Principles of Electric Machines and Power Electronicsli/slides/ch02-1Sli.pdf · Principles of...

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Chapter 2-1Transformers

Third Edition

P. C. Sen

Principles of Electric Machines

and

Power Electronics


Transformer application 1: power transmission


Ideal Transformer

1 1 1

2 2 2

1 1

2 2

dv e N

dt

dv e N

dt

v Na

v N

1, step-down transformer

1, step-up transformer

a

a

Assumptions:

1. Negligible winding resistance

2. No leakage flux

3. Infinite permeability of core

4. Zero core loss

𝑣1

𝑉1 𝜔 = 𝑗𝜔𝐿 ∗ 𝐼1(𝜔)

Voltage relation:

Excitation current:

⇒ 𝐼1 𝜔 =𝑉1 𝜔

𝑗𝜔𝐿

𝐿 =𝑁12𝜇0𝜇𝑟𝐴

𝑙

⇒ 𝐼1 𝜔 = 0


Basic equations- general definition

1 1 2 2

1 1 2 2

1 2

2 1

net mmf=0

1

N i N i

N i N i

i N

i N a

1 1 2 2v i v i

Current relationship: Instantaneous power:

𝑁1𝑖1 𝑁2𝑖2

ℜ =𝑙

𝐴𝜇


Polarity marking:

Like polarity: two entering

currents produce the same

direction of magnetic flux


Voltage polarity 2

Parallel operation of two single-phase transformer


Impedance Reflection:

𝑍2′ =

𝑉1𝐼1=𝑎𝑉2𝐼2𝑎

= 𝑎2𝑉2𝐼2= 𝑎2𝑍2

' 11 2

ZZ

a


Example: Determine the primary and secondary currents for

the ideal transformer below if Zs = (18-j4) Ωand Z2 = (2+j1) Ω.


Other single-phase transformers

• Single primary multiple secondary

windings

𝑉𝑖 =𝑉1𝑎𝑖

𝐼1 =𝑎𝑖𝐼𝑖


Non- ideal transformer: model development 1

• Ideal transformer assumptions:1. Negligible winding resistance

2. No leakage flux

• Practical transformer: ideal transformer+ external

impedance


• Ideal transformer assumptions:3. Infinite permeability of core

4. No core loss

• Practical transformer: Magnetizing reactance

Core resistance




𝑉2′ = 𝑎𝑉2

𝐼2′ =

𝐼2𝑎= 𝐼1

𝑅𝑤2′ = 𝑎2𝑅𝑤2

𝑋𝑙2′ = 𝑎2𝑋𝑙2

𝑍2′ = 𝑎2𝑍2


Approximate equivalent circuit

Voltage drops across

primary winding resistance

and reactance is quite small

Neglect excitation branch


Determine equivalent circuit parameters- No load test

Procedure:• Apply rated voltage to either high-voltage or low-voltage side

• Primary current: exciting current

• Loss: core loss (the same for applying either high-voltage or low-voltage side)

Parameters obtained:

• Magnetizing reactance

• Core resistance


Determine equivalent circuit parameters – short circuit

test

Procedure:• Short-circuiting one winding

• Apply rated current to the other winding

Parameters obtained:

• Primary and secondary resistance

• Primary and secondary leakage reactance

Current

source


Transformer ratings: power, primary/secondary voltages

→ turn ratio, current ratings

Example: single-phase transformer, 10 kVA, 2200/220V, 60 Hz

(1) Determine core loss resistance and magnetizing inductance from no load

test.

(2) Derive the parameters for the approximate equivalent circuits referred to

the High voltage side

(3) Derive the parameter when refer to low voltage side


Solution (practice): single-phase transformer,

10 kVA, 2200/220V, 60 Hz


Voltage regulation

2 2

2

NL L

L

V VVR

V

' '

2 2

'

2

NL L

L

V VVR

V

Refer to the primary

Basic definition


Load voltage is normally taken as rated voltage

'

1 2 rated

'

2 rated

100%V V

VRV

Voltage regulation


Example: single-phase transformer,

10 kVA, 2200/220V, 60 Hz

Determine voltage regulation in percent for

(a) 75% full load, 0.6 power factor lagging

(b) 75% full load, 0.6 power factor leading

(c) Draw the phasor diagram for (a) and (b)


Solution (practice)


Efficiency of a transformer

out out out

in out loss out c cu

100%P P P

P P P P P P

2 2 2

cu 1 1 2 2 2 2w w eqP I R I R I R

•Loss of transformerCore loss (hysteresis and eddy current)

Winding (copper) resistive loss

•Core loss Almost constant

Obtain from no-load test


Example (practice)

Example (practice): single-phase transformer,

10 kVA, 2200/220V, 60 Hz

Determine

Efficiency at 75% rated output and 0.6 PF

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