Fault Analysis Using z Bus

Fault Analysis Using Z-bus 1.0 1.0 Intr Introd oduc ucti tion on

The The previo evious us notes otes on Z-bus -bus resu esulte lted in some some usef useful ul know knowle ledg dge: e: diag diagon onal al elem elemen entt Zkk  of   of the Z-bus is the Thevenin impedance seen looking into the network from bus k. Combining this knowledge with equation (4 from notes called !"#mmetrical $aults %&' which was V   I ′′ = (  Z  + Z  enab enable less us to eff efficie icien ntl# tl# obta obtain in the the fault ault current for an# bus in the network.   f  

  f  

  f  

Thev

This is eas# if we have Z-bus and )f . *owever' one other thing that we will frequentl# need is the line currents' because the circuit breakers are going to be in series with the lines' not with the faults.



2.0 Fault Fault calcula calculation tionss using using Zbus Zbus

+ecall that (% "ince (% represents a set of linear equations' superposition holds' and we ma# write: (, This sa#s that the change in voltage at all  buses ) ma# be computed if the change in inections at all buses / are known. 0e can write eq. (, in e1panded form as: V  

 Z  I 

∆ V   = Z ∆  I 

 ∆V (   Z ((          ∆V k   =  Z k (         ∆V  N    Z  N (



 Z (k 











 Z kk 









  Z  Nk 



 Z ( N    ∆ I ( 

      Z kN    ∆ I k          Z  NN   ∆ I  N   

(4

 2ow consider a fault fault at bus k' where the prefault voltage at bus k is ) f . 3et the fault curr curren entt be / /f ' and assume tha that the fault impedance Zf 56 56 (this is t#picall# worst-case scenario. "ince a fault is a short circuit' then )k 5-) 5-)f .

%

7lso' since the fault current is out of bus k' then /k 55- /f . "ubstituting these into eq. (4 results in  ∆ V (   Z ((           row  k →  − V  f   =  Z k (         ∆ V  N   Z  N (

  Z (k    Z ( N    6

            Z kk    Z kN    − I  f ′             Z  Nk    Z  NN   6 

(8  2oting that the right-hand-side results in' for each row ' onl# the Z k  being   being multiplied b# a non-9ero current. Therefore:  ∆ V   − Z  I k   f ′              row  →  − V  f  =  − Z kk  I  f ′  (           ∆ V  N   − Z  Nk  I  f ′  0e observe from row k that: V   Z  I  (; "olving (; for /f  results  results in V   I ′′ = (<  Z  (

k

−   f   = −

′′ kk    f  

  f  

  f  

kk 

,

(

 2otice that eq. (< is consistent with eq. ( when Zf 56. 56.  2ow substitute eq. (< into eq. eq. ( to get:  ∆ V   ( −  Z k  = Z kk ) V  f               row  →  − V  f   =  − V  f   (>           ∆ V  N   ( −  Z  Nk  = Z kk ) V  f    (

(

k

 2ow eq. (> provides the change in the bus voltages due to the fault. Change from what? /t is the change from the voltage without the fault' i.e.' it is the pre-fault voltage. Consider an# bus' lets sa# bus ' with a prefault voltage of ) . Then we can compute the bus  voltage under the faulted condition as V  V  V  (6   jf  

=

  j

4

+∆

  j

$rom eq. (>' we know that  Z   jk 

∆V   j = −

 Z kk 

V   f  

(

"ubstitution of ( into (6 results in  Z   jk 

(%  2ow eq. (% is useful for computing fault currents in the circuits. Consider $ig. . V   jf   = V   j −

 Z kk 

V   f  

Z b  bus i

bus .

$ig.  0e can use eq. (% to write down the volt voltag ages es und under the the fault aulted ed condi ondittion ion for for  buses i and ' as  Z  V  = V  − V  (,  Z  ik 

if  

i

  f  

kk 

 Z   jk 

(4  2ow we can compute the subtransient current flowing from bus i to bus  under the fault condition as V  −V   I ′′ = (8  Z  V   jf  

V   j

=

−

 Z kk 

if  

  jf  

ij

b

8

V   f  

"ubst ubstit itut utin ing g eqs. eqs. (, (, and (4 (4 into into (8 (8 results in V i −  I ij′′ =

=

 Z ik   Z kk 

V   f   −V   j −

 Z   jk   Z kk 

 Z b V i −V   j  Z b

−V   f  

 Z ik  − Z   jk 

V   f  

(

 Z b Z kk 

0e can use eq. ( to get the fault current in the circuits. These values provide us with the appropriate information for selecting the circuit breakers in the lines. 3.0 Some importan importantt comments comments Zbus should be developed using • subtransient reactances in generator=motor models. • @ecause fault currents are t#picall# much larger than load currents' it ma# be assumed that there are no loads. o 7ll pre-faults currents are 6. o 7ll buses have voltage (pre-fault equal to )f . o Aquation ( becomes:  I ij′′ = −V   f  



 Z ik  − Z   jk   Z b Z kk 

(;

$rom ( and (;' we see that onl# the k th column of the Z-bus is required to anal#9e a fault at bus k.

•

The last observation can be utili9ed in an effe effect ctiv ivee fashi ashio on when hen per perform formin ing g faul faultt anal#sis. 3ets assume that we want to compute the short circuit currents for a fault at onl# one bus k. "o we ust want to get the k th column of Z-bus' but we do not need the entire Z-bus. There is an efficient wa# to get the k th column of Z-bus. 3ets stud# it. Consider that the Z-bus and the B-bus are inverses of each other' i.e.'  Z  Y  (< This means that their product gives the identit# matri1. Y  Z   I  (> where / is given b# a matri1 of 9eros e1cept the diagonal which contains all ones' i.e.' =

−(

=

;

( 6  I  =    6

6



(







6



6

    ( 6

(%6

ur approach will depend on two ideas: . Column of Z: 0e can ust consider a single column of Z' instead of the entire matri1. Call it Zk D it is a column vector. The right hand side of (> will ust be a colum column n of /. Call Call it /k . /t will also be ust a vector and will contain 9eros in ever# row e1cept for row k. The resulting relation is: Y  Z  = I  (% k 

k 

%. 3E-F 3E-Fec ecom ompo possitio ition n: /f #ou #ou hav have taken aken AA 48 or a linear algebra course in math' then #ou are familiar with 3E decomposition. 3E decomposition  provides a wa# to solve for the vector 1 in the matri1 relation  A x b (%% where 7 is a nGn square matri1' 1 is an unknown nG vector' and b is an known =

<

nG vector. The advantage to 3E decomposition is it does not require inverting the matri1 7. The basis of 3Edecomposition is that we ma# factor 7 into a matri1 product 3E' i.e.'  A  LU  (%, where where 3 is a lower lower diagon diagonal al nGn nGn matri matri1 1 of the form =

l (( 6   L = l %( l %%  l ,( l ,%

6

  l ,,  6

(%4

and E is an upper diagonal nGn matri1 of the form ( u(%  U  = 6 (  6 6

u(, 

  (  

u%,

(%8

"ubstitution of eq. (%, in (%% #ields  LU  x =b

(%

Fefining U  x

(%;

=w

 provides that eq. (%8 (%8 becomes: (%< /f we have have 3 and and E' then then (%< (%< is easi easil# l# solved solved for for w (witho (without ut invert inverting ing 3 using using forward substitution' and then (%; can  L w

=b

>

 be easil# solved for 1 (without inverting E usin using g back backwa ward rdss subs substit titut utio ion. n. Hore Hore details on 3E decomposition ma# be found in the notes called !3EFecomposition.doc.&  2ow observe that eq. eq. (% and (%% are in the the same form. Therefore we want to solve the foll follow owin ing g equa equati tion onss in the the orde orderr the# the# are are given:  L w = I  (%> U  Z  = w (,6 k 

k 

*omework I,: Fue $rida#' Januar# %;. Cons Consid ider er the the 4-bu 4-buss s#st s#stem em show shown n belo below. w. @oth machines have subtransient reactances of 6.%6 pu (#ou can combine the machine subtransient reactance with the transformer impedance to get a single reactance con connectin cting g the mach achine ine inter ntern nal volta oltag ge with the network.

6

@us %

 6.%8  6.%8

6.(%8 6.(%8

 6.%8  6.%8 @us ,

6.%6 6.%6

 6.46  6.46 @us 4

@us (  6.(6  6.(6

6.(6 6.(6

a. Construct Construct the the B-bus B-bus for this this network network (should be a 4G4 matri1.  b. Consider that there there is a three-phase (s#mmetrical fault at bus %. • Ese 3E decomposition to obtain the %nd column of the Z-bus. • Compute the subtransient fault current. Ese eq. (% to find the voltages • during the fault. Ese eq. (; to find the subtransient • currents in lines ,-%' -%' and 4-%.

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