1. ENERGY CONVERSION ONE (Course 25741)
Chapter Two
TRANSFORMERS
…continued

2. 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

3. 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:

4. 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:

5. Transformer Voltage Regulation and Efficiency
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

6. 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 > VS
V.R. > 0
Unity P.F.
VP / a > VS
V.R. >0 (smaller)
Leading P.F.
VS > VP/ a
V.R. < 0

7. 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

8. 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:

9. 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

10. 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:

11. Transformer Voltage Regulation and Efficiency
Impedance of excitation branch referred to primary:
Short Circuit Impedance angle:
Equivalent series Impedance:
Req=4.45 Ω, Xeq=6.45 Ω

12. Transformer Voltage Regulation and Efficiency
The equivalent circuits shown below:

13. Transformer Voltage Regulation and Efficiency
(b) To find eq. cct. Referred to L.V. side,
impedances divided by a²=NP/NS=10
RC=1050 Ω , XM=110 Ω
Req= Ω , Xeq= 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

14. Transformer Voltage Regulation and Efficiency
Therefore:
VP / a =
V.R.=( )/230 x 100 %=2.1 % for 0.8 lagging
At PF=0.8 leading  IS=65.2/_36.9◦ A

15. Transformer Voltage Regulation and Efficiency
V.R. = ( )/230 x 100%= %
At PF=1.0 , IS= 65.2 /_0◦ A
VP/a=
V.R. = ( )/230 x 100% = 1.28 % for PF=1

16. Transformer Voltage Regulation and Efficiency
Example: Phasor Diagrams …

17. 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

18. 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 / 1050=52.5 W
output power:
Pout=VSIS cosθ=230x65.2xcos36.9◦=12000 W
η= VSIS cosθ / [PCu+PCore+VSIS cosθ] x 100%=
12000/ [ ] = %

19. Efficiency of Distribution Transformers

20. 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 = 10^9 kWh)
it is further estimated [2] that the losses in all of the world’s 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 TWh.

21. Electrical Energy Losses in Distribution Networks

22. 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%

23. 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 /480 V
+2.5% tap /480 V
Nominal rating /480 V
-2.5% tap /480 V
-5% tap /480 V

24. 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

25. 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

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

27. AUTO TRANSFORMER
A step-down auto-transformer :
IH=ISE
IL=ISE+IC

28. 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:
VL=VC (3)
VH=VC+VSE (4)
current in coils are related to terminal currents:
IL=IC+ISE (5)
IH=ISE (6)

29. 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 . IH 
IL / IH = (NSE + NC)/NC (8)

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

31. 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)

32. 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:
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.

33. 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 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

34. 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 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

35. 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.)

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

37. AUTO TRANSFORMER 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= j pu
Power advantage of autotransformer is 11, so its per unit impedance would be:
Zeq=(0.01+j0.08)/11= j pu

38. Example of Variable Auto-Transformer