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					             ECE 476
POWER SYSTEM ANALYSIS

              Lecture 6
Development of Transmission Line Models



        Professor Tom Overbye
        Department of Electrical and
          Computer Engineering
Announcements

   For lectures 5 through 7 please be reading Chapter 4
    –   we will not be covering sections 4.7, 4.11, and 4.12 in detail
   HW 2 is 4.10 (positive sequence is the same here as per
    phase), 4.18, 4.19, 4.23. Use Table A.4 values to determine
    the Geometric Mean Radius of the wires (i.e., the ninth
    column). Due September 15 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
    marcott1@illinois.edu for more information or to sign up.

                                                                         1
Two Conductor Line Inductance

Key problem with the previous derivation is we
assumed no return path for the current. Now consider
the case of two wires, each carrying the same current
I, but in opposite directions; assume the wires are
separated by distance R.
             R                       To determine the
                                     inductance of each
                                     conductor we integrate
                                     as before. However
                                     now we get some
Creates counter-   Creates a         field cancellation
clockwise field    clockwise field
                                                        2
 Two Conductor Case, cont’d

              R                   R
                                                             Rp

                                             Direction of integration
Key Point: As we integrate for the left line, at distance 2R from
the left line the net flux linked due to the Right line is zero!
Use superposition to get total flux linkage.
For distance Rp, greater than 2R, from left line
          0      Rp 0        Rp  R 
left       I ln       I ln         
          2      r ' 2       R 
         Left Current        Right Current
                                                                   3
Two Conductor Inductance
Simplifying (with equal and opposite currents)
          0  Rp            Rp  R  
left       I  ln     ln           
          2  r '               R 
          0
            I  ln Rp  ln r ' ln( Rp  R)  ln R 
          2
          0  R              Rp 
            I  ln  ln
          2  r '         Rp  R   
          0  R 
            I  ln  as Rp  
          2  r ' 
          0  R 
Lleft        ln  H/m
          2  r ' 
                                                        4
Many-Conductor Case

Now assume we now have k conductors, each with
current ik, arranged in some specified geometry.
We’d like to find flux linkages of each conductor.
                               Each conductor’s flux
                               linkage, k, depends upon
                               its own current and the
                               current in all the other
                               conductors.
To derive 1 we’ll be integrating from conductor 1 (at origin)
to the right along the x-axis.

                                                                 5
Many-Conductor Case, cont’d


                                                        Rk is the
                                                        distance
                                                        from con-
                                                        ductor k
                                                        to point
                                                        c.


At point b the net   We’d like to integrate the flux crossing
contribution to 1   between b to c. But the flux crossing
from ik , 1k, is    between a and c is easier to calculate and
zero.                provides a very good approximation of 1k.
                     Point a is at distance d1k from conductor k.
                                                                6
Many-Conductor Case, cont’d

     0      R1         R2                Rn 
1     i1 ln '  i2 ln d         in ln     
     2      r1          12               d1n 
     0      1           1                1 
1     i1 ln '  i2 ln d        in ln     
     2      r1          12              d1n 
      0
         i1 ln R1  i2 ln R2     in ln Rn 
      2
As R1 goes to infinity R1  R2  Rn so the second
                                  0  n 
term from above can be written =   i j  ln R1
                                  2  j 1 
                                                    7
Many-Conductor Case, cont’d
               n
Therefore if    i j  0, which is true in a balanced
               j 1
three phase system, then the second term is zero and
     0      1           1                 1 
1     i1 ln '  i2 ln d         in ln     
     2      r1          12               d1n 
1  L11i1  L12i2      L1nin

System has self and mutual inductance. However
the mutual inductance can be canceled for
balanced 3 systems with symmetry.
                                                        8
Symmetric Line Spacing – 69 kV




                                 9
Birds Do Not Sit on the Conductors




                                 10
Line Inductance Example

Calculate the reactance for a balanced 3, 60Hz
transmission line with a conductor geometry of an
equilateral triangle with D = 5m, r = 1.24cm (Rook
conductor) and a length of 5 miles.
                         Since system is assumed
                         balanced
                         ia  ib  ic

      0        1             1            1 
 a     ia ln( r ')  ib ln( D )  ic ln( D ) 
      2                                       
                                                     11
Line Inductance Example, cont’d
 Substituting
 ia  ib  ic
 Hence
      0         1   i ln  1  
 a      ia ln  r '  a  D  
      2                    
      0        D
        ia ln  
      2        r'
      0  D  4  107         5      
 La    ln           ln           3 
      2  r '   2        9.67  10 
                  6
     1.25  10        H/m
                                             12
Line Inductance Example, cont’d
                 6
La  1.25  10        H/m
Converting to reactance
X a  2  60  1.25  106
    4.71  104 /m
    0.768 /mile
X Total for 5 mile line  3.79 
(this is the total per phase)
The reason we did NOT have mutual inductance
was because of the symmetric conductor spacing
                                                 13
Conductor Bundling

To increase the capacity of high voltage transmission
lines it is very common to use a number of
conductors per phase. This is known as conductor
bundling. Typical values are two conductors for
345 kV lines, three for 500 kV and four for 765 kV.
                                         Fourth edition
                                         book cover
                                         had a
                                         transmission
                                         line with
                                         two conductor
                                         bundling
                                                        14
Bundled Conductor Pictures




                            The AEP Wyoming-Jackson
                            Ferry 765 kV line uses
                            6-bundle conductors.
                            Conductors in a bundle are
                            at the same voltage!
      Photo Source: BPA and American Electric Power   15
Bundled Conductor Flux Linkages

                               For the line shown on the left,
                               define dij as the distance bet-
                               ween conductors i and j. We
                               can then determine  for each
        i  1           1         1          1        
         a  ln  ln         ln       ln      
        4  r'        d12        d13       d14        
                                                 
     0  ib      1        1         1         1
1           ln    ln       ln        ln      
     2  4  d15
                        d16        d17       d18  
                                                  
                                                       
         ic  1           1           1          1 
          ln       ln         ln        ln       
         4  d19
                        d1,10       d1,11      d1,12  
                                                        
                                                             16
Bundled Conductors, cont’d

 Simplifying
                                            
         i ln              1           
         a                        1
                                      4       
                 (r ' d12 d13d14 )          
                                         
      0                   1             
 1       ib ln
      2         ( d d d d ) 14  
                 15 16 17 18             
                                              
                                           
                              1             
         ic ln                           1 
         
                 (d19 d1,10 d1,11d1,12 ) 4  
                                               
                                                   17
Bundled Conductors, cont’d

Rb       geometric mean radius (GMR) of bundle
                           1
      (r ' d12 d13d14 )       4   for our example
                          1
      (r ' d12    d1b   ) b       in general
D1b         geometric mean distance (GMD) of
            conductor 1 to phase b.
                                    1
         (d15d16 d17 d18 )             4    D2b  D3b  D4b  Dab
                                            1
D1c       (d19 d1,10 d1,11d1,12 )              4    D2c  D3c  D4c  Dac

                                                                          18
Inductance of Bundle

If D ab  Dac  Dbc  D and ia  ib  ic
Then
     0        1           1 
1     ia ln  R   ia ln  D  
     2        b           
     0        D        0         D
       I a ln            4 I1 ln  
     2         Rb      2          Rb 
     0          D
L1      4  ln  
     2           Rb 

                                              19
Inductance of Bundle, cont’d

But remember each bundle has b conductors
in parallel (4 in this example). So
              0  D 
La  L1 / b    ln  
              2  Rb 




                                            20
 Bundle Inductance Example
  Consider the previous example of the three phases
  symmetrically spaced 5 meters apart using wire
  with a radius of r = 1.24 cm. Except now assume
  each phase has 4 conductors in a square bundle,
  spaced 0.25 meters apart. What is the new inductance
  per meter?
                           r  1.24  102 m   r '  9.67  103 m
         0.25 M
                                                                     
                                                                          1
                                          3
                           R b  9.67  10  0.25  0.25  2  0.25           4

0.25 M            0.25 M       0.12 m (ten times bigger!)
                                0    5
                           La     ln    7.46  107 H/m
                                2 0.12
                                                                      21
Transmission Tower Configurations

The problem with the line analysis we’ve done so far is
we have assumed a symmetrical tower configuration.
Such a tower figuration is seldom practical.

                                    Therefore in
                                    general Dab 
                                    Dac  Dbc

                                    Unless something
                                    was done this would
  Typical Transmission Tower        result in unbalanced
         Configuration              phases
                                                       22
Transposition

   To keep system balanced, over the length of a
    transmission line the conductors are rotated so each
    phase occupies each position on tower for an equal
    distance. This is known as transposition.




    Aerial or side view of conductor positions over the length
                     of the transmission line.
                                                                 23
Line Transposition Example




                             24
Line Transposition Example




                             25
Transposition Impact on Flux Linkages

 For a uniformly transposed line we can
 calculate the flux linkage for phase "a"
      1 0            1             1          1 
 a           I a ln r '  I b ln d  I c ln d     “a” phase in
      3 2                          12         13    position “1”

       1 0           1             1          1 
               I a ln r '  I b ln d  I c ln d  
                                                       “a” phase in
       3 2                         13         23    position “3”

       1 0           1             1          1 
               I a ln r '  I b ln d  I c ln d 
                                                       “a” phase in
       3 2                         23         12    position “2”


                                                                 26
Transposition Impact, cont’d

Recognizing that
1                               1
  (ln a  ln b  ln c)  ln(abc) 3
3
We can simplify so
         I ln 1  I ln              1            
                                                 
         a r' b                            1
     0                      d12 d13d 23  3    
a 
     2                 1                        
         I c ln                1                 
        
                 d12 d13d 23  3                
                                                  

                                                      27
Inductance of Transposed Line

Define the geometric mean distance (GMD)
                                1
   Dm          d12 d13d 23     3

Then for a balanced 3 system ( I a  - I b - I c )
        0         1             1  0        Dm
   a      I a ln r '  I a ln D   2 I a ln r '
        2                        m
Hence
        0 Dm            7  Dm
   La     ln     2  10 ln    H/m
        2    r'             r'

                                                       28
Inductance with Bundling

 If the line is bundled with a geometric mean
 radius, R b , then
         0        Dm
    a     I a ln
         2        Rb
         0 Dm              7  Dm
    La     ln       2  10 ln    H/m
         2 Rb                  Rb




                                                29
Inductance Example

   Calculate the per phase inductance and reactance of
    a balanced 3, 60 Hz, line with horizontal phase
    spacing of 10m using three conductor bundling with
    a spacing between conductors in the bundle of
    0.3m. Assume the line is uniformly transposed and
    the conductors have a 1cm radius.




Answer: Dm = 12.6 m, Rb= 0.0889 m
Inductance = 9.9 x 10-7 H/m, Reactance = 0.6 /Mile
                                                      30
Grid Weakness




                31
New Southwest Campus 138 kV Line

•   In 2010 Ameren proposed building a new 138 kV
    transmission between the Southwest Campus and
    Bondville Substations
•   Project cost is estimated to be about $14 million
    using the “preferred” nine mile route. An
    alternative seventeen mile route would cost about
    $22 million.
•   Project should begin in 2012 and finish by 2014.
•   Campus also wants to install at 138 kV line
    between Southwest Campus and North Champaign
    substations.
                                                   32
The Local Grid




                 33
Proposed and Alternative Route




                                 34

				
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