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Unformatted text preview: Chapter 3: 1'" wrightmen 31. 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 = DEN5.53.1" + 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: Eﬂxtﬂﬂth: Em w norm =93.1%
PM 853' W
A Eﬂk‘u'ﬂi SDDD‘ETTV 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 highvoltage 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 = RUDDQT? = 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 ﬁllW'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 petunit 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— 35.8?!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 tatrates) = 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 EDGEVA 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 talren from the side ofthe trmfornier.
(J) Find the equivalent circuit ofthis transformer referred to the lowvoltage 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 Laostrams;
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: usesﬂag}
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: X1UU%=l.l‘}fu [5} 0.3 PT Leadinw: 1;. = 1'3 + amtY =11szoc 1; +{o14r} + ﬁliﬂ rijﬁsjzsssrc A} T}. =113'r3ﬁl2ilzI ‘LT
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= KID0'!" =—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 HitINA 14‘2415.’
transformer through a feeder impedance of 33.2 +j14ﬂ $1. The transmrmer's equivalent series impedance referred to its low—voltage side is 0.12 +_;'C_5 £1. The load on the transformer is 9U kW at 1135 P1: lagging
and 2300 11'. “Hilll r'Jtlilﬂ nun _.:sﬂ Loud
‘.. HP:
(mi l'l l:I.:QRII'IR 1,.. unm ‘I'IIIIII.'$ looIsl
{Itlmmlsuun IIMI 'l'm mtnn'mr _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 {lowvoltage) 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—melﬂﬂﬁa = 914%
9323:1553? LN EL lika‘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 = 31331230 = 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+j3UUﬂl+{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 theyEmmy 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]=—j423a1ﬂ
The referred secondary ciu'rent is
'= :ssxgo v = :ssxgo 1 =2025ﬂEEcﬁ
{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“ = —'.'.[}'.'% ...
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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.
 Fall '08
 Staff

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