Transformer
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Transformer Professor Mohamed A. ElSharkawi
Why do we need transformers? • Increase voltage of generator’s output – Transmit high power at low current – Reduce cost of transmission system
• • • •
Adjust voltage to a usable level Create electrical isolation Match load impedance Filters El-Sharkawi@University of Washington
2
220kV-750kV Distribution Transformer
15 kV- 25kV
Service Transformer
Transmission Transformer 208V- 416V
El-Sharkawi@University of Washington
3
Transmission Transformer
El-Sharkawi@University of Washington
4
Distribution Transformer
El-Sharkawi@University of Washington
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Distribution Transformer
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Service Transformer
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7
Service Transformer bank
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Service Transformer bank
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9
Service Transformer
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Service Transformer
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11
Service Transformer
El-Sharkawi@University of Washington
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Service Transformer
El-Sharkawi@University of Washington
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Service Transformer
El-Sharkawi@University of Washington
14
Low power Transformer
El-Sharkawi@University of Washington
15
Basic Components Iron Core
Insulated Copper Wire
El-Sharkawi@University of Washington
16
Basic Components Laminated iron core
Insulated copper wire
El-Sharkawi@University of Washington
17
El-Sharkawi@University of Washington
18
El-Sharkawi@University of Washington
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El-Sharkawi@University of Washington
20
dφ e1 = N1 dt 1 φ = e1 dt ∫ N1
dφ e2 = N 2 dt
i1
e1
i2 N1
N2
Primary
e2
Secondary El-Sharkawi@University of Washington
21
Basic Analysis:Voltage dφ N1 e1 ( t ) N1 dt = = e2 ( t ) N dφ N 2 2 dt
E1 N1 = E2 N 2
i1
e1
+ _
N1
N2
i2 + _
e2
E1 E2 = N1 N 2
• Volts/turn is constant • Voltages are in phase (no phase shift) • Voltage magnitudes vary with turns ratio. El-Sharkawi@University of Washington
22
Basic Analysis: Power and current i1
S1 = S 2
EI =E I * 1 1
* 1 * 2
* 2 2
I E2 N 2 = = I E1 N1
e1
+ _
N1
I1 N 2 = I 2 N1
N2
i2 + _
e2
N1 I 1 = N 2 I 2
• Currents are in phase. • Current ratio is opposite to the voltage ratio El-Sharkawi@University of Washington
23
Basic Analysis: Reflected impedance Flux I2
I1
Load
Source
E1
N1
N2
E2
Zload
Secondary
Primary
Z load El-Sharkawi@University of Washington
E2 = I2
24
Basic Analysis: Reflected impedance I1 Source
Z
E1 Primary
Z
' load
' load
E1 = I1
Z E1 I 2 N1 = = Z load E2 I1 N 2 ' load
Z
' load
N1 = Z load N2
El-Sharkawi@University of Washington
2
2
25
Single-Phase, Ideal Transformer Ratings
I1
Apparent Power 2 KVA, 120/240 V Primary Voltage
+ V1
N1
I2
N2
+ V2
Secondary Voltage
El-Sharkawi@University of Washington
26
Rated Values • Rated voltage: The device can continuously operate at the rated voltage without being damaged due to insulation failure • Rated current: The device can continuously operate at the rated current without being damaged due to thermal destruction El-Sharkawi@University of Washington
27
Example I1 Transformer rating: 2 KVA, 240/120 V Compute the currents
S = V1 I1 = V2 I 2 = 2 KVA
+ V1
N1
I2 N2
+ V2
S
2 KVA I2 = = = 16.67 A V2 120 V
S
2 KVA I1 = = = 8.33 A V1 240 V El-Sharkawi@University of Washington
28
Multi-Secondary Transformer
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29
Multi-secondary windings
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I3 N3
I1 E1
I2
N1 N2
Primary
E3
E2
E1 N1 = E2 N 2 E1 N1 = E N 3 3 El-Sharkawi@University of Washington
31
Current ratio: superposition
I12
I12 E1
N2 = I2 N1
I2
N1 N2
E2
Primary
El-Sharkawi@University of Washington
32
Current ratio: superposition
I3 N3
I13 E1
N1
E3
N3 I13 = I 3 N1
Primary
El-Sharkawi@University of Washington
33
I3
N3
I1 E1
Superposition
E3
I2
N1 N2
E2
Primary
N3 N2 I1 = I12 + I13 = I 2 + I3 N1 N1 I 1 N1 = I 2 N 2 + I 3 N 3 El-Sharkawi@University of Washington
34
I3 N3
I1 E1
Superposition
E3 I2
N1 N2
E2
Primary
S1 = S 3 + S 3 EI =E I +EI * 1 1
* 2 2
El-Sharkawi@University of Washington
* 3 3 35
Example •
The transformer consists of one primary winding and two secondary windings. The number of turns is each winding is
N1 = 4000; N 2 =1000; N 3 = 500 A voltage source of 120V is applied to the primary winding, and purely resistive loads are connected across the secondary windings. A wattmeter placed in the primary circuit measures 300W. Another wattmeter placed in the secondary winding N2 measures 90W. Compute the following: • • •
The voltages of the secondary windings The currents in N3 The power consumed by the load connected across N3
El-Sharkawi@University of Washington
36
Solution N2 1000 E2 = E1 =120 = 30 N1 4000 N3 500 E3 = E1 =120 = 15 N1 4000
P1 300 I1 = = = 2.5 E1 cos θ1 120
P2 90 I2 = = =3 E 2 cos θ 2 30
El-Sharkawi@University of Washington
37
Solution I1 N1 = I 2 N 2 + I 3 N 3 2.5 * 4000 = 1000 * 3 + 500 I 3
I 3 = 14 E1 I1 = E 2 I 2 + E3 I 3 120 * 2.5 = 30 * 3 +15 I 3
I 3 = 14 El-Sharkawi@University of Washington
38
Solution P1 = P2 + P3 300 = 90 + P3
P3 = 210 El-Sharkawi@University of Washington
39
Autotransformer
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I1
V1
A1 E1
I2
B1 N1
E2
N2
V2
A2 B2
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41
Autotransformer: Voltage and current Is
I load = I1 + I 2
A1
N1
I1
E1
V1 = E1 + E2
Iload
A2
V1
B1
E2
N2
I2
V2
B2 El-Sharkawi@University of Washington
42
I1 + E1
N1
I2 N2
Autotransformer + E2
Is
E1
I1 Iload
A2
V1
S A = E1 I1 = E2 I 2
N1
A1
E2
N2
B1
I2
V2
B2
S B = V1 I s = V2 I load El-Sharkawi@University of Washington
43
Autotransformer: Power S B = V1 I s = ( E1 + E2 ) I1 = E1 I1 + E2 I1 S B = S A + E 2 I1
SB > S A El-Sharkawi@University of Washington
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Example Ratings of regular transformer: 10 kVA, 400/200 V New voltage ratio: 600/200 V Compute the new ratings
Solution
Is N1 E1
10 I2 = = 50 A 0.2 10 I1 = = 25 A 0.4
E2
El-Sharkawi@University of Washington
I1 Iload
A2
V1
S B = V1 I1 = 600 × 25 = 15 kVA
A1
N2
B1
I2
V2
B2
45
VARIC: Variable Auto-Transformer Z
Is
N3 Y
V1
N1
N2
Sliding terminal
I1
Iload I2
V2
El-Sharkawi@University of Washington
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Output Voltage Vload
N2 = Vs N1 + N 2
Z Is
N3 Y
At Y
Vload
Sliding terminal
I1 N1
N1 + N 2 = Vs = Vs N1 + N 2
Iload
V1 N2
I2
V2
At Z
Vload
N1 + N 2 + N 3 = Vs N1 + N 2 El-Sharkawi@University of Washington
47
Three-Phase Transformer
El-Sharkawi@University of Washington
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3-phase transformer
El-Sharkawi@University of Washington
49
3-phase transformer Y-Y connection. Also known as star-star connection A
a
N1
N2
c
Van N1 = VAN N2
N
b C Vac = VAC
B 3 Van N1 = N2 3 V AN
El-Sharkawi@University of Washington
o Ve ni L f o oit a R
l o Ves a h P f o oit a R
n
50
3-phase transformer (∆ -∆ ) A
a
N2
N1
c
B
b
oit a R
C
Vac N1 = VAC N2
El-Sharkawi@University of Washington
51
3-phase transformer (Y-∆ ) Also known as star-delta connection a
Ves a h P f o oit a R
n
c
Van N1 = VAC N2
N2
B
b
C
Vac 3 Van = = VAC VAC El-Sharkawi@University of Washington
3 N1 N2
tl o Ve ni L f o oit a R
N1
A
52
3-phase transformer bank (Y∆ ) a V AB N2 = Van N1
Vab
Van
N1
N2
A
VAB
B
b
V AB V AB = = Vab 3 Van
N2 3 N1
N1
N2
C
c N1 El-Sharkawi@University of Washington
N2 53
El-Sharkawi@University of Washington
54
Ratings of Ideal 3-phase Transformer Apparent Power (3-phase) 100 MVA, 13.8/138 KV Primary Voltage line-to-line
Secondary Voltage line-to-line
El-Sharkawi@University of Washington
55
Example • Three single-phase transformers are used to form a threephase transformer bank. Each single-phase transformer is rated at 10 kVA, 13.8 KV/240 V. One side of the transformer bank is connected to a three-phase, 13.8 kV transmission line. The other side of the transformer is connected to a three-phase residential load of 415.7V, 9kVA at 0.8 power factor lagging. Determine the connection of the transformer bank, the voltage ratio of the transformer bank, and the line current of the bank at the 13.8 kV side El-Sharkawi@University of Washington
56
Solution • Secondary voltage (Low voltage side) should be in Y to provide the needed residential voltage 415.7 = 240 V 3
• The high voltage side must be Delta-connection – The line-to-line voltage of the supply is 13.8 kV. Same as the transformer rating of the primary. – If the primary is connected in Y, the voltage of the load would be lower than 240 V. El-Sharkawi@University of Washington
57
Solution
Van = 240 V
A
VAB = 13.8 kV
N1
N2
B
Van
a Vab
b N1
N2
C
c El-Sharkawi@University of Washington
58
VAB VAB = = Vab 3 Van
Solution 13,800 3 240
Phase current of the load
I2 =
9000 3 240
Van = 240 V
A
VAB = 13.8 kV
N1
Van V ab
B
b N1
= 12.5 A
N2
a
N2
C
c
Phase current of the Transformer primary
N2 240 I1 = I 2 = 12.5 = 0.2174 A N1 13,800
Line Current in primary
I A = 3 I1 = 0.377 A
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Actual Transformer • Windings:
i2
i1
– Resistance – Inductance
e1
+ _
N1
N2
+ _
e2
• Core: – Eddy Current – Hysteresis
El-Sharkawi@University of Washington
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Windings Impedance R1
X1
N1
R2
X2
N2
Ideal Transformer El-Sharkawi@University of Washington
61
Core Hysteresis i
B
+ e _
N H
B= f
( ∫ e dt )
H = f (i)
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∫e
i
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i
Core Model Let e = Emax sin ωt
e
Emax ∫ e dt = − ω cos ωt
e Emax i= = sin ωt R R El-Sharkawi@University of Washington
R ∫e
i
64
i
Core Model Let
e
e = Emax sin ωt Emax ∫ e dt = − ω cos ωt
di e=L dt 1 Emax i = ∫ e dt = − cos ωt L ωL El-Sharkawi@University of Washington
Xl ∫e
i
65
i e
R
Xl
∫e
∫e
i
El-Sharkawi@University of Washington
i
66
Equivalent Circuit ' 2
N1 N2
R2
Io
I1 Ro
Xo
E1 N1 V1 = ≠ E2 N 2 V2
E1
E2
X2
I2
V2
' I2
N2 I1 = ≠ I2 N1 I2
El-Sharkawi@University of Washington
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load
R1
V1
I
X1
Referred impedance
V1
I
X1
R1
' 2
N1 N2
R2
Io
I1 Ro
Xo
E1
E2
X2
I2
V2
N1 N1 [ E1 = E2 = I 2 ( R2 + jX 2 ) + V2 ] N2 N2 El-Sharkawi@University of Washington
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Referred impedance
V1
I
X1
R1
' 2
N1 N2
R2
Io
I1 Ro
Xo
E1
E2
X2
I2
V2
N1 N1 ' N1 ( R2 + jX 2 ) + V2 E1 = E2 = I 2 N2 N2 N2 El-Sharkawi@University of Washington
69
N1 N1 ' N1 ( R2 + jX 2 ) + V2 E1 = E2 = I 2 N2 N2 N2 2
Define: R'2 =
N1 N1 ( R2 + jX 2 ) + E1 = I V2 N2 N2 ' 2
2
N1 R2 N2 2
N1 X 2 N2 ' N1 V2 = V2 N2 X '2 =
Then:
(
)
E1 = I 2' R2' + jX 2' + V2' El-Sharkawi@University of Washington
70
Equivalent Circuit Referred to Source Side
X1
R1
Ro
R2
N1 N2 E1
Xo
R'2
Io
I1 Ro
Xo
E1
I 2'
El-Sharkawi@University of Washington
X2
I2
E2
X1
R1
V1
' 2
Io
I1
V1
I
E1 = I 2' ( R2' + jX 2' ) + V2'
V2 X '2
V2' 71
Practical Considerations '
X1
R1
V1
R2
Io
I1 Ro
' R1 << Ro >> R2 ' X 1 << X o >> X 2
Xo
E1
I
' 2
I1 =
V2'
' I2 + Io
' I 2 >> I o
El-Sharkawi@University of Washington
X '2
<< I1 72
X1
R1
V1
Io
I1 Ro R1
V1
X '2
R'2
I
' 2
V2'
Xo
X1
R'2
I1
X '2
I 2' Io
Ro
El-Sharkawi@University of Washington
Xo
V2'
73
R1
V1
Define:
X1
R'2
X '2
I 1 ≈ I '2
V2'
' Req = R1 + R2 ' X eq = X 1 + X 2 El-Sharkawi@University of Washington
74
Analysis of Transformer X eq
Req
' I1 ≈ I 2
V1
' V1 = V2
+
' I2
V2'
Z
( Req + jX eq )
El-Sharkawi@University of Washington
75
V2 N1 V = V2 N2 ' 2
Terminologies Load Voltage Load Voltage referred to Source side Impedance referred to Source side 2
Load Current
I2
Load current referred to Source side
N2 I = I2 N1 ' 2
El-Sharkawi@University of Washington
N1 R2 R = N2 ' 2
2
N1 X 2 X = N2 ' 2
76
Analysis of Transformer X eq
Req
' I1 ≈ I 2
V1
' V1 = V2
+
' I2
V2'
Z’
( Req + jX eq )
El-Sharkawi@University of Washington
77
Req
X eq
' I1 ≈ I 2
V1
V2'
Z’
' I 2 X eq
' I 2 Req
' I2
El-Sharkawi@University of Washington
78
Req
V1
X eq
' I1 ≈ I 2
V2'
Z
V1 = V2' + I 2' Req + j I 2' X eq
δ
θ
V1 I '2 Z eq
V2'
I '2 X eq
I '2 Req
' I2 El-Sharkawi@University of Washington
79
Ratings of Actual 3phase Transformer Apparent Power (3-phase) 100 MVA, 13.8/138 KV ' 2
V
V2
line-to-line
line-to-line
El-Sharkawi@University of Washington
80
Example A transformer has the following parameters:
N1 =10 N2 Req = R1 + R2' =1Ω;
Z l = 0.5∠30 Ω o
X eq = X 1 + X 2' =10Ω; Ro =1000Ω;
X 0 = 5000Ω
The rated voltage of the primary winding is 1000V. Compute the load voltage.
El-Sharkawi@University of Washington
81
Solution
' I1 ≈ I 2
2
N1 = 50 Ω Z = Z L N2 ' L
X eq
Req
V1
V2'
Z’
0 V 1000 ∠ 0 o 1 I 2' = = = 17 . 7 ∠ − 38 . 31 A ' o ( Req + j X eq ) + Z L (1+ j 10) + 50 ∠30
V = I Z = (17.7 ∠ − 38.31 ' 2
' 2
' L
o
) (50 ∠30 ) = 885 ∠ − 8.31 o
0
V
N2 1 V2 = V = 885 = 88.5 V N1 10 ' 2
El-Sharkawi@University of Washington
82
Req
V1
eq
' I1 ≈ I 2
Vno load − V full load VR ≡ V full load Measured at the load side
V2'
da o L
Voltage Regulation VR X
V1 − V2' VR ≡ ' V2
El-Sharkawi@University of Washington
83
Example Calculate the voltage regulation of the transformer in the previous problem.
Solution:
VR ≡
' V1 − V2 ' V2
7209.5 − 7200 VR = × 100 = 0.14 % 7200 El-Sharkawi@University of Washington
84
Efficiency
η
Xeq
Req
V1
Io
I1 Ro
Output Power Pout η = = Input Power Pin
Pin = Pout + Losses
I
Xo
Pcu =
' 2
( )
' 2 I2
V2'
Req
V12 Piron = Ro
Plosses = Pcu + Piron
Pout = V2' I '2 cos θ El-Sharkawi@University of Washington
85
Example A 10 kVA, 2300/230 V, single phase distribution transformer has the following parameters:
R1 = 5.8 Ω ; X 1 = X '2 = 12 Ω ; R'2 = 6.05 Ω Ro = 75.6 kΩ ; X o = 69.4 kΩ At full load and 0.8 power factor lagging, compute the efficiency of the transformer.
El-Sharkawi@University of Washington
86
Solution Pout = V2' I '2 cos θ = S cos θ = 10 × 0.8 = 8 kW ' I2
Pcu =
( )
' 2 I2
Piron
S
10 ,000 = = = 4.35 A ' 2300 V2
Req = ( 4.35 )2 × ( 5.8 + 6.05 ) = 224.23 W
( )
( V V 2300 ) = ≈ = = 70 W Ro Ro 75,600 2 1
' 2 2
2
Pout 8000 η = = × 100 = 96.45 % Pin 8000 + 224.23 + 70 El-Sharkawi@University of Washington
87
Why do we need transformers? • Increase voltage of generator’s output – Transmit high power at low current – Reduce cost of transmission system
• • • •
Adjust voltage to a usable level Create electrical isolation Match load impedance Filters El-Sharkawi@University of Washington
2
220kV-750kV Distribution Transformer
15 kV- 25kV
Service Transformer
Transmission Transformer 208V- 416V
El-Sharkawi@University of Washington
3
Transmission Transformer
El-Sharkawi@University of Washington
4
Distribution Transformer
El-Sharkawi@University of Washington
5
Distribution Transformer
El-Sharkawi@University of Washington
6
Service Transformer
El-Sharkawi@University of Washington
7
Service Transformer bank
El-Sharkawi@University of Washington
8
Service Transformer bank
El-Sharkawi@University of Washington
9
Service Transformer
El-Sharkawi@University of Washington
10
Service Transformer
El-Sharkawi@University of Washington
11
Service Transformer
El-Sharkawi@University of Washington
12
Service Transformer
El-Sharkawi@University of Washington
13
Service Transformer
El-Sharkawi@University of Washington
14
Low power Transformer
El-Sharkawi@University of Washington
15
Basic Components Iron Core
Insulated Copper Wire
El-Sharkawi@University of Washington
16
Basic Components Laminated iron core
Insulated copper wire
El-Sharkawi@University of Washington
17
El-Sharkawi@University of Washington
18
El-Sharkawi@University of Washington
19
El-Sharkawi@University of Washington
20
dφ e1 = N1 dt 1 φ = e1 dt ∫ N1
dφ e2 = N 2 dt
i1
e1
i2 N1
N2
Primary
e2
Secondary El-Sharkawi@University of Washington
21
Basic Analysis:Voltage dφ N1 e1 ( t ) N1 dt = = e2 ( t ) N dφ N 2 2 dt
E1 N1 = E2 N 2
i1
e1
+ _
N1
N2
i2 + _
e2
E1 E2 = N1 N 2
• Volts/turn is constant • Voltages are in phase (no phase shift) • Voltage magnitudes vary with turns ratio. El-Sharkawi@University of Washington
22
Basic Analysis: Power and current i1
S1 = S 2
EI =E I * 1 1
* 1 * 2
* 2 2
I E2 N 2 = = I E1 N1
e1
+ _
N1
I1 N 2 = I 2 N1
N2
i2 + _
e2
N1 I 1 = N 2 I 2
• Currents are in phase. • Current ratio is opposite to the voltage ratio El-Sharkawi@University of Washington
23
Basic Analysis: Reflected impedance Flux I2
I1
Load
Source
E1
N1
N2
E2
Zload
Secondary
Primary
Z load El-Sharkawi@University of Washington
E2 = I2
24
Basic Analysis: Reflected impedance I1 Source
Z
E1 Primary
Z
' load
' load
E1 = I1
Z E1 I 2 N1 = = Z load E2 I1 N 2 ' load
Z
' load
N1 = Z load N2
El-Sharkawi@University of Washington
2
2
25
Single-Phase, Ideal Transformer Ratings
I1
Apparent Power 2 KVA, 120/240 V Primary Voltage
+ V1
N1
I2
N2
+ V2
Secondary Voltage
El-Sharkawi@University of Washington
26
Rated Values • Rated voltage: The device can continuously operate at the rated voltage without being damaged due to insulation failure • Rated current: The device can continuously operate at the rated current without being damaged due to thermal destruction El-Sharkawi@University of Washington
27
Example I1 Transformer rating: 2 KVA, 240/120 V Compute the currents
S = V1 I1 = V2 I 2 = 2 KVA
+ V1
N1
I2 N2
+ V2
S
2 KVA I2 = = = 16.67 A V2 120 V
S
2 KVA I1 = = = 8.33 A V1 240 V El-Sharkawi@University of Washington
28
Multi-Secondary Transformer
El-Sharkawi@University of Washington
29
Multi-secondary windings
El-Sharkawi@University of Washington
30
I3 N3
I1 E1
I2
N1 N2
Primary
E3
E2
E1 N1 = E2 N 2 E1 N1 = E N 3 3 El-Sharkawi@University of Washington
31
Current ratio: superposition
I12
I12 E1
N2 = I2 N1
I2
N1 N2
E2
Primary
El-Sharkawi@University of Washington
32
Current ratio: superposition
I3 N3
I13 E1
N1
E3
N3 I13 = I 3 N1
Primary
El-Sharkawi@University of Washington
33
I3
N3
I1 E1
Superposition
E3
I2
N1 N2
E2
Primary
N3 N2 I1 = I12 + I13 = I 2 + I3 N1 N1 I 1 N1 = I 2 N 2 + I 3 N 3 El-Sharkawi@University of Washington
34
I3 N3
I1 E1
Superposition
E3 I2
N1 N2
E2
Primary
S1 = S 3 + S 3 EI =E I +EI * 1 1
* 2 2
El-Sharkawi@University of Washington
* 3 3 35
Example •
The transformer consists of one primary winding and two secondary windings. The number of turns is each winding is
N1 = 4000; N 2 =1000; N 3 = 500 A voltage source of 120V is applied to the primary winding, and purely resistive loads are connected across the secondary windings. A wattmeter placed in the primary circuit measures 300W. Another wattmeter placed in the secondary winding N2 measures 90W. Compute the following: • • •
The voltages of the secondary windings The currents in N3 The power consumed by the load connected across N3
El-Sharkawi@University of Washington
36
Solution N2 1000 E2 = E1 =120 = 30 N1 4000 N3 500 E3 = E1 =120 = 15 N1 4000
P1 300 I1 = = = 2.5 E1 cos θ1 120
P2 90 I2 = = =3 E 2 cos θ 2 30
El-Sharkawi@University of Washington
37
Solution I1 N1 = I 2 N 2 + I 3 N 3 2.5 * 4000 = 1000 * 3 + 500 I 3
I 3 = 14 E1 I1 = E 2 I 2 + E3 I 3 120 * 2.5 = 30 * 3 +15 I 3
I 3 = 14 El-Sharkawi@University of Washington
38
Solution P1 = P2 + P3 300 = 90 + P3
P3 = 210 El-Sharkawi@University of Washington
39
Autotransformer
El-Sharkawi@University of Washington
40
I1
V1
A1 E1
I2
B1 N1
E2
N2
V2
A2 B2
El-Sharkawi@University of Washington
41
Autotransformer: Voltage and current Is
I load = I1 + I 2
A1
N1
I1
E1
V1 = E1 + E2
Iload
A2
V1
B1
E2
N2
I2
V2
B2 El-Sharkawi@University of Washington
42
I1 + E1
N1
I2 N2
Autotransformer + E2
Is
E1
I1 Iload
A2
V1
S A = E1 I1 = E2 I 2
N1
A1
E2
N2
B1
I2
V2
B2
S B = V1 I s = V2 I load El-Sharkawi@University of Washington
43
Autotransformer: Power S B = V1 I s = ( E1 + E2 ) I1 = E1 I1 + E2 I1 S B = S A + E 2 I1
SB > S A El-Sharkawi@University of Washington
44
Example Ratings of regular transformer: 10 kVA, 400/200 V New voltage ratio: 600/200 V Compute the new ratings
Solution
Is N1 E1
10 I2 = = 50 A 0.2 10 I1 = = 25 A 0.4
E2
El-Sharkawi@University of Washington
I1 Iload
A2
V1
S B = V1 I1 = 600 × 25 = 15 kVA
A1
N2
B1
I2
V2
B2
45
VARIC: Variable Auto-Transformer Z
Is
N3 Y
V1
N1
N2
Sliding terminal
I1
Iload I2
V2
El-Sharkawi@University of Washington
46
Output Voltage Vload
N2 = Vs N1 + N 2
Z Is
N3 Y
At Y
Vload
Sliding terminal
I1 N1
N1 + N 2 = Vs = Vs N1 + N 2
Iload
V1 N2
I2
V2
At Z
Vload
N1 + N 2 + N 3 = Vs N1 + N 2 El-Sharkawi@University of Washington
47
Three-Phase Transformer
El-Sharkawi@University of Washington
48
3-phase transformer
El-Sharkawi@University of Washington
49
3-phase transformer Y-Y connection. Also known as star-star connection A
a
N1
N2
c
Van N1 = VAN N2
N
b C Vac = VAC
B 3 Van N1 = N2 3 V AN
El-Sharkawi@University of Washington
o Ve ni L f o oit a R
l o Ves a h P f o oit a R
n
50
3-phase transformer (∆ -∆ ) A
a
N2
N1
c
B
b
oit a R
C
Vac N1 = VAC N2
El-Sharkawi@University of Washington
51
3-phase transformer (Y-∆ ) Also known as star-delta connection a
Ves a h P f o oit a R
n
c
Van N1 = VAC N2
N2
B
b
C
Vac 3 Van = = VAC VAC El-Sharkawi@University of Washington
3 N1 N2
tl o Ve ni L f o oit a R
N1
A
52
3-phase transformer bank (Y∆ ) a V AB N2 = Van N1
Vab
Van
N1
N2
A
VAB
B
b
V AB V AB = = Vab 3 Van
N2 3 N1
N1
N2
C
c N1 El-Sharkawi@University of Washington
N2 53
El-Sharkawi@University of Washington
54
Ratings of Ideal 3-phase Transformer Apparent Power (3-phase) 100 MVA, 13.8/138 KV Primary Voltage line-to-line
Secondary Voltage line-to-line
El-Sharkawi@University of Washington
55
Example • Three single-phase transformers are used to form a threephase transformer bank. Each single-phase transformer is rated at 10 kVA, 13.8 KV/240 V. One side of the transformer bank is connected to a three-phase, 13.8 kV transmission line. The other side of the transformer is connected to a three-phase residential load of 415.7V, 9kVA at 0.8 power factor lagging. Determine the connection of the transformer bank, the voltage ratio of the transformer bank, and the line current of the bank at the 13.8 kV side El-Sharkawi@University of Washington
56
Solution • Secondary voltage (Low voltage side) should be in Y to provide the needed residential voltage 415.7 = 240 V 3
• The high voltage side must be Delta-connection – The line-to-line voltage of the supply is 13.8 kV. Same as the transformer rating of the primary. – If the primary is connected in Y, the voltage of the load would be lower than 240 V. El-Sharkawi@University of Washington
57
Solution
Van = 240 V
A
VAB = 13.8 kV
N1
N2
B
Van
a Vab
b N1
N2
C
c El-Sharkawi@University of Washington
58
VAB VAB = = Vab 3 Van
Solution 13,800 3 240
Phase current of the load
I2 =
9000 3 240
Van = 240 V
A
VAB = 13.8 kV
N1
Van V ab
B
b N1
= 12.5 A
N2
a
N2
C
c
Phase current of the Transformer primary
N2 240 I1 = I 2 = 12.5 = 0.2174 A N1 13,800
Line Current in primary
I A = 3 I1 = 0.377 A
El-Sharkawi@University of Washington
59
Actual Transformer • Windings:
i2
i1
– Resistance – Inductance
e1
+ _
N1
N2
+ _
e2
• Core: – Eddy Current – Hysteresis
El-Sharkawi@University of Washington
60
Windings Impedance R1
X1
N1
R2
X2
N2
Ideal Transformer El-Sharkawi@University of Washington
61
Core Hysteresis i
B
+ e _
N H
B= f
( ∫ e dt )
H = f (i)
El-Sharkawi@University of Washington
62
∫e
i
El-Sharkawi@University of Washington
63
i
Core Model Let e = Emax sin ωt
e
Emax ∫ e dt = − ω cos ωt
e Emax i= = sin ωt R R El-Sharkawi@University of Washington
R ∫e
i
64
i
Core Model Let
e
e = Emax sin ωt Emax ∫ e dt = − ω cos ωt
di e=L dt 1 Emax i = ∫ e dt = − cos ωt L ωL El-Sharkawi@University of Washington
Xl ∫e
i
65
i e
R
Xl
∫e
∫e
i
El-Sharkawi@University of Washington
i
66
Equivalent Circuit ' 2
N1 N2
R2
Io
I1 Ro
Xo
E1 N1 V1 = ≠ E2 N 2 V2
E1
E2
X2
I2
V2
' I2
N2 I1 = ≠ I2 N1 I2
El-Sharkawi@University of Washington
67
load
R1
V1
I
X1
Referred impedance
V1
I
X1
R1
' 2
N1 N2
R2
Io
I1 Ro
Xo
E1
E2
X2
I2
V2
N1 N1 [ E1 = E2 = I 2 ( R2 + jX 2 ) + V2 ] N2 N2 El-Sharkawi@University of Washington
68
Referred impedance
V1
I
X1
R1
' 2
N1 N2
R2
Io
I1 Ro
Xo
E1
E2
X2
I2
V2
N1 N1 ' N1 ( R2 + jX 2 ) + V2 E1 = E2 = I 2 N2 N2 N2 El-Sharkawi@University of Washington
69
N1 N1 ' N1 ( R2 + jX 2 ) + V2 E1 = E2 = I 2 N2 N2 N2 2
Define: R'2 =
N1 N1 ( R2 + jX 2 ) + E1 = I V2 N2 N2 ' 2
2
N1 R2 N2 2
N1 X 2 N2 ' N1 V2 = V2 N2 X '2 =
Then:
(
)
E1 = I 2' R2' + jX 2' + V2' El-Sharkawi@University of Washington
70
Equivalent Circuit Referred to Source Side
X1
R1
Ro
R2
N1 N2 E1
Xo
R'2
Io
I1 Ro
Xo
E1
I 2'
El-Sharkawi@University of Washington
X2
I2
E2
X1
R1
V1
' 2
Io
I1
V1
I
E1 = I 2' ( R2' + jX 2' ) + V2'
V2 X '2
V2' 71
Practical Considerations '
X1
R1
V1
R2
Io
I1 Ro
' R1 << Ro >> R2 ' X 1 << X o >> X 2
Xo
E1
I
' 2
I1 =
V2'
' I2 + Io
' I 2 >> I o
El-Sharkawi@University of Washington
X '2
<< I1 72
X1
R1
V1
Io
I1 Ro R1
V1
X '2
R'2
I
' 2
V2'
Xo
X1
R'2
I1
X '2
I 2' Io
Ro
El-Sharkawi@University of Washington
Xo
V2'
73
R1
V1
Define:
X1
R'2
X '2
I 1 ≈ I '2
V2'
' Req = R1 + R2 ' X eq = X 1 + X 2 El-Sharkawi@University of Washington
74
Analysis of Transformer X eq
Req
' I1 ≈ I 2
V1
' V1 = V2
+
' I2
V2'
Z
( Req + jX eq )
El-Sharkawi@University of Washington
75
V2 N1 V = V2 N2 ' 2
Terminologies Load Voltage Load Voltage referred to Source side Impedance referred to Source side 2
Load Current
I2
Load current referred to Source side
N2 I = I2 N1 ' 2
El-Sharkawi@University of Washington
N1 R2 R = N2 ' 2
2
N1 X 2 X = N2 ' 2
76
Analysis of Transformer X eq
Req
' I1 ≈ I 2
V1
' V1 = V2
+
' I2
V2'
Z’
( Req + jX eq )
El-Sharkawi@University of Washington
77
Req
X eq
' I1 ≈ I 2
V1
V2'
Z’
' I 2 X eq
' I 2 Req
' I2
El-Sharkawi@University of Washington
78
Req
V1
X eq
' I1 ≈ I 2
V2'
Z
V1 = V2' + I 2' Req + j I 2' X eq
δ
θ
V1 I '2 Z eq
V2'
I '2 X eq
I '2 Req
' I2 El-Sharkawi@University of Washington
79
Ratings of Actual 3phase Transformer Apparent Power (3-phase) 100 MVA, 13.8/138 KV ' 2
V
V2
line-to-line
line-to-line
El-Sharkawi@University of Washington
80
Example A transformer has the following parameters:
N1 =10 N2 Req = R1 + R2' =1Ω;
Z l = 0.5∠30 Ω o
X eq = X 1 + X 2' =10Ω; Ro =1000Ω;
X 0 = 5000Ω
The rated voltage of the primary winding is 1000V. Compute the load voltage.
El-Sharkawi@University of Washington
81
Solution
' I1 ≈ I 2
2
N1 = 50 Ω Z = Z L N2 ' L
X eq
Req
V1
V2'
Z’
0 V 1000 ∠ 0 o 1 I 2' = = = 17 . 7 ∠ − 38 . 31 A ' o ( Req + j X eq ) + Z L (1+ j 10) + 50 ∠30
V = I Z = (17.7 ∠ − 38.31 ' 2
' 2
' L
o
) (50 ∠30 ) = 885 ∠ − 8.31 o
0
V
N2 1 V2 = V = 885 = 88.5 V N1 10 ' 2
El-Sharkawi@University of Washington
82
Req
V1
eq
' I1 ≈ I 2
Vno load − V full load VR ≡ V full load Measured at the load side
V2'
da o L
Voltage Regulation VR X
V1 − V2' VR ≡ ' V2
El-Sharkawi@University of Washington
83
Example Calculate the voltage regulation of the transformer in the previous problem.
Solution:
VR ≡
' V1 − V2 ' V2
7209.5 − 7200 VR = × 100 = 0.14 % 7200 El-Sharkawi@University of Washington
84
Efficiency
η
Xeq
Req
V1
Io
I1 Ro
Output Power Pout η = = Input Power Pin
Pin = Pout + Losses
I
Xo
Pcu =
' 2
( )
' 2 I2
V2'
Req
V12 Piron = Ro
Plosses = Pcu + Piron
Pout = V2' I '2 cos θ El-Sharkawi@University of Washington
85
Example A 10 kVA, 2300/230 V, single phase distribution transformer has the following parameters:
R1 = 5.8 Ω ; X 1 = X '2 = 12 Ω ; R'2 = 6.05 Ω Ro = 75.6 kΩ ; X o = 69.4 kΩ At full load and 0.8 power factor lagging, compute the efficiency of the transformer.
El-Sharkawi@University of Washington
86
Solution Pout = V2' I '2 cos θ = S cos θ = 10 × 0.8 = 8 kW ' I2
Pcu =
( )
' 2 I2
Piron
S
10 ,000 = = = 4.35 A ' 2300 V2
Req = ( 4.35 )2 × ( 5.8 + 6.05 ) = 224.23 W
( )
( V V 2300 ) = ≈ = = 70 W Ro Ro 75,600 2 1
' 2 2
2
Pout 8000 η = = × 100 = 96.45 % Pin 8000 + 224.23 + 70 El-Sharkawi@University of Washington
87
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