auto transformer energy conversion 6

7/22/2019 Auto Transformer Energy Conversion 6

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ENERGY CONVERSION ONE

(Course25741)

Chapter Two

TRANSFORMERScontinued


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Transformer Voltage Regulation

and Efficiency Output Voltage of Transformer Varies with Load

Due to Voltage Drop on Series Impedance of TransformerEquivalent Model

Full Load Regulation Parameter, compares output no-load

Voltage with its Full Load Voltage:V.R. =

At no load VS= VP / a thus :

V.R.=

in per unit: V.R. =

For Ideal Transformer V.R.=0

%100..,

..,..,

LFS

LFSLNS

V

VV

%100)/(

..

..

LF

LFP

V

VaV

%100,,

,,,

puFLS

puFLSpuP

V

VV


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Transformer Voltage Regulation and

Efficiency

The transformer phasor diagram

To determine the voltage regulation of a transformer:

The voltage drops should be determined

In below a Transformer equivalent circuit referred to

the secondary side shown:


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and Efficiency

since current which flow in magnetizing branch is small

can be ignored

Assuming secondary phasor voltage as reference VS with

an angle of 0

Writing the KVL equation:

From this equation the phasor diagram can be shown: At lagging power factor:

SeqSeqSP IjXIRVa

V


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Efficiency

If power factor is unity, VS is lower than VP soV.R. > 0

V.R. is smaller for lagging P.F.

With a leading P.F., VS is larger VP V.R.


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and Efficiency

Table Summarize possible Value for V.R. vs Load P.F.:

Since transformer usually operate at lagging P.F., asimplified method is introduced

Lagging P.F. VP/ a > VS V.R. > 0

Unity P.F. VP / a > VS V.R. >0 (smaller)

Leading P.F. VS > VP/ a V.R. < 0


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Efficiency

Simplified Voltage Regulation Calculation

For lagging loads: the vertical components

related to voltage drop on Req & Xeq partially

cancel each otherangle of VP/a very small


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and Efficiency

Transformer Efficiency (as applied to motors, generators and motors)

Losses in Transformer:

1- Copper IR losses

2- Core Hysteresis losses

3- Core Eddy current losses

Transformer efficiency may be determined as follows:

%100xP

P

in

out %100x

PP

P

lossout

out

%100cos

cos

xIVPP

IV

SScoreCu

SS


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and Efficiency

Example:

A 15kVA, 2300/230 V transformer tested to determine

1- its excitation branch components, 2- its series

impedances, and 3- its voltage regulation Following data taken from the primary side of the transformer:

Open Circuit Test Short Circuit Test

VOC=2300 V VSC=47 V

IOC=0.21A ISC=6 A

POC= 50 W PSC= 160 W


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and Efficiency

(a) Find the equivalent circuit referred to H.V. side

(b) Find the equivalent circuit referred to L. V. side

(c) Calculate the full-load voltage regulation at 0.8 lagging PF,

1.0 PF, and at 0.8 leading PF(d) Find the efficiency at full load with PF 0.8 lagging

SOLUTION:

Open circuit impedance angle is:

Excitation admittance is:

8421.02300

50coscos 11

OCOC

OC

OC

IVP

0000908.00000095.0

841013.9842300

21.084

5

j

V

IY

OC

OC

E


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and Efficiency

Impedance of excitation branch referred to primary:

Short Circuit Impedance angle:

Equivalent series Impedance:

Req=4.45 , Xeq=6.45

kX

kR

M

C

110000908.0

1

1050000095.0

1

4.55647

160coscos 11

SCSC

SCSC

IV

P

45.645.4

4.55833.74.556

47

j

I

VZ SC

SC

SCSE


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Efficiency

The equivalent circuits shown below:


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Efficiency

(b) To find eq. cct. Referred to L.V. side,

impedances divided by a=NP/NS=10

RC=1050 , XM=110 Req=0.0445 , Xeq=0.0645

(c) full load current on secondary side:

IS,rated=Srated/ VS,rated=15000/230 =65.2 ATo determine V.R., VP/ a is needed

VP/a = VS + Req IS + j Xeq IS , and:

IS=65.2/_-36.9 A , at PF=0.8 lagging


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and Efficiency

Therefore:

VP / a =

V.R.=(234.85-230)/230 x 100 %=2.1 % for 0.8 lagging

At PF=0.8 leading IS=65.2/_36.9 A

VP / a =

Vj

jj

j

4.085.23462.184.234

36.352.274.132.2230

1.5321.49.369.20230

9.362.650645.0)9.362.65)(0445.0(0230

Vj

jj

j

27.185.22910.58.229

36.352.274.132.2230

9.12621.49.369.20230

9.362.650645.0)9.362.65)(0445.0(0230


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and Efficiency

V.R. = (229.85-230)/230 x 100%= -0.062%

At PF=1.0 , IS= 65.2 /_0 A

VP/a=

V.R. = (232.94-230)/230 x 100% = 1.28 % for PF=1

Vj

j

j

04.194.23221.49.232

21.49.22309021.409.20230

)02.65)(0645.0()02.65)(0445.0(0230


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Efficiency

Example: Phasor Diagrams


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Efficiency

(d) to plot V.R. as a function of load is by

repeating the calculations of part c for many

different loads using MATLAB


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Efficiency

(e) Efficiency of Transformer:

- Copper losses:

PCu=(IS)Req =(65.2) (0.0445)=189 W- Core losses:

PCore= (VP/a) / RC= (234.85) / 1050=52.5 W

output power:Pout=VSIS cos=230x65.2xcos36.9=12000 W

= VSIS cos / [PCu+PCore+VSIS cos] x 100%=

12000/ [189+52.5+12000] = 98.03 %


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Efficiency of Distribution Transformers


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Energy Losses in Electrical Energy

Systems

The total electrical energy use per annum of the worldis estimated as 13,934

TeraWatthours [TWh] (1 TWh = 10^9 kWh)

it is further estimated [2] that the losses in all of theworlds electrical distribution systemstotal about1215 TWh or

about 8.8% of the total electrical energy consumed.

About 30-35% of these losses are generated in theTransformersin the Distribution systems.

Studies estimate that some 40-80% of thesetransformer losses are potentially saveable by

increasing transformer efficiencies, i.e. 145-290 TWh.


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Electrical Energy Losses in Distribution

Networks


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Transformer Taps & Voltage Regulation

Distribution Transformers have a series taps in

windings which permit small changes in turn

ratio of transformer after leaving factory

A typical distribution transformer has four taps

in addition to nominal setting, each has a 2.5%

of full load voltage with the adjacent tap

This provides possibility for voltage adjustment

below or above nominal setting by 5%


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Transformer Taps & Voltage

Regulation

Example: A 500 kVA, 13200/480 V distribution

transformer has 4, 2.5 % taps on primary

winding. What are voltage ratios?

Five possible voltage ratings are:

+5% tap 13860/480 V

+2.5% tap 13530/480 V

Nominal rating 13200/480 V

-2.5% tap 12870/480 V

-5% tap 12540/480 V


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Transformer Taps & Voltage Regulation

Taps on transformer permit transformer to be adjustedin field to accommodate variations in tap voltages

While this tap can not be changed when power isapplied to transformer

Some times voltage varies widely with load, i.e. whenhigh line impedance exist between generators &particular load;while normal loads should be suppliedby an essentially constant voltage

One solutionis using special transformer called: tapchanging under load transformer

A voltage regulatoris a tap changing under loadtransformer with built-in voltage sensing circuitry thatautomatically changes taps to preserve systemvoltage constant


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AUTO TRANSFORMER

some occasions it is desirable to changevoltage level only by a small amount

i.e. may need to increase voltage from 110 to

120 V or from 13.2 to 13.8 kV This may be due to small increase in voltage

drop that occur in a power system with long

lines In such cases it is very expensive to hire a two

full winding transformer, however a specialtransformer called: auto-transformer can be

used


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AUTO TRANSFORMER

Diagram of a step-up auto-transformer shown in

figure below:

C: common, SE: series


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AUTO TRANSFORMER

A step-down auto-transformer :

IH=ISE IL=ISE+IC


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AUTO TRANSFORMER

In step-up autotransformer:

VC / VSE = NC / NSE (1)

NC IC = NSE ISE (2)

voltages in coils are related to terminal voltagesas follows:

VL=VC (3)

VH=VC+VSE (4)

current in coils are related to terminal currents:

IL=IC+ISE (5)

IH=ISE (6)


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AUTO TRANSFORMER

Voltage & Current Relations in Autotransformer

VH=VC+VSE

since VC/VSE=NC/NSE VH=VC+ NSE/NC . VC

Noting that: VL=VC VH=VL+ NSE/NC . VL= (NSE+NC)/NC . VL

VL/ VH = NC/ (NSE+NC) (7)

Current relations:

IL=IC+ISE employing Eq.(2) IC=(NSE / NC)ISE

IL= (NSE / NC)ISE + ISE, since ISE=IH

IL= (NSE / NC)IH +IH = (NSE + NC)/NC . IHIL/IH = (NSE + NC)/NC (8)


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AUTO TRANSFORMER

Apparent Power Rating Advantage of Autotransformer

Note :not all power transferring from primary to

secondary in autotransformer pass through windings

Therefore if a conventional transformer bereconnected as an autotransformer, it can handle

much more power than its original rating

The input apparent power to the step-up

autotransformer is : Sin=VLIL

And the output apparent power is:

Sout=VH IH


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AUTO TRANSFORMER

And :

Sin=Sout=SIO

Apparent power of transformer windings:

SW= VCIC=VSE ISE This apparent power can be reformulated:

SW= VCIC=VL(IL-IH) =VLIL-VLIH

employing Eq.(8) SW= VLIL-VLIL NC/(NSE+NC)

=VLIL [(NSE+NC)-NC]/(NSE+NC)=SIO NSE /(NSE+NC)

SIO / SW = (NSE+NC) / NSE (9)


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AUTO TRANSFORMER

Eq.(9); describes apparent power rating advantage of

autotransformer over a conventional transformer

smaller the series winding the greater the advantage

Example one: A 5000 kVA autotransformer connectinga 110 kV system to a 138 kV system has an NC/NSE of

110/28

for this autotransformer actual winding rating is:

SW=SIO NSE/(NSE+NC)=5000 x 28/ (28+110)=1015 kVA

Example Two: A 100 VA 120/12 V transformer is

connected as a step-up autotransformer, and primary

voltage of 120 applied to transformer.


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AUTO TRANSFORMER

(a) what is the secondary voltage of transformer

(b) what is its maximum voltampere rating in this

mode of operation

(c) determine the rating advantage of this

autotransformer connection over transformers

rating of conventional 120/12 V operation

Solution: NC/NSE= 120/12 (or 10:1)

(a) using Eq.(7),VH= (12+120)/120 x 120 = 132 V

(b)maximum VA rating 100 VA

ISE,max=100/12=8.33 A


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AUTO TRANSFORMER

Sout=VSIS=VHIH= 132 x 8.33 = 1100 VA = Sin(c) rating advantage:

SIO/SW=(NSE+NC)/NSE=(12+120)/12=11 or:

SIO/SW= 1100/100 = 11

It is not normally possible to reconnect an ordinary transformeras an autotransformer due to the fact that insulation of L.V. sidemay not withstand full output voltage of autotransformerconnection

Common practice: to use autotransformer when two voltages

fairly close Also used as variable transformers, where L.V. tap moves up &

down the winding

Disadvantage:direct physical connection between primary &secondary circuits, and electrical isolation of two sides is lost


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AUTO TRANSFORMER

Internal Impedance of an Autotransformer

Another disadvantage: effective per unit

impedance of an autotransformer w.r.t. the

related conventional transformer is the

reciprocal of power advantage

This is a disadvantage where the series

impedance is required to limit current flowsduring power system faults (S.C.)


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AUTO TRANSFORMER

Example three:

A transformer rated 1000 kVA, 12/1.2 kV, 60 Hz

when used as a two winding conventionaltransformer and its series resistance &

reactance are 1 and 8 percent per unit

It is used as a 13.2/12 kV autotransformer

(a) what is now the transformers rating ?

(b) what is the transformers series impedance

in per unit?


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AUTO TRANSFORMER

Solution:

(a) NC/NSE= 12/1.2 (or 10:1) the voltage ratio ofautotransformer is 13.2/12 kV & VA rating :

SIO=(1+10)/1 x 1000 kVA=11000 kVA(b) transformers impedance in per-unit whenconnected as conventional transformer:

Zeq=0.01 + j 0.08 puPower advantage of autotransformer is 11, soits per unit impedance would be:

Zeq

=(0.01+j0.08)/11=0.00091+j0.00727 pu


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Example of Variable Auto-Transformer

auto transformer energy conversion 6

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