2/7/2013. Topics. 15-System Model Text: One-Line Diagram. One-Line Diagram
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1 /7/013 Topics 15-ystem Model Text: One-line Diagram ystem Modeling Regulating Transformers ECEGR 451 Power ystems Dr. Henry Louie 1 Dr. Henry Louie Generator us Transformer Transmission line Circuit breaker Load One-Line Diagram One-Line Diagram imple generator model: V G : terminal voltage E G : open circuit voltage G : generator current G E G V G 3 4 ystem Model We ve discussed transmission lines, transformers, per unit, one line diagrams Now we put them all together to model the system ystem Model G 1 : 50 MVA, 1. kv, X = 0.15 p.u. G : 0 MVA, 13.8 kv, X = 0.15 p.u. T 1 : 80 MVA 1./161 kv, X = 0.10 p.u. T : 40 MVA 13.8/161 kv, X = 0.10 p.u. Load: 50 MVA, 0.80 PF (lagging), operating at 154 kv T j T G j 0 80 j G Dr. Henry Louie 5 Dr. Henry Louie 6 1
2 /7/ Pick a power base for the system Common to select power base equal to or near the largest generator in the system Let = 100 MVA (three phase) T 1 : 80 MVA 1./161 kv, X = 0.10 p.u. T : 40 MVA 13.8/161 kv, X = 0.10 p.u T j T a. elect voltage base Common to select voltage base equal to or near the line-line voltage at any section Must keep track of voltage base and section ections are separated by transformers Let V = 13 kv (line-line) at the transmission line section three sections T j T G j 0 80 j G G j 0 80 j G Dr. Henry Louie 7 Dr. Henry Louie 8 b. Compute voltage bases for all sections Use transformer ratios (line-line) to relate base voltages between sections G1 section: V 1 = 13 x (1./161) = kv G section: V 13 x (13.8/161) = kv 3. Express all impedances in consistent p.u. terms All sections have same power bases, but different voltage bases mpedances are given with different power and voltage bases V Convert using: Zp.u. Zp.u. V T j T T j T G j 0 80 j G G j 0 80 j G Dr. Henry Louie 9 Dr. Henry Louie Express all impedances in consistent p.u. terms G 1 : X = 0.15 x (100/50) x (1./10.00) = p.u. G : X = 0.15 x (100/0) x (13.8/11.31) = p.u. T 1 : X = 0.1 x (100/80) x (1./10.00) = 0.1 x (100/80) x (161/13) = p.u. T = 0.1 x (100/40) x (13.8/11.31) = p.u. Z V Z V p.u. p.u. 3. Express all impedances in consistent p.u. terms The transmission line and : Z 3 = (13) /100 = p.u. Z line,a = (40 + j160)/174.4 = j Z line,b = (0 + j80)/174.4 = j p.u. and for the : = 50(.8 + j.6) = j MVA Z = {(154) /( j)} * = j84.59 =.18 + j1.63 p.u. T j T T j T G j 0 80 j G G j 0 80 j G Dr. Henry Louie 11 Dr. Henry Louie 1
3 /7/013 4a. Draw the impedance diagram Redrawing the system 4b. olve for desired quantities Use per-phase analysis T j T G j 0 80 j G j j j j j j j j j j1.63 j G 1 G j j j j1.63 G 1 G j impedance diagram Dr. Henry Louie 13 Dr. Henry Louie Convert back to actual quantities, if needed j j j j j j j1.63 G 1 G j pecifications G 1 : 75 MVA, 10 kv, X = 0.10 p.u. G : 75 MVA, kv, X = 0.08 p.u. T 1 : 75 MVA 10/365 kv, X = 0.1 p.u. T : 80 MVA 4/380 kv, X = 0.14 p.u. Load: Load A E G 50kV Choose V = 365kV at the transmission line and = 75MVA impedance diagram T 1 6+j178 T G 1 j35 G Dr. Henry Louie 15 Dr. Henry Louie 16 ection base voltages: V V V kV 10kV 3.05kV T 1 T 6+j178 New generator impedances Z Zp.u. V V p.u XG XG G 1 j35 G Dr. Henry Louie 17 Dr. Henry Louie 18 3
4 /7/013 New transformer impedances Z Zp.u. V V p.u XT XT Transmission line impedance in p.u. Z V 365k 75M j178 ZL j j35 ZL j Dr. Henry Louie 19 Dr. Henry Louie 0 Load current and generator voltage 118.6A E Load G 3V mpedance diagram with sources T 1 T 6+j178 E G1 G 1 j35 G j0.1 j0.1 j0.1 j0.073 j0.0 j Dr. Henry Louie 1 Dr. Henry Louie Regulating Transformers We can now describe the system model in per unit with the impedance diagram We have seen that using per unit on normal systems, we have eliminated the transformers However, this is not a general result as there are some transformers that do not disappear when per unit normalization is used regulating and off-nominal transformers Regulating Transformers Transformers used to adjust voltage magnitude or phase are called regulating transformers They do this by adding a small amount (+ or 10 %) of voltage to the line or phase voltages Voltage can also be changed by adjusting the turns ratio of the transformer (called tap changing) Tap changing may be automatic and may occur while the transformer is energized (-tapchanging) Dr. Henry Louie 3 Dr. Henry Louie 4 4
5 /7/013 Regulating Transformers Regulating Transformer a b c a b c V V V an an an V V V b n bn bn V V V c n cn cn V an a b c n a b c V an n voltage magnitude increase V cn V bn V cn V bn Dr. Henry Louie 5 Dr. Henry Louie 6 Phase-hifting Transformer asic idea: add a voltage that is 90 degrees out of phase j j Vbc 3Vane Van Van 1 p 3e Van 1 jp 3 Phase shift, and small voltage magnitude change occur f: phase shift V an V an Consider two transformers in parallel with different ratios Assume system is per unit normalized to T 1 ratio, n T 1 disappears from the impedance circuit assume T has the ratio n (off-nominal) define: n n n T 1 : X = 0. T 1 T : X = 0.4, n is such that n Load: V 10 T V 1 V V cn V cn V bn V bn Dr. Henry Louie 7 Dr. Henry Louie 8 T does not disappear from the circuit, we must include it using a transformer model How does arrangement affect power flows through each transformer? T T V 1 V + V : n 1.05 V Using KVL and KCL: V1 10 j j = j0.1 j j j V : n 1.05 V Dr. Henry Louie 9 Dr. Henry Louie 30 5
6 /7/013 ummary Finding the power: V j0.46 T1 V j T f the turns ratios were the same: T j note the large T j0.475 affect on VAR V 1-1: n 1.05 V The system model procedure discussed takes a one line diagram of a power system and produces an impedance diagram per unit is convenient and lends itself to three phase or single phase quantities most transformers disappear from the system Regulating transformers can be used to adjust real and reactive power flows through the system Next lecture we begin to focus on two other representations of the system using network matrices Dr. Henry Louie 31 Dr. Henry Louie 3 6
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