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Transformer Professor Mohamed A. El-Sharkawi

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Presentation on theme: "Transformer Professor Mohamed A. El-Sharkawi"— Presentation transcript:

1 Transformer Professor Mohamed A. El-Sharkawi

2 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 of Washington

3 El-Sharkawi@University of Washington
220kV-750kV Distribution Transformer Transmission Transformer 15 kV- 25kV Service Transformer 208V- 416V of Washington

4 Transmission Transformer
of Washington

5 Distribution Transformer
of Washington

6 Distribution Transformer
of Washington

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Service Transformer of Washington

8 Service Transformer bank
of Washington

9 Service Transformer bank
of Washington

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Service Transformer of Washington

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Service Transformer of Washington

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Service Transformer of Washington

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Service Transformer of Washington

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Service Transformer of Washington

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Low power Transformer of Washington

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Basic Components Iron Core Insulated Copper Wire of Washington

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Basic Components Laminated iron core Insulated copper wire of Washington

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Secondary Primary of Washington

22 Basic Analysis: Voltage
+ _ N2 e2 Volts/turn is constant Voltages are in phase (no phase shift) Voltage magnitudes vary with turns ratio. of Washington

23 Basic Analysis: Power and current i2
+ _ N2 e2 Ampere turn is constant Currents are in phase. Current ratio is opposite to the voltage ratio of Washington

24 Basic Analysis: Reflected impedance
Primary Secondary Source Load Flux Zload of Washington

25 Basic Analysis: Reflected impedance
+ _ N2 E2 Source E1 Primary of Washington

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Key relationships Constant voltage per turn I1 I2 N1 E1 + _ N2 E2 Constant Ampere turn Reflected impedance of Washington

27 Single-Phase, Ideal Transformer Ratings
+ - V 1 I 2 Apparent Power (S) 2 KVA, 120/240 V Primary Voltage (V1) Secondary Voltage (V2) of Washington

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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 of Washington

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Example N2 N1 + - V 1 I 2 Transformer rating: 2 KVA, 240/120 V Compute the currents of Washington

30 Multi-Secondary Transformer
of Washington

31 Multi-secondary windings
of Washington

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Voltage/Turn I3 I1 N3 E3 E1 I2 N1 E2 N2 Primary of Washington

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Primary I3 N3 E3 Ampere’s law of Washington

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Primary I3 N3 E3 Power of Washington

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Example The transformer consists of one primary winding and two secondary windings. The number of turns is each winding is 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 of Washington

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Solution of Washington

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Solution of Washington

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Solution of Washington

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Autotransformer of Washington

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B1 V1 E1 E2 N1 N2 V2 A2 B2 of Washington

41 Autotransformer: Voltage and current
I2 N2 V1 N1 I1 Is Iload V2 A1 A2 B1 B2 E1 E2 of Washington

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Autotransformer N2 N1 + - E 1 I 2 I2 N2 V1 N1 I1 Is Iload V2 A1 A2 B1 B2 E1 E2 of Washington

43 Autotransformer: Power
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 I2 N2 V1 N1 I1 Is Iload V2 A1 A2 B1 B2 E1 E2 of Washington

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VARIAC: Variable Auto-Transformer I2 N2 V1 N1 I1 Is Iload V2 N3 Z Y Sliding terminal of Washington

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Output Voltage I2 N2 Vs N1 I1 Is Iload Vload N3 Z Y Sliding terminal At Y At Z of Washington

47 Three-Phase Transformer
of Washington

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3-phase transformer of Washington

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3-phase transformer of Washington

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3-phase transformer of Washington

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single-phase transformer of Washington

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3-phase transformer bank of Washington

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3-phase transformer of Washington

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3-Phase Transformer of Washington

55 3-phase transformer Y-Y connection. Also known as star-star connection
of Washington

56 3-phase transformer Y-Y connection. Also known as star-star connection
b A B C N1 N2 Voltage per turn Ratio of Line Voltage n N of Washington

57 3-phase transformer ( -)
of Washington

58 3-phase transformer ( -)
B C a b c N 2 N1 Voltage per turn of Washington

59 3-phase transformer (Y-) Also known as star-delta connection
of Washington

60 3-phase transformer (Y-) Also known as star-delta connection
Ratio of Line Voltage N 1 Voltage per turn N 2 B n c b C of Washington

61 3-phase transformer bank (Y-)
2 1 V an V ab V AB B b N 2 1 C c N 2 1 of Washington

62 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 of Washington

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Example Three single-phase transformers are used to form a three-phase 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 of Washington

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Solution Secondary voltage (Low voltage side) should be in Y to provide the needed residential voltage 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. of Washington

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Solution Van= 240 V A a N2 N1 Van VAB= 13.8 kV Vab B b N2 N1 C c of Washington

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Solution N2 N1 Van a c A B C VAB= 13.8 kV Vab b Van= 240 V Phase current of the load Phase current of the Transformer primary Line Current in primary of Washington

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Actual Transformer Windings: Resistance Inductance Core: Eddy Current Hysteresis I1 I2 N1 E1 + _ N2 E2 of Washington

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Windings Impedance R1 X1 R2 X2 N1 N2 Ideal Transformer of Washington

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Core Hysteresis i B H N e + _ of Washington

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Core Model e i R Let i of Washington

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Core Model i e Xl Let i of Washington

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Xl i i of Washington

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Equivalent Circuit X1 R1 R2 X2 N1 N2 Io load I1 I2 V1 E1 E2 V2 Ro Xo of Washington

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Key Relations Ro Xo R1 X1 R2 X2 V1 I1 Io I2 V2 E1 E2 N1 N2 load Voltage per turn Ampere turn of Washington

76 Referred (reflected) impedance
X1 R1 R2 X2 N1 N2 Io I1 I2 V1 E1 E2 V2 Ro Xo of Washington

77 Referred (reflected) impedance
X1 R1 R2 X2 N1 N2 Io I1 I2 V1 E1 E2 V2 Ro Xo of Washington

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Define: Then: of Washington

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Equivalent Circuit Referred to Source Side Ro Xo R1 X1 R2 X2 V1 I1 Io I2 E1 E2 N1 N2 V2 X1 R1 Io I1 V1 E1 Ro Xo of Washington

80 Practical Considerations
X1 R1 Ro Xo V1 I1 Io E1 of Washington

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Ro Xo Io R1 X1 OK V1 I1 Ro Xo Io R1 X1 OK of Washington

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X1 OK Define: of Washington

83 Analysis of Transformer
V1 Z’ of Washington

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Terminologies Load Voltage Load Voltage referred to Source side Impedance referred to Source side Load Current Load current referred to Source side of Washington

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Z’ of Washington

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Z of Washington

87 Ratings of Actual 3-phase Transformer
Apparent Power (3-phase) 100 MVA, 13.8/138 KV line-to-line line-to-line of Washington

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Example A transformer has the following parameters: The rated voltage of the primary winding (V1) is 1000V. Compute the load voltage. of Washington

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Solution V1 Z’ of Washington

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Voltage Regulation VR V1 Load Measured at the load side of Washington

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Example Calculate the voltage regulation of the transformer in the previous problem. Solution: of Washington

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Efficiency  V1 I1 Ro Xo Io Req Xeq of Washington

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Example A 10 kVA, 2300/230 V, single phase distribution transformer has the following parameters: At full load and 0.8 power factor lagging, compute the efficiency of the transformer. of Washington

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Solution of Washington


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