25471 Energy Conversion 5

8/6/2019 25471 Energy Conversion 5

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ENERGY CONVERSION ONE

(Course 25741)

Chapter Two

TRANSFORMERScontinued


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Equivalent Circuit of Transformer

Major Items to be considered in Construction of

Transformer Model:

Copper losses (in primary & Secondarywinding) ~ I

Eddy current losses (in core) ~ V

Hysteresis losses (in core) a complex nonlinearfunction of applied V

Leakage flux : LP & LS, these fluxes produce

self-inductance in primary & secondary coils


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Exact Eq. cct. Model for Real Transformer

Copper losses modeled by resistances Rp & Rs

As discussed before:

p=m+Lp p; total av. Primary flux

S=m+LS S; total av. Secondary flux

where m; flux linking both P & S

Lp; primary leakage fluxLS; secondary leakage flux

The average primary (& Secondary) flux, each, isdivided into two components as:

mutual flux & leakage flux


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Based on application of these components,Faradays law for primary circuit can beexpressed as:

Vp(t)=Np dp/dt = Np dM/dt + Np dLp/dt or:

Vp(t)=ep(t) + eLp(t) similarly for secondary:

Vs(t)=Ns ds/dt = Ns dM/dt + Ns dLs/dt or:

Vs(t)=es(t) + eLs(t)

primary & secondary voltages due to mutualflux :

ep(t) = Np dM/dt es(t)= Ns dM/dt


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Note : ep(t)/Np = dM/dt =es(t)/Ns

ep(t)/es(t) = Np / Ns =a

while eLp(t) = Np dLp/dt & eLs(t)= Ns dLs/dtif = permeance of leakage flux path

Lp=(p Np) ip & Ls=(p Ns) is

eLp(t) = Np d/dt (p Np) ip = Npp dip/dt

eLs(t) = Ns d/dt (p Ns) is = Nsp dis/dt

Defining:Lp = Npp primary leakage inductanc

Ls = Nsp secondary leakage inductance


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eLp(t)=Lp dip/dt

eLs(t)=Ls dis/dt

Therefore leakage flux can be modeled by primary & secondary

leakage inductances in equivalent electric circuit Core Excitation that is related to the flux linking both windings(m; flux linking both P & S) should also be realized in modeling

im (in unsaturated region) ~ e (voltage applied to core)

and lag applied voltage by 90 modeled by an inductance Lm

(reactance Xm) Core-loss current ie+h is ~ voltage applied & It can be modeled

by a resistance Rc across primary voltage source

Note: these currents nonlinear therefore: Xm & Rc are bestapproximation of real excitation


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The resulted equivalent circuit is shown:

Voltage applied to core = input voltage-internal

voltage drops of winding


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Equivalent Circuit of Transformer to analyze practical circuits including Transformers, it

is required to have equivalent cct. at a single voltage

Therefore circuit can be referred either to its primary

side or secondary side as shown:


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Approximate Equivalent Circuits of a Transformer

in practice in some studies these models are more

complex than necessary

i.e. the excitation branch add another node to

circuit, while in steady state study, current of this

branch is negligible

And cause negligible voltage drop in Rp & Xp Therefore approximate eq. model offered as:


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Approximate transformer models a- referred to primary b- referred to secondary

c- with no excitation branch referred to p

d- with no excitation branch referred to s


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Determination of Transformer Eq. cct.

parameters Approximation of inductances & resistances obtained

by two tests:

open circuit test & short circuit test

1- open circuit test : transformers secondary windingis open circuited, & primary connected to a full-rated

line voltage, Open-circuit test connections as below:


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parameters Input current, input voltage & input power measured From these can determine p.f., input current, and

consequently both magnitude & angle of excitationimpedance (RC, and XM)

First determining related admittance andSusceptance:

GC=1/RC & BM=1/XM YE=GC-jBM=1/RC -1/XM

Magnitude of excitation admittance referred to primarycircuit : |YE |=IOC/VOC

P.f. used to determine angle,

PF=cos=POC/[VOC . IOC]

=cos1 {POC/[VOC . IOC]}


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Determination of Transformer Eq.

cct. parameters Thus: YE = IOC/VOC = IOC / VOC

Using these equations RC & XM can be

determined from O.C. measurement

PF1cosU


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parameters Short-Circuit test : the secondary terminals of

transformer are short circuited, and primary

terminals connected to a low voltage source:

Input voltage adjusteduntilcurrent in s.c.

windings equal to its rated value


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cct. parameters The input voltage, current and power are again measured Since input voltage is so low during short-circuited test, negligible current

flows through excitation branch

Therefore, voltage drop in transformer attributed to series elements

Magnitude of series impedances referred to primary side of transformer is:|ZSE| = VSC/ ISC , PF=cos=PSC/[VSC ISC]

=cos1 {PSC/[VSC ISC]}

ZSE= VSC / ISC = VSC/ ISC

series impedance ZSE is equal to:

ZSE=Req+jXeq = (RP+ aRS) + j(XP+aXS)

It is possible to determine the total series impedance referred to primary side

, however difficult to split series impedance into primary & secondary

components although it is not necessary to solve problem

These same tests may also be performed on secondary side of transformer

U0 U


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cct. parameters Determine Equivalent cct.Impedances of a 20 kVA,8000/240 V, 60 Hz

transformer O.C. & S.C. measurements

shown

P.F. in O.C. is:

PF=cos=POC/[VOCIOC]=400 W/ [8000V x 0.214A]

=0.234 lagging

O.C. test(on primary)

S.C. test(on primary)

VOC=8000V VSC=489V

IOC=0.214A ISC=2.5 A

POC=400W PSC=240W


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cct. parameters excitation impedance:

YE=IOC/VOC

= 0.214 A / 8000 V= 0.0000268

= 0.0000063 j 0.0000261 = 1/RC- j 1/XM

Therefore: RC=1/ 0.0000063 = 159 k

XM= 1/0.0000261=38.4 k

PF1cos

234.0cos 1

; Q5.76


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cct. parameters PF in sc test:

PF=cos = PSC/[VSCISC]=240W/ [489x2.5]=0.196 lagging

Series impedance:

ZSE=VSC/ISC

= 489 V/ 2.5 A =195.6

=38.4 +j 192

The Eq. resistance & reactance are :

Req=38.4 , Xeq=192

PF1cos

Q7.78 Q7.78


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parameters The resulting Eq. circuit is shown below:


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The Per Unit System For Modeling

As seen in last Example, solving cct. containing

transformers requires tedious operation to refer all

voltages to a common level

In another approach, the need mentioned above iseliminated& impedance transformation is avoided

That method is known asper-unit system of measurement

there is also another advantage, in application ofper-unit: as size of machinery & Transformer varies its

internal impedances vary widely, thus a 0.1 cct.

Impedance may not be adequate & depends on

devices voltage and power ratings


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In per unit system the voltages, currents,

powers, impedance and other electrical

quantities not measured in SIunits system

However it is measured and define as a

decimal fraction of some base level

Any quantity can be expressed onpu basis

Quantity in p.u. =

Actual Value / base value of quantity


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Two base quantities selected & other base quantitiescan be determined from them

Usually; voltage, & power

Pbase,Qbase, or Sbase = Vbase Ibase

Zbase= Vbase/Ibase

Ybase=Ibase/Vbase

Zbase=(Vbase) / Sbase

In a power system, bases for power & voltageselected at a specific point, power base remainconstant, while voltage base changes at everytransformer


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Example: A simple power system shown inFigure below:

Contains a 480 V generator connected to anideal 1:10 step up transformer, a transmissionline, an ideal 20:1 step-down transformer, and a

load


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Example

Impedance of line 20+j60,impedance of load

Base values chosen as 480 V and 10 kVA at genertor

(a) Find bas voltage, current, impedance, and power

at every point in power system

(b) convert this system to its p.u. equivalent cct.

(c) Find power supplied to load in this system

(d) Find power lost in transmission line(a) At generator: Ibase=Sbase/Vbase 1=10000/480=20.83 A

Zbase1=Vbase1/Ibase1=480/20.83=23.04

Turn ratio of transformer T1 , a=1/10 =0.1 so base

voltage at line Vbase2=Vbase1/a=480/0.1=4800 V

; Q3010


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Example

Sbase2=10 kVA

Ibase2=10000/4800=2.083 A

Zbase2

=4800 V/ 2.083 A = 2304 Turn ratio of transformer T2 is a=20/1=20, sovoltage base at load is:

Vbase3=Vbase2/a =4800/20= 240 V

Other base quantities are:

Sbase3=10 kVA

Ibase3=10000/240=41.67 A

Zbase3=240/41.67 = 5.76


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Example

(b) to build the pu equivalent cct. Of power system, each

cct parameter divided by its base value

VG,pu=

Zline,pu=(20+j60)/2304=0.0087+j0.0260 pu

Zload,pu=

Per unit equivalent cct of PWR. SYS. Shown below:

puQQ

00.1480/0480 !

puQQ

0736.176.5/3010 !


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Example (c) current flowing in:

Ipu=Vpu/Ztot,pu=

pu Per unit power of load :

Pload,pu =IpuRpu=(0.569)(1.503)=0.487

actual power supplied to load:

Pload=Pload,puSbase=0.487 x 10000=4870 W(d) power loss in line:

Pline loss,pu =IpuRline,pu=(0.569)(0.0087)=0.00282

Pline= Pline loss,pu Sbase= (0.00282)(10000)=28.2 W

Q

Q

Q

Q

6.30569.0894.0512.1

01

)30736.1()026.00087.0(

01

!

!

j

j

25471 Energy Conversion 5

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