ELG4125: Lecture 2 Power

The Ideal The Real Transformer Equivalent Circuits for Practical Transformers The Per-Unit System Three-Phase Transformers Transformers in Power Systems

• Typically in power systems, voltages get transformed approximately five times between generation and delivery to the users. • Generation in power systems, primarily by synchronous generators, takes place at around 20-kV level. • Transmission voltages of 230 kV, 345 kV, 500 kV, and 765 kV is common. • At the load, these voltages are stepped down to 120/240 V single-phase in residential usage. • Another advantage of transformers in many applications is to provide electrical insulation for safety purposes. Real Transformers

Transformer Types Power Transformers Current Transformers Voltage Transformers Transformer Connections Series Transformers Each leg is a single phase transformer Transformer Purchasing Issues Y-Y connections (no phase shift) Efficiency D-D connections (no phase shift) Audible Noise Y-D connections (-30 degrees phase shift) Installation Costs D-Y connections (+30 degrees phase shift) Manufacturing Facilities Performance Record

3 Nameplate Information

• kVA is the apparent power. • Voltage ratings for high and low side are no-load values. The symbol between the values indicate how the voltages are related: • Long dash (—) From different windings. • Slant (/) …from same winding: • 240/120 is a 240 V winding with a . • Cross (X) Connect windings in series or parallel. Not used in wye-connected winding: • 240X120 two part winding connected in series for 240 or parallel for 120. • Wye (Y) A wye-connect winding. Basic Principles of Transformer Operation

• Transformers consist of two or more coupled windings where almost all of the flux produced by one winding links the other winding. Transformer Exciting Current

• Transformers use ferromagnetic materials that guide magnetic flux due to their high permeability, require small ampere- turns to produce the desired flux density. • The magnetising current is that current which flows in the primary winding when the primary voltage is applied with the secondary unloaded. It is the necessary current that satisfies the excitation condition as determined by the fundamental transformer equation. • In modern power transformer of large kVA ratings, the magnetizing currents in their normal operating region are very small, for example, well below 0.2% of the rated current at the rated voltage. Voltage Transformation

d d v  e  N vS  eS  NS P P P dt dt

• The EMF of a transformer at a given flux density increases with frequency. By operating at higher frequencies, transformers can be physically more compact because a given core is able to transfer more power without reaching saturation and fewer turns are needed to achieve the same impedance. However, core loss and conductor skin effect also increase with frequency. • Aircraft and military equipment employ 400 Hz power supplies which reduce core and winding weight. Conversely, frequencies used for some railway electrification systems were much lower (16.7 Hz and 25 Hz) than normal utility frequencies (50/60 Hz) for historical reasons concerned mainly with the limitations of early electric traction systems. • As such, the transformers used to step down the high over-head line voltages (15 kV) were much heavier for the same power rating than those designed only for the higher frequencies. 2π EP  f NP φmax  4.44 f NP φmax 2

VP max  4.44 f NP

• Where VP is the RMS value of the applied voltage with the assumption that eP = VP. The second Equation shows that the flux is determined by the applied voltage, the frequency of operation, and number of turns in the winding.

Example

Transformer Equivalent Circuit Viewed from the Primary Side

Modeling a Real Transformer

Equivalent Circuit with Parameters Moved to the Primary Side

Transformer Phasor Diagram

Transformer Tests

Test to Determine Transformer Parameters, we require to conduct

Open Circuit Test: Energize Low voltage winding at rated voltage, leaving other winding open

Measure Current (Ioc) and Power (Poc) into energized winding. Calculate parameters

Short Circuit Test: Energize Low current () winding at rated current with a solid short circuit applied across the other winding Measure Voltage and Power at terminals of energized winding Calculate parameters

15 Transformer Test

Open-Circuit Test Short-Circuit Test

S S PC I C  , and VS

S S 2 S 2 I C  I e  - I C 

2 V V  S S PSC SC 2 RC  , and R  X T    - RT I S T 2  I  C ISC  SC  S VS X m  S Im Example 1

Example 2

The Per unit system

• In the per-unit system, the voltages, currents, powers, impedances, and other electrical quantities are expressed on a per-unit basis by the equation:

Actual value Quantity per unit =

Base value of quantity

• It is customary to select two base quantities to define a given per-unit system. The ones usually selected are voltage and power.

S  S Vb Vrated b rated

Then compute base values for currents and impedances

S 2 b Vb Vb I b  Z   V b b I b Sb

Vactual Iactual Vp.u.  I p.u.  Vb Ib S Z S  actual actual p.u. Z p.u.  Sb Z b

Z%  Z p.u. 100% Example 1

Example 2 An electrical load is rated 120 volts, 500 watts. Compute the per-unit and percent impedance of the load. Give the per unit equivalent circuit.

2 2 V V (120)2 P   R    28.8 R P 500

Power factor = 1.0 Z  28.80

23 Select base quantities

Sb  500VA V 120V b Compute base impedance

2 2 Vb (120) Zb    28.8 Sb 500 The per-unit impedance is

Z 28.80 Z p.u.    10p.u. Zb 28.8 24 Percent impedance:

Z% 100% Per-unit equivalent circuit:

Z 10p.u. VS 10p.u.

25 Per-unit System for Single Phase Circuits One-phase circuits

Sb  S1- V I where V Vline-to-neutral

I  Iline-current

VbLV VLV VbHV VHV

Sb Sb IbLV  IbHV  VbLV VbHV 26 2 2 VbLV (VbLV ) VbHV (VbHV ) ZbLV   ZbHV   IbLV Sb IbHV Sb

S * S pu  VpuI pu Sb P Ppu  VpuI pu cos Sb Q Qpu  VpuI pu sin Sb 27 Transformation Between Bases

Sb1  SA Vb1 VA

2 Vb1 Z L Zb1  Z pu1  Sb1 Zb1 And second

Sb2  SB Vb2 VB

2 Vb2 Z L Zb2  Z pu2  Sb2 Zb2

28 Transformation Between Bases

2 Z pu2 ZL Zb1 Zb1 Vb1 Sb2      2 Z pu1 Zb2 ZL Zb2 Sb1 Vb2

2  V   S   b1   b2  Z pu2  Z pu1    Vb2   Sb1  “1” – old “2” - new 2  V   S  Z  Z  b,old   b,new  pu,new pu,old     Vb,new   Sb,old 

29 Transformation Between Bases Generally per-unit values given to another base can be converted to new base by the equations:

Sbase1 (P,Q, S) pu_on_base_ 2  (P,Q, S) pu_on_base_1 Sbase2

Vbase1 Vpu_on_base_ 2 Vpu_on_base_1 Vbase2

2 (Vbase1) Sbase2 (R, X , Z) pu_on_base_ 2  (R, X , Z) pu_on_base_1 2 (Vbase2 ) Sbase1

When performing calculations in a power system, every per-unit value must be converted to the same base.

30 Per-unit System for Single Phase Transformer Consider the equivalent circuit of transformer referred to LV side and HV side shown below:

RS X S 2  j 2 RS  jX S a a

VLV VHV V N N1 N2 Define a  LV  1 1 S VHV N2 Referred to LV side Referred to HV side

31 Per-unit System for Single Phase Transformer Choose

Vb1 VLV ,rated

Sb  Srated Compute

VHV 1 Vb2  Vb1  Vb1 VLV a 2 2 Vb1 Vb2 Zb1  Zb2  Sb Sb

2 2 Zb1 Vb1 Vb1 2  2   a Z V 1 2 b2 b2 ( V ) 32 a b1 Per-unit System for Single Phase Transformer Per-unit impedances are

RS  jX S Z p.u.1  Zb1 R jX R jX S  S S  S 2 2 2 2 R  jX Z  a a  a a  S S p.u.2 Z Zb2 b1 Zb1 a2

Z p.u.1  Z p.u.2

33 Voltage Regulation Voltage regulation is defined as V - V VR  no-load full-load 100% V full-load In per-unit system

V - V VR  pu,no-load pu, full-load 100% Vpu, full-load

Vfull-load: Desired load voltage at full load. It may be equal to, above, or below rated voltage.

Vno-load: The no load voltage when the primary voltage is the desired voltage in order the secondary voltage be at its desired value at full load.

34 Example A single-phase transformer rated 200-kVA, 200/400-V, and 10% short circuit reactance. Compute the voltage regulation when the transformer is fully loaded at unity power factor and rated voltage 400-V.

Vb2  400V VP j0.1 VS

Sb  200kVA

X S Sload Sload, pu 10pu

X S, pu  j0.1pu

35 Rated voltage

VS, pu 1.00pu

* *  Sload, pu  1.00  I     1.00pu load, pu      VS, pu  1.00 

VP, pu VS, pu  I puX S, pu 1.00 1.00 j0.1 1 j0.1 1.0015.7o pu

36 Secondary side

Vpu, full-load VS, pu 1.00pu

o Vpu,no-load VP, pu 1.0015.7 pu

Voltage regulation

V - V VR  pu,no-load pu, full-load 100% Vpu, full-load 1.001-1.0  100%  0.1% 1.0

37 Problem

j100 G

20 kV 22kV/220kV 220kV/20kV 50MVA 80MVA 50MVA 0.8 PF 14% 10% lagging

Select Vbase in generator circuit and Sb=100MVA, compute the per unit equivalent circuit.

38 Residential Distribution Transformers

Single phase transformers are commonly used in residential distribution systems. Most distribution systems are 4 wire, with a multi-grounded, common neutral.

39 Three-Phase Transformers

Per-unit System for Three Phase Circuits

Sb  S3-  3S1-  3V I

V Vline-to-neutral VL(line) / 3

I  Iline-current  IL

Sb  3VL IL

VbLV VL,LV VbHV VL,HV

Sb  3VbLV IbLV  3VbHV IbHV 41 Per-unit System for Three Phase Circuits

Sb Sb IbLV  IbHV  3VbLV 3VbHV

2 VLV VbLV 3VbLV (VbLV ) ZbLV    ILV 3 Sb Sb

2 (VbHV ) ZbHV  Sb * S3 3VL I L * S pu   VpuI pu Sb 3Vb Ib

42 Per-unit System for Three Phase Transformer Three 25-kVA, 34500/277-V transformers connected in D- Y. Short-circuit test on high voltage side:

VLine,SC  2010V

ILine,SC 1.26A

P3,SC  912W

Determine the per-unit equivalent circuit of the transformer.

43 Per-unit System for Three Phase Transformer Using Y-connection equivalent circuit

RS  jX S ISC 1.26 2010 V  Sb  25000VA SC 3 34500 277 3 2010 V  1160.47V SC 3 1160.47 Z   921.00 SC 1.26

44 Per-unit System for Three Phase Transformer

912 P 304 P   304W RS  2  2 191.48 3 ISC 1.26

2 2 2 2 X S  ZSC - RS  921 -191.48  900.86

ZSC 191.48  j900.86 34500 S  25000VA V  19918.58V b b,HV 3 19918.582 Z  15869.99 b,HV 25000 191.48  j900.86 Z   0.012  j0.0568pu SC, pu,Y 15869.99

45 Per-unit System for Three Phase Transformer Using D-connection equivalent circuit 1.26 I SC  3 ZSC,D

VSC  2010 Sb  25000VA

34500 277 1.26 VSC  2010V I   0.727A SC 3 2010 Z   2764.79 SC,D 0.727

46 Per-unit System for Three Phase Transformer

912 P 304 P   304W RS,D  2  2  575.18 3 ISC 0.727

2 2 2 2 X S,D  ZSC,D - RS,D  2764.79 -575.18  2704.30

ZSC 191.48  j900.86

Sb  25000VA Vb,HV  34500V 345002 Z   47610 b,HV 25000 575.18  j1704.30 Z   0.012  j0.0568pu SC, pu,D 47610

47 Transformer Efficiency

Output power P P    out  out Input power Pin Pout  Ploss

Pout  VS IS cos θ

VS IS cos θ    100  VS IS cos θ  PCu  PC Load Tap Changing Transformers

• LTC transformers have tap ratios that can be varied to regulate bus voltages.

• The typical range of variation is 10% from the nominal values, usually in 33 discrete steps (0.0625% per step).

• Because tap changing is a mechanical process, LTC transformers usually have a 30 second dead-band to avoid repeated changes to minimize wear and tear.

• Unbalanced tap positions can cause “circulating VARs;” that is, reactive power flowing from one winding to the next in a three phase transformer.

49 Phase Shifting Transformers

• Phase shifting transformers are used to control the phase angle across the transformer.

• Since power flow through the transformer depends upon phase angle, this allows the transformer to regulate the power flow through the transformer.

• Phase shifters can be used to prevent inadvertent "loop flow" and to prevent line overloads.

50 Phase Shifting Transformer Picture Costs about $7 million, weighs about 1.2 million pounds

230 kV 800 MVA Phase Shifting Transformer During factory testing

Source: Tom Ernst, Minnesota Power 51 Power Transformers from China

Autotransformers

• Autotransformers are transformers in which the primary and secondary windings are coupled magnetically and electrically. • This results in lower cost, and smaller size and weight. • The key disadvantage is loss of electrical isolation between the voltage levels. This can be an important safety consideration when a is large. For example in stepping down 7160/240 V we do not ever want 7160 on the low side!

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