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ECE 476 POWER SYSTEM ANALYSIS Lecture 8 Transmission Lines, Transformers, Per Unit Professor Tom Overbye Department of Electrical and Computer Engineering Announcements Start reading Chapter 3. HW 2 is due now. HW 3 is 4.32, 4.41, 5.1, 5.14. Due September 22 in class. “Energy Tour” opportunity on Oct 1 from 9am to 9pm. Visit a coal power plant, a coal mine, a wind farm and a bio-diesel processing plant. Sponsored by Students for Environmental Concerns. Cost isn’t finalized, but should be between $10 and $20. Contact Rebecca Marcotte at [email protected] for more information or to sign up. 1 V, I Relationships, cont’d Define the propagation constant as yz j where the attenuation constant the phase constant Use the Laplace Transform to solve. System has a characteristic equation ( s 2 2 ) ( s )( s ) 0 2 Equation for Voltage The general equation for V is V ( x) k1e x k2e x Which can be rewritten as e x e x e x e x V ( x) (k1 k2 )( ) (k1 k2 )( ) 2 2 Let K1 k1 k2 and K 2 k1 k2 . Then x x x e e e V ( x) K1 ( ) K2 ( 2 2 K1 cosh( x) K 2 sinh( x) e x ) 3 Real Hyperbolic Functions For real x the cosh and sinh functions have the following form: d cosh( x) sinh( x) dx d sinh( x) cosh( x) dx 4 Complex Hyperbolic Functions For x = + j the cosh and sinh functions have the following form cosh x cosh cos j sinh sin sinh x sinh cos j cosh sin 5 Determining Line Voltage The voltage along the line is determined based upon the current/voltage relationships at the terminals. Assuming we know V and I at one end (say the "receiving end" with VR and I R where x 0) we can determine the constants K1 and K 2 , and hence the voltage at any point on the line. 6 Determining Line Voltage, cont’d V ( x) K1 cosh( x) K 2 sinh( x) V (0) VR K1 cosh(0) K 2 sinh(0) Since cosh(0) 1 & sinh(0) 0 K1 VR dV ( x) dx zI K1 sinh( x) K 2 cosh( x) K2 zI R IR z z IR y yz V ( x) VR cosh( x) I R Z c sinh( x) where Zc z y characteristic impedance 7 Determining Line Current By similar reasoning we can determine I(x) VR I ( x) I R cosh( x) sinh( x) Zc where x is the distance along the line from the receiving end. Pout Define transmission efficiency as Pin 8 Transmission Line Example Assume we have a 765 kV transmission line with a receiving end voltage of 765 kV(line to line), a receiving end power SR 2000 j1000 MVA and z = 0.0201 + j0.535 = 0.53587.8 mile y = j7.75 10 6 = 7.75 10 6 90.0 mile Then zy 2.036 88.9 / mile c z y 262.7 -1.1 9 Transmission Line Example, cont’d Do per phase analysis, using single phase power and line to neutral voltages. Then VR 765 441.70 kV 3 6 * (2000 j1000) 10 IR 1688 26.6 A 3 3 441.70 10 V ( x) VR cosh( x) I R Z c sinh( x) 441, 7000 cosh( x 2.03688.9) 443, 440 27.7 sinh( x 2.03688.9) 10 Transmission Line Example, cont’d 11 Lossless Transmission Lines For a lossless line the characteristic impedance, Zc , is known as the surge impedance. Zc jwl l (a real value) jwc c If a lossless line is terminated in impedance VR Zc IR Then I R Z c VR so we get... 12 Lossless Transmission Lines V ( x) VR cosh x VR sinh x I ( x) I R cosh x I R sinh x V ( x) Zc I ( x) 2 V(x) Define as the surge impedance load (SIL). Zc Since the line is lossless this implies V ( x) VR I ( x) I R If P > SIL then line consumes vars; otherwise line generates vars. 13 Transmission Matrix Model Oftentimes we’re only interested in the terminal characteristics of the transmission line. Therefore we can model it as a “black box”. + VS IS IR Transmission Line - + VR - VS A B VR With I I C D R S 14 Transmission Matrix Model, cont’d VS A B VR With I I C D R S Use voltage/current relationships to solve for A,B,C,D VS VR cosh l Z c I R sinh l VR I S I R cosh l sinh l Zc cosh l A B 1 T sinh l C D Z c Z c sinh l cosh l 15 Equivalent Circuit Model The common representation is the equivalent circuit Next we’ll use the T matrix values to derive the parameters Z' and Y'. 16 Equivalent Circuit Parameters VS VR Y' VR I R Z' 2 Z 'Y ' VS 1 VR Z ' I R 2 Y' Y' I S VS VR I R 2 2 Z 'Y ' 1 Z 'Y ' I I S Y ' 1 V R R 4 2 1 Z 'Y ' Z ' VR VS 2 I Z 'Y ' Z 'Y ' I S Y ' 1 R 1 4 2 17 Equivalent circuit parameters We now need to solve for Z' and Y'. Using the B element solving for Z' is straightforward B ZC sinh l Z ' Then using A we can solve for Y' Z 'Y ' A = cosh l 1 2 Y' cosh l 1 1 l tanh 2 Z c sinh l Z c 2 18 Simplified Parameters These values can be simplified as follows: Z ' ZC sinh l zl z sinh l yl z sinh l Z with Z zl (recalling zy ) l Y' 1 l tanh 2 Zc 2 yl y l tanh zl y 2 l tanh Y 2 with Y 2 l 2 yl 19 Simplified Parameters For short lines make the following approximations: sinh l Z' Z (assumes 1) l Y' Y tanh( l / 2) (assumes 1) 2 2 l /2 sinhγl tanh(γl/2) Length γl γl/2 50 miles 0.9980.02 1.001 0.01 100 miles 0.9930.09 1.004 0.04 200 miles 0.9720.35 1.014 0.18 20 Medium Length Line Approximations For shorter lines we make the following approximations: sinh l Z' Z (assumes 1) l Y' Y tanh( l / 2) (assumes 1) 2 2 l /2 sinhγl tanh(γl/2) Length γl γl/2 50 miles 0.9980.02 1.001 0.01 100 miles 0.9930.09 1.004 0.04 200 miles 0.9720.35 1.014 0.18 21 Three Line Models Long Line Model (longer than 200 miles) l sinh l Y ' Y tanh 2 use Z ' Z , l 2 2 l 2 Medium Line Model (between 50 and 200 miles) Y use Z and 2 Short Line Model (less than 50 miles) use Z (i.e., assume Y is zero) 22 Power Transfer in Short Lines Often we'd like to know the maximum power that could be transferred through a short transmission line V1 + - I1 S12 I1 Transmission Line with Impedance Z S21 + - V2 * V1 V2 S12 V1 Z with V1 V1 1 , V2 V2 2 V1I1* Z Z Z 2 S12 V1 V1 V2 Z Z 12 Z Z 23 Power Transfer in Lossless Lines If we assume a line is lossless with impedance jX and are just interested in real power transfer then: 2 P12 jQ12 V1 V1 V2 90 90 12 Z Z Since - cos(90 12 ) sin 12 , we get V1 V2 P12 sin 12 X Hence the maximum power transfer is Max P12 V1 V2 X 24 Limits Affecting Max. Power Transfer Thermal limits – – – – limit is due to heating of conductor and hence depends heavily on ambient conditions. For many lines, sagging is the limiting constraint. Newer conductors limit can limit sag. For example, in 2004 ORNL working with 3M announced lines with a core consisting of ceramic Nextel fibers. These lines can operate at 200 degrees C. Trees grow, and will eventually hit lines if they are planted under the line. 25 Other Limits Affecting Power Transfer Angle limits – while the maximum power transfer occurs when line angle difference is 90 degrees, actual limit is substantially less due to multiple lines in the system Voltage stability limits – as power transfers increases, reactive losses increase as I2X. As reactive power increases the voltage falls, resulting in a potentially cascading voltage collapse. 26 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 we’ll development models for the transformer and discuss various ways of connecting three phase transformers. 27 Transmission to Distribution Transfomer 28 Transmission Level Transformer 29 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 We’ll define the “primary” side of the transformer as the side that usually takes power, and the secondary as the side that usually delivers power. – primary is usually the side with the higher voltage, but may be the low voltage side on a generator step-up transformer. 30 Ideal Transformer Relationships Assume we have flux m in magnetic material. Then 1 N1m d 1 2 N 2m d 2 d m d m v1 N1 v2 N2 dt dt dt dt d m v1 v2 v1 N1 a = turns ratio dt N1 N2 v2 N2 31 Current Relationships To get the current relationships use ampere's law mmf H dL N1i1 N 2i2' H length N1i1 N 2i2' B length N1i1 N 2i2' Assuming uniform flux density in the core length ' N1i1 N 2i2 area 32 Current/Voltage Relationships If is infinite then 0 N1i1 N 2i2' . Hence i1 N2 or ' N1 i2 i1 N2 1 i2 N1 a Then v1 i 1 a 0 v 2 1 0 i2 a 33 Impedance Transformation Example Example: Calculate the primary voltage and current for an impedance load on the secondary a v1 i 0 1 v1 a v2 i1 0 v2 1 v2 Z a 1 v2 aZ v1 a2 Z i1 34 Real Transformers Real transformers – – – have losses have leakage flux have finite permeability of magnetic core 1. Real power losses – – resistance in windings (i2 R) core losses due to eddy currents and hysteresis 35 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 36 Effect of Leakage Flux Not all flux is within the transformer core 1 l1 N1m 2 l 2 N 2m Assuming a linear magnetic medium we get l1 Ll1i1 l 2 Ll 2i 2' dm di1 v1 r1i1 Ll1 N1 dt dt v 2 r2i 2 Ll 2 ' di 2' dm N2 dt dt 37 Effect of Finite Core Permeability Finite core permeability means a non-zero mmf is required to maintain m in the core N1i1 N 2i2 m This value is usually modeled as a magnetizing current m N 2 i1 i2 N1 N1 i1 N2 im i2 N1 m where i m N1 38 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 r2' a 2 r2 re r1 r2' x2' a 2 x2 xe x1 x2' 39 Simplified Equivalent Circuit 40 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. 41 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. 42 Transformer Example, cont’d From the short circuit test 100 MVA 30 kV I sc 500 A, R e jX e 60 200kV 500 A 2 Psc Re I sc 500 kW R e 2 , Hence X e 602 22 60 From the open circuit test 200 kV 2 Rc 4M 10 kW 200 kV R e jX e jX m 10, 000 20 A X m 10, 000 43 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. 44 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 45 Per Unit Conversion Procedure, 1 1. 2. 3. 4. 5. Pick a 1 VA base for the entire system, SB Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral. Calculate the impedance base, ZB= (VB)2/SB Calculate the current base, IB = VB/ZB 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) 46 Per Unit Solution Procedure 1. 2. 3. Convert to per unit (p.u.) (many problems are already in per unit) Solve Convert back to actual as necessary 47 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 48 Per Unit Example, cont’d Z BLeft 8kV 2 0.64 100 MVA Z BMiddle Z BRight 80kV 2 64 100 MVA 2 16kV 2.56 100 MVA Same circuit, with values expressed in per unit. 49 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* 50 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 51