IGEE 401 Power Electronic Systems. Solution to Midterm Examination Fall 2004

Transcription

1 Jós, G GEE 401 wer Electrnc Systems Slutn t Mdterm Examnatn Fall 2004 Specal nstructns: - Duratn: 75 mnutes. - Materal allwed: a crb sheet (duble sded 8.5 x 11), calculatr. - Attempt all questns. Make any reasnable assumptn. - All questns and sub-questns carry equal weght. - Return the cmpleted answer sheet attached. 1. QUESTON 1 (25 pnts) A sngle-phase dde rectfer, fed frm a 240, 60 Hz mans, supples an R-L lad (seres cnnectn). The lad has a dc resstance f 10 Ω, and a tme cnstant f 300 ms. a) Draw the wavefrm f the utput vltage and current f the rectfer, ndcatng maxmum and mnmum values. Draw the wavefrm f the vltage acrss the resstance and the nductance. Gve the average value f the nductr vltage. b) Fnd the average rectfer utput vltage and current. Cmpute the pwer delvered t the lad. c) Draw the nput ac lne current wavefrm, ndcatng the peak value. Cmpute the rms value. Cmpute the apparent nput pwer and pwer factr. d) Draw the wavefrm f the current n each dde. Gven the vltage drp n the dde t be 2, cmpute the lsses n the rectfer, assumng swtchng lsses are neglgble. Cmpute the effcency f the rectfer. Slutn: L = τ R = = 3 H (a) Wavefrms f the utput vltage and current f the rectfer are:

2 GEE 401 wer Electrnc Systems Jós, G. Mdterm examnatn Fall 2004 max = = ; mn =0 ; Wavefrm f the vltage acrss the resstance: Wavefrm f the vltage acrss the nductance: L = 0 (b) = = 216 ; (c) = / R 216 /10 = 21.6 A; 0 = 0 = 0 = W. nput ac lne current wavefrm: = 21.6 A; peak RMS value f nput current: = = 21.6 A; peak age 2 f 9

3 GEE 401 wer Electrnc Systems Jós, G. Mdterm examnatn Fall 2004 Apparent nput pwer: S = = 5184 A; = wer factr: p. f. = = = 0.9. S S (d) Wavefrm f the current n ddes 1 and 4: Wavefrm f the current n ddes 2 and 3: Lss f dde: = / 2 = / 2 = 21.6 W; Lss f rectfer: d lss d d = 4 = 86.4 W r lss Effcency f the rectfer: = d lss / = /4665.6= 0.98 = 98%. 1 r lss 2. QUESTON 1 (25 pnts) A sngle-phase ac cntrller, fed frm a 575, 3 phase 60 Hz mans, s used t cntrl the current n an nductr. The nductr mpedance s 10 Ω, at 60 Hz. a) Draw the vltage and current wavefrm f the nductr fr a frng angle f 105 deg. Shw the extnctn angle f the current and gve the average value f the nductr vltage fr each half cycle. b) Draw the lne current wavefrm. Cmpute the rms value, assumng that the rms value fr 105 deg s 0.5 pu f the maxmum current. Cmpute the real and reactve pwer cnsumed. c) Gve the angles fr whch the current s maxmum and mnmum. Draw the theretcal relatnshp between the current n the nductr and the delay angle usng the 3 knwn pnts. d) ndcate the apprxmate harmnc spectrum f the lne current. Add the thyrstr cntrlled crcut requred t prvde a leadng pwer factr. ndcate dfferences wth the nductr crcut. Slutn: age 3 f 9

4 GEE 401 wer Electrnc Systems Jós, G. Mdterm examnatn Fall 2004 (a) Wavefrms f the nductr vltage and current are: (b) Extnctn angle: β = 360 α = = 255 degree; Average value f the nductr vltage fr each half cycle: L = 0. (Average perd starts frm cnductn pnt, namely α=105, and spans half cycle) Lne current wavefrm s: (c) The maxmum current acheves when α=90, and the value s: max = / X = 575/10 = 57.5 A; Thus, = 0.5 =0.5*57.5 = A; max = 0 W; Q = S = = 575*28.75 = ar. Angles fr whch the current s maxmum and mnmum: α max =90, α mn =180 ; Relatnshp usng the 3 knwn pnts: α (A) age 4 f 9

5 GEE 401 wer Electrnc Systems Jós, G. Mdterm examnatn Fall 2004 α=90 α=105 α=180 α (d) The apprxmate harmnc spectrum f the lne current: Addng a thyrstr cntrlled capactr can prvder a leadng pwer factr. Thus by adjustng the frng angle f thyrstrs, the devce can ether prvde reactve pwer t, r absrb reactve pwer frm the system. (lease refer t Lab 1.) 3. QUESTON 2 (25 pnts) A sngle-phase fully cntrlled thyrstr rectfer, fed frm a 240, 60 Hz ac mans, supples the feld wndng f a dc mtr. The wndng s rated 1.5 kw at 100. The tme cnstant s 300 ms. a) Cmpute the delay angle fr rated vltage. Draw the apprxmate utput vltage and lad current wavefrms. Fnd the average value f the utput vltage and current, ndcatng maxmum and mnmum values, and the pwer cnsumed by the lad. b) lt the average rectfer utput vltage as a functn f delay angle, and gve the value f the vltage fr angles f 0, 60, 90 and 150 deg. Draw the apprxmate vltage wavefrm fr these angles. c) Draw the nput lne current wavefrm fr rated wndng current, estmatng the peak value. Estmate the rms and peak values f the fundamental cmpnent. d) Fr rated vltage, cmpute the nput pwer, the apparent pwer, the pwer factr, the dsplacement factr and the reactve pwer absrbed. Draw the apprxmate Q- plt as a functn f the delay angle, fr angles f 0, 60, 90 and 150 deg. Slutn: r = r / r =1500 / 100 = 15A; R = r / r = 100 / 15 = hm; L = τ R = = 2 H (a) = 0.9 csα = r cs α =100/(0.9*240) =0.463 α=62.4 ; Apprxmate utput vltage and lad current wavefrms: age 5 f 9

6 GEE 401 wer Electrnc Systems Jós, G. Mdterm examnatn Fall 2004 Average value f the utput vltage and current are: = 100, 0 = 15 A; max = = 216 ; mn =0 ; = = 1500 W 0 0 (b) = 0.9 csα, α [0,90 ]; (Nte there s NO vltage surce n the DC sde. Thus f α excesses 90 degree, utput vltage s 0) α () Rectfer utput vltage wavefrm fr angles f 0 degree: Rectfer utput vltage wavefrm fr angles f 60 degree: age 6 f 9

7 GEE 401 wer Electrnc Systems Jós, G. Mdterm examnatn Fall 2004 Rectfer utput vltage wavefrm fr angles f 90 degree: Rectfer utput vltage wavefrm fr angles f 150 degree: (c) nput lne current wavefrm: = 15 A; peak RMS and peak values f the fundamental cmpnent: 1 = 0. 9 peak = 13.5 A; = = 19.1A; 1 peak 2 1 (d) Fr rated vltage, the nput pwer = = 1500 W; RMS value f nput current: = = 15 A; The apparent pwer, peak S = =240*15 = 3600 A; The pwer factr, p. f. = = 1500 / 3600 = 0.417; S The dsplacement factr: p. f. dsp. = csα =0.463; The reactve pwer absrbed, Q 2 2 = S = 3273 ar. age 7 f 9

8 GEE 401 wer Electrnc Systems Jós, G. Mdterm examnatn Fall 2004 Draw the apprxmate Q- plt as a functn f the delay angle, fr angles f 0, 60, 90 and 150 degree: α=90 α=150 Q α=60 α=0 4. QUESTON 3 (25 pnts) A three-phase thyrstr rectfer, fed frm a 575, 3 phase 60 Hz mans, s used t cntrl the feld f an ac generatr. The feld s rated at 200, 5 kw and the tme cnstant s equal t 300 ms. a) Draw the utput vltage wavefrm fr a delay angle f 0 deg. Gve the average value, the frequency f the frst 3 harmnc cmpnents and draw the apprxmate harmnc spectrum, ndcatng the dc cmpnent. b) Fnd the delay angle crrespndng t the rated vltage. Draw the rectfer utput vltage wavefrm, ndcatng the perd and peak t peak ampltude f the vltage rpple. c) Draw the nput lne current wavefrm fr rated dc current, ndcatng the peak and rms values. Cmpute the rms value f the fundamental cmpnent. d) Cmpute the nput pwer, the pwer factr, the dsplacement factr and the reactve pwer absrbed. Slutn: r = r / r =5000 / 200 = 25A; R = r / r = 200 / 25 = 8 hm; L = τ R = = 2.4 H (a) Wavefrms f the utput vltage and current f the rectfer are: = = 1.35*575 = The frequences f the frst 3 harmnc cmpnents are 360Hz, 720Hz, and 1080Hz. The apprxmate harmnc spectrum: age 8 f 9

9 GEE 401 wer Electrnc Systems Jós, G. Mdterm examnatn Fall 2004 (b) = 1.35 csα = cs α =200/(1.35*575) =0.258 α=75.1 ; Rectfer utput vltage wavefrm: r (c) erd: T= 1/60/6 = s = ms; v = 2 575sn(60 + α) = 574 p+ v p = sn( 60 + α + 60 ) = v p = vp+ vp = 574 ( 212) = 786 The nput lne current wavefrm fr rated dc current: = 25 A; peak RMS value f nput current: = 3 peak 2 = 20.4 A; RMS value f fundamental cmpnent: 1 = peak = 19.5 A; (d) Fr rated vltage, the nput pwer = = 5000 W; The apparent pwer, S = 3 =1.732*575*20.4 = A; The pwer factr, p. f. = = 5000 / = 0.246; S The dsplacement factr: p. f. dsp. = csα = 0.258; The reactve pwer absrbed, Q 2 2 = S = ar. age 9 f 9