9-transformers

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ELECTRIC CIRCUITS THEORY 1 These lecture slides have been compiled by Mohammed LECTURE 9 SalahUdDin Ayubi. Transformers 24 August 2005

Engineer M S Ayubi

1

Outline • Ideal Transformers • Transformer Equations – Voltage Transform – Current Transform – Resistance Transform

• Transformer Ratings • Real Transformers • Transformers and the Power Grid 24 August 2005

Engineer M S Ayubi

2

Physical Configuration

+

i1(t)

φ

V1(t)

+

V2(t)

-

-

n2 turns

n1 turns

24 August 2005

i2(t)

Engineer M S Ayubi

3

Transformer Symbol i1(t)

i2(t)

+

+

V1(t)

V2(t)

-

-

n1:n2

24 August 2005

Engineer M S Ayubi

4

Outline • Ideal Transformers • Transformer Equations – Voltage Transform – Current Transform – Resistance Transform

• Transformer Ratings • Real Transformers • Transformers and the Power Grid 24 August 2005

Engineer M S Ayubi

5

Transformer Equations di1 (t ) v1 (t ) = L1 dt di1 (t ) di2 (t ) v1 (t ) = L1 +M dt dt di1 (t ) di2 (t ) v2 (t ) = M + L2 dt dt

24 August 2005

Engineer M S Ayubi

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Transformer Equations v2 n2 = v1 n1 i2 n1 =− i1 n2

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Engineer M S Ayubi

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Transformer Equations i1(t)

i2(t)

+

+

V1(t)

Req

V2(t)

-

-

n1:n2

v2 n2 = v1 n1 i2 n =− 1 i1 n2

2

 n1  R Reqeq = ?  R  n2  24 August 2005

Engineer M S Ayubi

8

Example 1 • Find turns ratio for maximum power (i.e., match the impedances)

24 August 2005

n1 =5 n2

Engineer M S Ayubi

9

Example 2 • Find V2

i1(t)

i2(t)

+

+

V1(t)

V2(t)

-

-

2:1 V2 = 0 V 24 August 2005

Engineer M S Ayubi

10

Example 3 • Find V2

i1(t)

i2(t)

+

+

V1(t)

V2(t)

-

-

2:1 V2 = 5 sin2t V 24 August 2005

Engineer M S Ayubi

11

Example 4 • Find the magnitude of the secondary voltage, the primary current, and the secondary current • Find power delivered by the source

60∠ 0° Vrms

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Engineer M S Ayubi

12

Example 4 • Magnitude of secondary voltage: vˆ1 = 60∠0° v1 = 60

60∠ 0° Vrms

v2 n2 = v1 n1 v2 =

24 August 2005

n2 1 v1 = 60 = 20Vrms n1 3

Engineer M S Ayubi

13

Example 4 • Magnitude of primary current: 60∠ 0° Vrms

v1 i1 = Zˆ

eq 2

2

 n1   3 Z eq =   Z 2 =   2.2 2 + 12 = 21.75Ω 1  n2  60V i1 = = 2.76A rms 21.75Ω 24 August 2005

Engineer M S Ayubi

14

Example 4 • Magnitude of secondary current:

60∠ 0° Vrms

n1 3 i2 = − i1 = − 2.76 = −8.28A rms n2 1

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Engineer M S Ayubi

15

Example 4 • Power delivered by source:

60∠ 0° Vrms

Ps = PR = I 2 R = (8.28 A)(2.2Ω) = 151W

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Engineer M S Ayubi

16

Outline • Ideal Transformers • Transformer Equations – Voltage Transform – Current Transform – Resistance Transform

• Transformer Ratings • Real Transformers • Transformers and the Power Grid 24 August 2005

Engineer M S Ayubi

17

Transformer Ratings 0.125 KVA; 230/115 VAC; 50/60 HZ

24 August 2005

Engineer M S Ayubi

18

Outline • Ideal Transformers • Transformer Equations – Voltage Transform – Current Transform – Resistance Transform

• Transformer Ratings • Real Transformers • Transformers and the Power Grid 24 August 2005

Engineer M S Ayubi

19

Real Transformers • Internal power loss due to: – – – –

core loss core reluctance uncontained flux lines winding resistance

24 August 2005

Engineer M S Ayubi

20

Outline • Ideal Transformers • Transformer Equations – Voltage Transform – Current Transform – Resistance Transform

• Transformer Ratings • Real Transformers • Transformers and the Power Grid 24 August 2005

Engineer M S Ayubi

21

The Power Grid

Diagram courtesy howstuffworks.com

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Engineer M S Ayubi

22

Power Grid

3-phase industrial

Intertie Power Plants Generator 0.5 – 5kV

Step-up Transformer

Generator 0.5 – 5kV

Step-up Transformer

Step-down Transformer

Substation Step-down Transformer 100-750 kV

Step-down Transformer

35-70 kV

4-15 kV

Step-down Transformer 120/240V residential

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Transmission Tower

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Incoming Power at Substation

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Incoming Power Conditioning

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Step-Down Transformers

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Transformer

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Outgoing Switch Banks

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Outgoing 3-Phase Power

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Utility Pole Carrying Three Phase

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Three Phase Transformers

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