ELEC-E8403 Converter Techniques Exam

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1 ELEC-E8403 Converter Techniques Exam Remember to answer the course feedback questionnaire. You will receive an extra bonus point by doing this. 1. A six-pulse thyristor rectifier has a load of R = 50 Ω and the dc current can be assumed to be ideal dc (large inductance). The line-to-line supply voltage is 400 V, frequency 50 Hz and the delay angle of the rectifier is 45. a) Calculate active and reactive power taken from the supply system ( points) b) The inductance of the supply system is 1 mh per phase. What is the active power now? It can be assumed that commutation causes an additional delay equal to half of the commutation angle. How large is the reactive power in this case? (3 points). Three-phase squirrel cage motor is supplied from a pulse width modulated inverter. In one operating point the line-to-line voltage of the motor comprises of two pulses within a half cycle and the control principle is constant flux. Calculate the frequencies on which the minimum and maximum values of the 7 th harmonic occur, ƒ < 50Hz. The nominal line-to-line voltage of the motor is 400 V and th e intermediate dc-bus voltage is 540 V. Nominal frequency of the motor is 50 Hz. 3. Explain shortly the operating principle of the converter shown below. Draw the phasor diagram of the system and use it to calculate reference amplitude of v conv1 and its angle as a function of the line current, when the power taken from the power systems varies between 0-6 kw and reactive power is zero. Rms voltage of the power system is 30 V, inductance 0 mh and resistance 0,1. Explain without mathematics what changes if the converter supplies all the time 1 kva reactive power to the power system. 4. The line frequency apparent power of a six-pulse thyristor rectifier S = 0 MVA and the line-toline rms voltage is 110 kv. For compensation and filtering purposes there is a filter for the 5 th harmonic with parameters: X0 5L 480 and quality factor Q X 0 R f 50. Short circuit power of the power system Sk = 590 MVA and system can be assumed to be reactive. Calculate the 5 th harmonic voltage without and with the filter. What is the effect of the quality factor Q on the harmonic voltage? When calculating harmonics, the effect of commutation is neglected and dc current can be assumed ideal. 5. Pulse width modulation in power electronics. Explain the main principles of PWM and discuss advantages and disadvantages on using PWM.

2 Page S Suuntaajatekniikka (Old course lectured in Finnish till Spring 015) Tentti Oletetaan, että 6-pulssisen tasasuuntaajan tasavirtapiirissä on äärettömän suuri induktanssi. Sillan kuormituksena on resistanssi R = 50 Ω, verkon pääjännite on 400 V, taajuus 50 Hz ja sillan ohjauskulma 45. a) Laske verkosta otettu pätö- ja loisteho ( p.) b) Kussakin vaiheessa on 1 mh kommutointi-induktanssit. Kuinka suuri pätöteho on tässä tapauksessa? Laske myös likiarvo loistehotarpeelle. (3 p.). Oikosulkumoottoria syötetään pulssinleveysmoduloidusta (PWM) vaihtosuuntaajasta. Eräässä toimintapisteessä pääjännitteessä (symmetrinen 3-vaihejärjestelmä) on pulssia/puolijakso ja säätö perustuu vakiovuoperiaatteeseen. Laske millä taajuuksilla 7. yliaallon maksimi- ja minimiarvot esiintyvät, ƒ < 50Hz. Moottorin nimellinen pääjännite on 400 V, joka on myös taajuusmuuttajan kuusipulssista diodisiltaa syöttävän verkon jännite. Moottorin nimellistaajuus on 50 Hz. 3. Selitä seuraavan kuvan mukaisen suuntaajan periaatteellinen toiminta lyhyesti. Piirrä kytkentää vastaava osoitindiagrammi. Laske sen avulla tarvittavan vastajänniteohjeen perusaallon itseisarvo v conv1 ja kulma verkkovirran funktiona, kun verkosta otettu teho vaihtelee välillä 0-6 kw ja loisteho on nolla. Verkkojännitteen tehollisarvo on 30 V ja verkon induktanssi 0 mh sekä resistanssi 0,1. Selitä sanallisesti miten tilanne muuttuu jos verkkoon syötetään koko ajan lisäksi 1 kva loistehoa. 4. Kuusipulssinen tyristoritasasuuntaaja ottaa verkosta perustaajuisen näennäistehon S = 0 MVA verkon pääjännitteen tehollisarvon ollessa 110 kv. Kompensointiin ja suodatukseen on rakennettu 5. yliaallolle viritetty suodatin. Suodattimen arvot ovat: X0 5L 480 ja hyvyysluku Q X 0 R f 50. Verkon oikosulkuteho Sk = 590 MVA ja verkko voidaan olettaa reaktiiviseksi. Laske verkon 5. yliaallon jännite ilman suodatinta ja suodattimen kanssa. Miten suodattimen hyvyysluku vaikuttaa yliaaltojännitteeseen? Yliaaltojen laskennassa kommutoinnin vaikutusta ei oteta huomioon ja tasasuuntaajan tasavirta oletetaan täysin tasoittuneeksi. 5. Kytkentäsuojapiirien käytön pääperiaatteet suuntaajatekniikassa.

3 Page 3 Question 1. a) The ideal uncontrolled dc voltage is V o,dc = 3 π V LL,p = 540,18 V And it is reduced by using control angle. If the losses of the converter are zero, active power taken from the supply is P = V o,dc I d cos α f = V o,dc R cos α f,9 kw In diode rectifier power factor can be calculated as and it is constant. However, in thyristor rectifiers control angle is used to reduce the output voltage value. At the same time output power decreases too as was seen in the active power. Nevertheless, power factor and reactive power taken from the supply behaves differently. Control angle introduces a delay in the ac side between current and voltage. This is reflected in power factor, which reduces. When comparing to the diode rectifier, there is an additional factor cos (f), which means that there is a phase shift equal to f between the voltage and fundamental component of line current. Reactive power from the supply is therefore simply Q = 3V LL I L1 sin α f Where I L1 is the fundamental component of the line current. It could be calculated by Fourier-series from the line current waveform but it can be obtained from the previous active power too because P = 3V LL I L1 cos α f = V o,dc I d cos α f And the fundamental current component is 3 π I L1 = V o,dc 3V LL I d = 3 I d = 6 π I d Q = 3V LL 6 π I d sin α f = V o,dc I d sin α f,9 kvar Active and reactive power are equal because control angle is 45 o. b) When commutation is taken into account it has been shown in the textbook that There i k = V LL,p X s is the peak value of short circuit current when two line are connected together. Line reactance Xv 0,314 and i k 900,3 A. Further, it has been shown in the textbook that there is a voltage drop due to commutation and it can be modelled as resistive voltage drop

4 Page 4 where R r = 3 π X v and then the dc voltage can also be written in the form V o,dc = 3 π V LL,p cos α f 3 π X v V LL,p X v [cos α f cos(α f + μ)] = 3 π V LL,p[cos α f + cos(α f + μ)] The dc voltage can be calculated with the dc quantities to, i.e. it is equal to RId and Udi U cos cos ˆ d RId Rik cos cos 1 Riˆ k cos Udi cos o cos 0, 699 0, U ˆ di Rik Because of the commutation dc-voltage drops and power is reduced. I ˆ d ik cos cos 7, 59 A P RId,88 kw Consumption of reactive power increases as commutation increases the phase shift between the input voltage and current. Without commutation it was equal to and with commutation one often used approximation is + f/. Question. Symmetric three-phase system means that the center points of the pulses must be at 60 and 10 in line-to-line voltage but the width of the pulses is not known. We will mark the width of the pulse with. Marking the starting and ending points of the two pulses with unknown angles the Fourier-voltage of the voltage is U d 1 bn sin nx sin nx nπ 11 1 U d cos n11 cos n1 cos n 1 cos n, n 1,3,5, 7,... nπ 4U d sin n1 cos n1 sin n cos n, n 1,3,5, 7,... nπ k = 1 k k1 k = 1 k1 + k k = k = Here k = 60, when k = 1 ja k =10 when k = U n = b n => U n = 4 U d sin n sin n60 + sin n sin n10 n

5 U 7 = Zero values of U 7 are the minimum points of, i.e => U 7 = 0 => sin 7 = 0 => 7 = m 180, m = 0, 1,, 3... => = m Physical limitations for are 0 30 m = 0 => = 0, m = 1 => = 5,7, m = => = 51,4 > 30 Maximum values of the harmonics are obtained by derivation. du 7 d = 4 U d 3 cos 7 = 0 => = l 180 7, l = 0, ± 1, ±,... l = 0 => = 1,86 l = 1 => = 38,57 > 30 => 7. harmonic minima when = 0 ja = 5,7 and maximum when. Additionally should be studied from U 7 but it can be seen from that it is for sure smaller than at. As we are operating in the constant flux region Fundamental component: 4 U d n sin 7 sin sin 7 sin 710 = 8 U d 7 3 sin 7 U 7 U f = U N f N => f = f N U 1 U N In the question line-to-line voltage is 400 V and the previous values are,5 Hz ja 43,37 Hz. Page 5 U 1 = 4 U d (sin sin 60 + sin sin 10 ) = 4 U d : 3 sin = 0 => U 1 = 0 V => f = 0 Hz = 1,86 => U 1 = sin 1,86 = 178 V => f = = 3,4 Hz 380 = 5,7 => U 1 = sin 5,7 = 347 V => f = = 45,7 Hz 380 Question 3

6 Page 6 As we know when cos 1 = 1, so U v = R + jl I 1 + U s1 U v RI 1 = U s1 cos jli 1 = U s1 sin (1) () (3) U s1 = U v RI 1 + 1/ LI1 = arc tan LI 1 U v RI j (4) (5) By solving it: I 1 max = P max U v = 6 kw 80 V 1,4 A => U S1 min = U v = 80V; min = 0 U S1 max = 80 0,1 1, ,4 1/ 309 V => max = arctan , ,4 5,8 Question 4 I Line-current harmonics are at n = kp ± 1 = 5, 7, 11, 13, 17, 19, and amplitudes n = I 1 n where I 1 = is the fundamental component. Effect of commutation is neglected. Fifth harmonic is S = 3UI 1 => I 5 = S 3U5 = A 0,99 A Network is assumed to be reactive and its impedance is calculated from short circuit power. U 3 U U Sk 3UI k 3U X v 4, X X S v v k When operating without the filter, fifth harmonic line-to-line voltage is U 5 = 3X v5 I 5 = 35X v I V 4% With filter situation is as shown below

7 Page 7 At resonance frequency only the resistance of the filter remains. X Rf 9,6 Q 50 Fifth current harmonics sees the parallel connection of the network impedance and filter and in lineto-line voltage jx R X R 5X R U 3 I 3 I 3 I 191, 4V 0,000174% v5 f v5 f v f Rf jx v5 Rf X v5 Rf 5X v With the numerical values of this question, the result is 348 V and 0,3 %, i.e. much smaller than in the first case. Quality factor has an effect on the resistance and in an ideal case it is infinite (resistance zero) and the harmonic voltage would be zero. Question 5. See textbook