EE 369 POWER SYSTEM ANALYSIS Lecture 8 Transformers, Per Unit Tom Overbye and Ross Baldick 1 Announcements For lectures 8 to 10 read Chapter 3. HW 6 is problems 5.14, 5.16, 5.19, 5.26, 5.31, 5.32, 5.33, 5.36; case study questions chapter 5 a, b, c, d, is due Thursday, 10/6. Power plant tour is 10/6. Instead of coming to class, go to UT power plant. Turn in homework at beginning of tour. Homework 7 is 5.8, 5.15, 5.17, 5.24, 5.27, 5.28, 5.29, 5.34, 5.37, 5.38, 5.43, 5.45; due 10/20. 2

Transformers Overview Power systems are characterized by many different voltage levels, ranging from 765 kV down to 240/120 volts. Transformers are used to transfer power between different voltage levels. The ability to inexpensively change voltage levels is a key advantage of ac systems over dc systems. In this section well development models for the transformer and discuss various ways of connecting three phase transformers. 3 Ideal Transformer First we review the voltage/current relationships for an ideal transformer no real power losses magnetic core has infinite permeability

no leakage flux Well define the primary side of the transformer as the side that usually receives power from a line etc, and the secondary as the side that usually delivers power to a load etc: primary is usually the side with the higher voltage, but may be the low voltage side on a generator step-up transformer. 4 Ideal Transformer Relationships Note that I2 and I2 are in opposite directions Assume we have flux m in magnetic material. Then flux linking coil 1 having N1 turns is: 1 N1m , and similarly 2 N 2m dm dm

d 1 d 2 v1 N1 , v2 N2 dt dt dt dt dm v1 v2 V1 N1

a = turns ratio dt N1 N2 V2 N2 5 Current Relationships To get the current relationships use ampere's law with path around core having total length L: mmf HdL H L ' N1i1 N 2i2

' N1i1 N 2i2 B L ' N1i1 N 2i2 Assuming uniform flux density in the core having area A, then B A L N1i1 N 2i2' A 6 Current/Voltage Relationships

If is infinite then 0 N1i1 N 2i2' . Hence i1 N2 or ' N1 i2 i1 N2 1 , where i2 i2' i2 N1 a I1 1 Then: and: I2 a

V1 I 1 a 0 V 2 1 0 I2 a 7 Impedance Transformation Example

Example: Calculate the primary voltage and current for an impedance load Z on the secondary I 2 V2 / Z and substituting: V1 I 1 V1 aV2 1 V2 I1 aZ a 0 V2 1 V2 0

Z a V1 2 a Z primary referred value of I1 secondary load impedance 8 Real Transformers Real transformers have losses have leakage flux have finite permeability of magnetic core Real power losses

resistance in windings (I2 R) core losses due to eddy currents and hysteresis 9 Transformer Core losses Eddy currents arise because of changing flux in core. Eddy currents are reduced by laminating the core Hysteresis losses are proportional to area of BH curve and the frequency These losses are reduced by using material with a thin BH curve 10 Effect of Leakage Flux Not all flux is within the transformer core 1 l1 N1m , where l1 is the coil 1 leakage flux, 2 l 2 N 2m , where l 2 is the coil 2 leakage flux,

Assuming a linear magnetic medium we get l1 Ll1i1 l 2 Ll 2i 2' dm di1 v1 r1i1 Ll1 N1 , including winding dt dt resistance r1, ' v2 r2i 2 Ll 2 di 2' d m N2

, including resistance r2 . 11 dt dt Effect of Finite Core Permeability Finite core permeability means a non-zero mmf is required to maintain m in the core N1i1 N 2i2 R m , where R is the reluctance. This effect is usually modeled as a magnetizing current R m N 2 i1 i2 N1 N1 i1 N2

im i2 N1 R m where im , N1 modeled by resistance and inductance. 12 Transformer Equivalent Circuit Using the previous relationships, we can derive an equivalent circuit model for the real transformer This model is further simplified by referring all impedances to the primary side (and approximating by swapping the referred elements and the shunts): r2' a 2 r2

x2' a 2 x2 re r1 r2' xe x1 x2' 13 Simplified Equivalent Circuit 14 Calculation of Model Parameters The parameters of the model are determined based upon: nameplate data: gives the rated voltages and power open circuit test: rated voltage is applied to primary with secondary open; measure the primary current and losses (the test may also be done applying the rated voltage to the secondary, calculating the values, then referring the values back to the primary side).

short circuit test: with secondary shorted, apply (lower than rated) voltage to primary to get rated primary current to flow; measure voltage and losses. 15 Transformer Example Example: A single phase, 100 MVA, 200/80 kV transformer has the following test data: open circuit: 20 amps, with 10 kW losses short circuit: 30 kV, with 500 kW losses Determine the model parameters. 16 Transformer Example, contd From the short circuit test I sc 100MVA

30 kV 500 A, Re jX e 60 200kV 500 A 2 2 Psc Re I sc 500 kW Re Psc / I sc 500,000 /(500) 2 2 , Hence X e 602 2 2 60 From the open circuit test (Vrated )2 (200) 2 (kV) 2 Rc 4M Poc

10 kW Re jX e jX m Vrated 200 kV 10,000 I oc 20 A X m 10,000 17 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.

18 Per Unit Calculations A key problem in analyzing power systems is the large number of transformers. It would be very difficult to continually have to refer impedances to the different sides of the transformers This problem is avoided by a normalization of all variables. This normalization is known as per unit analysis. actual quantity quantity in per unit base value of quantity 19 Per Unit Conversion Procedure, 1 1. Pick a 1 VA base for the entire system, SB

2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral. 3. Calculate the impedance base, ZB= (VB)2/SB 4. Calculate the current base, IB = VB/ZB 5. Convert actual values to per unit Note, per unit conversion affects magnitudes, not the angles. Also, per unit quantities no longer have 20 units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts) Per Unit Solution Procedure 1. Convert to per unit (p.u.) (many problems are already in per unit) 2. Solve 3. Convert back to actual as necessary 21 Per Unit Example

Solve for the current, load voltage and load power in the circuit shown below using per unit analysis with an SB of 100 MVA, and voltage bases of 8 kV, 80 kV and 16 kV, respectively. Original Circuit 22 Per Unit Example, contd Z BLeft 82 (kV) 2 0.64 100MVA Middle ZB Z BRight

802 (kV) 2 64 100MVA 162 (kV) 2 2.56 100MVA Same circuit, with values expressed in per unit. 23 Per Unit Example, contd 1.00 I 0.22 30.8 p.u. (not amps)

3.91 j 2.327 VL 1.00 0.22 30.8 p.u. 2 VL SL 0.189 p.u. Z SG 1.00 0.2230.8 30.8p.u. VL I L* 24 Per Unit Example, contd To convert back to actual values just multiply the per unit values by their per unit base V LActual 0.859 30.8 16 kV 13.7 30.8 kV S LActual 0.1890 100 MVA 18.90 MVA

SGActual 0.2230.8 100 MVA 22.030.8 MVA I BMiddle 100 MVA 1250 Amps 80 kV Actual I Middle 0.22 30.8 275 30.8 25 Three Phase Per Unit Procedure is very similar to 1 except we use a 3 VA base, and use line to line voltage bases 1. Pick a 3 VA base for the entire system, S B3 2. Pick a voltage base for each different voltage

level, VB,LL. Voltages are line to line. 3. Calculate the impedance base VB2, LL ( 3 VB , LN )2 VB2, LN Z B 3 1 1 SB 3S B SB Exactly the same impedance bases as with single phase using the corresponding single phase VA base and voltage base!26 Three Phase Per Unit, cont'd 4. Calculate the current base, IB 3 IB S B3 3 S 1B

S 1B 1 I B 3 VB , LL 3 3 VB , LN VB , LN Exactly the same current bases as with single phase! 5. Convert actual values to per unit 27 Three Phase Per Unit Example Solve for the current, load voltage and load power in the previous circuit, assuming:

a 3 power base of 300 MVA, and line to line voltage bases of 13.8 kV, 138 kV and 27.6 kV (square root of 3 larger than the 1 example voltages) the generator is Y-connected so its line to line voltage is 13.8 kV. Convert to per unit as before. Note the system is exactly the same! 28 3 Per Unit Example, cont'd 1.00 I 0.22 30.8 p.u. (not amps) 3.91 j 2.327 VL 1.00 0.22 30.8 p.u. 2

VL SL 0.189 p.u. Z SG 1.00 0.2230.8 30.8p.u. * VL I L Again, analysis is exactly the same! 29 3 Per Unit Example, cont'd Differences appear when we convert back to actual values VLActual 0.859 30.8 27.6 kV 23.8 30.8 kV SLActual 0.1890 300 MVA 56.70 MVA SGActual 0.2230.8 300 MVA 66.030.8 MVA I BMiddle

300 MVA 1250 Amps 3 138 kV Actual I Middle 0.22 30.8 Amps 275 30.8 (same current!) 30