JEE3320-Topic 02-Power and Power Factor Correction

JEE3320-Topic 02-Power and Power Factor Correction - Power...

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Unformatted text preview: Power and Power Factor Correction 1. Phasors and Phasor Algebra-Required Knowledge 2. Phasor Representation of AC Waveforms t ‘ F0 V; L, WA 0 Polar and Rectangular FEW?“ Trl‘fi WWW 9 Polar Form 0 Rectangular Fen/HA Phasor Addition Phasor Subtraction Phasor Multiplication Phasor Division 9 Phasor Expotentiai Form 3. Phase Angle-Where does it come from? 4. Apparent, Real, and Reactive Power 5. Derivation of Power, p(t), Average Power and Power Factor 6 . Power in Complex Form 0 R o R-L o R-C 7. Combination Loads 8. Single Load-Finding a Capacitor 9.' Multiple Loads —Rmd.Cv-t £L. Calmitolr 10.Voltage ImprovementJ .00. T) hora:er Mir. Phage» I've/CL 'RQQUQVGQL [WM \1 3 ?M°V ' VbQMYI‘QkM o‘F W‘awe firm “14"?01‘14" i'fivm — . [email protected]fiu(av PW” 0.+&;{9 ‘ WMQW Alia-mm <1= o~I+gbl+aL+atoLsccwa+a .m‘) " Mariam v‘ gub'fwdnhx a : (QM m —~ (afifi‘ob : CQ1’&:>+éLb|“bzr) ° P hi! how WLLJWFY [JUIHZLW a '3 @fl‘fi [0'5 (afl‘éL’Q 7: @1211." 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QTILQC 2:. @4 =1 4:3.IKVA-a Mac:— wmtwm %Lv:”\fi I)“ “QC. ' @C-nflgmfirnvma 3 ' ' ' fin éorreqtmr aafiomim (Eu-m: mamean 7L1<e4erc Maw/d w-EfoEerJE/Ufié'figf’m m” (yaflfifs we flgéO/e& mu .Eu: «aw 3:) DC! DEERE . ’JUAHE 1:: mm on. L- . .1 5 5 "5 5 % 5m EHEE 5L! SHEETS - K t? . W - r: 1' & Harlan.) Brand _ An industrial load confirm of: One 50-hp induction motor; load, 30 hp; efficiency. 0.86; and 'pf, 0.70. One 100-bp induction motor; load, 75 hp; efficiency. 0.89; and pf, 0.80. . . . Two 15-hp induction motors; load, 15 hp; efficiency. 0.92; and pf, 0.35. . One 300-bp induction moror; loud, 310 hp; efficiency, 0.92; lilo-85. ... . . _ Lighting load, 33 kw. Synchronous motor (to be added): 500 hp, 0.80 pf, leading current; loud, 300 hp ; and efficiency. 0.925, exclusive of field loss. Determine for load, over-all, (a) kilowatts; (b) kva; (c) kilo- vars; (d) power factor. _________._._____2_———-—-—-——-—— (1) E (2) (a) (4) c5) (6). (7) ; (B?! ; ~19 ——e—§—-———— - I ' Loud Input Factors Input liming '-—--—"—" Em- eieucy _ . 5- hp ‘ kw kw PM.“ kva Kvar -renctance Given) ,. (em-n) 0-74“ (cranial/w (ii/(951mm) . X (2) :l'US 5 sin 9 _ l x _ fl_ 1—50 hpi 30 22.4 0.35 '2u.usru..-nn :15 3.52;: 2mm l~llltlhp , 75 50.0 0.80 03.000 HUI) m 75.75 ~l'.‘ ‘2“ 2-15 hp; 15 11.2 0.92 ' 12.200550 52: 14 35:9 7.5a; recoup; 310 231.5 0.92 :252.oo-o.35_u 527 29000' 150.00 1-33 kwl —~ 33.0 1.00 i33.000.000 33.00; 0 ' = " ; a9 5 437.35 3 i- l 4’ .3' §373.W0.80>D.00 455.0 : +280.00 1—500 hp : - - ———-—-— 1300 224 0.925 :242410- . . Tom r 5523.25; A 42.64 __________________———,——-——-——'— ';{{><0.746) 373 1.000 This combination can be udiusted to uuigf pf where kw = km. . Otherwise lava. = W/(kxv)! + Lunbalsnced liver)! The above figures indicate almost unity power factor so than kva in case will be very close to 629.5 kva. - - ({3quch ‘(br BMK S “(HERVE)” awh‘cafmm Vasfm‘ms memluofikm and Eyai‘m f‘e7uz'rafiéé'xfi‘i; 7% defer Mme .mfi eWed'ive, Isa 79%.”) Mum: meet/:5), Md Magma :7 Prof‘ed’fis-n . ' r3 " 0 Emdlflnemij Cafcmdwg are car/mac! ‘5 M #1 aubmfiw 1‘15 PM ( aisMVW Meg/s mwfeffigg flag}, 2) fflfw r1629»? MM! {Fm-15w" ~62; dainty-l oioflmum ,bwez47‘g €97“ WMLWM/u MQ/ Va/7/217¢, C’JWeaUCb—n ' 3) KL?" 7%; and m” 7% Me. ficfrérwé. mm‘mum W/éye @kmégfgggm ‘ ddrfictaLom awn/9a Mace! M pawl/M Md fie. l/téV-ZdrV/lm/d 30¢: Aw? oftij cit vaing Vwfiofieza. QC: 3%: 75%:er V17 diet 52: vi“ C a fédfi aoghdhm mm [ahead in safes. 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Java (Vex! {A sdechma Q {baa Q0 V g faéfwdud aoJFACL‘dLOV me’ixgcfilgfll . . . a ami‘t‘rwa 05 au WW v Wmfirfifi‘cuvwaflfi‘ J 'FMH' QUVVflwT" ' “12mg. wrafinre. cum/a fia-wdmai‘lgn “Vol QM M 6 CL‘QLUV‘S ‘5 Mm (Us; 6L t'/e1,.war‘7L *’ 0015/. us {aware E E E T61“ {break Gwd‘m U90 2, m WWW Qufabmx’fi 1L5 chm wit/d Hear grader m {554% 5P %_ miei ea Many? BEEF. rammed 5/5212; ,5 2902-3 4* ,o’, (my mad“: {Mfr/as lerK Kvgfiumrfl E ZLII‘JK— EVENIL 4.49 A two-element Mfr-volt 5-way pdyphnse watt—hour meter, having a. basic watt-hour constant of K5: = 9g, is con- nected to a. three~phase three—wire circuit through lOO/é—amp current transformers and 2,300/115-v01t patential transformers. A stop-watch check shows that the meter disk is making 15 revolutions in 50 sec. “M is the kilowatt load on the circuit? K (3.600x15x?5) __ _ _ 4.49 PM:3.6mXRX€-=—_T——=l20“att§ Puma = P1,: X (CT ratio):;PT ratio) = 720 X 20 X 20 = 233 kilowatts E l E PMN. MS 89. 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Mfm m5 )4», #fi G/(le'dcff‘fm 0Q" 5 W‘Oaaoa ado B aj_+LL su-loa‘faj-bm } 2—) MKWL pr-g'vavT 4.952.414,» ) ‘5) av (with , 57h auwwefl de/m 191 a. ch’ar me: 0-. Maj? flaw? flaflr aéiarmér We Mm"! ( 05mm adaééoag, V‘fieéFV/incffig Wt:an mQ-d'fle ijfficf‘fm“ bad: I ‘11) ‘VLL “gamma .7542, \er Hwa-c Nae, UJ EH be. 43¢; w} +1: “fig reach/mew aflfh‘; U V“ C L” *F Gama: +29 Harlem; m1) new:le :64 M cr—La. GaEMCL‘fi) r awmfl" (mums ic-dra dccfi WEE" ‘(fili tr Hem/or VG/QYL: 2114575? mwita LL, Fvelalrmucéémf a? ‘fLL (Odd, QEVCUWIML‘LJLS ffmfieg-r‘arf’ “I'LL hub/W OS; fw-L Cmrqoihvg cam—A Aficreagag Jm mu wee. 4. ‘Kz. ° Gmsfci-exr ‘H-Lttwwti" 5W M 3 Wfiaflanfleflmd IL50. cpx‘qic pm U?) 4' g M”: Mole—Q 5‘9 a. 25197.2qu 51/ Olin” VLrJ +Vn4‘23 rim-VFW NAM ' _ _ Faw- a. 3g! @Qffl’flérhfllf'. “ Pam. 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Solution: The inductive reactance of the transformer is 0.01244 0 as given in the problem statement. Direct application of Equation (15—19) results in 300,000 var (480 W ] x 100 = 1.6% gal/gm}; = PW W’ eqsummm — m Wai’fmwtew * "712w. WmHWEGFILGLh (‘5 am 2h? Mle mdmk Mé’cwri‘rfl r14 firfiflcfi YKSWf‘OkW T 1“ W '19».(’-,="“’-.; Hz . and MaUCLHe. am I .. gaggiwcggxauni [nefgbwrk $1 (3, NW. awé‘mi ‘fafiw‘fi m» 3...; ‘ v “1 wt. {harm 5‘ ‘1‘» C 4 fig “Pk‘hwth— «319 “Ht. mum/Ewe Gail dkfomds chm:ch M‘W'fe. M041} Lflafl MP“ l£fid Lab , fl. < t r I I ‘ l +— 4—9 w a a? T L ,. 1““) '7 “Wig-:5? “W flageflmm "‘3 Mfwm‘mf a ' VOW “Q” a “4&3” W {if-“Elfin :5 flaw“; gag. Elr'i.*x'§':”w 'fP‘V\§#‘T$"'-.‘=. QR“ Q‘m‘é’"m§§"&uy\i€‘g wafim-g G I I ’ M a 76“ M “C” Q‘l‘r‘wfil"; fi‘Q— W “1‘84; M‘si Ga’wremf “Emma raw? wagwiz: L-wf’é’ai». vi L" ltr. v\ :5 “Li‘i.m «4m "'2’ I J i . 3;. «~53 r- Wan-J 0% $555; {flag}. scinrfii’igr‘. 1%-F’jggwegw ,5 :_ gq-agr‘jn L EEG Vti’i‘akfi. 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JEE3320-Topic 02-Power and Power Factor Correction - Power...

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