Chapter 2 342 Transformer 4

7/28/2019 Chapter 2 342 Transformer 4

1/51


2/51

Fida Muhammad (Air University)

Induction heating Non-contact heating process uses high frequency electricity to heat materials that areelectrically conductive

1. What makes the conductive material to heat up?

Eddy current in conductive non-magnetic materials (Cu) & eddy& Hysterisis in conductive-magnetic materials (iron & steel)

2. What about plastics?


3/51


4/51


100

load_fullV

load_fullVload_noVVR%

s

ss

Transformer (T/F) Characteristics (distinctiveness)

Efficiency () Voltage Regulation (VR)

InputPower

LossesInputPower

InputPower

OutputPower

Eddy Current Losses

Hysterisis Losses

Copper Losses(RP& RS)

Voltage Drop

Across XP & XS

Voltage Drop

(RP& RS)

In seriesvoltage

drop

Question: What about XM

Not used for Efficiency. Why?

XM - Why not for VR ?


5/51


Page#90 Determining the Values of Components in the

Transformer Model

How to find Values of Components ?

Open Circuit Test Short Circuit Test

Do the Following Test

Reqp jXeqp

aVS

IS/a

VP

IO

IMIC

RC jXM

IP

find RC & XM find Reqp, Xeqp OR Reqs , Xeqs


6/51


IP(t)

Wattmeter

VP(t)

V

A

V(t)

Transformer

Open

Circuit

Ampere

Meter

Volt

Meter

Figure 2-19 Connection fortransformer open-circuit test

Open Circuit Test

Apply rated primary (input) Voltage

IO No-Load current flows primary

Wattmeter reads ..?

Core losses

small

Scale

(IO)

Why?

Large

Scale

Why ?


7/51


IP(t)

Wattmeter

VP(t)

V

A

V(t)

Transformer

Ampere

Meter

Volt

Meter

Figure 2-20 Connection for

transformer short-circuit test

A

Secondary

Short Circuit

Short Circuit Testlargescale

(IO)

Why?

small

Scale

Why ?

Apply rated current

Full load IP & IS current flows

Wattmeter reads ?.....I2R (Copper)


8/51


Page#90 Determining the Values of Components in the

Transformer ModelTests : Find Values of Components

Open Circuit Test Short Circuit Test

IP(t)Wattmeter

VP(t)V

A

V(t)

Transformer

Open

Circuit

Ampere

Meter

Volt

Meter


transformer open-circuit test

IP(t)Wattmeter

VP(t)V

A

V(t)

Transformer

Open

Circuit

Ampere

Meter

Volt

Meter


transformer open-circuit test

IP(t)Wattmeter

VP(t)V

A

V(t)

Transformer

Ampere

Meter

Volt

Meter



A

Secondary

Short

CircuitIP(t)Wattmeter

VP(t)V

A

V(t)

Transformer

Ampere

Meter

Volt

Meter



A

Secondary

Short

Circuit

Apply rated primary (input) Voltage

IO No-Load current flows primary

Wattmeter reads ..?Core losses

Apply rated current

Full load IP & IS current flows

Wattmeter reads I2R (Copper)


9/51


A 20-kVA, 8000/240-V, 60-Hz T/F. The open-circuit test (OCT) and

the short-circuit test (SCT) were performed on the primary side of

the T/F, and the following data were taken:

Find the impedances of the approximate equivalent circuit referred

to the primary side, and sketch the circuit.

OCT (on primary)

Voc= 8000 V

Ioc

= 0.214 A

Poc= 400 W

SCT (on primary)

Vsc= 489 V

Isc

= 2.5 A

Psc = 240 W

Example 2.2 (page 92)

IP(t)Wattmeter

VP(t)V

A

V(t)

IP(t)Wattmeter

VP(t)V

A

V(t)

IP(t)Wattmeter

VP(t)V

A

V(t)

A

IP(t)Wattmeter

VP(t)V

A

V(t)

A

W tt tW tt t


10/51


Answer

OCT (on primary)

Voc = 8000 V

Ioc =Io= 0.214 A

Poc = 400 W ??

IP(t)Wattmeter

VP(t)V

A

V(t)

IP(t)Wattmeter

VP(t)V

A

V(t)

IOC=I0=IP=0.214

VOC8000

IC IM

RC XM

POC=400W

VOC


11/51


OCT (on primary)

Voc= 8000 V

Ioc=Io= 0.214 A

Poc= 400 W ???

IOC=I0=0.214

VP=8000

IC IM

RC XM

POC=400W

ococococ cosIVP

YY

2336021408000

400.

.xcos oc

01

57623360 ..cosoc

mhox..

Y 510728000

2140 ococ

ococ

IV

Pcos

ocOC

.OCV

I

8000

2140 05

5761072 .x.

Y=1/Z


12/51


OCT (on primary)

Voc= 8000 V

Ioc=Io= 0.214 A

Poc= 400 W

IOC=I0=0.214

VP=8000

IC IM

RC XM

YY ocOC

.OC

V

I

8000

2140 05 5761072 .x.

oc=76.5o

jBM

Gc=Yoccosoc

=YocsinocYoc=2.7x10-5mho

)circuitsparallel(cetanAdmit

Yoc=Yoccosoc -jYoc sinoc

C

CG

R1

M

MB

X1

MCoc jBGY


13/51


R= Zcos

Z X= Zsin

G=Ycos

YB=Ysin

jBGY

Y=Ycos -jY sin

Y

I

R jX

V

jXRZ

Z=Zcos +jZ sin

Z

Z

Y=1/ZR

G1

jXjB

1

Why

Negative ?

Why

Negative ?


14/51


OCT (on primary)

Voc= 8000 V

Ioc=Io= 0.214 AP

oc= 400 W

IOC=I0=0.214

VP=8000

IC IM

RC XM

MCoc jBGY

Yoc=Yoccosoc -jYoc sin oc

MCoc Xj

RZ

111

Yoc=2.7x10-5 cos76.5 -J2.7x10-5 sin76.5

Y=6.25x10-6 -j 2.53x10-6

Z=159000 + j38400

RC=159kXM=

38.4k

M

M

C

C

oc

oc

X

B;

R

G;

Z

Y111

oc=76.5o

jBM

Gc

Yoc=2.7x10-5mho

W tt tW tt t


15/51


Answer IP(t)Wattmeter

VP(t)V

A

V(t)

A

IP(t)Wattmeter

VP(t)V

A

V(t)

A

SCT (on primary)

Vsc= 489 V

Isc

= 2.5 A

Psc

= 240 W ??

ISC=Irated=215

Reqp jXeqp

VSC=489

PSC=240W

VSC

R


16/51


SCT (on primary)

Vsc= 489 V

Isc = 2.5 A

Psc

= 240 W

ISC=Irated=215

Reqp jXeqp

VSC=489

scscscsc cosIVP

scscsc ZZ

196052489

240.

.xcos sc

01

68761960 ..cossc

ohms.

.

Zsc 619552

489

scsc

sc

sc IV

P

cos

sc.sc

scI

V

52

489 0

68766195 ..

R


17/51


SCT (on primary)

Vsc= 489 V

Isc = 2.5 A

Psc

= 240 W

ISC=Irated=215

Reqp jXeqp

VSC=489

scscsc ZZ 068766195 ..

=Zsc

cossc

)eqp(SC)eqp(SC)eqp(SC jXRZ oc=76.68

RSC

jXSC =jZscsinsc

ZSC

Zsc=Zsccossc+ jZsc sin sc

Zsc=195.6

cos76.68 +j195.6sin76.68

Zeqp=38.4

+j 192

=38.4 =j192

=195.6


18/51


Question:- For the same transformer given in the previous

example, if the OCT & SCT are performed on the secondary

side, (a) what will be the Voltage, current and power readings

for the OCT and SCT.

(b) find the impedances of the approximate equivalent circuit

referred to the secondary side. Compare the data with part (a)

and sketch the circuit diagram.

Assignment ( Due Friday for section B & Saturday for section A

Reqp jXeqp

aVS

IS/a

VP

IO

IMIC

RC jXM

IP

38.4 j192

159k 38.4k


19/51


100

loadfullVloadfullVloadnoV

/aVVload-noatsince

100loadfullV

loadfullVloadnoVRegulationVoltage%

s

sp/a

PS

s

ss

The voltage regulation of a transformer is the change in the

magnitude of the secondary terminal voltage from no-load to

full-load.

2.7 (page 100) Transformer Voltage Regulation(VR) & Efficiency

Transformer Voltage Reg lation & Vector Diagram


20/51


100loadfullV

loadfullVloadnoVRegulationVoltage%

S

SP/a

SSeqSeqa

VVIjXIRP

VReq VjXeq

SjXeqqReaV VVVP

=ReqIS =XeqIS

IS

Reqp jXeqp

L

O

Ad

VS

a

VP

Transformer Voltage Regulation & Vector Diagram



21/51


100loadfullV

loadfullVloadnoVRV%

S

SP/a


IS VReq VjXeq

=ReqIS =XeqIS

Reqp jXeqp

L

O

A

d

VSa

VP

SeqSeqSa

VIjXIRVP loadfullVloadnoV SP/a

100

loadfullVRV%

s

SeqSeq

IjXIR

The VR of the

T/F depends on

the voltage

magnitude of

these seriesparameters.

Fig 2 26 (page 101) Vector Dig


22/51


IS

VReq

IS

VReq VjXeq

=ReqIS =XeqIS

Reqp jXeqp

L

oa

d

VSa

VP

VS

SjXeqqReaV VVVP

VjXeq

VP/a

Phase relation

between IS &

VReq ?Phase relation

between IS &

VXeq ?

Fig 2-26 (page 101) Vector Dig

Lagging Power Factor

Lagging

Power FactorLoad

?100V

VVVR%

sfl

sflp/a

Positive

VP/a > VS

-

VS is assumed to

be at angle 0o

(

VS-). All othervoltages & currents

are compared to

this reference

IR

Vector Diagram


23/51


IS

VReq VjXeq

Reqp jXeqp

L

o

ad

VSa

VPVeq

:Angle between Vs & VP/a

Veq

: Angle between VReq & VXeq

: Power Factor angle (Between Vs & Is)

Vector Diagram

Fig 2 27(a) Vector Dig


24/51


-

VS

IS

VReq VjXeq

=ReqIS =XeqIS

Reqp jXeqp

L

oa

d

VSa

VP

SjXeqqReaV VVVP

VP/a

Fig 2-27(a) Vector Dig

Unity Power Factor

UnityPower Factor

Load

?100V

VV

VR%sfl

sflp/a

Positive

VP/a > VS

Fig 2 27(a) Vector Dig


25/51


VS

IS

VReq VjXeq

=ReqIS =XeqIS

Reqp jXeqp

L

oa

d

VSa

VP

SjXeqqReaV VVVP

VP/a

Fig 2-27(a) Vector Dig

Leading PF

VReq

LeadingPower Factor

Load

?100V

VV

VR%sfl

sflp/a

Negative

VP/a


26/51


InputPower

OutputPower

The efficiency for a power transformer is between 0.9 to 0.99.

The higher the rating of a transformer, the greater is its efficiency.

Transformer Efficiency

InputPower

Losses

1

InputPower

LossesInputPower

cosIVPP

PP

sslosscorelosscopper

losscorelosscopper1


27/51

A 15 kVA 2300/230 V 60 Hz T/F The open circuit test (OCT) andExample 2 5 (page 103)


28/51


A 15-kVA, 2300/230-V, 60-Hz T/F. The open-circuit test (OCT) and

the short-circuit test (SCT) were performed on the primary side of

the T/F, and the following data were taken:

OCT (on primary)

Voc= 2300 V

Ioc

= 0.21 A

Poc= 50 W

SCT (on primary)

Vsc= 47 V

Isc

= 6.0 A

Psc = 160 W

Example 2.5 (page 103)

IP(t)Wattmeter

VP(t)V

A

V(t)

IP(t)Wattmeter

VP(t)V

A

V(t)

IP(t)Wattmeter

VP(t)V

A

V(t)

A

IP(t)Wattmeter

VP(t)V

A

V(t)

A

(a) Find the impedances of the approximate equivalent circuit


29/51


(b) Find the impedances of the approximate equivalent circuit

referred to the LV-side. (You can do this part: refer to slide#9) .

(a) Find the impedances of the approximate equivalent circuit

referred to the HV-side (You can do this part: refer to slide#9) .

38.4 j192

159k

Reqp jXeqp

aVS

IS/a

VP

IO

IMIC

RC jXM

IPReqp jXeqp

aVS

IS/a

VP

IO

IMIC

RC jXM

IP

j11k105k

4.45 J6.45

Figure:2-29(a)

Reqs jXeqs

VS

IS

VP/a

IO

IMIC

RC jXM

aIPReqs jXeqs

VS

IS

VP/a

IO

IMIC

RC jXM

aIP

38.4

159k 38.4k

Figure:2-29(b)

j110k1050k

0.0445 J0.0645

( ) C l l t th f ll l d ( t d l d) lt l ti


30/51


(c) Calculate the full-load (rated-load) voltage regulation:

(i) at 0.8 lagging power factor

(ii) at unity (1) power factor

(iii) at 0.8 leading power factor

Reqs jXeqs

VS

IS

VP/a

IO

IMIC

RC jXM

aIPReqs jXeqs

VS

IS

VP/a

IO

IMIC

RC jXM

aIP

38.4

159k 38.4k

Figure:2-29(b)

j110k1050k

0.0445 J0.0645

Reqs jXeqs

VS

IS

VP/a

IO

IMIC

RC jXM

aIPReqs jXeqs

VS

IS

VP/a

IO

IMIC

RC jXM

aIP

38.4

159k 38.4k

Figure:2-29(b)

j110k1050k

0.0445 J0.0645

We can use any of the

referred circuits to find

VR. Its easy to use

figure referred to thesecondary side.

SjXeqqRea

VVVVP

SjeqSeqSaV IjXIRVP

?100

V

VVVR%

sfl

sflp/a

VS=Vsf=230V - Given

Vp/a=unknown

To get Vp/a find ReqIS & XeqIS

Req=0.0445; Xeq=0.0645

from SCT & OCT.

IS=IS(rated) un-known

dthtl dllFHow to find I I ?


31/51


Va

V oP0230

rateds,

ratedrated,S

V

SIisF/Tthisofside

ondarysectheoncurrentloadullF

How to find IS =IS(rated) ?

IS VReq VjXeq

ReqIS,rated XeqIS,rated

Reqp jXeqp

L

O

A

d

VS

a

VP

0.0445 0.0645

IS=IS,rated

Parallel circuit is omitted.

Why ??

26500015

.230

VA,I rated,S

lagging.PFat

..I rated,S

80

9362650

0936265 ..

)A..)(.( o93626504450 )A..)(.(j o93626506450 Va

V oP0230

SjeqSeqSa

VIjXIRVP

V.. o153214 V.. o93692

?100V

VVVR%

sfl

sflp/a

).()VA( 8015000

V


32/51


%1.23085.34

2100

230V

2VR%

sfl

VV Angles are not used in

VR has no unit why?

InputPowerOutputPower

What is the

problem with

this equation ?

(e) What is the efficiency of the T/F at full load with a PF=0.8

lagging?

coreCuSS

SS

PPcosIVcosIV

InputPower

LossesInputPower

Use this

equation

VS IS =S (apparent power) and cos=0.8 lagging are known

Power input is unknown. We can use input equation, if the T/F

circuit is referred to the primary side. - Try & find efficiency

Va

V oP0230 V..

o153214 V..

o93692 V.. o4085234

IV


33/51


PCu= (IS)2

Req = (65.2A)2

(0.0445) = 189 Watts

Watts.

V.

R

aV

PC

P

core 5521050

852342

2

000398100

5521898015000

8015000.x

W.W).()VA(

).()VA(

InputPower

OutputPower

coreCuSS

SS

PPcosIV

cosIV

Why the SCT cupper losses(160W) and OCT core losses(50W)are not used in efficiency calculation ?

Open circuit test at the primary is performed at 2300 V. Where

as with rate-load (PF=0.8 lagging) VP=234.85x10=2348.5V, i.e

48.5V more than the OCT.

SCT is performed

with ISC=6A, where

as IP(rated)= 6.52A


34/51


What is the condition for the maximum efficiency in the T/F?

Core Losses = Copper LossesPcore = Pcopper

In the previous problem find the current at maximum efficiency?

I2Rseq= 52.5 W

Amps..

.

Imax3434

04450

552

T/F will operate at almost half the rated load (half of T/F capacity)

Example 2-5 Page # 103


35/51


p g

Solution


36/51



37/51



38/51


38.4 j192

159k

Reqp jXeqp

aVS

IS/a

VP

IO

IMIC

RC jXM

IPReqp jXeqp

aVS

IS/a

VP

IO

IMIC

RC jXM

IP

j11k105k

4.45 J6.45

Figure:2-29(a)


39/51



40/51



41/51



42/51



43/51



44/51


Continued


45/51



46/51



47/51


Practice Problem


48/51


A single-phase, 100-kVA, 1000:100-V, 60-Hz transformer has thefollowing test results:

Open-circuit test (HV side open): 100 V, 6 A, 400 W

Short-circuit test (LV side shorted): 50 V, 100 A, 1800 W

Draw the equivalent circuit of the transformer referred to the high-voltage

side. Label impedances numerically in ohms and in per unit. Determine the voltage regulation at rated secondary current with 0.6

power factor lagging. Assume the primary is supplied with rated voltage

Determine the efficiency of the transformer when the secondary current is75% of its rated value and the power factor at the load is 0.8 lagging witha secondary voltage of 98 V across the load

Example 4


49/51


A 20-kVA, 8000:277-V distribution transformer has the followingresistances and reactances:

The excitation branch impedances are referred to the high-voltage side.

a) Find the equivalent circuit of the transformer referred to the high-voltageside.

b)

b) Find the per unit equivalent circuit of this transformer.

c) Assume that the transformer is supplying rated load at 277 V and 0.8power factor lagging. What is this transformers input voltage? What isits voltage regulation?

d) What is this transformers efficiency under the conditions of part (c)?

RP= 32 ohm

XP= 45 ohm

RC= 250,000 ohm

RS

= 0.05 ohm

XS= 0.06 ohm

XM

= 30,000 ohm

Neglect

PU S t


50/51


PU System

QuantityBase

QuantityActualValuePU

base

ohm

PU

base

basebase

base

base

base

base

base

base

basebasebase

basebasebasebasebasebase

base

basebase

Z

ZZ

V

IY

VA

V

S

V

I

V

ZXR

IVVASQP

V

VAI

22

Per unit system, a system of dimensionless parameters, is used for

computational convenience and for readily comparing the performance

of a set of transformers or a set of electrical machines.

Where actual quantity is a value in volts, amperes, ohms, etc.

[VA]base and [V]base are chosen first.

ratioturnsVV

VAVA

secbase

pr ibase

secbasepr ibase


51/51


Next Time

Chapter # 3

Power Electronics

Must Read

Chapter_2_342_transformer_5PPT presentation

Chapter 2 342 Transformer 4

Documents

Transcript of Chapter 2 342 Transformer 4