Chapter 4- Transformers

1

‫بسم ال الرحمن الرحيم‬

Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

2

Electrical Machines (EELE 4350)

By

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

[email protected]

Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

3

Chapter Four

Transformers Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

4

Introduction A 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

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5

Construction of a Transformer

Transformer core is built up of thin lamination of highly permeable ferromagnetic material such as silicon sheet steel The lamination’s thickness varies from 0.014 to 0.024 inch to keep core losses to a minimum. A thin coating varnish is applied to provide electrical Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

6

Φ=Φ

An Ideal Transformer m

sin ωt

e1 =N 1 ω Φ

m

cos ωt

1 E 1 = N 1 Φ∠ ω m 0o 2 o E 1 = 4.44fNΦ∠ 0 1 m o E 2 = 4.44fNΦ∠ 0 2 m dΦ N ν Ve11 = E 11 = N1 1 dt a = = = V 2 E 2 N 2d Φ N ν Ie22 = N 12 = dt2 = = a Iν N 1 1 N1 2e 1 === a * * Vν V22e I2 N or V 1I= 1I 2 1 = 1 V 2I 2 2 Dr. Assad Abu-Jasser, ECE-iugaza

An Ideal Transformer has the following properties: The core is highly permeable i.e. it requires a very small mmf to set up flux Φ The core does not have V eddy-current V1 2 Ni Ni = any or Z 2 11 = = 2 2 12 I 2hysteresis a I 1 loss

i 2 N1 The = core flux = 1exhibits no a Z = Z 12 i.e. the flux is i 12 leakage N 2 a

confined within the core

ν ν i = Z 11 aZ =resistance 2i 2 The of each 1

2 2

windingElectrical is negligible Machines (EELE 4350)

7

Transformer Polarity and Ratings Transformer Polarity Ratings The 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 Dr. Assad Abu-Jasser, ECE-iugaza Electrical Machines (EELE 4350)

Example 4.1

8

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

For t=0.1 -0-0 .0.1 62 ss

Φ Φ = − 0.1 0.4 55tW tW bb e aabb e aabb e ccdd e ccdd

ddΦ Φ = = −= e −= −N aabb b ba a − tt Φ = 0.009 W b dd = = − − 2 − 0 0e 0* 0*0 = = − 0 .1 .4 55 9 30 0VV V e2 == 0 For t=0.06-0.1s

ab

cd

ddΦ Φ = N =ccdd dd tt = =5 50 0 − 0* 0*0 = = 0 .1 − .4 55 7522 V 5 V

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Electrical Machines (EELE 4350)

Example 4.2

9

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 =150/7 = 50 0.2 I2 4 I1 = = = 20 A a 0.2 V1 240 V2 == = 1200 V a 0.2 PL = VI 2 2= cos θ12= 00*4*0.8 3840 W Φ= m

E1 240 = = 4.44 fN 14.44*50*150

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10

A Nonideal Transformer For Nonideal Transformer

E1 I 2 N 1 = = = a E2 I 1 N 2 Nonideal Transformer has the following V1 = E 1+ R+ 1( jX 1 I 1 ) parameters V2 = E 2− R+ 2( jX 2 I 2 ) Winding Resistances Leakage Fluxes Finite Core Permeability Core losses (Eddy-Current & Hysteresis)

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Example 4.3

11

A 23-kVA, 2300/230-V, 60-Hz, step-down transformer has the following resistance and leakage-reactance values: R1=4Ω, R2=0.04Ω, X1=12Ω, and X2=0.12Ω. 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.

I 223000 II1 2= = =7.5 30*0.75 °A= 75 A,I=2 75 ° 30 A a 230 o ZZ1 2= =RR jX 1 4=2= +12 j0.04 + jΩ 0.12 Ω 1 + 2 + jX

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

2

1

2

1

2

1

2

VE1 = 2269.578 4.72.33 °°V V = 228.287 2 * Po = Re[ V * I Re[230 * 75 30− ] °14938.94 = W 2 2 ] = 2300

a=

= * 10,E= 1 a *E=

2282.87 °2.33

V

Pin = Re[ V1 * I1 ] =Re[2269.578 4.7 * 7.5 ° 30 −] 15389.14 ° = W 230 Dr. Assad Abu-Jasser, ECE-iugaza

2

Electrical Machines (EELE 4350)

12

Finite Permeability

Transformer excitation cuincreases rrent When Unloaded the load on the draws transformer I m is the current magnetizingincreases current The Isecondary φ = I c + I m ,winding

The currentEsupplied by the source increases 1 Ic = core-loss current The voltage drop R c 1across primary impedance increases TheEinduced emf E1 drops X m 1 = 1 magnetizing reactance The mutual flux decreases because magnetizing current jI m drops Dr. 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 kΩ and 15kΩ, respectively. If the transformer delivers the same load, what is its efficiency?

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

V 2 = 230 VI a = 10

m

Rc1

=

20000

°

0.114 ° 2.33

75 = 30

I

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

°−0.152 °87.67 AE 2 228.287 = 2.33

2282.87 = A2.33

E

A

I φ =I c+ − 50.8 ° 1 I m= 0.19

Vp

°I

I =I + I = 7.5°+ 30 0.19 − 50.8 °= 7.53 °28.57 Po = 14938.94 W V =E + I Z = 2282.87 2.33 °+ 7.53 28.57 ° + (4 j 1

p

1

1

φ

1

1

V 1 = 2271.9 4.71 °

Dr. Assad Abu-Jasser, ECE-iugaza

Po 14938.94 = = Pin 15645.35

A

A

12)

V

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

η=

7.5 30=

W

0.955, or 95.5% Electrical Machines (EELE 4350)

V

°

14

Phasor Diagram

E 2 = V 2 + I2 R2 Dr. Assad Abu-Jasser, ECE-iugaza

+jI 2 Electrical Machines (EELE 4350)

15

Approximate Equivalent Circuit Ze2R =+ e 2 jX e 2 R e 1R =+ Re 1a

=R + jX e 1

Z

e2

2

/ 2

1

2

= X+aa R / R e 1 X= R+ 1 2

X

e2

2

Rc 2 R = c 1a

X

1

/

2

=X +a/ X

X me 12X= ma 11

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 shown Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

16

Example 4.5 Analyze 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.

' 2269.594.7 ° ° V aV 2== 1° 0*2 = 2.7 3000 I c 2== ° 300 0.1134 A V

20000 Ip = 7.53 °0 A ° 2269.594.7 Im = = − ° 0.151 85.3 A 2 2 j 1 5 0 0 0 Re1 = R 1+ a R=+ =Ω 4 10 *0.04 8 2 I1 = I ++ I m °2 7.5428.6 2A pI c=

X

e1

= X 1+ a X=2 +

Po =Re 2 *7.5 30 [3°000−°=

= 12 Ω 10 *0.12 24

14938.9 ]4 W

Z e1 = R e 1+X =+ Ω j 8 e1

24

Pin =Re 2 °94.7 −*7 °= .54 28.6 [' 269.5

15645.36 W ]

V1 = V 2+I Z=1p e°+ 2 °30 + 00 7.53 j0 (8 Po 14938.94 η= = = 0.955 or 95.5% V1 = 94 Pin2269.5 15° 645.7 .36 V

Dr. Assad Abu-Jasser, ECE-iugaza

24) Electrical Machines (EELE 4350)

17

Voltage Regulation

V 2 NL V − 2 FL VR % = *100 V 2 FL V 1 −aV 2 VR % = *100 aV 2 V1 −V 2 a VR % = *100 V2 Dr. 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 Ω, Xe1 =4 Ω, Rc1 =2.5 kΩ, and Xm1 =2 kΩ. The transformer is delivering full load at rated voltage with a power factor of 0.707 lagging. Determine the efficiency and the voltage regulation of the transformer. 464.7620 =.44 2 = V ° 220 V S A 2200 VA Iac = =440/220 = ° 0.1 860 .44 2= 2500 2200 −1 620.44 =464.7= 10° ° cos =0.2 0.707 − 10A 45 A IIm2 = =−− 32°89.56 j 2 0 0 0 220 I 1 =+ I pII += I m − ° 5.296 45.33 A c

10 − 45° 2 I = = = − °.6 Pop =Re 4[40* °= 545 1 555 3 W5 45 A ] a 2 Pin ' =Re 4[64.7 ° 620.44 ° *5 = .29645.33 1716.91 W ] V2 = aV 2= 2*220 = ° 4400 V

η=1555.63/1= 716.91 0.906

' 4 64.76 0 − V = V + R2+ (44 jX=1 2 I p1 VR1 % = *e

440

Dr. Assad Abu-Jasser, ECE-iugaza

90.6%

e

or

) 464.7620.44 ° V

100 5 = .63%

Electrical Machines (EELE 4350)

19

Maximum Efficiency Criterion

Po a = V I pP cos θ 2 Core Losses (eddy-current and I pη = 2 m Pcu I= Re Rare p 1 constant and called hysteresis) e1 2 Pin a = V I p fixed ++ P co slosses Iθ Re 2 m p 1 Pm Pm I pη = I pfl = I aV2I2 pflp cos θ I pare Pcufl 2 as the η= losses Copper varying flRe 1 aV I2 p cosP θ I+ R+ m p 1e square of the current and called P dη m 2 VA |max. VA = | =Variable →= 0 I R P losses eff . rated pηe 1 m P dI

p

cufl

Efficiency is zero at no load & it Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

Example 4.7

20

A 120-kVA, 2400/240-V, step-down transformer has the (a) following parameters: R1=0.75 Ω, X1=0.8 Ω, R2=0.01 Ω, X2=0.02 Ω. The transformer is designed to operate at maximum kVA 120000 efficiency at 70% of its rated load 50 withA 0.8 power factor lagging. (b) I = = = (c)pDetermine' (a) the kVA rating of the transformer at maximum V 2400 2 ' efficiency, (b) the maximum efficiency, (c) the efficiency at full ' P = V I * pf * Pload = V I * p f * o 2 p η p fl 0.8 power factor Iopη and =20.7* I p= 0.7*50 = lagging, 35 Aand (d) the equivalent coreloss resistance. Use approximate P = 2400*35*0.8 = W equivalent circuit. P =24 00*5 0*0 = .8 960 00 the W67200

(d)

o

VA Pcufl

2 2 V 2400 35*2400 1 2 2 |= = = 84 kV A R =+ == = 71487.5 2686.88W Ω 50 .7 5 10+ *0 1 ) 4 37 P = P P.0 = max. ceff 1.(0 in o +Pm cu η5 W 1000 P 2143.75 m Po 96000 o

2 P 2 67200 2 2 oa R η =η = = *100) 93.6% Pcu = Iη R + ( = ) + 35 (0.75 10 *0.01 *100 94% η p = =1 2 = Pin 96000 4+ 375 + 2143.75 Pin 71487.5 Pcu η = 2143.75 W

Pm = Pcu =η Dr. Assad Abu-Jasser, ECE-iugaza

2143.75 W Electrical Machines (EELE 4350)

21

Determination of Transformer Parameters The Open-Circuit Test One 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 = V oc I*oc

,= oc φ cos

I c =I oc cosocφ ,I=m I oc sin φ oc R cL

V oc V oc 2 = = X, = mL I c Poc

2 Qoc = S oc− Poc

−1

 Poc   S oc

V oc V oc 2 = I m Qoc

2

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Determination of Transformer Parameters The Short-Circuit Test

22

The test is designed to determine the winding −1  P sc resistances and leakage reactances S sc = V sc2 I * ,=sc φ cos   2 sc eH The test H is conducted L eH H sc circuitL S  by placing a short winding and exciting the other with Psc V sc 2 across one 2 R eH = Z= ,frequency H H L L I 2 ratedeH I

= R +a R

R

I R

=X +a X

, X

=I R

sc

sc

The2 applied voltage is carefully adjusted until 2 N 2 H flows LeH =current eHeH L Xrated Z eH−2R =a inHthe , windings

R H = a R = 0.5R

, X

= a X = 0.5X eH N

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

Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

23

Example 4.8 SC-Test OC-obtained from testing a 48-kVA, The following data were 4800/240-V. step-down transformer; Test Voltage (V)

Current (A)

Power (W)

--------- ------------------------------------------------------------------Open-circuit test: Short-circuit test:

240 150

2 10

120 600

150 S oc =PVsc oc *600 I oc = 240*2 = 480 VVA sc RReH == the =equivalent = 6Ω=circuit , Z eHof = the=transformer = Ω 15 as viewed Determine 2 2 0.5 R = 0.5*6 3 Ω 10 10low-voltage side 2eH 2 from H (a)Ithe high-voltage side andI (b) the sc sc V 240 a=0.5 = 4800 == 20 X RHcL= Xoc2 eH= =/ 240 0.5*13.75 = 6.88 Ω 480 2 2 Ω2 X eH = ZPeH − R eH = 215 − 6= 13.75 Ω oc 2 120 R0.5 =eH a RcL0.5*6 = 20 (480) = 192 k Ω cH R R L = R eH 2 2 =6 2 2 = 0.0075 Ω2 or 7.5 mΩ 2 2 2 P=oc20 =0.015 480 −or120 464.76k Ω VA R eLQ= ==oca− X Ω(123.94) 15 == mΩ ocX= 2aS = 20 49.58 2 mH mL a 20 0.5VX 2eH 240 2 0.5*13.75 X XL = X=eH 2oc =13.75 = = 0.017 Ω or 17 mΩ 2= 123.94 Ω mL a= X eL = = 20 0.034 Ω or 34 mΩ 2Q 2 464.76 a oc 20 Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

24

Per-Unit Computations

When an electric is designed or analyzed Sb the machine 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 I b = ,parameters Zb = =in per-unit shows I b operates Sb immediately howVthe machine around its b ratings

actual quantity Quantity, pu = Per-unit values of machines of the same type with itswithin bas value widely different ratings lie a narrow range V bH has four quantities of interest: An electric system a= for transformers voltage, current, apparent power, and impedance V bL If base (reference) values of any two of them are selected, the remaining two can be calculated Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

Example 4.9

25

A single-phase generator with an internal impedance of 23+j92 mΩ 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 Ω. The parameters of step-up and step-down transformers are: RH

XH

RL

XL

RcH

XmH

-------------------------------------------------------------------------Section BA Section  1 1 Section B C ISection = 1 .0 − 1 6 ° + 3 0 .3 1 ° 1 .1 + 8 87 .2 = 1 − ° 1 .0 36 30 .84 gp ,u   Step-up: 2.3Ω 6.9Ω 23mΩ 13.8kΩ 6.9kΩ V = 2 3 0 0V S = , 4 6 0 0 0 V A A t full load and 0.866 lagging pf 1 2 0 60j bA bA  69mΩ

V bA = 230 V,S bA =

46000 VA

17.25mΩ 11.5kΩ VI bC = = 46000/ 115 V, and S 46000 = VA 2.3 6.9 bC 200 A ,= Z 230/ == 200 Ω = 1.15 0.06 9.2kΩ V = 1 .31 3 ° 1 1 .0230 8j= 0.06)(1 bA = bA 2 .3 6 .9 R = = 0.02, X E 10 ° + (0.02 + − °= 30 ) 1.0482.29 ° H0.0 ,pu R H ,pu== X== 0.02, 6 V = 23 0*10.023 .3 1 ° = 3 1 1 .0 8 ° 3 0 1 .9 9 1 1 .0 8 V IDetermine = 46000 /115 = 400 A, Z bC 115 =generator / 400 current, 0.2875 = 1 1 5 1 1 5 +0.092 j (a) the generator (b) the 115 115 bC  1 voltage, 1 V = 1 .1 87 ° + .2 − ° 3 6 3 0 .8 4(0 + .0 j2 23 0 .0 800.0 j1 2 I bLgA,pu = 4 6 08 0 0 = /2 3 01.0 = 2 0 A Z =Ω ,+ 0 /2 15 0.06) V = 10 ° and I1 =− ° 1+ 30 2.3Ω 5.75mΩ p , u Step-down bA 6.9Ω L0 ,pu gp ,u

l , pu Hp ,u

Hp ,u

+4 − 0.02 j1.016 0.08 at full load and 0.866 IIZlg, pu =− 10= °+ 1.0482.29 ° + ° 30.31 g= ,pu 20 *30 1 − .03 6 ° = 3 0 .8 − 4 °efficiency 2 0= 7.2 = 30 A 1 3 8 0 0 6 90 0 (c) the overall of the system .8  80 j R cH pu, = = 1.15 X= = 2 ,100  60 mH1 pu , 0 1 5 Po p, u =0.8 66,115 Pin p,uRe1 = .31311[.08 °*11 .0 3 63 °= 0.84 1.012 power factor ] lagging. E g ,pu =1.0482.29 °+ 1.016 − °30.31+ 2(0.02 + j( 0.06) + 0.018j 0.03 6) 0.023 0.069 2 .0 7 + 4 .1 4 j 0 .8,6 6 LLcH ,,pu pu LmH , pu pu ,pu0.06 R = = 0.02, Xj L=,pu 0.01 = Z = = + 8 0 .0 3 6 η = = * 1 0 0 8 5 .6 % l , p u E g ,pu1.0 = 1.1887.21 ° 115 1.15 1.15 12 g

RR

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

Dr. Assad Abu-Jasser, ECE-iugaza

9200 0.01725 == = 0.06 80 = 0.2875 115

Electrical Machines (EELE 4350)

26

The Autotransformer When 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 Electricalcurrent Machines (EELE 4350) An autotransformer requires lower excitation

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27

The Autotransformer Connections

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Electrical Machines (EELE 4350)

28

The Autotransformer

a-ratio of Autotransformer

(N + N ) I = N I ( a + 1) 1   V = V = E = E + E 1 2 1 a 2 2 a 2 a 2 1 1 a 1 2 S =V I = S 1+

II = aN I+Na( 2= a1) + I a  V22aa == ET 2a1a = E=2 +1 =a 1a aT SIo1a≡ power N transfer through induction 2 VSSI 11aa ==VI 1EI1a E 1 + E 2 ooa = 2a transfer 2a = through conduction ≡ power VSa2oaa = V2VE 12I(a a 1)+ E 2 S oa =  1a  [aT I a1 =] V a1I a1 V 1a Na1T +(a N + 21)  1 1= a1S=+o aT =  S oa == V2 2I VS2oaa =S ina N a2 a  oa 2a

Dr. Assad Abu-Jasser, ECE-iugaza

2 2 1

o

Electrical Machines (EELE 4350)

29

Example 4.10

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

connected as an autotransformer. For each possible V combination, So primary =24 kVA, I 1 10 = A, I 2(b) 100 = A determine (a) the winding voltage, the 1 = 2400 V, V 2 =240 V, secondary winding voltage, (c) the ratio of transformation, and (d) the nominal V 1aa ==rating E 12 2240 +4=ofE0 02the 22400 4VV0 =0autotransformer. = 2240 4 0 +2+2640 6 4 0 VV= =

V 2aa == 24E +01 2V +4= E0E0222 2400 4V0 0= 2 240 4 0 + 2+ 2640 6 4 0 VV== aT =

VV11aa VV22aa

22640 24640 04 0 = 1011 0.091 .1 .9 1 == 22640 26440 40 0

S ooaa == V22 a 22I

a

=V1=212 aa2121II a

V1 =a 1I

S ooaa == 22640 6 4 0 ** 110 0 0 26400 2=6 =4 0 0 V 0 AV A 26.4 2 6 =4k 2266 4.4-k-kV V AA, , 224640 640 04 0 // 264 /224 64 4000-V -V Dr. 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.

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Electrical Machines (EELE 4350)

Example 4.11

31

A 720-VA, 360/120-V, two-winding transformer has the following I 2 a Ω, 2 4and 5° X =2.4 Ω, R =8.64 kΩ, constants: RH=18.9 Ω, XH=21.6 Ω,I RL==2.1 c°H = 8 L4 5 = A pa and XmH =6.84 kΩ. The transformer is connected aT 0.2 5 as a 120/480-V, step-up V 1 2 1 .5 1 3 1 3 .6 ° 3the full load at 0.707 autotransformer. If the autotransformer IIc a = = 1a I =−delivers 0 .1 2 7 ° 1 3 .6 =3 A I 6 45 = A power factor leading, determine its efficiency and the voltage com R pa 2 9 6 a0 cL regulation. E H = V 1aa E L 1 2 1 .5 1 3 1 3 .6 ° 3 I ma = = 0 .1 6 7 6 .3 7= A − 4E =jX mIL (R 7 6+j0jX ) V + I − (R jX L

aT =120/ = 0.25=a P 480 678.82

η=

360/120 = 3

= = 0.803 or 80.3% 8640 Pin 2= 845.3 R cL = Ω 960 3 V 1a 121.513 6840 VX 2mL = 2== Ω = 486.056 V 760 anL = 3 aT 0.25 720 I 2a =I H = = −V 2 A V 2anL 2a 360 VR % = *100= 1.26% V 2a I 2a = 2 45 ° A o

Dr. Assad Abu-Jasser, ECE-iugaza

2 a

H

H

2 a

com

L

L

)+

I4φ E = =I ca 2 + I4m5a 0 .2 0=8.9 4 3 8 j+ 2A1.6 − ) ° 480 a ° (1 45+° 6 − ( 2.1 L I 1a = I p a φ+ I a8 4 5= 0 .2 0 °4 + 3 8 8 .0−2 7 4 3° .5=6 A

E L = 119 .745 4.57

I 4 8 0 *= 2 4 5 V 1a = E L + I2com ( a R L jX +) L Po =

R e 

V *a 2

R e 

V* a 1

V°

*

6 7−8 .8 2 ° W=

2 11 3= 1 3 .6° 3 * 8 .0 2 7 − 4 3 .5 6 ° 8=4 5 1 .5 V 1a = 1 19.7 451 I4a.57 6° 45 + ( 2 .1 ° j2.4 )+ Pin =

*

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

32

Three-Phase Transformers Power 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

Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

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Three-Phase Transformers

Construction & Windings Connection

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

Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

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Analysis of a 3-Phase Transformer Under 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 Electrical Machines (EELE 4350) replaced by its equivalent Y-connected winding using

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Example 4.12

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

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

V 11LL = 360 V 3 =623.54 3 **360 =624 V V V 2 L == 120 *V 3 ==207.85 208 V V V 3 *120 The nominal nominalRatings Ratingsare are

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

Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

36

Example 4.13

ThreeY/∆ single-phase transformers,Primary each rated at 12-kVA, 120/240-V, 60Connection Secondary Hz, are connected to form a three-phase, step-up, Y-∆ connection. The parameters of each transformer are R Phase Voltage 120 240 mΩ, RL=39.5 H=133.5 mΩ, XH=201 mΩ, and XL=61.5 mΩ, RcH =240 Ω, and XmH =290Ω. What are the nominal Line Voltage 240 voltage, current, and power ratings of 208 the three-phase transformer. When it delivers the rated load at the rated voltage and 0.8 pf lagging, Phase Current 100 determine the line voltages, the line currents, and the50 efficiency of the transformer.

Line Current

100

86.6

* Y-Y For equivalent connection V = E + I (0.0395 + j Primary EY/Y = V + I (0.0445 j 0.067) 36.87   P = 3Re V * I = 3Re 138.564*86.6 ° 1 n 1 n pA [+0.0615) ] Connection Secondary 2n 2n 2 A2A  o 2n V 2n = 138.564 °− 0 V6.87 V = 125.7 30.92 °+ 100 ° −36.87 (0.0395 + °(0.0445 j 0.0615) 1n= 28800 P W or 28.8 kW E = 138.564 0 ° + 86.6 + j 0.067) o Phase Voltage 120 138.564 2n I 2 A = 86.6 =− 36.87 ° A V = 132.6131.97 ° V * V 1n *I 1A0.92 = 3Re 132.6131.97 P ° *100.72 6.88 ° ] Ein12nn= Voltage =3Re 145.147 Line 240 120  ° V [ 208 aV3 =1n 30 = 229.69 0.866 V °= 61.97 ° V 1L == P 100.72 − 6.88 A 138.564 Phase 86.6 Ein2 L = Current 3E 2 n 30 ° = 251.4 100 30.92 °V  − 1 86.6 36.87° 1 28.8 I = I + V + =  30 A IpA =1n= 30 °= − −100 °100.72 6.876.88 A ηE1A= *100 92.3% pA  Line Current 100 86.6 = a * E ° = 125.7 30.92 ° V 1n 31.2 2 n 0.866 240 j290  Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)

37

End of Chapter Four

Dr. Assad Abu-Jasser, ECE-iugaza

Electrical Machines (EELE 4350)