- 123 Views
- Uploaded on
- Presentation posted in: General

ENERGY CONVERSION ONE (Course 25741)

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

ENERGY CONVERSION ONE(Course25741)

Chapter Two

TRANSFORMERS

…continued

- 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

- 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:

- 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:

- If power factor is unity, VS is lower than VP so V.R. > 0
- V.R. is smaller for lagging P.F.
- With a leading P.F., VS is larger VP V.R.<0
- P.F. =1

- P.F. leading

Table Summarize possible Value for V.R. vs Load P.F.:

- Since transformer usually operate at lagging P.F., a simplified method is introduced

- 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

- 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:

- 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:

(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:

- Impedance of excitation branch referred to primary:
- Short Circuit Impedance angle:
- Equivalent series Impedance:
Req=4.45 Ω, Xeq=6.45 Ω

- The equivalent circuits shown below:

- (b) To find eq. cct. Referred to L.V. side,
impedances divided by a²=NP/NS=10

RC=1050 Ω , XM=110 Ω

Req=0.0445 Ω , Xeq=0.0645 Ω

- (c) full load current on secondary side:
IS,rated=Srated/ VS,rated=15000/230 =65.2 A

To 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

- 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 =

- 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

- Example: Phasor Diagrams …

- (d) to plot V.R. as a function of load is by repeating the calculations of part “c” for many different loads using MATLAB

- (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 / 1050=52.5 W

output power:

Pout=VSIS cosθ=230x65.2xcos36.9◦=12000 W

η= VSIS cosθ / [PCu+PCore+VSIS cosθ] x 100%=

12000/ [189+52.5+12000] = 98.03 %

- The total electrical energy use per annum of the world is estimated as 13,934
- TeraWatthours [TWh] (1 TWh = 10^9 kWh)
- it is further estimated [2] that the losses in all of theworld’s electrical distribution systemstotal about 1215 TWh or
- about 8.8% of the total electrical energy consumed. About 30-35% of these losses are generated in the Transformersin 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.

- 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%

- 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

- 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 solutionis using special transformer called: “tap changing under load transformer”
- A voltage regulatoris a tap changing under load transformer with built-in voltage sensing circuitry that automatically changes taps to preserve system voltage constant

- 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

- Diagram of a step-up auto-transformer shown in figure below:
- C: common, SE: series

- A step-down auto-transformer :
- IH=ISE
- IL=ISE+IC

- In step-up autotransformer:
- VC / VSE = NC / NSE (1)
- NC IC = NSE ISE(2)
- voltages in coils are related to terminal voltages as follows:
- VL=VC (3)
- VH=VC+VSE (4)
- current in coils are related to terminal currents:
- IL=IC+ISE (5)
- IH=ISE (6)

- 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 . IH

IL /IH = (NSE + NC)/NC (8)

- 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 : Sin=VLIL
- And the output apparent power is:
Sout=VH IH

- 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)

- 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:
- 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.

(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 transformer’s

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

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 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

- 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.)

- 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 transformer’s rating ?

(b) what is the transformer’s series impedance in per unit?

- Solution:
(a) NC/NSE= 12/1.2 (or 10:1) the voltage ratio of autotransformer is 13.2/12 kV & VA rating : SIO=(1+10)/1 x 1000 kVA=11000 kVA

(b) transformer’s impedance in per-unit when connected as conventional transformer:

Zeq=0.01 + j 0.08 pu

Power advantage of autotransformer is 11, so its per unit impedance would be:

Zeq=(0.01+j0.08)/11=0.00091+j0.00727 pu