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Energy conversion I

Lecture 7:

Topic 2: Transformers & its performance (S. Chapman, ch. 2)

- Introduction

- Types and Construction of Transformers.

- Ideal Transformer.- Theory of operation of real single-phase transformers.

- The Equivalent Circuit of a Transformer.

- The Per-Unit System of Measurement.

- Transformer voltage regulation and efficiency.- Autotransformers.

- Three phase transformers.

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Voltage regulation:

Series transformer impedance is the origin ofoutput voltage variations even if theinput voltage is constant (load regulation)

VR : Voltage regulation (percent)Vs,nl : No-load Secondary voltageVs,fl : full-load Secondary voltage (usually rated or 1 p.u)

Using Simple equivalent circuit and p.u values:

Transformer voltage regulation

%100,

,,x

V

VVVR

flS

flSnlS

%100,,

,,,x

V

VVVR

puflS

puflSpup = Series equivalent impedance

limits short circuit current

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Transformer voltage regulation

Load effect :

Using phasor diagram:

Vp,pu

= Vs,pu

+ Req

Is,pu

+jXeq

Is,pu

considering Vs as the reference phasor:

Vs

Vp

IsReqsIs

jXeqsIs

Vs

Vp

Is ReqsIs

jXeqsIs

Vs

Vp

Is

ReqsIs

jXeqs

Is

Lagging power factor

Leading power factor

Unity power factor

Note: Voltage regulation can benegative for capacitive loads.

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Transformer Tap changer

Transformer Tap changer is a device to change the voltage ratio forregulation ofthe output voltage:

On load tap changer: can change the voltage ratio while transformer is loaded

Off load tap Changer: changes the voltage ratio of a no-load transformer

Example: having 4 taps of 2.5% in HV for a 13200/480 V transformer means:

+5.0% tap : 13860/480 V+2.5% tap : 13530/480 V

Nominal : 13200/480 V

-2.5% tap : 12870/480 V-5.0% tap : 12540/480 V

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Transformer Efficiency

Transformer efficiency:

Considering power Losses:

Pcu: copper (winding RI2) Loss (proportional to load current).

Pcore: Core (Eddy current and Hystersis) loss (proportional to voltage).

Show that for a given output voltage and power factor, maximum efficiencyhappens if Pcu = Pcore

%100xP

P

in

out

%100cos

cos%100 x

IVPP

IVx

PP

P

SScoreCu

SS

lossout

out

++=

+=

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Transformer voltage regulation and efficiency

Example:

The equivalent Circuit parameters of a 15kVA, 2300/230 V transformer referred to low

voltage side are:

Req = 0.0445

, Xeq = 0.0645

, Rc = 1050

, XM = 110

A- Calculate the full load voltage regulation @ 0.8 lagging power factor, 1.0 power

factor and 0.8 leading power factor.

B- What is the efficiency at full load with 0.8 lagging power factor.

Solution:

The full-load current in the LV side is: Is,n = Sn / Vs,n = 15000/230 = 65.2 A

For 0.8 lagging power factor: Is

Vp = Vs + Is(Req+jXeq)=

o9.362.65 =

oj 4.085.234)0645.00445.0(9.362.650230 =++

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Therefore: VR=

For unity power factor: Is

Vp = Vs + Is(Req+jXeq) =

Therefore: VR=

For 0.8 leading power factor: Is

Vp = Vs + Is(Req+jXeq) =

Therefore: VR=

%1.2%100230

23085.234=

o02.65 =

oj 04.194.232)0645.00445.0(2.650230 =++

%28.1%100230

23094.232=

o9.362.65 =

oj 27.185.229)0645.00445.0(9.362.650230 =++

%062.0%100

230

23085.229=

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B- efficiency at full load with 0.8 lagging power factor.

Losse can be calculated as:

Pcu = Req Is2 = 0.0445(65.2)2 = 189 W

Pcore

= (Vp

)2 / Rc = (234.85)2 / 1050 = 52.5 W

Output power is : Pout = VsIscos = 230 65.2 cos (36.9o) = 12000 W

= Pout

/ (Pout

+ Ploss

) 100%= 12000 / (12000+189+52.5) 100%= 98.03%

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Autotransformers

For voltage ratios close to 1 it is usual to use autotransformer instead ofTransformer.

For voltage ratios close to one, autotransformer is much smaller in weightand volume compared to transformer.

For voltage ratios close to one, kVA rating of autotransformer is muchhigher than transformer used to make it.

Problem: no electrical isolation!

c

SE

SE

C

SE

c

SE

c

N

N

I

I

N

N

V

V

=

=

C: Common coilSE: Series coil

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Step-up Autotransformers

VL = VC

VH = VC + VSE

IL = IC + ISE

IH = ISE

SEC

C

SEC

C

H

L

NN

N

VV

V

V

V

+=

+=

STR = ISEVSE

SATR = IH VH = ISE VH

SE

SEC

SE

H

N

NN

V

V +==

TR

ATR

S

S>>1 If voltage

ratio is close to 1

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Autotransformer rating

C

SEC

L

H

SE

SEC

SE

H

N

NN

V

V

NNN

VV

+=

+==

TR

ATR

SS

STR : Transformer or each winding kVA (SW)

SATR : Autotransformer kVA (SIO)

Example: For a 5000 kVA, 110 kV / 138 kV we can use a 110 / 28 transformer:

NC: 110 , NSE = 28

STR = SW = 28 / (28 + 138) * 5000 = 1015 kVA

Think about Series impedance

of Autotransformer comparedto Transformer

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Variable Autotransformer (VARIAC)

Used as Variable AC voltage Source

I2No

V1

NC

IC

IL

Iload

V2

NSE

Z

Y

Sliding

terminal

Output voltage varies between 0 to (NSE+NC)/NCV1