HW3 - Chapter 3: 1'" wright-men 3-1. The secondary ofa...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 3: 1'" wright-men 3-1. The secondary ofa transformer has a terminal voltage of v1 {1} = 232.8 sin 37"?f'v'. The tums ratio of the transformer is iflflflfl (a = 0.25}. if the secondary current of the transformer is is {I} = Tl}? sin {3TH — 363?“) A , what is the primary current ofthis transformer? What are its voltage regulation and efi'iciency'? The impedances ofthis transformer referred to the side are Rq=fl05fl RC=TSQ Jr”1 = 0.225 n X“ = as o SOLUTION The equivalent circuit of this transformer is sheen below. (Since no particular equivalent circuit was specified we are using the approximate antivalflit circuit referred to the side.) 'P 5: pr. r' tilt REP - Ry. + “gm XE“. = If. 1- right..- The secondary voltage and current are T3 = 2813 50° V = 200513" V J? I5. = 5— 315.8?” A = 5i — 36.8?” A. The secondary voltage referredto the side is v; = avg = seam v The secondary current referred to the side is I; = % = ECU.” - 35.37” A The primary circuit voltage is given by n = Ts +15 {Rs + as} Tr = 502:}: v + [2:];— 35.8?” 310.05 r: + fitflfi n]: 53.15am“ v The excitation current of this transformer is _ 53.5.53? v Essen: v Tin Jase In data—turn In = Itr + rM = DEN-5.53.1" + 2.5?91— ass: 21' Lu !4 Therefore. thetotal primaryemrent ofthis transformer is II, =I_.I~ +121! = 201—36.8T3+2.TTZ—?1.9°= 22.31—41.03 A Thevoltage regulation ofthe transformeratthis load is I’}. — arrg 53.6 — so —-><1oo% = — off... so 1’R= x100%=?.2% The input power to this transformer is PH = r31! cos a: [53.5 turns a”; eos [3.2='— [41.03)] Pm = [53.6 1122.3 ajcos 44.23 = 35? w The output power fi'om this transformer is PM = 1ng cos a = [200 t-‘jlrfis A his [sears]: son or Therefore, the transformH‘s efiecieney is u: Eflxtflflt-h: Em w norm =93.-1% PM 853' W A Efl-k‘u'fli SDDD-‘ETT-V distribution transformer has the following resistanoes andreaetanees: Rp=32fl R3 =0_E}5fl IF =45fl Jr'_.‘.=[}.Ufifl Rt? = 250m I“ = SDI-:51 The excitation branch impedanees are given referred to the high—voltage side of the transformer. (it) Find the equivalflit circuit of this transformer referredto the high-voltage side. I'l‘r) Find the per—unit erpti‘ualent circuit of this transformer. I?) Assume that this transformer is supplying rated load at 21'? 1E and (1.3 PF lagging. What is this transformer’s input voltage? What is its voltage regulation?I {at} What is the transformer’s efiieienr'j; under the oonditions of part (s)? Summit (at) The turns ratio ofthis transformer is :1 = RUDD-QT? = 23.39. Therefore: the sewnde impedanees referred to the side are an. = airs? = {aassflooj r2}: 41m I3 = air? = {sassfloos :1]: 50.1 n The resulting equivalent circuit is It 'c 32:: ms 41:11 fjfllfl ? IO M fill-W's I W W a. V]. 330 kn I30 Ltfl "If, D 0 (1'?) The rated k‘iu'fil. ofthe transformer is 10 k‘u'fit: and the rated voltage on the side is Eflflfl V. so the rated current in the primary side is 2'1] kh’A-‘Etlfltl V = 2.3 A. Therefore, the base impedance on the side is 21m _ ifhm = EDflUE‘ _ = 32430;: 11“ 2.3a Since Elm = 2m Elm: the resulting pet-unit equivalent circuit is as shown below: 'x- out mom 0.013 men? 4. W... VJ, 738.125 19.3 7'5 N5 (c; To simplify the calculations: use the simplified equivalent circuit referred to the side of the transformer: T J Ir 2;; you 41:11 J'Sfllfl ? E. 350 km j3U kfl "I. D U The sec ondarj,’ current in this transformer is 2 I I_.I. = z — 36.8?”5. = Tl25— 3-5.8?!1 The secondary current referred to the side is I; = I_-= 150;; 3537': A ' :‘I 23.39 29 rJJ Therefore. the voltage on the transformer is I t}. = v3 +{Rm + JXEQ tr. 1}. = somzmv + [T33 + 193.1}[1joz - 3:315?D A] = 319435053: V The voltage regulation of the transformer under these conditions is '1 _ ‘R = wx 100%: 3.63% I’d) Under the conditions ofpart {c}. the transformers output power copper losses and core losses are: PM = 5 cos 9 = {so tat-rates) = 16 kw / .Ix': PE- : [In {J RR = [rsflrar]: 451w R = 3 = 32901 = “m 12., 250.001: The efficiency: of this transformer is 2'5 W 15.000 H Pour 15.0130 +461— 275 =—><10rrss= PUL1+PCU+PM K100?“ = 95.6% A EDGE-VA 13051 15-? transformer has been tested to determine its equivalent circuit. The results of the tesm are shown below. Eta—circuit teat Short—circuit test 12>; =13D'I.’ Ifsr=13.2"v' Ia: = 1143 A I“: = till] A Pa = 30 W PM: = 2111 ‘5." All data given were tal-ren from the side ofthe trmfornier. (J) Find the equivalent circuit ofthis transformer referred to the low-voltage side ofthe transformer. r1?) Find the transformers voltage regulation at rated conditions and (1] I113 PF lagging, {1) ID PF. {3113 PF leading. (c; Determine the transfornier‘s eficienev at rated conditions and 0.3 PF lagging. SoLL‘t'rorr fa) W _ 0.459. fl _ P’E~|=|G::-JBM|= 23W =o.uolsj.- s S=CD§1IPDC =c55'l#=1 : FDCIOC Laos-trams; Tn =Ggr—J'BM = onolssrz—rsije s = u.mugm__,-D_Dmg¢3 S Re: =i=lrsso I' . 1 A“ =—=534o Bill 31} SHORT CIRCUIT TEST: 13.21; 2: :3 + X = =1qu lml IE? J ml aoa P 2am: fl=cos'] 5’: =cos'J—v. =7 . ° F5315: uses-flag} em = am Harm = 12055.? o: 0.553 Homo. REQ=U558£1 rm: 12.1mm To convert the equivalent circuit to the seconden‘ side, divide each impedance by the square of the ruins ralio {a = Bil-'1 15 = 2:1. The resulting equivalent circuit is shown below: “I.” 3 ix '5 —- we: L' H _|.' —II- REQJ; = fl XEQJ- = 9 Re"? = 441 o XMIS = 134 :1 r1!) To find the required T.a'oltage regulation, we will use the equivalent circuit of the transformer referred to the secondary side. The rated secondary current is _1ooova 1151: We will now calculate the voltage referred to the secondary side and use the voltage regulalion equation for each power factor. (1} as H Laminar: T? = 1'3 + amt? =115mc 1.: +{o.1ao + jfl.532 okay—363?: A] I E =ssoa .- 1}. = 11831131”.I RR = $Xlflfl‘lfi = 3.3% [2} LI] PI: ‘5'}, = T3 + amt¢ =115auc 1: +{o.1ao + filfifl missz A] .- 1}. = 1 16.352 .2 8‘: V 31 1163—115 TR: X1U-U%=l.l‘}fu [5} 0.3 PT Leadinw: 1;. = 1'3 + amtY =11szoc 1; +{o14r} + filifl rijfisjzsssrc- A} T}. =113'r3fil2-il-zI ‘L-T 1133—115 115 r'r'; At rated conditions and I33 PF lagging, the output power ofthis Iransformer is PM =1'"I.I~I_.‘.eos 94115113.? Alas} = son or 's-‘R= KID-0'!" =—l.5% The oopper and core losses of this transformer are 2 PW = If em. = [so A] (a. 14s or): 10.5 w Fr ‘ .‘2 I; f] = {113.31%} Pm = = 32.0 W Rf. 441 [1 Therefore the efficiency of this transformer at these condilions is rg—PAKIUIH’ = 3mm =9¢.9% _Pm+Pw+Pm soownoswasow A. single—phase power system is shown in Figure PS—l. The power source feeds a Hit-INA 14-‘24-15.’ transformer through a feeder impedance of 33.2 +j14-fl $1. The transmrmer's equivalent series impedance referred to its low—voltage side is 0.12 +_;'C|_5 £1. The load on the transformer is 9U kW at 1135 P1: lagging and 2300 11'. “Hill-l- r'J-tlilfl nun _.-:-sfl Loud ‘-.||.| HP: (mi l'l- l:I.:QRII'IR 1,..- unm- -‘I'IIIIII.'$ loo-Isl {Itlmmlsuun IIMI 'l'm mtnn'm-r |_c.:..| (a) TWhat is the voltage at the power some of the system? r12) What is the 1.roltage reg'ulalion of the transformer? (c; How efficient is the overall power system? SoLLI'rroN To solve this problem we will refer the circuit to the secondary {low-voltage) side. The feeder‘s impedance referred to the secondary side is ’ {2.4- It? 1'“ = k 141a.I [3329+ moo =1_12+ 4.1151 s r 32 The secondary mirrent I3 is given by I _ «30 kW 3 [330:] visa] I... = 43.434 — 35.33 a =43.48A (a) The voltage at the power souree of this system l{refeued to the secondary side) is a Tm = v; + [32h + LizBEE Tm = 33310on v+(43.4se— 3333 A][1.12 +j4.11nj+{43.43z— 33.33 anoufios o] Tm = 244153.331! Therefore, the voltage at the power souree is - - 143:3.F v. = [34415333 v} m _.- = 1424/13.?" er r13) To find the voltage regulation of the transformer. we must find the voltage at the side of the transformer {referred to the secondary side] under filll load conditions: 3'? = v. +132m ‘3}. = 2300£0° V + {43.485— 25.8: £143.12 + j‘flj fl] = 23 1450.43" V There is a voltage drop of 14 1-“ under these load oondilions. Therefore the voltage regulalion of the transformeris 7‘ _ ‘R = w x1oo33= {1531: 3300 (r; The power supplied to the load is Pum= 90 kW. The power supplied by the souree is Pm = V. I3 cos a: {3441 3143.43 Aleos 33.33: 33.3? kw am Therefore, the efi'ieienev of the power system is If: FEE><1tlltilttia = W—melfl-flfia = 914% 9323:1553? LN EL li-ka-‘Pr SEMI-'23 EI-‘v' dislrihmr'on transformer has an impedance referred to the of Bi] +_r'3 I30 51. The components ofthe excitation branch referred to the side are RI- = 350 kn and XM : TU kfl. far} If the primary voltage is 796'." ‘5." and the load impedance is 2L = 3.2 +j1.5 [1. what is the secondary voltage of the transfornier'.’I What is the voltage regulafion of the transformer? r12) If the load is disconnected and a capacitor of—fij it is connected in its place. what is the secondary voltage of the transfor'rlier'.‘I What is its voltage regulaiion imder these condilions'.‘I SoLLIrroN (a) The easiest way to solve this problem is to refer all components to thepn'mmjv side of the transformer. The tlu'ns ratio is a = 31331-230 = 34.?3. Thus the load impedance referred to the side is 2; = {343333 [3.2 +J'1sr1): EST-'1 +1131er The referred secondary clu'rent is I ’ _ HEAT V _ MEAT V -" [80+j3U-Ufll+{3371+J1'1815fl] Mamaaam and the referred secondary voltage is = 1.?M—23.2c A a T3 = I_... 3L = [1385— 282“ AH38T1 +J1'1815 fl] = Tfilflz— 11" V The actual secondary voltage is Thus a 1;. _ asisz— sicv T9: F =213.E£—3.1‘=V ' or 34.98 The voltage regulation is 1.13 = MX 100% = 42% Tfilfl r12) As before: the easiest vray to solve this problem is to refer all components to they-Emmy side ofthe transformer. The turns ratio is again a = 34.73. Thus the load impedance referred to the side is z,_ = [34.73fl— 3'3.5£1]=—j423a1-fl The referred secondary ciu'rent is '= :ssxgo v = :ssxgo 1- =2-025flE-Ecfi {so+;3oooj+{—;423m} 39351—38_8°fl and the referred secondary voltage is I .i‘ T3 = In. 2L = {2.2558333 AH—j4234 fl] = SST-'31 — 1.2” V The actual secondary voltage is thus . Fl. _ 1c :" =L=w= ganja—1.1%" a 34.73 The voltage regulaliflll is T9 I5? - 35 "."3 Ti? E’R = X1009“ = —'.-'.[}'.-'% ...
View Full Document

This note was uploaded on 01/03/2012 for the course EEL 3211 taught by Professor Staff during the Fall '08 term at University of Florida.

Page1 / 8

HW3 - Chapter 3: 1'" wright-men 3-1. The secondary ofa...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online