. Since the phase angle is known to be 3 we can use the relationship between voltage and current to find B. G. 3 B tan3 B G or B G tan 3. tan Note that B is actually negativ e. The resulting circuit is +jbl. mhos -j.5 Having the line current being in phase means that the reactance is zero. This requires jbc + jbl BC - BL -- (. 5). 5 X C W for the desired power factor correction -5-

26 6. A parallel combination of a resistance (W), capacitance (88.5µf), and inductance (66.3 mh) has 6Hz, 3 volt (RMS) applied. Obtain the: (a) reactances of C and L. (b) admittance of each circuit element. (c) phasor diagram for the currents, using the applied voltage as the reference. (d) admittance diagram for the circuits, including the total admittance. (e) input current as a sine function, taking the applied voltage as a reference. (Is the circuit inductive or capacitive?) (f) power factor. (g) power triangle. The circuit is 3V,6Hz W 88.5µf 66.3mH The angular frequency is w p( 6) rad sec (a) X j L L j ( ) -3 w 377 ( ) j5 X C j j j C - ( )( ) w (b) BL -j4. mhos j5 B C + j. 334mhos - j3 G. mhos R (c) i 3 R 3 amps i 3 C amps - j3 i amps L j5-6-

27 ic7.67 ir3 il9. (d) BC.334 G. BL.4 Yt. + j j j (e) i YV ( ) ( 3 ) 3. - j The circuit is inductive since the angle is negative. Recall the phasor diagram ic V il (f) PF cos ( ) (g) p vi* ( 3 ) ( ) * j The power triangle will look like this 53.5VA 349.4VAr watts -7-

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