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ECE 476 Power System Analysis Lecture 10: Per Unit, Three-Phase Transformers, Load and Generators Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign [email protected] Announcements • Please read Chapter 2.4, Chapter 6 up to 6.6 • HW 4 is due now; HW 5 is 5.31, 5.43, 3.4, 3.10, 3.14, 3.19, 3.23, 3.60, 6.30 should be done before exam 1 • Exam 1 is Thursday Oct 6 in class • • Closed book, closed notes, but you may bring one 8.5 by 11 inch note sheet and standard calculators Last name A-M here, N to Z in ECEB 1013 • Power Area Scholarships, Awards (Oct 1 deadlines, except Nov 1 for Grainger) • • http://energy.ece.illinois.edu/ Turn both into Prof Sauer (4046 or 4060) 1 Service Entrance Grounding We’ll talk more about grounding later in the semester when we cover faults Image: www.osha.gov/dte/library/electrical/electrical_10.gif 2 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 3 Per Unit Conversion Procedure, 1f 1. Pick a 1f 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 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) 4 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 5 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 6 Per Unit Example, cont’d Z BLeft 8kV 2 0.64 100 MVA Z BMiddle Z BRight 80kV 2 64 100 MVA 16kV 2 2.56 100 MVA Same circuit, with values expressed in per unit. 7 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* 8 Per Unit Example, cont’d To convert back to actual values just multiply the per unit values by their per unit base Actual VL 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 Middle B 100 MVA 1250 Amps 80 kV I Actual Middle 0.22 30.8 Amps 275 30.8 9 Three Phase Per Unit Procedure is very similar to 1f except we use a 3f VA base, and use line to line voltage bases 1. Pick a 3f VA base for the entire system, S B3f 2. Pick a voltage base for each different voltage level, VB. Voltages are line to line. 3. Calculate the impedance base ZB VB2, LL S B3f ( 3 VB , LN ) 2 3S 1Bf VB2, LN S 1Bf Exactly the same impedance bases as with single phase! 10 Three Phase Per Unit, cont'd 4. Calculate the current base, IB I3Bf S B3f 3 S 1Bf S 1Bf I1Bf 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 11 Three Phase Per Unit Example Solve for the current, load voltage and load power in the previous circuit, assuming a 3f 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 1f 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! 12 3f 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! 13 3f Per Unit Example, cont'd Differences appear when we convert back to actual values V LActual 0.859 30.8 27.6 kV 23.8 30.8 kV S LActual 0.1890 300 MVA 56.70 MVA SGActual 0.2230.8 300 MVA 66.030.8 MVA I Middle B 300 MVA 1250 Amps (same current!) 3 138 kV I Actual Middle 0.22 30.8 Amps 275 30.8 14 3f Per Unit Example 2 •Assume a 3f 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 15 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 NewBase Z OriginalBase Z Z pu actual pu Hence ZOriginalBase pu OriginalBase Z pu 2 Vbase / OriginalBase S Base NewBase S Base OriginalBase S Base 2 Vbase NewBase S Base NewBase Z pu NewBase Z pu 16 Per Unit Change of Base Example •A 54 MVA transformer has a leakage reactance of 3.69%. What is the reactance on a 100 MVA base? 100 X e 0.0369 0.0683 p.u. 54 17 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 e 0.10 0.0286 p.u. 350 2 230 0.0286 15.1 100 18 Three Phase Transformers • There are 4 different ways to connect 3f transformers Y-Y D-D Usually 3f transformers are constructed so all windings share a common core 19 3f Transformer Interconnections D-Y Y-D 20 Y-Y Connection Magnetic coupling with An/an, Bn/bn & Cn/cn VAn VAB IA 1 a, a, Van Vab Ia a 21 Y-Y Connection: 3f Detailed Model 22 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. 23 D-D Connection Magnetic coupling with AB/ab, BC/bb & CA/ca VAB I AB 1 I A 1 a, , Vab I ab a I a a 24 D-D Connection: 3f Detailed Model To use the per phase equivalent we need to use the delta-wye load transformation 25 D-D Connection: Per Phase Model Per phase analysis similar to Y-Y except impedances are decreased by a factor of 3. Key disadvantage is D-D connections can not be grounded; not commonly used. 26 D-Y Connection Magnetic coupling with AB/an, BC/bn & CA/cn 27 D-Y Connection V/I Relationships VAB VAB a, Van Vab 3 Van30 Van a VAn30 VAB 30 Hence Vab 3 and Van 3 a a For current we get I AB 1 I a a I AB I ab a I A 3 I AB 30 I AB 1 a a I A30 3 1 I A30 3 28 D-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 29 Y-D Connection: Per Phase Model Exact opposite of the D-Y connection, now with a phase shift of -30 degrees. 30 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" 31 LTCs and Circulating Vars slack 64 MW 14 Mvar 1.00 pu 1 24.1 MW 12.8 Mvar 40.2 MW 1.7 Mvar 1.000 tap A A 80% 1.056 tap MVA MVA 40.0 MW -0.0 Mvar 24.0 MW -12.0 Mvar 0.98 pu 2 3 1.05 pu 0.0 Mvar 24 MW 12 Mvar 40 MW 0 Mvar 32 Phase Shifting Transformers • Phase shifting transformers are used to control the phase angle across the transformer – Also called phase angle regulators (PARs) or quadrature booster transformers • 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. Image Source: en.wikipedia.org/wiki/Quadrature_booster#/media/File:Qb-3ph.svg 33 Phase Shifter Example 3.13 345.00 kV 500 MW 341.87 kV 283.9 MW 39.0 Mvar 283.9 MW 6.2 Mvar slack 164 Mvar Phase Shifting Transformer 216.3 MW 125.0 Mvar 500 MW 100 Mvar 216.3 MW 0.0 deg 93.8 Mvar 1.05000 tap 34 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. Hence auto-transformers are not used when a is large. For example in stepping down 7160/240 V we do not ever want 7160 on the low side! 35