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Ece 476 Power System Analysis ECE 476 POWER SYSTEM ANALYSIS Lecture 9 Transformers, Per Unit Professor Tom Overbye Department of Electrical and Computer Engineering Announcements Be reading Chapter 3 HW 3 is 4.32, 4.41, 5.1, 5.14. Due September 22 in class. 1 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 '2 ' r22ar re r 12 r '2 ' xaxxxx22e 12 2 Simplified Equivalent Circuit 3 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 voltage to the secondary, calculating the values, then referring the values back to the primary side). – short circuit test: with secondary shorted, apply voltage to primary to get rated current to flow; measure voltage and losses. 4 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. 5 Transformer Example, cont’d From the short circuit test 100MVA 30 kV I 500AjX , R 60 sc 200kV e e 500 A 2 Psc RIesc 500 kW Re 2 , 22 Hence Xe 60 2 60 From the open circuit test 200 kV 2 RM4 c 10 kW 200 kV R jX jX 10,000 X 10,000 e em20 A m 6 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. 7 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 8 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. 2 3. Calculate the impedance base, ZB= (VB) /SB 4. Calculate the current base, IB = VB/ZB 5. Convert actual values to per unit Note, per unit conversion on affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts) 9 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 10 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. Original Circuit 11 Per Unit Example, cont’d 8kV 2 Z Left 0.64 B 100MVA 80kV 2 Z Middle 64 B 100MVA 16kV 2 Z Right 2.56 B 100MVA Same circuit, with values expressed in per unit. 12 Per Unit Example, cont’d 1.0 0 I 0.22 30.8 p.u. (not amps) 3.91 j 2.327 VL 1.0 0 0.22 30.8 p.u. V 2 SVI* L 0.189 p.u. LLLZ SG 1.0 0 0.22 30.8 30.8 p.u. 13 Per Unit Example, cont’d To convert back to actual values just multiply the per unit values by their per unit base V Actual 0.859 30.8 16 kV 13.7 30.8 kV L Actual SL 0.189 0 100 MVA 18.9 0 MVA Actual SG 0.22 30.8 100 MVA 22.0 30.8 MVA 100 MVA IMiddle 1250 Amps B 80 kV Actual I0.2230.8Middle Amps27530.8 14 Three Phase Per Unit Procedure is very similar to 1 except we use a 3 VA base, and use line to line voltage bases 3 1. Pick a 3 VA base for the entire system, SB 2. Pick a voltage base for each different voltage level, VB. Voltages are line to line. 3. Calculate the impedance base 2 22 VVVBLL,,,(3 BLN ) BLN ZB 311 SSSBBB3 Exactly the same impedance bases as with single phase! 15 Three Phase Per Unit, cont'd 4. Calculate the current base, IB 311 31SSSBBB3 IIBB 3VBLL, 33VVBLN,, BLN Exactly the same current bases as with single phase! 5. Convert actual values to per unit 16 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). Also assume 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! 17 3 Per Unit Example, cont'd 1.0 0 I 0.22 30.8 p.u. (not amps) 3.91 j 2.327 VL 1.0 0 0.22 30.8 p.u. V 2 SVI* L 0.189 p.u. LLLZ SG 1.0 0 0.22 30.8 30.8 p.u. Again, analysis is exactly the same! 18 3 Per Unit Example, cont'd Differences appear when we convert back to actual values V Actual 0.859 30.8 27.6 kV 23.8 30.8 kV L Actual SL 0.189 0 300 MVA 56.7 0 MVA Actual SG 0.22 30.8 300 MVA 66.0 30.8 MVA 300 MVA IMiddle 1250 Amps (same current!) B 3138 kV Actual I0.2230.Middle 8 Amps 275 30.8 19 3 Per Unit Example 2 Assume a 3 load of 100+j50 MVA with VLL of 69 kV is connected to a source through the below network: What is the supply current and complex power? Answer: I=467 amps, S = 103.3 + j76.0 MVA 20 Per Unit Change of MVA Base Parameters for equipment are often given using power rating of equipment as the MVA base To analyze a system all per unit data must be on a common power base OriginalBase NewBase ZZZpu actual pu VV22 OriginalBase base base NewBase Hence Z pu OriginalBase/ NewBase Z pu SBase SBase NewBase OriginalBase SBase NewBase Z pu OriginalBase Z pu SBase 21 Per Unit Change of Base Example A 54 MVA transformer has a leakage reactance or 3.69%. What is the reactance on a 100 MVA base? 100 X 0.0369 0.0683 p.u. e 54 22 Transformer Reactance Transformer reactance is often specified as a percentage, say 10%. This is a per unit value (divide by 100) on the power base of the transformer. Example: A 350 MVA, 230/20 kV transformer has leakage reactance of 10%. What is p.u. value on 100 MVA base? What is value in ohms (230 kV)? 100 X 0.10 0.0286 p.u. e 350 2302 0.0286 15.1 100 23 Three Phase Transformers There are 4 different ways to connect 3 transformers Y-Y - Usually 3 transformers are constructed so all windings share a common core 24 3 Transformer Interconnections -Y Y- 25 Y-Y Connection Magnetic coupling with An/an, Bn/bn & Cn/cn VVI1 Anaa,, AB A VVIaan ab a 26 Y-Y Connection: 3 Detailed Model 27 Y-Y Connection: Per Phase Model Per phase analysis of Y-Y connections is exactly the same as analysis of a single phase transformer. Y-Y connections are common in transmission systems. Key advantages are the ability to ground each side and there is no phase shift is introduced. 28 - Connection Magnetic coupling with AB/ab, BC/bb & CA/ca VI11 I ABa,, AB A VIaIaab ab a 29 - Connection: 3 Detailed Model To use the per phase equivalent we need to use the delta-wye load transformation 30 - Connection: Per Phase Model Per phase analysis similar to Y-Y except impedances are decreased by a factor of 3. Key disadvantage is - connections can not be grounded; not commonly used. 31 -Y Connection Magnetic coupling with AB/an, BC/bn & CA/cn 32 -Y Connection V/I Relationships VVAB AB aVVV,an ab3 an 30 Vaan VV30 30 Hence VV 3AB and 3 An ab aaan For current we get I AB 1 IaIaAB Iaab 1 II3 30 I I30 AAB AB3 A 1 aI 30 aA3 33 -Y Connection: Per Phase Model Note: Connection introduces a 30 degree phase shift! Common for transmission/distribution step-down since there is a neutral on the low voltage side. Even if a = 1 there is a sqrt(3) step-up ratio 34 Y- Connection: Per Phase Model Exact opposite of the -Y connection, now with a phase shift of -30 degrees. 35 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 deadband to avoid repeated changes. Unbalanced tap positions can cause "circulating vars" 36 LTCs and Circulating Vars 64 MW slack 14 Mvar 1 1.00 pu 24.1 MW 40.2 MW 12.8 Mvar 1.7 Mvar A A 1.056 tap 1.000 tap 80% MVA MVA 40.0 MW 24.0 MW -0.0 Mvar -12.0 Mvar 2 0.98 pu 3 1.05 pu 0.0 Mvar 24 MW 40 MW 12 Mvar 0 Mvar 37 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.
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