Basic Transformer Theory

7/31/2019 Basic Transformer Theory

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BASIC TRANSFORMER

THEORYW. H. Kersting

New Mexico State [email protected]

(505) 646-2434
mailto:[email protected]:[email protected]


2/43

Ideal Transformer Windings

Subtractive Polarity

N 1 N2


3/43

Subtractive Polarity Observations

Primary winding (#1) is wound such thatthe top of the winding is brought over thetop of the core

Secondary winding (#2) is wound such thatthe top of the winding is brought over the

top of the core


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Primary Connected to a Source

-

+I1E1 N2N1

01

01


5/43

Primary Connected to a Source

Basics Exciting current will flow in the primary

winding The flow of current will cause a magnetic

flux to flow in the core of the transformer The magnetic flux will link with the

secondary coil


6/43

Primary EquationsThe primary current I 1 is sinusoidal such that:

1 max1 cos( )i I t

This creates a sinusoidal flux in the coil:

1 max cos( )t

Applying Faradayss Law to coil #1:

11 1 1 maxsin( ) 1 cos( 90)

d e N N t E t

dt

This demonstrates that the primary current will lag theprimary ideal voltage by 90 degrees.


7/43


8/43

Secondary Induced Voltage

The flux changing with time is linked to thesecondary coil

The changing flux will induce a voltage onthe secondary coil

The polarity of the secondary voltage is

such that a secondary current will flow insuch a direction as to oppose the changingflux from the primary


9/43

Secondary Polarity

-

+I1E1 N2N1

0201

I2E2

+

-01


10/43

Secondary Polarity

For the windings specified, the secondary currentmust set up a flux to oppose the primary flux

Apply the right hand rule (wrap fingers in thedirection of the current flow and the thumb willpoint in the direction of the flux)

For this case the current must flow out of the topof the coil so that the plus sign of the secondaryvoltage will be on the top


11/43

Polarity Marks

-

+I1E1 N2N1

0201

I2E2

+

-01


12/43

Meaning of Polarity Marks

Polarity marks are place on the positiveterminal of the primary and secondary

voltages Primary current flows into the primary

polarity mark and secondary current flowsout of the polarity mark

The primary coil looks like a load while thesecondary coil looks like a source


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14/43

Complex PowerFor the ideal transformer, the complex power in will equal the complex power out

*1 1 1

*2 2 2

* *1 1 2 2

S E I

S E I

E I E I

But:2

2 11

N E E

N

Therefore:* *2

1 1 1 21

* *21 2

1

1 1 2 2

N E I E I

N

N I I

N

N I N I


15/43

Always Equations Subtractive Polarity

+ +

- -

I1

E1 E2

I2

No matter how the transformer is connected the relationshipbetween the primary and secondary voltages and currents willalways be:

1 2

1 2

1 1 2 2

E E N N

N I N I

This says that the voltages with polarities as shown will alwaysbe in phase and the currents with directions as shown willalways be in phase.


16/43

Ideal Transformer Windings

Additive Polarity

-

+I1E1 N2N1

0201

E2

01

-

+ I2


17/43

Additive Polarity Observations

Primary winding (#1) is wound such that the topof the winding is brought over the top of the core

Secondary winding (#2) is wound such that the topof the winding is brought behind the top of thecore

Position of the polarity marks on the secondarychange, same theory and equations apply


18/43

Always Equations Additive Polarity

+

-

I1

E1 E2

-

+I2

No matter how the transformer is connected the relationshipbetween the primary and secondary voltages and currents willalways be:

1 2

1 2

1 1 2 2

E E N N

N I N I

This says that the voltages with polarities as shown will alwaysbe in phase and the currents with directions as shown will

always be in phase.


19/43

Referring Impedances+ +

- -

I1E1 E2

I2Z

N 1 N 2

2

Secondary Equations:

2 2 2 E Z I

But: 12 12

N I I

N and 11 2

2

N E E

N

Therefore:

221

1 2 1 2 1 1 12

t N

E Z I n Z I Z I N

Where: The turns ratio =1

2t

N n

N

And: The referred impedance =

21 2t Z n Z


20/43

Application of Referred

Impedance+ +

- -

I1

E1 E2

I2Z

N 1 N 2

2Z

1

I1+

-

E1

Step 1: 1

2t

N n

N

Step 2: 111

E I

Z

Step 3: 2 1t I n I


21/43

Exact Equivalent Circuit

G -jB

Z

+

-

+

-

II

E 1 E 2

-

V

++

-

Vp

p

p Z s

sc m

I2 I s=

1n t :

1I ex

+

-

Vm


22/43

General Solution

G -jB

Z

+

-

+

-

II

E 1E

2

-

V

++

-

Vp

p

p Z s

sc m

I2 I s=

1n t :

1I ex

+

-

Vm

Given: V s and Is Required: V p and I p

Solution: 2 s s s E V Z I and 2 s I I

1 2t E n E and 1 21

t I I n

1mV E and ex c m m I G jX V

1 p ex I I I and p m s pV V Z I


23/43

Referred Z Solution

G -jB

+

-

+

-

II

E 1E

2

-

V

++

-

Vp

p

Z p

c m

1

n t2

Z s n t : 1

I2 = I s

I ex

s

+

-

Vm

Given: V s and Is Required: V p and I p

Solution: 2 s E V and 2 s I I

1 2t E n E

and 1 21

t I I

n

21 1m t sV E n Z I

and ex c m m I G jX V

1 p ex I I I and p m s pV V Z I


24/43


25/43

Other Operating Parameters

Compute input complex power:*

1000s s

p p pV I

S P jQ kW + j kvar

Compute output complex power:*

1000s ss s sV I S P jQ kW + j kvar

Compute efficiency: 100s

p

P Eff

P %

Compute voltage regulation: 100 100

ps

NL FL t reg

FL s

V V

V V nV

V V %


26/43

Short Circuit Test

+

-

+

-

-

++

-

V

I

E Esc

sc

1 2

PscZ T = RT+ j X T

Isc

Vs = 0

n t : 1

n t

1. Assumption: neglect shunt admittance branch2. Connect primary to an adjustable source and adjust so that I sc is equal to rated

primary current3. Measure: V sc, Isc and P sc 4. Then:

2sc

T sc

P R

I and scT

sc

V Z

I

1cos T T

R Z

and sinT m X Z

/ T T T T Z Z R jX


27/43

Impedance Diagram

XT

RT

0

Z T


28/43

Open Circuit Test

-jB

+

-

+

-

I

-

V

++

-

V G EE

I

p c m

1

1 2

oc

oc

Iex

Pocn t : 1

1. Assumption: neglect series impedance2. Connect the secondary to an adjustable source and adjust voltage to rated

secondary voltage3. Measure: V oc, Ioc and P oc 4. Then:

2

occ

t oc

PG

n V and

2

oc

t ocm

t oc t oc

I n I

Y n V n V

1cos cm

GY

and sinm m B Y

/ m m c mY Y G jB


29/43

Admittance DiagramG c

BmYm

0


30/43

Exact Equivalent Circuit

G -jB

Z

+

-

+

-

II

E 1E

2

-

V

++

-

Vp

p

p Z s

sc m

I2 I s=

1n t :

1I ex

+

-

Vm

With Z T and Y m known from the short-circuit and open-circuit tests, if necessary, the parameters of the exact equivalent circuit can be approximated by:

12 p T Z Z and 2

12

T s

t

Z Z n

m mY Y


31/43

Example

Given: The Rating of a Single -Phase Transformer

kVA 100 V hi 7200 V low 240

Compute the turns ra tio :

n tV hi

V lown t 30

Compute the rated pri mary current :

I ratedkVA 1000

V hiI rated 13.9 A


32/43

Short Circuit TestPerform the shor t-circui t test:

+

-

+

-

-

++

-

V

I

E Esc

sc

1 2

PscZ T = R T+ j X T

Isc

Vs = 0

nt : 1

nt

Measurements: V sc 250 I sc 13.9 P sc 127


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Calculate Series Impedance

Compute resistance: R TP sc

I sc2

R T 6.599

Compute impedance magnitude: Z magV sc

I scZ mag 17.9856

Compute the impedance angle: acosR T

Zmag

deg68.475 degrees

Def ine series impedance: Z T Z mag e j Z T 6.599 16.7313j


34/43

Open Circuit Text

Perform the open-ci rcuit test:

-jB

+

-

+

-

I

-

V

++

-

V G EE

I

p c m

1

1 2

oc

oc

Iex

Pocn t : 1

Measurements: V oc 240 I oc 13.5 P oc 355


35/43

Calculate Shunt Admittance

Compute conductance: G cP oc

n t V oc2

G c 6.848106 S

Compute admittance magnitude: Y magI oc

n t2 V oc

Y mag 6.25105 S

Compute the admittance angle: acosG c

Y mag deg83.7096 degrees

Define the s hunt admittance: Y m Y mag e j Y m 6.84810

6 6.212410 5 j S


36/43

Load Analysis

Transformer analysis us ing the approximate equivalent circuit:

-jB

+

-

+

-

I

-

V

++

-

V

Z

I

G EE

= II

p

p

c m

T

1

1 2

2 s

s

Iex

Load

De fine "load": kVA load 95 V load 240 PF load .9 lag


37/43

Load Voltage and Current

Compute load vol tage and current:

V s V load e j 0 V s 240

arg V s

deg0

I skVA load 1000

V loade

j acos PF load I s 395.8333arg I s

deg25.8419


38/43

Compute Primary V and I

-jB

+

-

+

-

I

-

V

++

-

V

Z

I

G EE

= II

p

p

c m

T

1

1 2

2 s

s

Iex

Load

I 2 I s I 2 395.8333arg I 2

deg25.8419

E 2 V s E 2 240

arg E 2

deg 0

E 1 n t E 2 E 1 7200arg E 1

deg0

I 1

I 2

n t I 1 13.1944

arg I 1

deg 25.8419


39/43

Compute Input V and I

-jB

+

-

+

-

I

-

V

++

-

V

Z

I

G EE

= II

p

p

c m

T

1

1 2

2 s

s

Iex

Load

V p E 1 Z T I 1 V p 7376.3418arg V p

deg1.2486

I ex Y m V p I ex 0.461 arg I exdeg82.461

I p I 1 I ex I p 13.4536arg I p

deg27.4816


40/43

Compute Operating

Characteristics

-jB

+

-

+

-

I

-

V

++

-

V

Z

I

GE E

= II

p

p

c m

T

1

12

2 s

s

Iex

Load

Compute complex power input: S pV p I p

1000S p 87.0215 47.7025j kVA

Compute complex power output: S sV s I s

1000S s 85.5 41.4095j kVA


41/43

(continued)

Compute efficiency: Eff Re S s

Re S p100 Eff 98.2516 %

Compute voltage regulation: V NL

V p

n t V NL 245.8781

V FL V s V FL 240

V regV NL V FL

V FL

100 V reg 2.4492


42/43

Homework Problem

A 50 kVA, 2400-240 volt transformer has thefollowing test results :

Short Circuit: V=60.0 V, I=20.8 A, P=750 W Open Circuit: V=240 V, I=5.97A, P=213 W Determine the parameters of the approximate

equivalent circuit


43/43

HW (continued)

The transformer serves a load of 45 kVA, at 235volts and 0.85 power factor lagging

Compute the input and output voltages andcurrents Compute the input and output complex powers Compute efficiency Compute voltage regulation

Basic Transformer Theory

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