ECE 325 Electric Energy System Components 5 Transmission Lines. Instructor: Kai Sun Fall 2015

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1 ECE 325 Electric Energy System Components 5 Transmission Lines Instructor: Kai Sun Fall

2 Content (Materials are from Chapter 25) Overview of power lines Equivalent circuit of a line Voltage regulation and power transmission of transmission lines 2

Overview Types of transmission lines Overhead lines Underground Cables (less than 1%) Properties Series Resistance (stranding and skin effect) Series Inductance (magnetic & electric fields; flux

3 Overview Types of transmission lines Overhead lines Underground Cables (less than 1%) Properties Series Resistance (stranding and skin effect) Series Inductance (magnetic & electric fields; flux linkages within the conductor cross section and external flux linkages) Shunt Capacitance (magnetic & electric fields; charge and discharge due to potential difference between conductors) Shunt Conductance (due to leakage currents along insulators or corona discharge caused by ionization of air) Line-to-line voltage levels 69kV, 115kV, 138kV and 161kV (sub-transmission) 230kV, 345kV, and 500kV (EHV) 765kV (UHV) Corona discharge on insulator string of a 500 kv line (source: wikipedia.org) 3

4 Overhead Transmission Lines Shield wires (ground wires) are ground conductors used to protect the transmission lines from lightning strikes (Source: wikipedia.org and EPRI dynamic tutorial) 4

Overhead Transmission Lines Materials AAC (All Aluminum Conductor), AAAC (All Aluminum

Alloy Reinforced) ACCC (Aluminum Conductor Composite Core) Why not copper?

5 Overhead Transmission Lines Materials AAC (All Aluminum Conductor), AAAC (All Aluminum Alloy Conductor) ACSR (Aluminum Conductor Steel Reinforced) ACAR (Aluminum Conductor Alloy Reinforced) ACCC (Aluminum Conductor Composite Core) Why not copper? Relative lower costs and higher strength-toweight ratios than copper Bundle conductors Preferred for high voltages, e.g. 2-conductor bundles for 230kV, 3-4 for kV, and 6 for 765kV ACSR (7 steel and 24 aluminum strands) 24/7 ACSR and modern ACCC conductors A bundle of 4 conductor 5

6 Equivalent circuit of a transmission line GMR L or GMR C GMD GMD GMD GMD L ln GMR L 1 GMD ln C GMR C R=rN, X L =x L N, X C =x C /N With the increase of the length of the line, R and X L increase but X C decreases For transmission lines, R<<X L 6

7 Simplifying the equivalent circuit 7

Example 25 3 X L =0.5 50=25 X C =300,000/50=6k, so 2X C =12k R=0.065 50=3.25 Note X C =480X L and X L =7.

8 Example 25 3 X L =0.5 50=25 X C =300,000/50=6k, so 2X C =12k R= =3.25 Note X C =480X L and X L =7.7R P=300/3=100MW E =230/3=133kV I =100MW/133kV=750A Loss I 2 R=1.83MW=0.0183P Q L = I 2 X L =14.1Mvar Q C = E 2 /X C =3Mvar <<Q L 8

9 Voltage regulation I 2 E2, NL E2, FL Voltage Regulation = 100% E (if ignoring X C ) E 2, FL 1 2, FL E E 2, FL 100% VR is a measure of line voltage drop and usually should not exceed 5% (or 10%) VR depends on the load power factor: VR is low for a low lagging power factor Perhaps, VR<0 for a leading power factor (i.e. E 1 < E 2 ). If ignore X C, three typical loads with lagging, unity and leading power factors E 1 E 1 I 2 E 1 ji 2 X L I 2 E 2 ji 2 X L I 2 R E 2 I 2 ji 2 X L I 2 R Lagging PF Unity PF Leading PF E 2 I 2 R 9

10 Resistive line There is an upper limit to the power the line can transmit to the load 2 ES P ER I I ( kr) kr R kr 2 2 ES ES R(1 / kk2) 4R 2 P-V curve PF=1 Let R R =kr k: 0 P=P max = E S 2 /4R when k=1, i.e. E R =E S /2 If we allow a maximum VR of 5%, i.e. E R =0.95E S, the line can support a load that is only 19% of P max The total power from the sender is P+ I 2 R VR is a key factor that limit the power transmission capacity 10

11 Inductive line 2 2 ES P I ( kx) kx jx kx ES E X k1/ k2 j 2X 2 2 S P max = E S 2 /2X when k=1, E R = E S /2 =0.707 E S A inductive line can deliver twice as much power as a resistive line (if X=R) If we allow a maximum VR of 5%, the line can support a load that is 60% of P max, i.e. 6x as much as power as a resistive line VR is a key factor that limit the power transmission capacity The total power from the sender is P+j I 2 X P-V curve PF=1 Let R R =kx k: 0 11

12 Inductive line connecting two systems 2 ES ES ER * * ES ER ES ER 0 SESI ES ES jx X90 90 ( 90 ) X X ES ER cos( 90 P ) X ES ER ES ER sin (in rad) X X 2 2 ES ES ER ES ES ER Q sin( 90 ) cos X X X X ES ES ES ER cos ES ER X X If E S = E R =E, Q0, i.e. almost no reactive flow E E P sin X X

13 Compensated inductive line To have E R = E S, the line can fully be compensated by adding a shunt capacitor to the receiving end whose X C is adjustable so that E S 2 /X C =0.5 I 2 X while the other 0.5 I 2 X is provided by the source or another capacitor with X C at the sending end. Thus, P max = E S 2 /X 13

14 Increasing the power transmission capacity To increase P max = E S 2 /X, an approach is to reduce line reactance X Use parallel lines: XX/2 or X/n, so P max 2P max or np max Improving security against a line trip. Use a series capacitor P max = E S 2 /(X-X CS ) It may cause sub-synchronous resonance (SSR) X CS 1 fssr f f X LC -jx CS S E S E R If f=60hz, f SSR = 30Hz for 25% compensation (X CS /X=1/4) 14

15 Voltage Regulation for EHV lines A 3-phase 735kV 60Hz 600km line, operated at 727kV, has inductive reactance of 0.5 /km and capacitive reactance of 300k/km. At no-load (open-circuit) conditions, for each phase, E S =727/3=420kV, X L =0.5600=300, X C =300k/600=500, X C1 =X C2 =2X C =1000 E R =E S (-jx C2 )/(jx L -jx C2 ) =4200 o 1000/( )=6000 o kv To bring E R back to E S, add a shunt reactor of X L2 at the receiving end: If X L2 =X C2, then -jx C2 //jx L2 = and E R = E S. The reactive power generated by X C2 is entirely absorbed by X L2 (cancelling each other) 15

16 Surge impedance load (SIL) When connected to a gradually increasing load with PF=1, E R decreases from E R,NL (open-circuit) to 0 (short-circuit). When E R = E S, the amount of load is called the surge-impedance load (SIL) and the corresponding load impedance is called the surge impedance, which has Z Y 400 for aerial lines SIL = E L2 /Z Y =E L2 /400 (MW) E L : 3-phase line voltage in kv E R (V) No-load 420 SIL = /400=1320MW Rated load 0 Distance to E S (km) Short-circuit

17 Inter region power exchange: Example 25 8 Calculate (1) the power transmitted by the line and (2) the required phase-shift enabling transmitting 70MW from E a to E b (1) P ab =(E 2 /X)sin( a - b )=(100 2 /20) sin(-11 o )=-95.4MW (E a E b ) (2) P ab =70=(100 2 /20) sin( a - b ), so a - b =8 o, i.e. 8-(-11)=19 o phase-shift of E a 17

18 Homework Assignment #4 Read Chapters 11, 12 &25 Questions: 11-7, 11-8, 11-9, 12-2, 12-7, 12-8, 25-11, 25-12, 25-13, 25-14, 25-15, 25-16, 25-17, 25-18, Due date: hand in your solution to Denis at MK 205 or by before the end of 10/14 (Wed) 18

Calculations of Capacitance for Transposed Bundled Conductor Transmission Lines

Calculations of Capacitance for Transposed Bundled Conductor Transmission Lines Multi-conductor Lines. An example with a conductor bundle r: conductor radius, d: distance between conductors of the same