(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 4 pot

55 The resulting equivalent circuit referred to the primary at 50 Hz is shown below: + - V S V P I S + - I P R C jX M R EQ jX EQ 1.31 MΩ j473 kΩ 260 Ω j917 Ω 2-20. Prove that the three-phase system of voltages on the secondary of the Y- ∆ transformer shown in Figure 2- 38b lags the three-phase system of voltages on the primary of the transformer by 30°. S OLUTION The figure is reproduced below: V C V B V A V C ' V A ' V B ' ++ + + + + - - - - + + ++ + + V A ' V B ' V C ' V A V B V C 56 Assume that the phase voltages on the primary side are given by °∠= 0 PA V φ V °−∠= 120 PB V φ V °∠= 120 PC V φ V Then the phase voltages on the secondary side are given by °∠= ′ 0 SA V φ V °−∠= ′ 120 SB V φ V °∠= ′ 120 SC V φ V where aVV PS / φφ = . Since this is a Y- ∆ transformer bank, the line voltage ab V on the primary side is °∠=°−∠−°∠=−= 3031200 PPPBAab VVV φφφ VVV and the voltage °∠= ′ = ′′ 0 SAba V φ VV . Note that the line voltage on the secondary side lags the line voltage on the primary side by 30 ° . 2-21. Prove that the three-phase system of voltages on the secondary of the ∆ -Y transformer shown in Figure 2- 38c lags the three-phase system of voltages on the primary of the transformer by 30°. S OLUTION The figure is reproduced below: ++ ++ V A V A ' V B V B ' V C V C ' Assume that the phase voltages on the primary side are given by °∠= 0 PA V φ V °−∠= 120 PB V φ V °∠= 120 PC V φ V 57 Then the phase voltages on the secondary side are given by °∠= ′ 0 SA V φ V °−∠= ′ 120 SB V φ V °∠= ′ 120 SC V φ V where aVV PS / φφ = . Since this is a ∆ -Y transformer bank, the line voltage ab V on the primary side is just equal to °∠= 0 PA V φ V . The line voltage on the secondary side is given by °−∠=°∠−°∠=−= ′′ 3031200 PPPCAba VVV φφφ VVV Note that the line voltage on the secondary side lags the line voltage on the primary side by 30 ° . 2-22. A single-phase 10-kVA 480/120-V transformer is to be used as an autotransformer tying a 600-V distribution line to a 480-V load. When it is tested as a conventional transformer, the following values are measured on the primary (480-V) side of the transformer: Open-circuit test Short-circuit test V OC = 480 V V SC = 10.0 V I OC = 0.41 A I SC = 10.6 A V OC = 38 W P SC = 26 W (a) Find the per-unit equivalent circuit of this transformer when it is connected in the conventional manner. What is the efficiency of the transformer at rated conditions and unity power factor? What is the voltage regulation at those conditions? (b) Sketch the transformer connections when it is used as a 600/480-V step-down autotransformer. (c) What is the kilovoltampere rating of this transformer when it is used in the autotransformer connection? (d) Answer the questions in (a) for the autotransformer connection. S OLUTION (a) The base impedance of this transformer referred to the primary side is () () 2 2 base, 480 V 23.04 10 kVA P P V Z S == =Ω The open circuit test yields the values for the excitation branch (referred to the primary side): , , 0.41 A 0.000854 S 480 V OC EX OC I Y V φ φ == = ()( ) 11 38 W cos cos 78.87 480 V 0.41 A OC OC OC P VI θ −− =− =− =− ° 0.000854 78.87 0.000165 0.000838 EX C M YGjB j=− = ∠− °= − 1/ 6063 CC RG==Ω 1/ 1193 MM XB==Ω The excitation branch elements can be expressed in per-unit as 6063 263 pu 23.04 C R Ω == Ω 1193 51.8 pu 23.04 M X Ω == Ω The short circuit test yields the values for the series impedances (referred to the primary side): 10.0 V 0.943 10.6 A SC EQ SC V Z I == = Ω 58 ()() 11 26 W cos cos 75.8 10.0 V 10.6 A SC SC SC P VI θ −− == =° 0.943 75.8 0.231 0.915 EQ EQ EQ ZRjX j=+ = ∠°= + Ω The resulting per-unit impedances are 0.231 0.010 pu 23.04 EQ R Ω == Ω 0.915 0.0397 pu 23.04 EQ X Ω == Ω The per-unit equivalent circuit is + - V S V P I S + - I P R C jX M R EQ jX EQ 0.010 j 0.0397 263 j 51.8 At rated conditions and unity power factor, the input power to this transformer would be IN P = 1.0 pu. The core losses (in resistor C R ) would be ( ) 2 2 core 1.0 0.00380 pu 263 C V P R == = The copper losses (in resistor EQ R ) would be ()( ) 2 2 CU EQ 1.0 0.010 0.010 puPIR== = The output power of the transformer would be OUT OUT CU core 1.0 0.010 0.0038 0.986PPPP=−−=− − = and the transformer efficiency would be OUT IN 0.986 100% 100% 98.6% 1.0 P P η =× = × = The output voltage of this transformer is ()( ) OUT IN EQ 1.0 1.0 0 0.01 0.0397 0.991 2.3Zj=− =−∠° + = ∠−°VVI The voltage regulation of the transformer is 1.0 0.991 VR 100% 0.9% 0.991 − =×= (b) The autotransformer connection for 600/480 V stepdown operation is 59 + - + - 480 V 600 V N C N SE V SE V C + + - - (c) When used as an autotransformer, the kVA rating of this transformer becomes: () SE IO SE 41 10 kVA 50 kVA 1 C W NN SS N ++ == = (d) As an autotransformer, the per-unit series impedance EQ Z is decreased by the reciprocal of the power advantage, so the series impedance becomes 0.010 0.002 pu 5 EQ R == 0.0397 0.00794 pu 5 EQ X == while the magnetization branch elements are basically unchanged. At rated conditions and unity power factor, the input power to this transformer would be IN P = 1.0 pu. The core losses (in resistor C R ) would be ( ) 2 2 core 1.0 0.00380 pu 263 C V P R == = The copper losses (in resistor EQ R ) would be ()( ) 2 2 CU EQ 1.0 0.002 0.002 puPIR== = The output power of the transformer would be OUT OUT CU core 1.0 0.002 0.0038 0.994PPPP=−−=− − = and the transformer efficiency would be OUT IN 0.994 100% 100% 99.4% 1.0 P P η =× = × = The output voltage of this transformer is ()( ) OUT IN EQ 1.0 1.0 0 0.002 0.00794 0.998 0.5Zj=− =−∠° + = ∠−°VVI The voltage regulation of the transformer is 1.0 0.998 VR 100% 0.2% 0.998 − =×= 2-23. Figure P2-4 shows a power system consisting of a three-phase 480-V 60-Hz generator supplying two loads through a transmission line with a pair of transformers at either end. (a) Sketch the per-phase equivalent circuit of this power system. 60 (b) With the switch opened, find the real power P, reactive power Q, and apparent power S supplied by the generator. What is the power factor of the generator? (c) With the switch closed, find the real power P, reactive power Q, and apparent power S supplied by the generator. What is the power factor of the generator? (d) What are the transmission losses (transformer plus transmission line losses) in this system with the switch open? With the switch closed? What is the effect of adding Load 2 to the system? Region 1 Region 2 Region 3 base1 S = 1000 kVA base2 S = 1000 kVA base3 S = 1000 kVA base2,L V = 480 V base2,L V = 14,400 V base3,L V = 480 V S OLUTION This problem can best be solved using the per-unit system of measurements. The power system can be divided into three regions by the two transformers. If the per-unit base quantities in Region 1 are chosen to be base1 S = 1000 kVA and base1,L V = 480 V, then the base quantities in Regions 2 and 3 will be as shown above. The base impedances of each region will be: ( ) 2 2 1 base1 base1 3 3 277 V 0.238 1000 kVA V Z S φ == = Ω () 2 2 2 base2 base2 3 38314 V 207.4 1000 kVA V Z S φ == =Ω ( ) 2 2 3 base3 base3 3 3277 V 0.238 1000 kVA V Z S φ == =Ω (a) To get the per-unit, per-phase equivalent circuit, we must convert each impedance in the system to per-unit on the base of the region in which it is located. The impedance of transformer 1 T is already in per- unit to the proper base, so we don’t have to do anything to it: 010.0 pu,1 =R 040.0 pu,1 =X The impedance of transformer 2 T is already in per-unit, but it is per-unit to the base of transformer 2 T , so it must be converted to the base of the power system. ()() ()() 2 base 1 base 2 pu on base 2 pu on base 1 2 base 2 base 1 ( , , ) ( , , ) VS RX Z RX Z VS = (2-60) ()( ) ()( ) 2 2,pu 2 8314 V 1000 kVA 0.020 0.040 8314 V 500 kVA R == ()( ) ()( ) 2 2,pu 2 8314 V 1000 kVA 0.085 0.170 8314 V 500 kVA X == 61 The per-unit impedance of the transmission line is line line,pu base2 1.5 10 0.00723 0.0482 207.4 Zj Zj Z +Ω == = + Ω The per-unit impedance of Load 1 is load1 load1,pu base3 0.45 36.87 1.513 1.134 0.238 Z Zj Z ∠°Ω == =+ Ω The per-unit impedance of Load 2 is load2 load2,pu base3 0.8 3.36 0.238 Zj Zj Z −Ω == =− Ω The resulting per-unit, per-phase equivalent circuit is shown below: + - 1∠0° T 1 T 2 Line L 1 L 2 0.010 j0.040 0.00723 j0.0482 0.040 j0.170 1.513 j1.134 -j3.36 (b) With the switch opened, the equivalent impedance of this circuit is EQ 0.010 0.040 0.00723 0.0482 0.040 0.170 1.513 1.134Zj j j j=+ + + ++ ++ EQ 1.5702 1.3922 2.099 41.6Zj=+ =∠° The resulting current is 10 0.4765 41.6 2.099 41.6 ∠° ==∠−° ∠° I The load voltage under these conditions would be ()() Load,pu Load 0.4765 41.6 1.513 1.134 0.901 4.7Zj== ∠−°+ =∠−°VI ()( ) Load Load,pu base3 0.901 480 V 432 VVVV== = The power supplied to the load is ()() 2 2 Load,pu Load 0.4765 1.513 0.344PIR== = ()( ) Load Load,pu base 0.344 1000 kVA 344 kWPPS== = The power supplied by the generator is ()( ) ,pu cos 1 0.4765 cos41.6 0.356 G PVI θ == °= ()( ) ,pu sin 1 0.4765 sin 41.6 0.316 G QVI θ == °= ()( ) ,pu 1 0.4765 0.4765 G SVI== = ()( ) ,pu base 0.356 1000 kVA 356 kW GG PPS== = ()( ) ,pu base 0.316 1000 kVA 316 kVAR GG QQS== = ()( ) ,pu base 0.4765 1000 kVA 476.5 kVA GG SSS== = The power factor of the generator is 62 PF cos 41.6 0.748 lagging=°= (c) With the switch closed, the equivalent impedance of this circuit is ()() ()() EQ 1.513 1.134 3.36 0.010 0.040 0.00723 0.0482 0.040 0.170 1.513 1.134 3.36 jj Zj j j jj +− =+ + + ++ + ++− EQ 0.010 0.040 0.00788 0.0525 0.040 0.170 (2.358 0.109)Zj j j j=+++++++ EQ 2.415 0.367 2.443 8.65Zj=+ =∠° The resulting current is 10 0.409 8.65 2.443 8.65 ∠° ==∠−° ∠° I The load voltage under these conditions would be ()() Load,pu Load 0.409 8.65 2.358 0.109 0.966 6.0Zj==∠−°+ =∠−°VI ()( ) Load Load,pu base3 0.966 480 V 464 VVVV== = The power supplied to the two loads is the power supplied to the resistive component of the parallel combination of the two loads: 2.358 pu. ()() 2 2 Load,pu Load 0.409 2.358 0.394PIR== = ()( ) Load Load,pu base 0.394 1000 kVA 394 kWPPS== = The power supplied by the generator is ()( ) ,pu cos 1 0.409 cos6.0 0.407 G PVI θ == °= ()( ) ,pu sin 1 0.409 sin 6.0 0.0428 G QVI θ == °= ()( ) ,pu 1 0.409 0.409 G SVI== = ()( ) ,pu base 0.407 1000 kVA 407 kW GG PPS== = ()( ) ,pu base 0.0428 1000 kVA 42.8 kVAR GG QQS== = ()( ) ,pu base 0.409 1000 kVA 409 kVA GG SSS== = The power factor of the generator is PF cos 6.0 0.995 lagging=°= (d) The transmission losses with the switch open are: ()( ) 2 2 line,pu line 0.4765 0.00723 0.00164PIR== = ()( ) line l ,pu base 0.00164 1000 kVA 1.64 kW ine PPS== = The transmission losses with the switch closed are: ()( ) 2 2 line,pu line 0.409 0.00723 0.00121PIR== = ()( ) line l ,pu base 0.00121 1000 kVA 1.21 kW ine PPS== = Load 2 improved the power factor of the system, increasing the load voltage and the total power supplied to the loads, while simultaneously decreasing the current in the transmission line and the transmission line losses. This problem is a good example of the advantages of power factor correction in power systems. 63 Chapter 3: Introduction to Power Electronics 3-1. Calculate the ripple factor of a three-phase half-wave rectifier circuit, both analytically and using MATLAB. S OLUTION A three-phase half-wave rectifier and its output voltage are shown below π /6 5 π /6 2 π /3 () sin AM vt V t ω = () ( ) sin 2 /3 BM vt V t ωπ =− () ( ) sin 2 / 3 CM vt V t ωπ =+ S OLUTION If we find the average and rms values over the interval from π /6 to 5 π /6 (one period), these values will be the same as the average and rms values of the entire waveform, and they can be used to calculate the ripple factor. The average voltage is () 5/6 /6 13 () sin 2 DC M VvtdtVtdt T π π ωω π == 5 6 6 333333 cos 0.8270 22222 MM DC MM VV Vt VV π π ω ππ π =− =− − − = = The rms voltage is () 5/6 222 rms /6 13 () sin 2 M VvtdtVtdt T π π ωω π == 5/6 2 rms / 6 311 sin 2 22 4 M V Vtt π π ωω π =− 64 2 rms 315 15 sin sin 22664 3 3 M V V ππ π π π =−−− 22 rms 315 3133 sin sin 234 3 3 23422 MM VV V πππ π ππ =−−=−−− 22 rms 31333 3 0.8407 23422 234 MM M VV VV ππ ππ =−−−=+= The resulting ripple factor is 22 rms DC 0.8407 1 100% 1 100% 18.3% 0.8270 M M VV r VV =−×= −×= The ripple can be calculated with MATLAB using the ripple function developed in the text. We must right a new function halfwave3 to simulate the output of a three-phase half-wave rectifier. This output is just the largest voltage of () tv A , () tv B , and () tv C at any particular time. The function is shown below: function volts = halfwave3(wt) % Function to simulate the output of a three-phase % half-wave rectifier. % wt = Phase in radians (=omega x time) % Convert input to the range 0 <= wt < 2*pi while wt >= 2*pi wt = wt - 2*pi; end while wt < 0 wt = wt + 2*pi; end % Simulate the output of the rectifier. a = sin(wt); b = sin(wt - 2*pi/3); c = sin(wt + 2*pi/3); volts = max( [ a b c ] ); The function ripple is reproduced below. It is identical to the one in the textbook. function r = ripple(waveform) % Function to calculate the ripple on an input waveform. % Calculate the average value of the waveform nvals = size(waveform,2); temp = 0; for ii = 1:nvals temp = temp + waveform(ii); end average = temp/nvals; % Calculate rms value of waveform [...]... ( B1 sin ω t + B2 cos ωt ) = M sin ω t RC RC cosine equation: ω B1 + 1 B2 = 0 ⇒ RC B2 = −ω RC B1 sine equation: −ω B2 + 1 V B1 = M RC RC ω 2 RC B1 + ω 2 RC + 1 V B1 = M RC RC 1 V B1 = M RC RC 1 + ω 2 R 2C 2 V B1 = M RC RC Finally, 71 B1 = VM and 1 + ω 2 R 2C 2 B2 = −ω RC VM 1 + ω 2 R 2C 2 Therefore, the forced solution to the equation is vC , f ( t ) = VM ω RC VM sin ωt − cos ωt 1 + ω 2 R 2C 2 1 +... be 1 ω = v (t ) dt = T π /6 π T Vave π /ω π /6 1 π /ω VM sin ω t dt = − VM cos ω t π π /6 1 3 2+ 3 = Vave = − VM 1 − VM = (0.5 94 ) (16 9.5 V ) = 10 1 V 2 2π π (c) For a firing angle of 90°, the average voltage will be 1 ω = v (t ) dt = T π /2 π T Vave π /ω π /2 1 π /ω VM sin ω t dt = − VM cos ω t π 69 π /2 1 1 Vave = − VM [ 1] = VM = ( 0. 318 ) (16 9.5 V ) = 54 V π 3-5 π For the circuit in Figure P3 -1, ... = 2π / ω , so T /12 is π / 6ω The average voltage over the interval from 0 to T /12 is VDC = VDC = 6ω π / 6ω 1 T v (t ) dt = 3 3 VM = 1. 6 540 VM π π 3VM cos ω t dt = 0 6 3 π The rms voltage is Vrms = 1 T v (t ) 2 dt = 6ω π π / 6ω 3VM 2 cos 2 ω t dt 0 66 π / 6ω VM sin ω t 0 Vrms = 18 ω π Vrms = VM VM 2 1 1 t+ sin 2ω t 2 4 18 ω π π 12 ω + π / 6ω 0 1 π 3 9 3 sin = VM + = 1. 65 54 VM 4 3 2 4 The resulting... t 1 + ω 2 R 2C 2 1 + ω 2 R 2C 2 If we substitute the known values for R, C, ω, and VM, this equation becomes vC (t ) = 42 e − 10 0t + 11 . 14 sin ωt − 42 cos ωt This equation is plotted below: It reaches a voltage of 30 V at a time of 3.50 ms Since the frequency of the waveform is 60 Hz, the waveform there are 360° in 1/ 60 s, and the firing angle α is 72 α = ( 3.50 ms ) 360° = 75.6° or 1. 319 radians 1/ ... resulting ripple factor is r= Vrms VDC 2 − 1 × 10 0% = 1. 65 54 VM 1. 6 540 VM 2 − 1 × 10 0% = 4. 2% The ripple can be calculated with MATLAB using the ripple function developed in the text We must right a new function fullwave3 to simulate the output of a three-phase half-wave rectifier This output is just the largest voltage of v A (t ) , v B (t ) , and vC (t ) at any particular time The function is shown below:... Laplace Transforms, if desired (b) The rms voltage applied to the load is Vrms = Vrms = Vrms = 1 T v (t ) 2 dt = ω π π /ω VM 2 sin 2 ω t dt α VM 2 1 1 ωt − sin 2ω t 4 π 2 π /ω α VM 2 1 1 (π − α ) − (sin 2π − sin 2α ) 4 π 2 Vrms = VM 0.32 84 = 0.573 VM = 97 .1 V 3-6 One problem with the circuit shown in Figure P3 -1 is that it is very sensitive to variations in the input voltage v ac ( t ) For example, suppose... d 1 V vC + vC = M sin ωt dt RC RC It must have a form similar to the forcing function, so the solution will be of the form vC , f ( t ) = B1 sin ω t + B2 cos ω t where the constants B1 and B2 must be determined by substitution into the original equation Solving for B1 and B2 yields: d 1 V ( B1 sin ωt + B2 cos ω t ) + ( B1 sin ω t + B2 cos ωt ) = M sin ω t dt RC RC (ω B1cos ω t − ω B2 sin ω t ) + 1. .. transformer will be vac (t ) = 16 9.5 sin ωt (a) The average voltage applied to the load will be the integral over the conducting portion of the half cycle divided by π/ω, the period of a half cycle For a firing angle of 0°, the average voltage will be 1 ω v (t ) dt = T 0 π T Vave = π /ω 1 VM sin ω t dt = − VM cos ω t 0 π π /ω 0 1 2 Vave = − VM [ 1 − 1] = VM = (0.637 ) (16 9.5 V ) = 10 8 V π (b) π For a firing... output of a three-phase full-wave % rectifier waveform = zeros (1, 128); for ii = 1: 128 waveform(ii) = fullwave3(ii*pi/ 64) ; end % Now calculate the ripple factor r = ripple(waveform); % Print out the result string = ['The ripple is ' num2str(r) '%.']; disp(string); 67 When this program is executed, the results are » test_fullwave3 The ripple is 4. 2 017 % This answer agrees with the analytical solution above... t 1 + ω 2 R 2C 2 1 + ω 2 R 2C 2 The initial condition for this problem is that the voltage on the capacitor is zero at the beginning of the halfcycle: vC (0 ) = Ae − 0 RC + VM ω RC VM sin 0 − cos 0 = 0 2 2 2 1+ ω R C 1 + ω 2 R 2C 2 A− ω RC VM =0 1 + ω 2 R 2C 2 A= ω RC VM 1 + ω 2 R 2C 2 Therefore, the voltage across the capacitor as a function of time before the DIAC fires is vC (t ) = ω RC VM − e 1 . EQ 0. 010 0. 040 0.00723 0. 048 2 0. 040 0 .17 0 1. 513 1. 134Zj j j j=+ + + ++ ++ EQ 1. 5702 1. 3922 2.099 41 . 6Zj=+ =∠° The resulting current is 10 0 .47 65 41 . 6 2.099 41 . 6 ∠° ==∠−° ∠° I The. Load 0 .47 65 41 . 6 1. 513 1. 1 34 0.9 01 4. 7Zj== ∠−°+ =∠−°VI ()( ) Load Load,pu base3 0.9 01 48 0 V 43 2 VVVV== = The power supplied to the load is ()() 2 2 Load,pu Load 0 .47 65 1. 513 0. 344 PIR==. PF cos 41 . 6 0. 748 lagging=°= (c) With the switch closed, the equivalent impedance of this circuit is ()() ()() EQ 1. 513 1. 1 34 3.36 0. 010 0. 040 0.00723 0. 048 2 0. 040 0 .17 0 1. 513 1. 1 34 3.36 jj Zj