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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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Equivalent Circuit of Transformer
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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Equivalent Circuit of Transformer
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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Equivalent Circuit of Transformer
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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Equivalent Circuit of Transformer
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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Equivalent Circuit of Transformer
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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Equivalent Circuit of Transformer
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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Equivalent Circuit of Transformer
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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Determination of Transformer Eq. cct.
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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Determination of Transformer Eq. cct.
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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Determination of Transformer Eq.
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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Determination of Transformer Eq.
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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Determination of Transformer Eq.
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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Determination of Transformer Eq.
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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Determination of Transformer Eq. cct.
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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The Per Unit System For Modeling
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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The Per Unit System For Modeling
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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The Per Unit System For Modeling
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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The Per Unit System For Modeling
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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The Per Unit System For Modeling
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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The Per Unit System For Modeling
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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The Per Unit System For Modeling
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