ECE 325 – Electric Energy System Components
5‐ Transmission Lines
Instructor:
Kai Sun
Fall 2015
1
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
Corona discharge on insulator string of a 500 kV line
(source: wikipedia.org)
– 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)
3
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 Conductor)
ACSR (Aluminum Conductor Steel Reinforced)
ACAR (Aluminum Conductor Alloy Reinforced)
ACCC (Aluminum Conductor Composite Core)
ACSR (7 steel and 24 aluminum strands)
• Why not copper?
– Relative lower costs and higher strength-toweight ratios than copper
• Bundle conductors
24/7 ACSR and modern ACCC conductors
– Preferred for high voltages, e.g. 2-conductor
bundles for 230kV, 3-4 for 345-500kV, and 6 for
765kV
5
A bundle of 4 conductor
Equivalent circuit of a transmission line
GMRL or GMRC
L  ln
GMD
GMD
GMD
GMRL
1
GMD
 ln
C
GMRC
GMD
• R=rN, XL=xLN, XC=xC/N
• With the increase of the length of
the line, R and XL increase but XC
decreases
• For transmission lines, R<<XL
6
Simplifying the equivalent circuit
7
Example 25‐3
XL=0.5 50=25
XC=300,000/50=6k, so 2XC=12k 
R=0.065 50=3.25
Note XC=480XL and XL=7.7R
P=300/3=100MW
|E|=230/3=133kV
|I|=100MW/133kV=750A
Loss |I|2R=1.83MW=0.0183P
QL=|I|2XL=14.1Mvar
QC=|E|2/XC=3Mvar <<QL
8
I2
Voltage regulation
Voltage Regulation =
(if ignoring XC)

E2, NL  E2, FL
E2, FL
E1  E2, FL
E2, FL
 100%
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. |E1|<|E2|).
– If ignore XC, three typical loads with lagging, unity and leading power factors
E1
E2
I2
I2
E1
I2R
Lagging PF
jI2XL
jI2XL
E2
E1
jI2XL
E2
I2R
I2
Unity PF
I2R
Leading PF
9
Resistive line
• There is an upper limit to the power the
line can transmit to the load
2
ES
2
P | ER |  | I || I | (kR) 
kR
R  kR
| ES |2
| ES |2


R(1/ k  k  2)
4R
PF=1
Let RR=kR
k:   0
P-V curve
P=Pmax=|ES|2/4R when k=1, i.e. ER=ES/2
• If we allow a maximum VR of 5%, i.e.
ER=0.95ES, the line can support a load that
is only 19% of Pmax
• The total power from the sender is P+|I|2R
• VR is a key factor that limit the power
transmission capacity
10
Inductive line
2
ES
2
P | I | (kX ) 
kX
jX  kX
PF=1
Let RR=kX
k:   0
| ES |2
| ES |2


2X
X | k  1/ k  2 j |
Pmax=|ES|2/2X
•
•
•
•
when k=1, |ER|=|ES|/2 =0.707|ES|
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 Pmax, 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|2X
P-V curve
11
Inductive line connecting two systems
*









E
E
|
E
|

|
E
|
0
*
S
R
S
R
S  ES I  ES 

 | ES |  


jX
X
90




| ES |2
| E || E |

90  S R (  90 )
X
X
| E || E |
P   S R cos(  90 )
X
| E || E |
| E || E |
 S R sin   S R  (in rad)
X
X
| ES |2 | ES || ER |
| ES |2 | ES || ER |

Q


sin(  90 ) 
cos 
X
X
X
X
|E |
|E |
 S | ES |  | ER | cos    S | ES |  | ER |
X
X
• If |ES|=|ER|=E, Q0, i.e. almost no reactive flow
E2
E2
P  sin  
X
X

12
Compensated inductive line
• To have |ER|=|ES|, the line can fully be compensated by adding a
shunt capacitor to the receiving end whose XC is adjustable so that
|ES|2/XC=0.5|I|2X
while the other 0.5|I|2X is provided by the source or another
capacitor with XC at the sending end.
• Thus, Pmax=|ES|2/X
13
Increasing the power transmission capacity
To increase Pmax=|ES|2/X, an approach is to reduce line reactance X
• Use parallel lines:
– XX/2 or X/n, so Pmax 2Pmax or nPmax
– Improving security against a line trip.
• Use a series capacitor
Pmax=|ES|2/(X-XCS)
– It may cause sub-synchronous resonance (SSR)
f SSR  f
-jXCS
ES
ER
X CS
 f
X
1
LCS
If f=60Hz, fSSR= 30Hz for 25%
compensation (XCS/X=1/4)
14
Voltage Regulation for EHV lines
A 3-phase 735kV 60Hz 600km line, operated To bring |ER| back to |ES|, add a
at 727kV, has inductive reactance of 0.5 /km shunt reactor of XL2 at the
and capacitive reactance of 300k/km.
receiving end:
• At no-load (open-circuit) conditions,
If XL2=XC2, then -jXC2//jXL2=
for each phase,
and |ER|=|ES|.
|ES|=727/3=420kV,
The reactive power generated by
XC2 is entirely absorbed by XL2
XL=0.5600=300, XC=300k/600=500,
(cancelling each other)
X =X =2X =1000 
C1
C2
C
ER=ES(-jXC2) /(jXL-jXC2)
=4200o 1000/(1000-300)=6000o kV
15
Surge‐impedance load (SIL)
• When connected to a gradually increasing load with PF=1, |ER| decreases from
|ER,NL| (open-circuit) to 0 (short-circuit). When |ER|=|ES|, the amount of load is
called the surge-impedance load (SIL) and the corresponding load impedance is
called the surge impedance, which has ZY400 for aerial lines
SIL = EL2/ZY=EL2/400 (MW)
|ER| (V)
EL: 3-phase line voltage in kV
No-load
SIL = 7272/400=1320MW
420
Rated load
Short-circuit
0
Distance to ES (km)
600
16
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 Ea to Eb
(1) Pab=(E2/X)sin(a-b)=(1002/20) sin(-11o)=-95.4MW (EaEb)
(2) Pab=70=(1002/20) sin(a-b), so a-b=8o, i.e. 8-(-11)=19o phase-shift of Ea
17
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, 25-21
• Due date:
– hand in your solution to Denis at MK 205 or by email
(dosipov@vols.utk.edu) before the end of 10/14 (Wed)
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