Electric Machinery Fundamentals Power & Energy_4 pptx

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 65 temp = 0; for ii = 1:nvals temp = temp + waveform(ii)^2; end rms = sqrt(temp/nvals); % Calculate ripple factor r = sqrt((rms / average)^2 - 1) * 100; Finally, the test driver program is shown below. % M-file: test_halfwave3.m % M-file to calculate the ripple on the output of a % three phase half-wave rectifier. % First, generate the output of a three-phase half-wave % rectifier waveform = zeros(1,128); for ii = 1:128 waveform(ii) = half wave3(ii*pi/64); end % Now calculate the ripple factor r = ripple(waveform); % Print out the result string = ['The ripple is ' num2str(r) '%.']; disp(string); When this program is executed, the results are » test_halfwave3 The ripple is 18.2759%. This answer agrees with the analytical solution above. 3-2. Calculate the ripple factor of a three-phase full-wave rectifier circuit, both analytically and using MATLAB. S OLUTION A three-phase half-wave rectifier and its output voltage are shown below 66 T /12 () sin AM vt V t ω = () ( ) sin 2 / 3 BM vt V t ωπ =− () ( ) sin 2 /3 CM vt V t ωπ =+ S OLUTION By symmetry, the rms voltage over the interval from 0 to T/12 will be the same as the rms voltage over the whole interval. Over that interval, the output voltage is: () () () 22 sin sin 33 CB M M vt vt vt V t V t ππ ωω =−= +− − () 22 22 sin cos cos sin sin cos cos sin 33 33 MM vt V t t V t t ππ ππ ωω ωω =+−− () 2 2cos sin 3cos 3 M vt V t t π ωω == Note that the period of the waveform is 2/T πω = , so T/12 is / 6 πω . The average voltage over the interval from 0 to T/12 is /6 / 6 0 0 16 63 () 3 cos sin DC M M Vvtdt VtdtVt T πω πω ω ωω ππ == = 33 1.6540 DC M M VV V π == The rms voltage is /6 222 rms 0 16 () 3 cos M Vvtdt Vtdt T πω ω ω π == 67 / 6 2 rms 0 18 1 1 sin 2 24 M VVt t πω ω ω πω =+ rms 18 1 3 9 3 sin 1.6554 12 4 3 2 4 MMM VV V V ωπ π πωω π =+=+= The resulting ripple factor is 2 2 rms DC 1.6554 1 100% 1 100% 4.2% 1.6540 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 fullwave3 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 = fullwave3(wt) % Function to simulate the output of a three-phase % full-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 ] ) - min( [ a b c ] ); The test driver program is shown below. % M-file: test_fullwave3.m % M-file to calculate the ripple on the output of a % three phase full-wave rectifier. % First, generate the 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); 68 When this program is executed, the results are » test_fullwave3 The ripple is 4.2017%. This answer agrees with the analytical solution above. 3-3. Explain the operation of the circuit shown in Figure P3-1. What would happen in this circuit if switch S 1 were closed? S OLUTION Diode D 1 and D 2 together with the transformer form a full-wave rectifier. Therefore, a voltage oriented positive-to-negative as shown will be applied to the SCR and the control circuit on each half cycle. (1) Initially, the SCR is an open circuit, since v 1 < V BO for the SCR. Therefore, no current flows to the load and v LOAD = 0. (2) Voltage v 1 is applied to the control circuit, charging capacitor C 1 with time constant RC 1 . (3) When v C > V BO for the DIAC, it conducts, supplying a gate current to the SCR. (4) The gate current in the SCR lowers its breakover voltage, and the SCR fires. When the SCR fires, current flows through the SCR and the load. (5) The current flow continues until i D falls below I H for the SCR (at the end of the half cycle). The process starts over in the next half cycle. 69 If switch S 1 is shut, the charging time constant is increased, and the DIAC fires later in each half cycle. Therefore, less power is supplied to the load. 3-4. What would the rms voltage on the load in the circuit in Figure P3-1 be if the firing angle of the SCR were (a) 0°, (b) 30°, (c) 90°? S OLUTION The input voltage to the circuit of Figure P3-1 is () ttv ω sin339 ac = , where rad/s 377= ω Therefore, the voltage on the secondary of the transformer will be () ttv ω sin5.169 ac = (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 / / ave 0 00 11 () sin cos T MM VvtdtVtdtVt T πω πω ω ωω ππ == =− [] ()( ) ave 12 1 1 0.637 169.5 V 108 V MM VV V ππ =− − − = = = (b) For a firing angle of 30°, the average voltage will be / / ave / 6 /6 /6 11 () sin cos T MM VvtdtVtdtVt T πω πω π ππ ω ωω ππ == =− ()( ) ave 1323 1 0.594 169.5 V 101 V 22 MM VV V ππ + =− − − = = = (c) For a firing angle of 90°, the average voltage will be / / ave / 2 /2 /2 11 () sin cos T MM VvtdtVtdtVt T πω πω π ππ ω ωω ππ == =− 70 [] ()( ) ave 11 1 0.318 169.5 V 54 V MM VV V ππ =− − = = = 3-5. For the circuit in Figure P3-1, assume that V BO for the DIAC is 30 V, C 1 is 1 µF, R is adjustable in the range 1-20 kΩ, and that switch S 1 is open. What is the firing angle of the circuit when R is 10 kΩ? What is the rms voltage on the load under these conditions? Note: Problem 3-5 is significantly harder for many students, since it involves solving a differential equation with a forcing function. This problem should only be assigned if the class has the mathematical sophistication to handle it. S OLUTION At the beginning of each half cycle, the voltages across the DIAC and the SCR will both be smaller then their respective breakover voltages, so no current will flow to the load (except for the very tiny current charging capacitor C), and v load (t) will be 0 volts. However, capacitor C charges up through resistor R, and when the voltage v C (t) builds up to the breakover voltage of D 1 , the DIAC will start to conduct. This current flows through the gate of SCR 1 , turning the SCR ON. When it turns ON, the voltage across the SCR will drop to 0, and the full source voltage v S (t) will be applied to the load, producing a current flow through the load. The SCR continues to conduct until the current through it falls below I H , which happens at the very end of the half cycle. Note that after D 1 turns on, capacitor C discharges through it and the gate of the SCR. At the end of the half cycle, the voltage on the capacitor is again essentially 0 volts, and the whole process is ready to start over again at the beginning of the next half cycle. To determine when the DIAC and the SCR fire in this circuit, we must determine when v C (t) exceeds V BO for D 1 . This calculation is much harder than in the examples in the book, because in the previous problems the source was a simple DC voltage source, while here the voltage source is sinusoidal. However, the principles are identical. (a) To determine when the SCR will turn ON, we must calculate the voltage v C (t), and then solve for the time at which v C (t) exceeds V BO for D 1 . At the beginning of the half cycle, D 1 and SCR 1 are OFF, and the voltage across the load is essentially 0, so the entire source voltage v S (t) is applied to the series RC circuit. To determine the voltage v C (t) on the capacitor, we can write a Kirchhoff's Current Law equation at the node above the capacitor and solve the resulting equation for v C (t). 0 21 =+ ii (since the DIAC is an open circuit at this time) 0 1 =+ − C C v dt d C R vv [...]... charging is τ = RLOADC = (250 Ω)(15 µF) = 0.00375 s and SCR2 will turn off in a few milliseconds (d) In this circuit, once SCR1 fires, a substantial period of time must pass before the power to the load can be turned off If the power to the load must be turned on and off rapidly, this circuit could not do the job 81 (e) This problem can be eliminated by using one of the more complex parallel commutation... purpose of SCR 2 ? What is the purpose of D2? (This chopper circuit arrangement is known as a Jones circuit.) SOLUTION First, assume that SCR1 is triggered When that happens, current will flow from the power supply through SCR1 and the bottom portion of transformer T1 to the load At that time, a voltage will be applied to the bottom part of the transformer which is positive at the top of the winding... cut off when the capacitor is fully charged Alternately, it will be cut off by the voltage across the capacitor if SCR1 is triggered before it would otherwise cut off In this circuit, SCR1 controls the power supplied to the load, while SCR2 controls when SCR1 will be turned off Diode D2 in this circuit is a free-wheeling diode, which allows the current in the load to continue flowing for a short time... three time constants Since τ = RC = (20 kΩ)(150 µF) = 3 s, the SCR will be ready to fire again after 9 s (c) In this circuit, the ON time of the SCR is much shorter than the reset time for the SCR, so power can flow to the load only a very small fraction of the time (This effect would be less exaggerated if the ratio of R to RLOAD were smaller.) (d) This problem can be eliminated by using one of the... voltage greater than VBO for the DIAC across the RC charging circuit This diode holds the voltage across the RC circuit constant, so that the capacitor charging time is not much affected by changes in the power supply voltage R vC 73 3-7 Explain the operation of the circuit shown in Figure P3-2, and sketch the output voltage from the circuit SOLUTION This circuit is a single-phase voltage source inverter . base3 0.901 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== = ()( ) Load Load,pu base 0. 344 1000 kVA 344 kWPPS== = The power supplied. 2.099 41 .6Zj=+ =∠° The resulting current is 10 0 .47 65 41 .6 2.099 41 .6 ∠° ==∠−° ∠° I The load voltage under these conditions would be ()() Load,pu Load 0 .47 65 41 .6 1.513 1.1 34 0.901 4. 7Zj==. j0. 040 0.00723 j0. 048 2 0. 040 j0.170 1.513 j1.1 34 -j3.36 (b) With the switch opened, the equivalent impedance of this circuit is EQ 0.010 0. 040 0.00723 0. 048 2 0. 040 0.170 1.513 1.134Zj j