Ideal Transformer Faraday’s Law - Department of ...ece.citadel.edu/barsanti/elec316/L4_Tranformers_slide.pdf · Open circuit (oc) test data to find Rc and Xm PFoc = cos( θoc)

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1

Ideal Transformer

Properties

• High permeability of the core

• No Leakage Flux

• No winding resistances.

• Ideal core has no reluctance.

• No Core losses.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.2

The negative sign is the statement of the Lenz’s law

stating that the polarity of the induced voltage should

be such that a current produced by it produces a flux in the opposite of the original flux. This is illustrated

below

Faraday’s Law

If a flux φ passes through N turns of a coil, the induced

voltage in the coil is given by

ind

de N

dt

ϕ= −

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

3

Relationships

• From Faraday’s Law

• Since there is no magnetic potential drop in

the ideal core,

1

( ) ( )

( )

( )

s p s s

p s

s p

N I t N I t

I t N

I t N a

=

= =


4

Relationships

• Since

• then

Power in equals power out.

No power loss in ideal transformer!

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Relationships

• Reflected Impedance

=

=

=

=

6

Dot Convention

• Help to determine the polarity of the voltage and direction of the current in the secondary winding.

• Voltages at the dots are in phase.

• When the primary current flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the

secondary winding.

Compare the following distribution schemes

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same

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Non-Ideal Single-Phase Transformers

Non-ideal facts:

• Winding resistances modeled as series resistors Rp and Rs Also called Copper Losses.

• Leakage flux modeled as series inductances Xp and Xs

• Core Losses (Eddy Current and Hysteresis Losses) produce heating losses modeled as a shunt resistor RC in the primary winding.

•Magnetizing Current flows in the primary to establish the flux in the core. Modeled as a shunt inductance Xm in the primary winding.

Mutual (M) and Leakage (L) Flux

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Excitation CurrentMagnetization Current (IM) is not sinusoidal

because of the non-linear relation between the current and flux (magnetization curve)

Total Excitation Current

In a well designed transformer Iex is small.

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17

The equivalent circuit for a single-phase non-ideal

transformer is shown below:


18

• The equivalent circuit may be simplified by reflectingimpedances, voltages, and currents from the secondary

to the primary as shown below:

• Below is the transformer model referred to Secondary Side.


19

• Simplified equivalent circuit referred to primary side:

• Simplified equivalent circuit referred to secondary side:


Transformer Equivalent Circuit without

Excitation Branch

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Transformer Voltage Regulation

• Because a real transformer has series impedances

within it, the output voltage will vary with the load

even if the input voltage remains constant.

• Voltage Regulation compares the no load to full load

voltage.

= |,||,|

|,|x100% =

|| |,|

|,|x 100%

Transformer Phasor Diagram

• ApplyingKirchhoff‘slaw0 = 1 + 3451 + 673451

• For a Lagging PF (I lags V)

Unity PF

Leading PF

24

Transformer Efficiency

• Pout = Ps = VsIsCos(θs)

• Pin = Ps + PLosses = VsIsCos(θs)+ Pcore+ Pcu

( ) 100%

( )out S S S

in S S S core cu

P V I Cosx

P V I Cos P P

θη

θ• = =

+ +


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Ex.. Given: 2.2 kVA, 440/220V transformer. Equiv.

circuit parameters referred to the primary are :

Req = 3Ω, Xeq = 4Ω, Rc = 2.5kΩ, Xm = 2 KΩ

Find : VR and η. Assume transformer is

delivering rated current and voltage at full load

and 0.707 pf.

Solution:

1. V s,fl = 220 V <0°

This is the full load secondary voltage.

We must now find the secondary voltage assuming the load was removed V s,nl . This is found by adding in the

voltage drop across Req + j Xeq that occurs at full load.

1,9: =0

= 1,;: + 51(=> + 67=>)

2. Is = 2200 VA / 220V = 10 A

3. θ = cosA 0.707 = 45°4. Using the primary referred circuit we get Vp,nl

0,9: =

=> + 67=> + 1

=(10<-45°)/2*(3+j4)+2(220)=464.8<0.4°

So,

1,9: =,

=232.4<0.4

=|,||,|

|,|x100% =

P.QR

R= 5.63%

• To find η we need Pin and Pout

1. Pin = Vp ∙ Ip cos(∅) = ReVp ∙Ip*

= Re(464.8<0.4°)(QUQ.VWR.Q

XRR+

QUQ.VWR.Q

YRRR+ARWQX°

)*

=Re(464.8<0.4°)(5.29<45.3°) = 1,717 W

2. Pout = Vs Is cos(∅) = |S| pf = 2200*0.707 = 1,555 W

3. η = Z,[[[

Z,\Z\= 90.6%

4. Note: Pin = Pout + Pcopper loss + Pcore loss

= Pout +]^_

`abc +

de_

af= 1555 +(5)23 +(464.8)2/2500

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Determining the Values of Components in the Transformer Model

Transformer impedances may be obtained from two tests:

• Open-circuit test: to determine core losses and

magnetizing reactance (Rc and Xm)

• Short-circuit test: to determine equivalent Series

Impedance Req. and equivalent leakage reactances, Xeq)


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Open-Circuit Test to Determine Rc and Xm

With secondary open, Voc, Ioc, and Poc are measured on the primary side.


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Short-Circuit Test to Determine Req and Xeq

With secondary shorted, a reduced voltage is

applied to primary such that rated current flows

in the primary. VSC, ISC, and PSC are measured on

the primary side.

1

ecos Rsc

eqq eq

sc

VZ PF jX

I

−= − = +


Req

Rc

Xeq1 2

Xm

1

2

Ip

Vp a Vs

Is/a

+

-

+

-

------>------>

Lab 3

1. Open circuit (oc) test data to find Rc and Xm

PFoc = cos(θoc) = Poc/(Voc Ioc) = 2.0 / (120.0)(0.02) = 0.8333

θoc = arcos(0.833) = 33.6 deg.

Y= 1/Rc – j/Xm = Ioc/Voc <- θoc = 0.02/120.0 <-33.6

= 1.6667 x 10-4 <-33.6 =(1.3882 – j 0.92232) x 10-4

Rc = 1/1.3882 x 10-4 = 7,204 ΩXm = 1/0.92232 x 10-4 = 10,842 Ω

Pre- lab 3 Solution

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2. Short circuit (sc) test data to find Req and Xeq

PFsc = cos(θsc)= Psc/(Vsc Isc) = 3.0 / (10.0)(0.4) = 0.75

θsc = arcos(0.75) = 41.4 deg.

Zeq = Req + j Xeq = Vsc/Isc < θsc

= 10.0/0.4 < 41.4 = 25.0 < 41.4 = 18.8 +j16.5 Ω

Req = 18.8 ΩXeq = 16.5 Ω

Pre- lab 3 Solution (continued) Pre- lab 3 Solution (continued)

VRc and ηc

1. Find secondary voltage using input voltage and voltage divider w/ ZL = 300 Ω

voltsj

VZZ

ZV

P

eqL

L

s8.1120.120

5.168.18300

300=

++=

+=

2. Find secondary current in load

AZ

VI

L

s

s 376.0300

8.112===

3. Find losses using eqv. circuit values

Pcopper = (Is)2 Req = (0.376)2 18.8 = 2.66 wPcore = (Vprimary/a)2/ Rc = (120.0)2/ 7204 = 2.0 W (note Pcor = Poc)

%38.6%1008.112

8.1120.120%100%100

,

,,=

−=

−=

−= xx

V

VVx

V

VVVR

s

sp

Loadfulls

LoadfullsLoadnos

c

%1.90%1000.266.24.42

4.42%100%100 =

++=

++== xx

PPP

Px

P

P

corecopout

out

in

out

cη

4. Find efficiency ηc and VRc using eqv. circuit values

Pout = Vs Is PF = (112.8)(0.376)(1) = 42.4 w since PF =1 for Z =300 Ω

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Three-Phase Transformers

Figure 2-35

A three-phase transformer bank composed of independent transformers


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Figure 2-36

A three-phase transformer wound on a single three-legged core.


Three-Phase Transformers Three-Phase Transformers

39

Y-Y Transformer Connection

gh

=3∅g3∅h

=

∅g

∅h=

Problems: 1. If loads are unbalanced, transformer phase voltageswill be severely unbalanced. Use neutral line

2. Large third harmonic voltages since they add.Use a third ∆ connected winding to cancel

40

Note: 1. Secondary phase voltage shifted by 30°due to connection.


Y-∆ Transformer Connection

gh

=3∅g∅h

= 3

∅g

∅h=

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41

Note: 1. Secondary phase voltage shifted by 30°due to connection.

∆-Y Transformer Connection

∅g∅h

=

g

h=

∅g

3∅h

=

3

42

Note: No issues.


∆- ∆ Transformer Connection

∅g∅h

=

g

h=

3∅g

3∅h

=

Transformer Ratings

Voltage and Frequency rating

- Protects the winding insulation from excessive

voltage.

- Protects winding from large magnetization

currents resulting from large voltages or low

frequencies. Since

∅ i = A

jkl i mi =

A

jknopωi mi

= −

sjcosti

Voltage and Frequency rating

∅u v =u vtw0

Φ increases as V increases

Φ increases as ω decreases

In sat. region large Φ causesextremely large ox

For f change from 60 to 50 Hz

ox can increase 60%.

Documents

Ideal Transformer Faraday’s Law - Department of ...ece.citadel.edu/barsanti/elec316/L4_Tranformers_slide.pdf · Open circuit (oc) test data to find Rc and Xm PFoc = cos( θoc)