ENERGY CONVERSION ONE (Course 25741)

1
ENERGY CONVERSION ONE (Course 25741)

• Chapter Two
• TRANSFORMERS
• continued

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Transformer Voltage Regulation and Efficiency

• Output Voltage of Transformer Varies with Load
• Due to Voltage Drop on Series Impedance of
  Transformer Equivalent 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

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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

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Transformer Voltage Regulation and Efficiency

• If power factor is unity, VS is lower than VP so
  V.R. gt 0
• V.R. is smaller for lagging P.F.
• With a leading P.F., VS is larger VP ? V.R.lt0
• P.F. 1
• ?
• P.F. leading
• ?

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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., a simplified method is introduced

Lagging P.F. VP/ a gt VS V.R. gt 0
Unity P.F. VP / a gt VS V.R. gt0 (smaller)
Leading P.F. VS gt VP/ a V.R. lt 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 other ?angle 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 I²R losses
• 2- Core Hysteresis losses
• 3- Core Eddy current losses
• Transformer efficiency may be determined as
  follows

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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
VOC2300 V VSC47 V
IOC0.21A ISC6 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

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Transformer Voltage Regulation and Efficiency

• Impedance of excitation branch referred to
  primary
• Short Circuit Impedance angle
• Equivalent series Impedance
• 
  Req4.45 O, Xeq6.45 O

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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 a²NP/NS10
• RC1050 O , XM110 O
• Req0.0445 O , Xeq0.0645 O
• (c) full load current on secondary side
• IS,ratedSrated/ VS,rated15000/230
  65.2 A
• To determine V.R., VP/ a is needed
• VP/a VS Req IS j Xeq IS , and
• IS65.2/_-36.9? A , at PF0.8 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 PF0.8 leading ? IS65.2/_36.9? A
• VP / a

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Transformer Voltage Regulation and Efficiency

• V.R. (229.85-230)/230 x 100 -0.062
• At PF1.0 , IS 65.2 /_0? A
• VP/a
• V.R. (232.94-230)/230 x 100 1.28 for PF1

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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)²Req (65.2)² (0.0445)189 W
• - Core losses
• PCore (VP/a)² / RC (234.85)² / 105052.5 W
• output power
• PoutVSIS cos?230x65.2xcos36.9?12000 W
• ? VSIS cos? / PCuPCoreVSIS cos? x 100
• 12000/ 18952.512000 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
  world is estimated as 13,934
• TeraWatthours TWh (1 TWh 109 kWh)
• it is further estimated 2 that the losses in
  all of the worlds electrical distribution
  systems total about 1215 TWh or
• about 8.8 of the total electrical energy
  consumed. About 30-35 of these losses are
  generated in the Transformers in the Distribution
  systems.
• Studies estimate that some 40-80 of these
  transformer 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
  adjusted in field to accommodate variations in
  tap voltages
• While this tap can not be changed when power is
  applied to transformer
• Some times voltage varies widely with load, i.e.
  when high line impedance exist between
  generators particular load while normal loads
  should be supplied by an essentially constant
  voltage
• One solution is using special transformer called
  tap changing under load transformer
• A voltage regulator is a tap changing under load
  transformer with built-in voltage sensing
  circuitry that automatically changes taps to
  preserve system voltage constant

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AUTO TRANSFORMER

• some occasions it is desirable to change voltage
  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 special
  transformer 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
• IHISE
• ILISEIC

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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
  voltages as follows
• VLVC
  (3)
• VHVCVSE
  (4)
• current in coils are related to terminal
  currents
• ILICISE
  (5)
• IHISE
  (6)

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AUTO TRANSFORMER

• Voltage Current Relations in Autotransformer
• VHVCVSE
• since VC/VSENC/NSE ? VHVC NSE/NC . VC
• Noting that VLVC ?
• VHVL NSE/NC . VL (NSENC)/NC . VL
• VL / VH NC / (NSENC)
  (7)
• Current relations
• ILICISE employing Eq.(2) ? IC(NSE / NC)ISE
• IL (NSE / NC)ISE ISE, since ISEIH ?
• IL (NSE / NC)IH IH (NSE NC)/NC .
  IH ?
• IL / 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 be
  reconnected as an autotransformer, it can handle
  much more power than its original rating
• The input apparent power to the step-up
  autotransformer is SinVLIL
• And the output apparent power is
• SoutVH IH

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AUTO TRANSFORMER

• And
• SinSoutSIO
• Apparent power of transformer windings
• SW VCICVSE ISE
• This apparent power can be reformulated
• SW VCICVL(IL-IH) VLIL-VLIH
• employing Eq.(8) ? SW VLIL-VLIL NC/(NSENC)
• VLIL (NSENC)-NC /(NSENC)SIO NSE
  /(NSENC)
• SIO / SW (NSENC) / 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
  connecting a 110 kV system to a 138 kV system has
  an NC/NSE of 110/28
• for this autotransformer actual winding rating
  is
• SWSIO NSE/(NSENC)5000 x 28/ (28110)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 101)
• (a) using Eq.(7),VH (12120)/120 x 120 132 V
  
• (b) maximum VA rating 100 VA?
• ISE,max100/128.33 A

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AUTO TRANSFORMER

• SoutVSISVHIH 132 x 8.33 1100 VA Sin
• (c) rating advantage
• SIO/SW(NSENC)/NSE(12120)/1211 or
• SIO/SW 1100/100 11
• It is not normally possible to reconnect an
  ordinary transformer as an autotransformer due to
  the fact that insulation of L.V. side may not
  withstand full output voltage of autotransformer
  connection
• 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 flows during 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 conventional
  transformer 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 101) the voltage ratio
  of autotransformer is 13.2/12 kV VA rating
  SIO(110)/1 x 1000 kVA11000 kVA
• (b) transformers impedance in per-unit when
  connected as conventional transformer
• Zeq0.01 j 0.08 pu
• Power advantage of autotransformer is 11, so
  its per unit impedance would be
• Zeq(0.01j0.08)/110.00091j0.00727 pu

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Example of Variable Auto-Transformer