Chapter 4- Transformers

1

Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

2


By

Assad Abu-Jasser, Ph. D.Electric Power Engineeringwww.iugaza.edu/homes/ajass er

[email protected]



3

Chapter Four

TransformersDr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

4

IntroductionA Transformer is a device that involves two electrically isolated but magnetically strong coupled coils These are primary winding connected to the source and secondary winding connected to the load Induced emf is proportional to the number of turns in the coil. If the secondary voltage is higher than the source, the transformer is called step-up transformer. On the other hand, a step-down transformer has higher source voltage than the load voltage. One-to-One ratio transformer is called isolation



5

Construction of a Transformer

Transformer core is built up of thin lamination of highly permeable ferromagnetic material such as silicon sheet steel The laminations thickness varies from 0.014 to 0.024 inch to keep core losses to a minimum. A thin coating varnish is applied to provide electricalDr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

6

An Ideal Transformerm

=

sin tm

e1 =N 1

cos t

An Ideal Transformer has the following properties: The core is highly permeable i.e. it requires a very small mmf to set up flux The core does not have V= 1 Ni11 Ni2 2V 1 or any eddy-current Z2 = 2 = 2 hysteresis loss I2 a I1

1 E 1 = N 1 m 0o 2 E 1 = 4.44 m 0 o fN 1 E 2 = 4.44 m 0 o fN 2 d e N1 N V 11 = = 1 E1 1d = = =t a V 2 E 2 N 2d e N2 I 22 = = d2 N1 = = a t I1 N 2e N 1 1 = * 1 == a * V 1I 2 =2e2 N 1I= V 2I 2 V V 2I 2or 1 1Dr. Assad Abu-Jasser, ECE-iugaza

aZresistance of each 2i i= 2 Z 11 = The1 2 2

i 2 N1 The = core exhibits no a flux = 1 Z 2 leakage i.e. the flux is Z i 1 =21 Na 2

confined within the core windingElectrical Machines (EELE 4350) is negligible

7

Transformer Polarity and RatingsTransformer Polarity RatingsThe nameplate of a transformer provides information on power and voltage-handling capacity of each winding A 5-kVA, 500/250-V, step-down transformer has the following Full-load power rating is 5 kVA or the transformer can deliver 5 kVA on a continuous basis Nominal Primary voltage V1=500 V and nominal secondary voltage V2=250 V Full-load primary current I1=5000/500=10 A and fullload secondary current I2=5000/250=20 A

The transformation ratio is usually not given by manufacturer but it can be calculated Machines (EELE 4350) Dr. Assad Abu-Jasser, ECE-iugaza Electrical

8

Example 4.1The core of the two-winding transformer shown is subjected to the magnetic flux variation as indicated. What is the induced emf in each winding?

F r t= .1.0 2s o 0-0 .1s -06

0055tW b .1 = .4 tW b e ab a b e ab a b e cd c d e cd c d d d = ba ab =ba ab = N e = t 0 0 W dt = .0 9 b d = 20 =.1 5 90 V = 2e =.4 0 V 5 3 V 0== 0*0 0 e 0* 0F r t= .0 -0 s o 0 6 .1

d d =cd N cd = dt d t =50 = 5 7 2V V =5 = 5 5 2 00*0 0* 0 .1 .4 5

a b

c d



9

Example 4.2

An ideal transformer has a 150-turn primary and 750-turn secondary. The primary is connected to a 240-V, 50-Hz source. The secondary winding supplies a load of 4 A at a lagging power factor of 0.8. Determine (a) the a-ratio, (b) the current in the primary, (c) the power supplied to the load, and (d) the flux in the core.

a =1 0/7 0 0 5 = 5 .2 I2 4 I1 = = = 2 A 0 a 0 .2 V1 20 4 V2 == = 10 V 20 a 0 .2 PL =I 2 2 = s 1 = V c o 2 0*4*0 3 4 W 0 .8 8 0 = m E1 20 4 = = 4 4 fN 14 4*5 1 0 .4 .4 0* 5

7 1m b .2 WElectrical Machines (EELE 4350)


10

A Nonideal TransformerF N nideal T or o ransform er

E1 I 2 N 1 = = = a E2 I 1 N 2 Nonideal Transformer has the following V 1 = 1parameters E + R 1( jX 1 I 1 ) + V 2 = 2 R 2( jX 2 I 2 ) E + Winding ResistancesLeakage Fluxes Finite Core Permeability Core losses (Eddy-Current & Hysteresis)



11

Example 4.3

A 23-kVA, 2300/230-V, 60-Hz, step-down transformer has the following resistance and leakage-reactance values: R1=4, R2=0.04, X1=12, and X2=0.12. The transformer is operating at 75% of its rated load. If the power factor of the load is 0.866 leading, determine the efficiency of the transformer.

P 14938.94 = = 0.971, or 97.1% = VE = = Pin * Z 15389.14 30+)(0.04 j 0.12) EV ++ I * 2282.87 2.33 (7.5 30 )(4 12) I j + Z = 230 (75 = + +1

I 223000 II 2= = =7.5 30 = 75 A,I= A *0.75 75 30 A 1 2 a 230 o Z 1 2 =R + + jX 4= +j0.04 j 0.12 Z = R1 2 jX 1 = 12 + 221

VE = 2269.578 4.7 V V 1 2 = 228.287 2.33 * Po = Re[V2 * I2 ] = Re[230 * 75 30 ] 14938.94 W = 2300

2

1

2

1

2

Pin = Re[ V * I ] = Re[2269.578 4.7 * 7.5 30 ] 15389.14 W = 230 1 12Dr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

a=

= * 10,E= 1 a * = E

2282.87 2.33

V

12

Finite Permeability

When Unloaded Transformer draws excitation cuincreases the load on the transformer rrent The I = I c + I m , I m is the current increases secondary winding magnetizing current The current supplied by the source increases E1 Ic = core-loss current The voltage dropc 1 R across primary impedance increases TheE induced emf E1 drops X m 1 = 1 magnetizing reactance The mutual flux decreases because magnetizing current jI m dropsDr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

13

Example 4.4

The equivalent core-loss resistance and the magnetizing reactance on the primary side of the transformer of example 4.3 are 20 k and 15k, respectively. If the transformer delivers the same load, what is its efficiency?

E 2282.87 From Example 4.3, we have2.33 I c = 1= =

V 2 = 230 V I a = 10

Rc1

m

=

E 1 2 2282.87 2.33 A = = jX m 1 j 15000

I

75 30 =

20000

0.114 2.33

A

I =I c+ I m= 0.19 50.8 11 p

E

2282.87 A2.33 =V W

87.67 A 2 228.287 2.33 E = 0.152V p

V

I

7.5 30 =

A

I =I = + 0.19 = 30 28.57 Po = 14938.94+ II Z = 7.52282.872.3350.8 7.53+ (4 j W V =E + + 7.53 28.57 1 1 1 1

A

12)

V 1 = 2271.9 4.71

Pin = Re [ 1I 1*] = V 15645.35

=Dr. Assad Abu-Jasser, ECE-iugaza

Po 14938.94 = = Pin 15645.35

0.955, or 95.5%Electrical Machines (EELE 4350)

14

Phasor Diagram

E 2 = V 2 + I2 R2Dr. Assad Abu-Jasser, ECE-iugaza

+2 jIElectrical Machines (EELE 4350)

15

Approximate Equivalent CircuitZ e 2 R e 2 jX e 2 =+ R e 1R + 1 = Re a

Z

=R + jX e 12 1

X

Rc 2 Rc 1 =a

= R e 1 X RX+a R 2/ =+ a 1e2 2 1

e2

/ 2

2 2

X m 1X m a1 e 2 =1

X

=X +a X /2

/

2

2

2

The low core loss implies high core loss resistance The high permeability of the core ensures high magnetizing reactance The impedance of the parallel branch across the primary is very high compared with Z1 and Z2 The high impedance of the parallel branch assures low excitation current and thus it can be moved as shownDr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

16

Example 4.5Analyze the transformer discussed in examples 4.3 and 4.4 using the approximate equivalent circuit as viewed from the primary side. Also sketch its phasor diagram.

2 = 2 V 2' =69.594.7 1 = 3 134.7 00 V V 0*2 00 2 0 A 3 I c =a 2 = 0 .1

200 00 I p =26994.7A 7 0 .53 2 .5 Im = = 0 52 8 .3 A .1 1 5 j a 0=+ 1 R0 2 5 02 R e 1 =1 R+ = *0 4 8 4 1 0 .0 I 1 = ++c= m I pI I 7 42 .6 2 .5 8 A 2 Po =R 2 00 *7 3 e [3 = 0 0 .5

X

e1

= 1 aX 2+ X + =

Pin =R 2 6 .5 .7 *7 4 2 .6 e [2 9 = 8 94 .5 '

Z e 1 =e 1 X =+ 8 R + e1 j

1 9 8 ]4 W 4 3 .9

= *0 2 2 1 1 2 0 .1 4 2 4165 6W 5 4 .3 ]

V 1 =2+ Z 1p e V I = + 20 + 3 00 7 j0 (8 .53 Po 198 4 4 3 .9 = = = 0 5 o 9 .5 .9 5 r 5 % V 1 =in 2 9 5645.36 V 94 P 2 6 .5 .7 1

2) 4Electrical Machines (EELE 4350)


17

Voltage Regulation

V 2 NL V 2 FL VR % = *100 V 2 FL V1 2 aV VR % = *100 aV 2 V1 2 V a *100 VR % = V2Dr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

18

Example 4.6

A 2.2-kVA, 440/220-V, 50-Hz, step-down transformer has the following parameters referred to the primary side: Re1 =3 , Xe1 =4 , Rc1 =2.5 k, and Xm =2 k. The transformer is 1 delivering full load at rated voltage with a power factor of 0.707 lagging. Determine the efficiency and the voltage regulation of the transformer.6 .7 20 .4 a =4 4 6= 4 220 V4 A 2200 V = = V A I c = 440/220 2 02= .4S .1 60 8 20 50 2200 .4 1 6 .7 20 4 6 I 2 =4 4 = 10 =0 0.707.510 45 A 8 Im = = cos .2 2 9 6 A 3 j 00 220 2 0 I 1 =+ c m I p+= I I 5 9 4 .3 A .2 6 5 3 I

0 V 1 % =4 4 6+4* 1 = 2+I .7 2 1 4 jX V ' 6 Rp ( e = V R 40 4Dr. Assad Abu-Jasser, ECE-iugaza

= 5 5 3/1 1 .9 0 0 1 5 .6 =6 1 .9 6 7

10 45 2 I op =R =0*54 = 1 5 3 W 45 A = e 4 5 5] 5 P 4 5 .6 [ = a 2 Pin ' =R 4 4 .4 .2 64 .3 e [ .7 20 4 *=9 5 3 1 1 .9 W ] 6 6 5 76 1 V 2 = 2= a V 2*220 4400 V = 9 .6 0 % o re

10 5 3 0 =% .6

) 464.7620.44 V Electrical Machines (EELE 4350)

19

Maximum Efficiency Criterion

Po = V o 2 Core a I pPm (eddy-current and Losses c s I p = 2 Pcu I p e are = R hysteresis)R e 11 constant and called 2 Pin a I p fixedslosses1 = V + Pm I e +o p c R 2 Pm P I p =fl Ip =2II pfl c s m a 2 p o V I pare 1 Pcufl 2 = Copper losses flRe varying as the a I2 p c s +p 1 V oP m R e I + square of the current and called P d | 2 VA ax. Variable e Pm VA | ratedRlosses m = I p 1 = 0 m =ff e. P d Ip

c fl u

Efficiency is zero at no load & itDr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

20

Example 4.7

A 120-kVA, 2400/240-V, step-down transformer has the (a) following parameters: R1=0.75 , X1=0.8 , R2=0.01 , X2=0.02 . The transformer is designed to operate at maximum kVA 120000 efficiency at 70% of its rated load 50 A power factor lagging. with (b)= Ip = =kVA rating of the 0.8 (c) Determine (a) the transformer at maximum V 2' the maximum efficiency, (c) the efficiency at full 2400 efficiency, (b) I * pf Popfl =f 2' p * Po = 2' * V * Vand 0.8 p I load power factor I p =0.7*p I = 0.7*50lagging, and (d) the equivalent core= 35 A loss resistance. Use W P =2P0*5 = .8 9 0 0 the 67200 W equivalent circuit. 4 = 2400*35*0.8 approximate 0 0*0 6= 0

(d)o

o

VA Pcufl

2 cu = = +a1R2 67200 2 (0.75 10 *0.01 93.6% * ) 1 P= =p R Po = 2 = I 2 (= ) = *100 94% 00 + 35 96000 4375 +143.75 Pin + 2 Pin 71487.5 Pcu = 2143.75 W

2 2 V35*2400 2400 1 2 2 |=501.(0.75 Pm== 1 4375 W2686.88 = = .0 ) A R eff + 1+ Pcu84 kV= 71487.5 W P mPc =o + 0 *0= ax. in = Pm 10006000 2143.75 Po 9

Pm =cu = P Dr. Assad Abu-Jasser, ECE-iugaza

2143.75 WElectrical Machines (EELE 4350)

21

Determination of Transformer Parameters The Open-Circuit TestOne winding of the transformer is left open while the other is excited by rated voltage and rated frequency It does not matter which side is excited, however it is safer to perform this test in the low-voltage side

S oc = oc I* V oc

, = cos oc

1

I c = oc cosoc ,I m I oc sinoc I = R cL V oc V oc 2 = = X= , mL I c Poc2

Poc S oc

V oc V oc 2 = I m Qoc

2 Qoc = S oc Poc



22

Determination of Transformer Parameters The Short-Circuit TestThe test is designed to determine the winding P 1 sc resistances and leakage reactances 2 S sc = sc I * V 2 sc , sc cos = eH The test is conducted by placing a short circuit H L eH H S L sc Psc V sc 2 across one winding and exciting the other with 2 R eH = 2 rated= Z eH ,frequency H H L L I I

R

= R +a R =I R

, X

=X +a X

I R

R H = a R = 0.5R

The applied voltage is carefully adjusted until 2 2 NH 2 2 rated current flows inH the windings eH L X L =Z eHR eH = a , eH

sc

sc

, X

= a X = 0.5X eHN

L It does not matter on which side the test is conducted. But for safety it is conducted on the high-voltage side



23

Example 4.8SC-Test OCThe following data were obtained from testing a 48-kVA, 4800/240-V. step-down transformer; TestVoltage (V) Open-circuit test: Short-circuit test: 240 150 Current (A) 2 10 Power (W) 120 600 --------- -------------------------------------------------------------------

Psc S oc =V oc *600 = 240*2 = 480 VVA 150 I oc R eH = 0.5 equivalent circuit of sctransformer as viewed = 15 Determine the= =2 = 6 = 3, Z eH= the= R H = I 2 R2 10 0.5*6 10 from (a) the eH sc sc V ochigh-voltage side andI (b) the low-voltage side 240 2 = 20 a 0.5 2 eH / 2 = X = 0.5*13.75 2 X R cL== 4800= 240 = 480 6.88 = H 2 X eH = Z eH 2R120 6 13.75 = Poc eH = 215 R0.5 R eH RcL0.5*6 (480) = 192 k = 20 cH = a R L = R eH 2 2 =6 2 2 = 0.0075 or 7.5 m Q=X= 2aS= a 2X oc20= 2022 1202= 49.58 k P 0.015 (123.94) m R eL oc =oc 2= = 480 or 15 = 464.76 VA mH a X 2 20 mL 2 0.5 eH 240 0.5*13.75 V2oc 13.75 2 = X X = X eH = 0.017 L = a = = = 20= 123.94 or 17 m mL X eL = 2Q 0.034 or 34 m 2 464.76 a oc 20Dr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

24

Per-Unit Computations

When an electric machine is designed or analyzed Sb the apparent power base using actual values of its parameters, it is not immediately obvious how its performance compare Vb the base voltage with similar-type machine 2

Sb Vb Vb Expressing machine ,parameters = per-unit shows in Ib = Zb = I b Sb immediately howV b machine operates around its the actual quantity Quantity, pu = Per-unit values of machines of the same type with its bas value widely different ratings lie within a narrow range V bH An electric system has four quantities of interest: a= for transformers voltage, current, apparent power, and impedance V bLIf base (reference) values of any two of them are selected, the remaining two can be calculated ratings



25

Example 4.9

A single-phase generator with an internal impedance of 23+j92 m is connected to a load via a 46-kVA, 230/2300-V, step-up transformer, a short transmission line, and a 46-kVA, 2300/115V, step down transformer. The impedance of the transmission line is 2.07+j4.14 . The parameters of step-up and step-down transformers are: RH XH RL XL RcH Xm H

V bC = 46000/V, and S bCZ 46000 200 1.15 = 115 2.3 200 A,= = 230/ 17.25m 11.5k VA 6.9 I bA =10.3= 1 230= = 30 )6 1.0482.29= = 0.06)(10.02, X V 1 R ===.08X== .9 bA H.06= = 0.06 9.2k E H ,pu1 1(0.02j 3.3 + 2+ ,pu R 0 2 .0 , V =246000 the3 1 9 1 81Vvoltage, (b) the = / 400 current, 115 0 1 generator IDetermine151.0/115.9 1= 115 A,0 Z bC 115 = 30*1.31= 8+0.092 .0 400 115 = 13 0.023 j (a) generator 0.2875 bC 1V L ,pu = 1087I 10 1 = A 4(0 j2 3.0 0 .01 .0 ) =6 0= ,pu +0 b + + 6 Z I g , =10 80 .2L .02.3+ 6.9 0 8 0 j2 5 4.1 3 0 2 30 A .0 /2 , /2 1 V bApu Step-down=30 3 =205.75m 0 6 and 1 .8l , pu Hp , ug gp ,u

-------------------------------------------------------------------------Sc n B e tio Section A CB 1 1 Section + 1 + .2 ISection .0 6 0 1 .1 87= = 1 3 .3 1 8 1 16 3 4 .0 3 0 .8 g, 2.3 A 13.8k 6.9k V pu Step-up: = 3 0V 2 0 lagging = 60 A full load and 0.866S ,A pf 4 0 0 V6.9 23m 6 j t b 1 0 69m 2 0 A b

V bA = 230 V, bA S=

46000 VA

Z (c)0*.0 6 80 efficiency6 0the 30.31 + of1.016 0.08 I lg, pu=21=+11.0482.290 .2 =.8 90.02 system at full load and 0.866 = 30 0.8 2 = 3 4 A j I g ,pu 0 the3330 + 1 overall 7 = 4 0 0 80 R cH pu = = 1.15H1 0 100 j X =pu , = 2 6 0 , m , Po pu =0 6 ,1 5in pu e1 1 1 [ 8 * 13 .8 .8 6 1 , R = 3 1 1 163 4 1 1 power factor lagging. P .3 .0 .05 = 0 .0 2 ] , E g ,pu =1.0482.29 1.01630.31 2(0.02j( 0.06) 0.018j 0.03 + + + + 6) 0.023 +4 0.069 2 7= 4 j 0.02, = .0 .1 0 6 pu = 1 0 5 % R LcH.8pu * 0 = + Xj L ,pu 0 1 0, pu,pu = = mH Z L ,,pu , 6 .0 8 L.0 6 0.06 3 =pu=1.1887.21 8 .6 = l , E g ,pu1 1 1 1.15 .0 2 1.15 1 5

Hp , u

R R

11500 0.00575 = 100,X = = 0.02, X = 0.2875 115

9200 0.01725 == = 0.06 80 = 0.2875 115



26

The AutotransformerWhen the two windings of a transformer are interconnected electrically, it is called an autotransformer The direct electrical connection between the windings ensures that a part of the energy is transferred by conduction in addition to the part transferred by the magnetic induction Autotransformer is cheaper in first cost than a conventional two-winding transformer of a similar rating Autotransformer delivers more power than a twowinding transformer of similar physical dimensions For a similar power rating, an autotransformer is more efficient than a two-winding transformer


Electrical Machines (EELE An autotransformer requires lower excitation current 4350)

27

The AutotransformerConnections



28

The Autotransformer

a-ratio of Autotransformer

a I = a I+N 2= a1) + I a N I = T ( = 1 a V22aa = E 2a1a = E 2 + = 1a aT SIo1a power N 2 through induction transfer I V11aa ==VI 1 I1a E 1 + E 2 S ooa = E 2a = S 2 a power transfer through conduction V 2a = VE 1I(a a 1)+ E 2 S oa 2 2 a V 1a S oa = [ T I a1 = V a1I a1 a ] V 1a a1 +a N 21) NT ( + 1 = 2 I 1= a1=o a = + S oa = V 2 S T V 2oa =S ina N a Sa a 2oa 2a 2 2 1 oDr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

( 2a = V )I 1 a= + 2 +E 1)1 I VN 1=V N 22a(a=+EN 2S Ea21 1 S 1 = I1 = +

29

Example 4.10

connected as an autotransformer. For each possible V combination, V 2 =240 V, the primary winding 10 A, I 2(b) 100 A = kVA, I 1 voltage, 24 = = the 1 = 2400 V, determine (a) S o secondary winding voltage, (c) the ratio of transformation, and (d) the nominal=rating24002400 2240 +22640 V= = V 1aa = E 12 2+of 2the == 4 E 2 4V 0 = 0 V0 autotransformer.V 40 + 640

A 24-kVA, 2400/240-V distribution transformer is to be For the 2-winding transformer

V 2aa = 24 + 2V E 02 2400 2 240 2 2640 VV= = E 0 + 0 22 4V 0 4E 0= = 40+ + 640 = 1 aT = V 11aa V V 22aa V 22640 2 40 40 640 = 111 1 == 00.091 .1 .9 22640 2 40 64 400a

S ooaa = V a 22I = 22

=12 aa2121II a V = 1 2

V = 1I 1 a

S ooaa = 2 6 4 0 * 1 0 0 264000 V A A 26.4 =k = 2640 * 10 26= 0 0 V = 4 264 2 6 4 -k V A , , 226640/ 264000-V 26 .4 -kV A 4400 / /2244 -V 0 4 6 4Dr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

30

A Nonideal Autotransformer

An equivalent circuit of a nonideal autotransformer can be obtained by including the winding resistances, the leakage reactances, the core-loss resistance, and the magnetizing reactance.



31

Example 4.11

A 720-VA, 360/120-V, two-winding transformer has the following I 2 a 2 and X constants: RH=18.9 , XH=21.6 , RL=2.1 , 4 5 8 L=2.4 , RcH =8.64 k, I pa = = 45 = A and Xm =6.84 k. The transformer is connected as a 120/480-V, step-up aT 0.2 5 H V 1 2 delivers autotransformer. If the autotransformer=1 .5 1 3 1 3 .6 3 0 .1 2 7 1 3 .6 = A 0.707 IIc a = = 1a I I 6 45 the full load at 3 = A power factor leading, determine its 6 efficiency and the voltage com R pa 2 9 a 0 cL regulation. E H = V 1a E L 1 2 1 .5 1 3 1 3 .6 3 I ma = a = 0 .1 6 7 6 .3 7= A jX j jX ) V + I (R 4E = mIL (R 7 6+0 jXL 2 a H H 2 a com L

L

)+

= = 0.803 or 80.3% 8640 P R cL = in 2= 845.3 960 3 V 1a 6840 121.513 V 2mL = 2== = 486.056 V X anL = 760 3 aT 0.25 720 I 2a = H = = 2 A I V V 360 2a *100= 1.26% VR % = 2anL I 2a = 2 45 A 2a VoDr. Assad Abu-Jasser, ECE-iugaza

=

aT =120/ 480 0.25 a = = P 678.82

360/120 3 =

I4 E= =I ca 2 + I m50 .2 0=8.9 3 8 j+ 2 1.6 ) 480 6 ( 2.1 A a 4 a (1 4 45 + L I 1a = I p a + I a8 4 5= 0 .2 0 4 + 3 8 8 .0 7 4 3 .5 6 A 2 = I 480 * 2 45 = V 1a = E L + I2coma L ( R jX L +) Po = V *a 2*

E L = 119 .745 4.57R e R e

V

6 78 .8 2 W =

4 1 2 = 45 * 8 1 V 1a = 1 19.7 45 I4a.57.511 3 61 3+.6 3( 2 .1.02 7 j2.4 3+.5 6 8=4 5 ) Pin = V* a 1*

V 1a = 121 .513 13.6 V 3Electrical Machines (EELE 4350)

32

Three-Phase TransformersPower is generated, transmitted, and consumed in three-phase form. 3-phase transformers are used in such systems Three exactly alike single-phase transformers are used to form a single three-phase transformer For economic reasons, a three-phase transformer is designed to have all six windings on a common magnetic core A common magnetic core of a three-phase transformer can be either a core type or a shell type Shell-type transformer exhibits less waveform distortion than core-type and this makes it preferable over the core-type


A three-phase winding on either side can be connected


33

Three-Phase Transformers

Construction & Windings Connection

Y-Y Connection -Y Y- - Core Type Shell-Type



34

Analysis of a 3-Phase TransformerUnder steady-state conditions, a single 3-phase transformer operates exactly the same as 3 singlephase transformers In our analysis we assume that we have 3 identical 1phase transformers connected to form a single 3-phase transformer Such an understanding allows the development of the per-phase equivalent circuit of a three-phase transformer It is also assumed that the 3-phase transformers delivers a balanced load and the waveforms are pure sinusoidal

This enables us employ the per-phase equivalent circuit of a transformer. A -connected winding can be Dr. Assad Abu-Jasser, ECE-iugaza its equivalent Y-connected winding using 4350) Electrical Machines (EELE replaced by

35

Example 4.12

transformer are RH=18.9 , XH=21.6 , RL=2.1 , and XL=2.4 , RcH =8.64 k, and Xm =6.84 k. For each of the four configurations, determine the H nominal voltage and power ratings of the three-phase transformer. Draw the winding arrangements and the per-phase equivalent circuit for each (c) For connection configuration. (a) For Y- (d) For Y-Y connection -Y (b) -

S 3Athree-phase transformer is assembled by orThe constants for each = 3* 720single-phase transformers. 2.16 kVA = 2160 VA connecting three 720-VA, 360/120-V,

V 1L = 360 * 3 =623.54 V V 3 * 360 = 624 V 1L V = 3 *120 207.85 V V 2 L = 120 * 3 ==208 V VThe nominal Ratings are nominal Ratings are

360/120-V, 2.16-kVA, 624/120-V, Y- connection 2.16-kVA, 624/208-V,Y-Y connection 360/208-V, -Y



36

Example 4.13

ThreeY/ Connection single-phase transformers, each rated at 12-kVA, 120/240-V, 60Primary Secondary Hz, are connected to form a three-phase, step-up, Y- connection. The parameters of each transformer are RH=133.5 m, XH=201 m, RL=39.5 Phase Voltage 120 240 m, and XL=61.5 m, RcH =240 , and Xm =290. What are the nominal H Line Voltage 240 voltage, current, and power ratings of 208 the three-phase transformer. When it delivers the rated rated Phase Current load at thecurrents, and the efficiency of the 100 voltage and 0.8 pf lagging, 50 determine the line voltages, the line transformer.

Line Current

100

86.6

* equivalent Y-Y connection Vo12== Connection 3Re Primary + 0.0615) E =For+ 2npAI 2 2A(0.0445 [+ j 0.067) Secondary = P nn 3Re2 I+ *I A(0.0395j 138.564*86.6 36.87 ] 1 V Y/YEVn n V 2 n = + 100 V 0 V 1n== 125.7W138.564kW 36.87 (0.0445 + j 0.067) 30.92 28.8 6.87 (0.0395 j 0.0615) + Po 2 n 28800 or E = 138.564 0 + 86.6 120 PhaseI Voltage 36.87 A 138.564 = 86.6 = 2A V 1n = 3Re *I * =V = 132.6131.97 3Re 132.6131.97 *100.72 6.88 Pin2 n = 145.147 A Vn ] E Line Voltage 10.92 V [ 208 240 1 120 = V 1L == aV 1n 30 =A 0.866 3 Pin 100.72 = 229.69 61.97 V 6.88 138.564 = 251.4 100 V Phase Current 86.6 E 2 L = 3E 2 n 30 30.92 1 36.87 1 28.8 +V 86.6+ = 100 IEA =Current 30 = 125.7 30 V6.876.88 A I I pA E n= 92.3% = 30.92 1 = a *100 1 Line = pA * =2 n 0.866 j 100 100.72 A 86.6 1n 31.2 240 290 Dr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

37

End of Chapter Four



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Chapter 4- Transformers