1
ا الرحن الرحيم بس م ا
2
Electrical Machines (EELE 4350)
By
Assad Abu-Jasser, Ph. D. Electric Power Engineering www.iugaza.edu/homes/ajass er
[email protected]
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Chapter Four
Transformers
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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 One-to-On e 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 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
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An Ideal Transformer
Φ=Φ
m
e1 =N 1 ω Φ E 1 =
An Ideal Transformer has the following properties:
sin ω t m
cω ost
1
N 1 Φ ω ∠ m 2
0
N∠ m0 1E= 4.44 fΦ 1 N m0 2E= 4.44 fΦ 2 ∠ d Φ e1 = N = 1 V E N 1 1 1 1 = = =dt a V 2 E 2 N 2d Φ = = 2 Ie22 N N 12 = = a dt Iν N 1 1 N 1 2e 1 * ν 1 2 1
VI
The core is highly permeable i.e. it requires a very small mmf to set up flux Φ
o
o o
ν
ν
a == = * oV r 1I= = V22e I2 2 N 1 V 2I 2
The core does not have 1 V 2 i V 1 N i N any eddy-current or 22 Z 2 1 1 I 2hysteresis a I 1 2 loss i 2 N 1 The core flux 1exhibits no a Z Z 12 i.e. the flux is 2 leakage i1 N 2 a confined within the core 2 ν ν i i Z1 a Z 1 2 1 The 2 2 resistance of each
=== = = = ==
winding is negligible
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Transformer Polarity and Ratings Transformer Polarity Ratings The nameplate of a transformer t ransformer 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 g iven by
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Example 4.1 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= t=0.1 .-1 .6 12 0-.0 .0.1 ss Φ = − Φ =
e aabb e aabb eccdd eccd d
45t W 0.0 1.5 bb t W
Fort= t=0.0 .06-0.1 .1s = − = − = − = − e N ba a ab b b a .009 Φ = 0.0
d Φ Φ d
t t W b dd = − − = 0 0* .4 V 0 0* 00 .1 55 9 30 0V V = −2 = − e ab2 e 0 == cd d Φ d Φ = N =ccd d d d t t 0 *= 0 .4 5 V = 50 0 − 0*0 = .− 1 55 7522 V =5
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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
=150/ 7 = 50
0.2
4 I 1 = = = 20 A a 0.2 V 1 240 V 2 = = = 1200 V a 0.2 P L = V I 2 2 = cos θ 12= 00*4*0 0*4*0.8 3840 W I 2
Φ = m
240 = = 4.44 fN 14.44*5 4*50*150 E 1
7.2 .21m 1mWb
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A Nonideal Transformer For Nonid ideal Transf sformer
E1 E2
I2
N1
I1
N2
= =
=
a
Nonideal Transformer has the following V1 = E 1+ R+ ) 1( jX 1I 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
A 23-kVA, 2300/230-V, 60-Hz, step-down transformer has the following resistance and leakage-reactance values: R 1=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 30* 0.°A I ° 30 7=5 75 A,I =2 75 30 A I1 2= = =7 .5 30 a 230 o η =4== +102j .+04 Ωj 0Ω Z1Z===R1 R+ +j X1jX .12 = 2
P
14938.94
P
15389.14
2
2
0.971, or 97.1%
V 1E= = E 1V+in I 1* I Z 1*Z2== 2 8 2 .283 702 .3(7 375°3(°0 7 .+5)(30 30+ 0.0)4( 4 °j 01j.21+)2) ( + + 2 2 2 2 V 4 .72.3°°3V V 1 = 2 =2 6292.85.72887 E 2
Po = R2300 e [ V2 * *2I ] =R e [ 2 3 0 * 7 5 3 0− ] °1 4=9 3 8 .9 4 W = * 10,E= 1 a *E= 2 2282.87°2.33 V a= Pin = R2e3[ 0V R e [ 2 2 6 9 .5 7 8 4 .7 * 7°.5 3 0 −] 1°5 3=8 9 .1 4 W 1 * 1I ] =
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Finite Permeability
Unloa Transformer Transform er draws excitation current cuincreases rrent When Unloaded the ded load on the transformer
Iφ = Ic + Im ,winding I m is the current magnetizingincreases current The secondary The current by the source increases E supplied 1 core-loss current I c = The voltage drop Rc 1across primary impedance increases The E induced emf E1 drops 1
X m 1 = magnetizing reactance The mutual flux decreases because magnetizing current jI m drops
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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?
2282 2282.87 .87 2.33 2.33 ° E 3, 1 From Exam I ple= 4.3, 4. = we have = c
V 2 = 230 V I
m
a = 10
R c 1
=
20000
75 30 =
2282 22= 82.87 .87 2.33 .33 A°
I
E 1 2 = m1
jX
15000 j
0.19 0.= 19 50 − 50.8 ° .8
I1 = I p+ I = φ
°+ 7.530
P o = 14938.94W
V 1 =E 1 + I 1Z 1=
°0 =.8 0.−19 50 5
2282.87 .87 2°.33 .3 +3
2271.9 .9 4.71 4.7°1 V 1 = 2271
η =
P o P in
=
14938.94
=
15645.35
E 2 2 2 8 . 2=8 7 2 . 3 3
V p
A
7 . 5 3 =0
° I
°8.57 7.53 2
A
7.53 .53 2°8.57 .5+7 (4 j 12)
V
* 15645.35 P in = Re [V 1I 1 ] =15
A
°−0.152 0.152 °87.67 87.67
2 2 8 2 . 8 7 A2 . 3 3
Iφ = I c+ 1 I m =
E
0.114 0.114 2.33 ° 2.33
W
0.95 .955, or 95.5% .5%
A
V
°
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Phasor Diagram
E2 = V2 + I2 R2
+j I2
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Approximate Equivalent Circuit +jXe 2 Ze 2 = Re 2 e1 e1 =2+ Re 2 R R 1a
Z = R + jXe 1
R
2
=R + a R 2/
=2+ X e2X X 1a e1 1 = Rc 2 R / c 1a
X
/ 2
2
2
=X + a/ X 2
= Xm a1 e 12 X m1
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
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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.
' 2269.594 594.7 1° ° V aV 2= 0= *2 30° 2.7 300A 0 V I c 2== = ° 00 .1134
20000 °0 A I p = 7.53 2269.594 9.594.7 I m = = − 2 j 1 5 0 02+ = Re 1 = R1+a0R I1 = I +pI+ c= I
°
° °
= Xe1 = X 1+a m X 2+ 2
P o
=Re
P in
=Re
*0.04 8
= 12 Ω 10
° +
14938.9e4 η = = = V 1 = 9 45.7 P i2269.5 15° 64 .36 V
*0.12 24
14938.9 ]4 W
Ω j 8
24
2 °94.7 −*7 ° .5 = 4 28.6 [' 269.5
V1 = V 2+I Z=1 p P o
= 4Ω 10
7.5428.6 2A
°.5 = 30 2 000− *7 [3°
Ze 1 = Re 1+X= e 1+
0.151 85.3 A 2
° + 2 30 00
15645.36 W ]
7.53 j0 (8
0.955 or 95.5%
24)
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Voltage Regulation
V2
VR % =
−2 V
NL
V 2 FL
*F1L00
V1 − aV 2 VR % = *100 aV 2 V 1 VR % =
−V 2 a *100 V 2
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Example 4.6
A 2.2-kVA, 440/220-V, 50-Hz, step-down transformer has the following parameters referred to the primary side: R e1 =3 Ω, Xe1 =4 Ω, Rc1 =2.5 kΩ, and X m1 =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. ac I
464.762= 00 .44 2 = 2 = V° 2 = V S A 2200 VA 0.2 1 860 .44 =440/ 2 = ° 20
I m2 I
2500 2206 020.44 =464.7= 10° = =−− j 2000 220
I1 = I +I I += p c
I 2
−1 co= s 0.2 0 − .7 °879.5610A 45 A 3 20
°
10 − 45°5.296
− ° m
45.33 A
I p = =Re= − °.63 W5 440* 545= 1 555 P ° = o P in
V
45 A [ ] a 2 e4 *5 .29645.33 1716.91 W ] = [64.7°620.44 ° ' =R aV = =
2*2 2* =220° 4400 V
2 1555.62 3/1 /1 716.91 0.906 = η =
4I 64.7R 62 V = V= (440− jX + p1 *e =1 2+ '
VR1 %
440
90.6%
or
) 45° 6 4.7620.44 V .63% =
e 100
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Maximum Efficiency Criterion
cos θ P aV I p P o = 2 Core Losses (eddy-current and I pη = 2 m = Pcu I pRe R 1 constant and called hysteresis) are e1 2 c o slosses Pin = aV Ip fixed ++ P I θ m pR e 2 1 P P m m I η p= I pfl = fo aV2I 2 I p c p l s θ R P c2u flas η = lossesI are Copper varying the pfl e1 + aV I2 p cosP θ I R+ m p 1e square of the current and called P d η m |max= | raIted2pη VA V A= → = 0 R P . Variable e f.f losses e 1 m P cufl dI p
Efficiency is zero at no load & it
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Example 4.7
A 120-kVA, 2400/240-V, step-down transformer has the (following a) parameters: R 1=0.75 Ω, X 1=0.8 Ω, R2=0.01 Ω, X2=0.02 Ω. The transformer is designed to operate at maximum k V A70% of1its 200 00 load with 0.8 power factor lagging. efficiency at rated = (b)=' (a) the =kVA rating of50 A ( I c) pDetermine the transformer at maximum V 2 2 4 0 0 ' maximum efficiency, (c) the efficiency at full efficiency, (b) the ' P = V p * *f P V I p f * = η 2I o* p o p f l 2 and 0.8 power0.factor lagging, and (d) the equivalent core Iload = I = = 0 . 7 * 7 * 5 0 3 5 A pη p loss resistance. Use the approximate equivalent circuit. =0*0 = 24 0*35*0 0*3 5*0.8 P =2 P = 400*50* 0*5 00 .8 96 0 00.8 W6 2 27200 W
(d)
o
o
=
V 35*2400 2400 1 2
=
=
Ω
2 Vcufl A |= = = 8 )4 4 k3V AW R P 5 (0 .7 5+1 0 *0 .0 1 7 5 + = P = P P + P = 72686.88 1487.5 W m ax .0 . e f f η icn1 o m cu 1000 P 2143.75 m P o 92 60 0 02 2 P 2 67 2 0 0 oa R η = = * = + = + Pc= I R ( ) 3 5 ( 0 . 7 5 1 0 * 01 .00 1) 93.6% * 1 0 0 9 4 % u η η p = =1 η 2 = P in 96000 4+ 375 + 2143.75
P in 71487.5 P cu η = 2143.75 W Pm = P cu =η
2143.75 W
Determination of Transformer Parameters The Open-Circuit Test
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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
Ic =Ioc cosocφ V oc V oc
R cL =
=
Ic
,= Im 2
X , = mL
Poc
2 Q oc = S oc− P oc
2
−1
P o c c S o
Ioc sφ inoc V oc V oc 2
=
Im
Qoc
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Determination of Transformer Parameters The Short-Circuit Test
The test is designed to determine the winding −1 P sc sc resistances resistance s and leakage reactances 2 S sc = V 2sc I *sc ,= sc φ cos ReH TheRtest a R , X X a X S sc H L e H H L s c is conducted by placing a short circuit across one winding and exciting the other with P V 2 2 sc sc Z = ,frequency I H R H RIeHL R=L 2 rated eH I I sc sc The2 applied voltage is carefully adjusted until 2 N 2 2 X R H a X Rrated 0 . 5 R , a X H L 0.5X current flows in the windings L e H H = − = Z R a , eH eH eH N It does not matter on which side the L test is conducted. But for safety it is conducted on the high-voltage side
=
+
=
+
=
=
=
=
=
eH
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Example 4.8 SC-Test OC-obtained from testing a 48-kVA, The following data were 4800/240-V. 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
PV oc *6I 0oc0 =sc 240 * 2= 480V VA 150 sc S oc = =equivalent = 6Ω=circuit , Z eHof = the =transformer = Ω 15 as viewed R Determine the 25R 2 0.5 * 6 = = Ω 0 . 3 ReH = H I H 1e0 10low-voltage 2high-voltage from (a) the side andI (b) the side sc 2 sc V 240 24 0 oc a 80=0= / 204 .05= 0 =.54X = Ω X RcL= 0 * .720 5 Ω62 .88 48 =123480 H= eH 2 2 15 5 X eH = Z = − 6 = 13.7Ω P eocH −2R120 12 eH 0 2 R0cH (480)= 192 k Ω .5 R =eH a RcL0.5*6 = 20 (48 02.00 .0075 Ω R L = R 2 2 =6 2 2 = 2 or 7.5 mΩ eH 4082 0(1− 1.9 24 0)= 4.876k Ω VA = S=oc a−2 2P =00.015Ω = 4496.58 =oc2 or 15 mΩ R eLQ= oc X 2a= X 2 2 3 .5 = mH mL 2 0 a 0.5 X 2eH 2 0.5*13.75 V 240 24 0 = = 0.01 .017 Ω or 17 mΩ mΩ X L = X 2oc 13.75 2 eH X = = = Ω 123.94 mL 20.034Ω 0 a= = or 34 mΩ X eL = 2Q 2 464.76 a oc 20
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Per-Unit Computations When an electric designed or analyzed appa apparrent ent powe pois wer r base base S b the machine using actual values of its parameters, it is not immediately obvious how its performance compare voltage V b the base with similar-type machine 2
Sb
Vb
V b
Expressing machine I b = Z b = ,parameters =in per-unit shows I b operates S b 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 a re selected, the remaining two can be calculated
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Example 4.9
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, 2300/1152300/115V, 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
-----------------------------------------------------------------------------------------------------------------------------------------------
Sec tio ion BA Section Se ction Section B°+ S e o n.− C I g p,uc t i= + = . ° . . −° S t e p u p : 2 . 3 Ω 6.9Ω = 2 3 0 0 V = , 4 6 0 0 0 V A V S A t full full load and . lagging pf bA bA
. . 23 m Ω 6 9 jmΩ 13.8kΩ 6.9kΩ 46000 VA V bA = 230 V,S b= A ° + ° + j 23 j1 .d . . ( . . 1 = 4 6 0 0 /w 2 3 0 0 = 2 0 A = ,+ Ω 0 / 0 5 .) I Z V b gAL,p, upuS I ° = − ° . a= n + t= e p -0d o n− 5 .7 5. m Ω 17.25mΩ 11.5kΩ == =2 , pu 2.3Ω bA 6.9Ω L
V bC 1 1 5 V S bC 4 6 0 0 0 V6A.9 , a n d 2 . 3 4°6 0 0 0/ 2= 30 . 2 20 0 0.0 A ,= 2 =30/ 2Ω 00 = 1.15 0.06 9.2kΩ I Z V bA = . . = ° b 2 .3 .9 = R 0 2 , X E = + + − ° = ° ( . j = )( )6 . A. = pu = X H ,H = , 6 pu = 0.0 = 2, 0.0 R =*.1 ° = 0 the 0.0 0(a) /51+°1 j 5.92=.410150Vvoltage, A 0 .2=8 7 5 IV Determine 1 5 .02 2 3.1 0generator .09 .0 generator 1 1111=5 b C = 4 60 C the 5 / 4 0 0 current, , Z b(b) ,u g p
l , pu pu
,
H pu
,
H pu
= = 0 .. 02 system .08 at full load and 0.866 Zl g , pu j − ° + + ° I 8 . .0 ,(c) g p= u − ° = ° I * . .− ° . . = − A 1 3 0 0 6 90 0 the overall efficiency of the p u= .+ j = = 1.15 = 20, 60 R cH pu, X = , mH1 pu 1 1 5 1 1 5 P o p, u =. , P in p,uRe .. . . power factor lagging. = °*. ° = E == j ) . j + + . .0 ° . − .°= .+(.+ . . ) 0 2 3+ 0.(06 9 g ,pu = = 2 . 0 7 4 . 1 4 j L. L6,pu u0.06 ,H c,,H u *= m.0 ,up = 0.0X 2j, =,L pu0.01 = R = = + 8 0 3 Z L p = = η . % l p u , = ° E g pu. . . 11 5 115 1 15 g
R R
0 .0 1 0 5175500 [ = = 0 .0 21,00,X X 0 .2 8171 55
0 .0 19722050]
=
= 0 .0860
0 .2 817155
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The Autotransforme Autotransformerr When the two windings of a transformer are interconnected 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 Autotransformer is cheaper in first cost than a conventional two-winding transformer of a similar rating Autotransformer delivers more power Autotransformer p ower than a twowinding transformer of similar physical dimensions For a similar power rating, an autotransformer is more efficient than a two-winding transformer
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The Autotransforme Autotransformerr Connections
The Autotransforme Autotransformerr a-ratio of Autotransform Autotransformer er =E ( N 1==+ V N )(Ia1=+ N 2+IE a 1 1 ) V E a 2 2 1= S 2 1+ S oa21a =V 2I 212a o a2 a I 2a N 1 +N I ( =E VI22aa === aE a I1a a T =2a1) + 1 += T 2a1a = S power N transfer transfer through through induction induction Io1a≡ power 2 I 11aa = I 1E 1a E 1 + E 2 V S Sooa = V 2aI 2a = = ≡ power transfer through conduction VSa2oa = V E E + I a ( 1) 2 2 a 2 1 V 1a
S oa = [aT I a1 =] V a1I a1 V 1a Na1T +(a N + 21)
1
VS2oa =S ina N a2
a
S o a == V2 2I
1= a1S=+o aT =
29
Example 4.10
Ading 24-kVA, distribution transformer is to be For the the 2-win 2-windi ng trans transff2400/240-V ormer ormer
connected as an autotransformer. For each possible V combination, So primary , the =2 4 k V A , I 1 1voltage, =0 A , I 2(b) 1the =0 0 A determine (a) winding 1 = 2400 V , V 2 =2 4 0 V secondary winding voltage, (c) the ratio of transformation, and =4E 00 000the 224V4V00=0autotransformer. (d) the nominal V E E V= = =0 224400 +2 +2664400 V V11aa ==rating E112 22++4=of E E 22 V V 22aa V a aT T
= 2 EE 0E 0222 24V 4 +101 2V 4000==0 2 24 40 0+ 2+ 26 64 40 0VV== +4 = E E
=
=
=
V V 11aa
24640400 22640
V 22aa V
22640 26444000
==
= VV22 a 22II
S ooaa S
=
S ooaa S
=
a
aa
1010.091 . 19 1 ==
=VV1=212 aa2121IIII a V1 =a 1I aa
= 2264 664400 **110 000 226= 64 64=4000 000 V 0A V A 2 62. 464 6 4= 100
226.4-kVA 6 4 - k V A , , 22640 46 04 0 // 2640 64 40 0--VV 240 /2240-V
30
A Nonideal Autotransfor Autotransformer mer
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 2a Ω, 2 4and 5° X =2.4 Ω, R =8.64 kΩ, constants: RH=18.9 Ω, XH=21.6 Ω, RL=2.1 c°H I pa = = 8 L4 5 = A and XmH =6.84 kΩ. The transformer is connected aT 0.25 as a 120/480-V, step-up V 1a 121.513 ° the full load at 0.707 13.63 autotransformer. If the autotransformer delivers I ca = 0 .1 .1 2 7 1 3 .6 .6=3 A Ic o m R = Ip a =−its I = ° 6 4 5 A power factor leading, determine efficiency and the voltage a0 2 96 cL regulation. E H = V 1aa EL121.513 ° 13.63 I m a = = − 0 . 1 6 7 6 . 3 7= A j X L ) 4 E L = m2LIj Xa ( R H 7 6+0j Xj H ) 2V +a I c o m− ( R L 120/ 48= 0 0.25=a 360/12 0/1 = 20 3 aT = P 678.82 o = = 0.803 or 80.3% η = 8640 P in 2= 845.3 Ω 960 R cL = 3 V 1a 121.513 6840 = 486.056 V V 760 X 2maLnL == 2== Ω 3 0.25 aT 720 I 2a =I H = −V 22a A V 2an= L 3 6 0 VR % = *100= 1.26% V 2a 24°5 A I 2a = 245
I a = Ic a + Im a 0.204 = 38 A − ° 448 5 = ° j + 4φ E 2 4 5 ( 1 8 . 9 2 1 . 6 ) 0+° 6 − ( 2 . L I1a = Ip a φ + I a8 4 5= 0 . 2 0°4 + 3 8 8 . 0−2 7 4 3° . 5= 6 A E = ° 1 19 1 9 .7 . 7 45 4 5 4 .5 . 5 7 V L * R e = = P V I 4 8 0 *= 2 4 5 * 6 7−8 . 8 2 ° W Vo1a = E2L a+ I2c(o ma RL j+X) L P = R e in
*
I 1 .513 2 1 = 1 3 .6 .6° 3 * 8 . 0 2 7 − 4 3 . 5 6
V 1a = 1 1 91 . 7 4 51 4a . 57 V *a
V 1a = 121 .513 13.6 3° V
6° 4+5 ( 2. 2.1 °
j2 . 4 )+
° 8=4
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 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 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 Y-∆ ∆-Y ∆-∆ Connection Core Type Shell-Type
34
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 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
35
Example 4.12
S 3Aφ three-phase = 3360/120-V, = 2 1is6assembled * 72 7transformer 2 0single-phase 0transformers. V Aby oconnecting rThe constants 2.1three 6 for k720-VA, Veach A transformer 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 ((a) cb)) For FFor o rr∆Y-Y Y∆-Y -∆ connection cconnection oonne n nnec ection configuration. (d) Fo on c ti on
V 11 L L == 3630**3V6 03 =6=62234.5V 4V V 2 L L == 1230*1 **1V2 03 ==220078.8V 5V V The nominal nominal n ominal Ratingsare nom inal Ratings arare e
.16-kVA, 24/1 /120 /20V, V, - connection connection nnection ∆Y-Y 2.16-kVA, 2.16624/208connect ion 2.16 2. kVA, kVA, 624/208-V, 6624 --V, Yconn co nnec ecti tion on ∆∆-Y 2.16-kVA, 360/208-V, 360/208∆ connection connect ion
36
Example 4.13
ThreeY single-phase 60/∆ Connetransformers, ction Peach rimarated ry at 12-kVA, Secon120/240-V, dary Hz, are connected to form a three-phase, step-up, Y-∆ connection. The parameters transformer are R mΩ, XH=201 Phasof e each Volta ge 1H2=133.5 0 240 mΩ, RL=39.5 mΩ, and XL=61.5 mΩ, RcH =240 Ω, and XmH =290Ω. What are the nominal Licurrent, ne Voland tagpower e 08three-phase transformer. 240 voltage, ratings of 2 the When it delivers the rated load at the rated voltage and 0.8 pf lagging, P h ase Current 100 50 determine the line voltages, the line currents, and the efficiency of the transformer.
Line Current
100
86.6
*Y-Y For equivalent connection VY E I (0.0 (0 .03 3 9 0r8 .06 .0 66 5r6 )y78)6.6 3S = + + E V I ( 0 . 0 454je5P .1a04*8 = + P V I = = 3Re 3R e * 3Re 3R 1[+3 .j0 5 4* 6.e 8°7 n n pA p A 1 1 / Y C o n n e c t i o n im con]dary 2n 2n 2 A2A o 2n 1.9 382.5 .56 64° 00 V6.8 = V 2n V 1 2 5 .7 3 0 .92 1 0 (0.0 (0 .06 .0 = ° + − + Eho12na=n s28800 38o.W 5ta6g 4ore028.8 6.6°.87 62 .0 8.03 73°9(50.0j 4041 53618 =e1V ° + 8kW −73 +5.j)506.0467 ) P P l 1 I = 86.6 =− 36°.87 A °*7 =V3Re V 1n =132 A 2.613 .61 3 1.97 1.9 P V I 3Re 3R e * 2.0 6131 13 = E = ° Ve 1[ 32 1 4 5 . 1 4 7 Linin2en Voltag1n1e201A0.923R 8 1.9°7 *100.722460.8° 8 ] aV 0.866 =1n 3°0 = 229.6 = = V 3 9 6° 1.97 V 1 L P in = 100. 100.72 72 6.88 6.88 A − 138.564 P Eh2a Ls=e C3uE rr2e nt 100 86.6 n 3 0 ° = 2 5 1 .4 3 0 . 9 2 °V − 1 86.6 36.8 .87 7° 1 28.8 I I u n t 3902+.3% .72 6.8 86A == + == −1 3 0 0°00 6.876.88 A = ° − °1 = *r1V 010 .3 % 1 p pA pAA Lη IiE n=1Ae C r e n 1 0 0 8 .6 a E * 1 2 5 . 7 3 0 . 9 2 V ° = 2 n 0.866 n 31.2 240 j290
37
End of Chapter Four
