Auto Transformer Energy Conversion 6
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Transcript of 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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Transformer Voltage Regulation
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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Transformer Voltage Regulation and
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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Transformer Voltage Regulation
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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Transformer Voltage Regulation and
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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Transformer Voltage Regulation
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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Transformer Voltage Regulation
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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Transformer Voltage Regulation
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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Transformer Voltage Regulation
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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Transformer Voltage Regulation and
Efficiency
The equivalent circuits shown below:
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Transformer Voltage Regulation and
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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Transformer Voltage Regulation
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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Transformer Voltage Regulation
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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Transformer Voltage Regulation and
Efficiency
Example: Phasor Diagrams
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Transformer Voltage Regulation and
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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Transformer Voltage Regulation and
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