EE239 Exam Formula Sheet - 'J' W ‘ .Llucceo ,4 2: VOL/U...

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Unformatted text preview: 'J' W ‘ .Llucceo ,4 2: VOL/U {o :— ‘WE- id” item” #707 Z 4 Ms J __i\.’V\/\-——~ L ,_ L 2> 2 7 e e )1 L». i4 r _ 4 e w — LIE ’ a 5 Y E 0&0“ Q saw/L" 1 5) 3:0 v; : 1) : “mm? A}: .1? i/M‘L' '“JL Izaq'wfl lg .iicx; \ : l 7 WE” , H2, 3 Ola ‘ a / + / 2 nuivwifi’is (41$ D 10m H5 eulka 2 7» 13+”, looiJ(1q|77zu L)=mo+j '00: fL/lfi‘Jl ice/w he») 2/: it: [La e goo-7034115,, 2 MIKE 5) if: y: 64;)de ZOVDEII+J(.‘J<MK’CI_()ifllLJCB-B/‘Vhs f L ‘ I: 73 .7 fl N 0‘, we "‘/‘ "WI" 5 lilvovolé huge-3‘3 g: 425;; 23;: (o-‘fljo-H) K’Z Lt \/:\'];,;(';I‘J:_ILl-’ejidzg [ i- laws (L: éjol #«K " ‘ 5' ~ . 0 ~ - A»; Kw) - U714 SE»- 943%, f , \ r—f’ / "a Q‘ 2 _ \/ P 1 U4 ‘9 9 ;7 Yv L) L ZR ? (w) LDC, ; \ 7i”: 7 \Irmi : \rf? E, :W? M]? 5 P Case : ; : P? PARAMEI'ER SVMBOL UNITS pARAMETER SYMBOL UNITS Impedance 2 g Admiuance Y =% 5 (Siemens; Resislance R n Conduclance G =1fi s (Siemensjr Reactance x 9 Suscepiance B =)1( 5 (Siemens; 2 = Fl + Jx I Y = G + 13 r L; r 12 v = l 2 I = 1 V = I i i i i z 7 I = YV ZR=R zL=0+ij Z°=° VR=1 VL:;L Vc=+imc H 02L xe : xL = mL lmL BL = we ADMI'ITANCES IN SERIES ‘ I. : (1- H 3 ("lo’éé A IMPEDANCES lN SERIES z=zi+22+23+___ Y=L+ 1 + 1 +._. Y1 V—z Y3 L LOC IMPEDANCES IN PARALLEL Ht 7‘ ‘00? s $1 l ADMI‘ITANCES IN PARALLE L H 11 L , IA 2 = 1 + 1_ + 1 + / Z1 22 23 Y = Y1 1. Y2 + Y: + it]; _ P = 12 2 ‘7 R P = v G M") ‘7 1" ,7; M u f n 2 j 7 _ I _ f e B~i$u 16W {6” )b"‘(v—H0.la(y—C‘I.3‘/ \ , If the voltage phase diagram ie tnullipllcd by 1 results in a poxvcr triangle as illustrated luelowv 7L; \Ym/n: g V ‘p 4 \‘ {‘4’ L V. I are RMS vames v. 1 are PEAK values ?\“ MMMLA we? , g E Or. be» Kata/13 2009 EE239 EXTRA AC CIRCUIT PROBLEM; 3/ "‘ i E POWER TRIANGLE POWER TRIANGLE The above power triangle can be aim be expretsed as t’oiiowez P= S 0036 e, Q=Ssin6 ° ta‘ne=Q REAL P P S=1/P2+Qz POWER TRIA NGLE LOAD A Load A is an induction motor with an efficiency of 85 percent, a powerfadur of 0.75 lagging, NOTE: ALL POWER QUANTITIES ARE GIVEN IN RMS VALUES OF VOLTAGE AND CURRENT. and an output load of 30.0 LAGGING POWER FACTOR LEADING POWER FACTOR ' LOAD - net inductive load — net capacitive load Lnad Bis a synchronous motor drawing 22D kVA and its load current has a leading phase _ e is positive — 6 is negative ,dkplaremenr with rpq-ial‘t to the line Voltage of 255 electrical degrees. LOAD C Load C is a lighting load of 13.2 kW at unity power factor. 3?:Um‘718f‘? " Q2000 cesgzsfa) : [LI/<50”! w Fox/oer drawn bv’ gyko. QC) Li) 9 _ O . P C050“ COS(2§<63:0.7603 I OE SYMIQ'ILAcHE Muirgr ‘45 We rm,“ ‘lO Nev L47: I’K 5)". 53m: .‘Kmuzt V52 2009 EE 239 FUNDAMENTAL$ OF ELECTRICAL ENGINEERING 1: S1, pf1 P2, pf: 5:, pr: Load #1: 51, kVAI, power factor p11 lagging (e.g., induction motor) Load #2: P2, kW, power factor p12 = unity (e.g., heater) Load #3: 53, power factor p15 lagging (e.g., fluorescent lights) METHOD CONSTRUCT THE POWER TRIANGLE PART BY PART AS ILLUSTRATED BELOW. "5332“ SPECIFICATIONS REAL POWEFI REACTIVE POWER S1, p11 P1 = 81 x pt1 Q1 : 81 7x sin(é13 7 2 P2, p12=1 P2 02:0 3 7 $3, ptai P3 : 53 x pm 03 = 83 x sin(93) 2Pi=PT=PI+P2+P3 ZQi=GT=Q1+02+Qa ST = [Pr2 + 072]°'5 = VRMS x IRMS e = tan'1 (QT/PT) Power Factor - pr = «5(6) = PT/ ST NOTE: If 01' is positive, then the power factor is lagging If QT is negative, then the power factor is leading It the power factor of the composite load is lagging and excessive (i.e., below 0.90 pf lagging), the composite load can be corrected by adding capacitors to the circuit to mak‘ the power factor unity (i.e., the most emcienI load), that is, Q(capacitors added to circuit) : GT so the net reactive power of the circuit is zero and the power factor at the composile load is unity. EXAMPLE #2 — composite loads A 300 kVA motor (p1 = 0.5 lagging) is wired in parallel with a 200 kW heating load (pf = 1.0). - I leads V - Q is positive - Q is negative Reactlve power is CONSUMED 0R ABSORBED Reactive power is SUPFLIED DR GENERATi P = Iamsz R (Watts) (1 = IRMS2 x (VAR) P = IRMSZ R (Watts) EXAMPLE PROBLEM #1 Draw the power Iriangle tor a 4 kw load with a lagging power factor at 0.30. Given: source voltage 120 VHMs Find the load current. P = Real Power = 4,000w or 4 kw Case = 0.50 lagging therefore a s cos"(o.so) = 1-36.49 Q = Ptane = 4 lan(36.9§) = 3 kVAR S=P/CosB=4/O.B=5kVA Q [load = S / VRMS = 5,000 / 120 = 41.67 A HMS A P=4kW ‘7) @We : LOSS/1C0 :Ctri—fl/ (ctJC\‘/\KJML‘_IV\I'Q:3 Firefliflefp a? m. Car-elem z’15>,‘l,\8:’7”{l3 {fwd/s): ‘QTBGS'C? UAR gem/5] {Ale o? Peacil‘v't Vow? 15% 4m focLomfimacdeol Ly Ct (Aft 6% Q 1 C wwwew (4 WM PF. ' ‘ 2 3» s— ; t T 7V1” 1 $! A v (1‘10 ‘v flee SENS ““fl 0f: Laid-t3” ,‘V‘ Ode; hi CtLLu’ix/i UV.\'+)/ ?(.I I («)L \ , . > C I M {7374 (75W; Q0 5-15316 Lictwvirwc “Add *0 waged Mvv'i/ PF Par "HQ, Q/Mvvx MOI}; me 95:1..‘7Ktyual a) :2rm9 EE 239 FUNDAMENTALS OF ELECTRICAL ENGINEERING sm e aim KVA he zoo rw mm: in 5 Inu‘lnil’ln pm : 1.n (a) Draw the power triangle (b) What is the total current required if the supply voltage Is 10 kv RMS? Considering the motor load Sm s 300 kVA pfm = 0.5 lagging (l.e., consumes +0 VAR) cos(6m) s 0.5 em oos"(pfm) = cos"‘(o.5) = soE Pm = Sm x cos(em) s 300 x 103 x cos(6clQ — 300 x 10" x 0.5 = 150 x 103: 150 kW em = Sm x sin(em) = 300 x 103 x sln (605) = 259.8 x 103 s 259.3 kVAR Considering the heating load Pvt = 200 kw pfm = 1.0 (i.e., consumes or supplies no VARs) cos(6h) s 1.0 an s cos"(pth) = cos"(1.o) s oQ Considering the composite load Pr=Pm+Ph=150kW+200kW =300kW Q'r = Qm + Qt. = 259.8 kVAR + o = 259.8 kVAFt ST = [PT2 + QT2]°'5 : [3002 .2593 2]“5 = 435.9 kVA Power Factor — pr PT 1 ST = 300 /435.9 = 0.803 lagging at = cos" (0.803) = 36.65 m E a :5 .. 5 g i as g u e a P», 2 200 kw P.“ : 150 kW PT = 350 kw (b) Calculaling [he lolal load current IRMS ans x [HMS e 435.9 x 103 IRMS: 435.9 x 103 e 43.59 A 10 x 103 SIGNIFICANCE VARIABLES IN POWER TRIANGLE POSITIVE 0 |$| = 1/P2 + Q2 a . E <3 5 P = VRMisMsCOSe g + + Q = VmIRmssine LAGGING PF =_P s NEGATIVE e lsl = 1/P2 + Q2 P = VRMSIRMSCOS(-e) POSITIVE REACTIVE e = tan'1 [ 'ulo _. - Q = VHMisMsSiI’K-G) LEADING PF = E S NEGATIVE REACTIVE ole: - When 9 = 0 then Ihe power laclor PF = 1 (i.e., unin powei lacloi) - When 9 is not equal to zero, the the PF will be less Ihan 1. Q is considered In be posilive for a predominanlly INDUCTIVE ciruil ( i.e., l lagging V) ...
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This note was uploaded on 12/13/2009 for the course ENGG ENGL 1199 taught by Professor Bruseski during the Spring '09 term at University of the Bío-Bío.

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EE239 Exam Formula Sheet - 'J' W ‘ .Llucceo ,4 2: VOL/U...

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