Courses

# Z ?BUS formation considering mutual coupling between elements - Chapter Notes Notes | EduRev

## : Z ?BUS formation considering mutual coupling between elements - Chapter Notes Notes | EduRev

Page 1

4.3
¯
Z
BUS
formation considering mutual coupling between
elements
Assume that the bus impedance matrix 
¯
Z
m
BUS
 is known for a partial network of ‘m’ nodes and
a reference node. The bus voltage and bus current relation for the partial network, shown in Fig.
4.26, can be expressed as:
Figure 4.26: Partial Network with m-buses

¯
V
BUS
= 
¯
Z
m
BUS

¯
I
BUS
 (4.33)
In equation (4.33),
¯
V
BUS
is (m× 1) bus voltage vector
¯
I
BUS
is (m× 1) bus current vector
¯
Z
m
BUS
is (m×m) bus impedance matrix
The new added element ‘p-q’ may be a branch or may be a link as discussed in the previous
algorithm.
128
Page 2

4.3
¯
Z
BUS
formation considering mutual coupling between
elements
Assume that the bus impedance matrix 
¯
Z
m
BUS
 is known for a partial network of ‘m’ nodes and
a reference node. The bus voltage and bus current relation for the partial network, shown in Fig.
4.26, can be expressed as:
Figure 4.26: Partial Network with m-buses

¯
V
BUS
= 
¯
Z
m
BUS

¯
I
BUS
 (4.33)
In equation (4.33),
¯
V
BUS
is (m× 1) bus voltage vector
¯
I
BUS
is (m× 1) bus current vector
¯
Z
m
BUS
is (m×m) bus impedance matrix
The new added element ‘p-q’ may be a branch or may be a link as discussed in the previous
algorithm.
128
4.3.1 Addition of a branch to this partial network:
The performance equation of the network with an added branch ‘p-q’ is
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
V
1
¯
V
2
?
¯
V
p
?
¯
V
m
¯
V
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
Z
11
¯
Z
12
?
¯
Z
1p
?
¯
Z
1m
¯
Z
1q
¯
Z
21
¯
Z
22
?
¯
Z
2p
?
¯
Z
2m
¯
Z
2q
?
¯
Z
p1
¯
Z
p2
?
¯
Z
pp
?
¯
Z
pm
¯
Z
pq
?
¯
Z
m1
¯
Z
m2
?
¯
Z
mp
?
¯
Z
mm
¯
Z
mq
¯
Z
q1
¯
Z
q2
?
¯
Z
qp
?
¯
Z
qm
¯
Z
qq
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
I
1
¯
I
2
?
¯
I
p
?
¯
I
m
¯
I
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
(4.34)
The network is assumed to contain bilateral passive elements and hence,
¯
Z
qi
=
¯
Z
iq
for i =
1, 2,?m, i ? q. The added branch ‘p-q’ is assumed to be mutually coupled with one or more
elements of the partial network.
To determine element
¯
Z
qi
, inject a current ati
th
node and calculate the voltage atq
th
node with
respect to reference, as shown in Fig. 4.27.
Figure 4.27: Calculation of
¯
Z
qi
Calculation of
¯
Z
qi
As all other bus currents are zero, bus voltages can be written as,
129
Page 3

4.3
¯
Z
BUS
formation considering mutual coupling between
elements
Assume that the bus impedance matrix 
¯
Z
m
BUS
 is known for a partial network of ‘m’ nodes and
a reference node. The bus voltage and bus current relation for the partial network, shown in Fig.
4.26, can be expressed as:
Figure 4.26: Partial Network with m-buses

¯
V
BUS
= 
¯
Z
m
BUS

¯
I
BUS
 (4.33)
In equation (4.33),
¯
V
BUS
is (m× 1) bus voltage vector
¯
I
BUS
is (m× 1) bus current vector
¯
Z
m
BUS
is (m×m) bus impedance matrix
The new added element ‘p-q’ may be a branch or may be a link as discussed in the previous
algorithm.
128
4.3.1 Addition of a branch to this partial network:
The performance equation of the network with an added branch ‘p-q’ is
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
V
1
¯
V
2
?
¯
V
p
?
¯
V
m
¯
V
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
Z
11
¯
Z
12
?
¯
Z
1p
?
¯
Z
1m
¯
Z
1q
¯
Z
21
¯
Z
22
?
¯
Z
2p
?
¯
Z
2m
¯
Z
2q
?
¯
Z
p1
¯
Z
p2
?
¯
Z
pp
?
¯
Z
pm
¯
Z
pq
?
¯
Z
m1
¯
Z
m2
?
¯
Z
mp
?
¯
Z
mm
¯
Z
mq
¯
Z
q1
¯
Z
q2
?
¯
Z
qp
?
¯
Z
qm
¯
Z
qq
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
I
1
¯
I
2
?
¯
I
p
?
¯
I
m
¯
I
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
(4.34)
The network is assumed to contain bilateral passive elements and hence,
¯
Z
qi
=
¯
Z
iq
for i =
1, 2,?m, i ? q. The added branch ‘p-q’ is assumed to be mutually coupled with one or more
elements of the partial network.
To determine element
¯
Z
qi
, inject a current ati
th
node and calculate the voltage atq
th
node with
respect to reference, as shown in Fig. 4.27.
Figure 4.27: Calculation of
¯
Z
qi
Calculation of
¯
Z
qi
As all other bus currents are zero, bus voltages can be written as,
129
¯
V
1
=
¯
Z
1i
¯
I
i
¯
V
2
=
¯
Z
2i
¯
I
i
?
¯
V
p
=
¯
Z
pi
¯
I
i
?
¯
V
m
=
¯
Z
mi
I
i
¯
V
q
=
¯
Z
qi
¯
I
i
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
(4.35)
Also from Fig. 4.27,
¯
V
p
and
¯
V
q
can be related as,
¯
V
q
=
¯
V
p
- ¯ v
pq
(4.36)
Where ¯ v
pq
is the voltage across the added element ‘p-q’. Also, the currents in the elements of
the network can be related to the voltages across the elements as,

¯
i
pq
¯
i
?s
= 
¯ y
pq,pq
¯ y
pq,?s
¯ y
?s,pq
¯ y
?s,?s

¯ v
pq
¯ v
?s
(4.37)
Where,
¯
i
pq
= the current through the added element ‘p-q’.
¯
i
?s
= (m× 1) current vector of the elements of the partial network.
¯ v
?s
= (m× 1) voltage vector of the elements of the partial network.
¯ y
pq,pq
¯ y
pq,?s
= (m× 1) vector of mutual admittances between the added element ‘p-q’ and the
elements ‘?-s’ of the partial network.
¯ y
?s,?s
= (m×m) primitive admittance matrix of the partial network.
¯ y
?s,pq
= ¯ y
pq,?s

T
The diagonal elements of primitive impedance matrix [¯ z] are the self impedance of the individual
elements while the o?-diagonal elements are mutual impedances between the elements. The inverse
of primitive impedance matrix is the primitive admittance matrix [¯ y]. This can be explained with
the help of an illustrative example.
A single line diagram of a power system is shown in Fig. 4.28. The self-impedances of lines are
written by the side of the line and are in p.u.. The two lines between nodes 1 and 3 are mutually
coupled with a mutual impedance ofj0.10. The primitive impedance matrix for the network can be
written as,
¯ z=
?
?
?
?
?
?
?
?
?
?
?
?
?
j0.60 0 0 0 0
0 j0.50 0 0 0
0 0 j0.50 0 0
0 0 0 j0.25 j1.0
0 0 0 j1.0 j0.20
?
?
?
?
?
?
?
?
?
?
?
?
?
130
Page 4

4.3
¯
Z
BUS
formation considering mutual coupling between
elements
Assume that the bus impedance matrix 
¯
Z
m
BUS
 is known for a partial network of ‘m’ nodes and
a reference node. The bus voltage and bus current relation for the partial network, shown in Fig.
4.26, can be expressed as:
Figure 4.26: Partial Network with m-buses

¯
V
BUS
= 
¯
Z
m
BUS

¯
I
BUS
 (4.33)
In equation (4.33),
¯
V
BUS
is (m× 1) bus voltage vector
¯
I
BUS
is (m× 1) bus current vector
¯
Z
m
BUS
is (m×m) bus impedance matrix
The new added element ‘p-q’ may be a branch or may be a link as discussed in the previous
algorithm.
128
4.3.1 Addition of a branch to this partial network:
The performance equation of the network with an added branch ‘p-q’ is
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
V
1
¯
V
2
?
¯
V
p
?
¯
V
m
¯
V
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
Z
11
¯
Z
12
?
¯
Z
1p
?
¯
Z
1m
¯
Z
1q
¯
Z
21
¯
Z
22
?
¯
Z
2p
?
¯
Z
2m
¯
Z
2q
?
¯
Z
p1
¯
Z
p2
?
¯
Z
pp
?
¯
Z
pm
¯
Z
pq
?
¯
Z
m1
¯
Z
m2
?
¯
Z
mp
?
¯
Z
mm
¯
Z
mq
¯
Z
q1
¯
Z
q2
?
¯
Z
qp
?
¯
Z
qm
¯
Z
qq
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
I
1
¯
I
2
?
¯
I
p
?
¯
I
m
¯
I
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
(4.34)
The network is assumed to contain bilateral passive elements and hence,
¯
Z
qi
=
¯
Z
iq
for i =
1, 2,?m, i ? q. The added branch ‘p-q’ is assumed to be mutually coupled with one or more
elements of the partial network.
To determine element
¯
Z
qi
, inject a current ati
th
node and calculate the voltage atq
th
node with
respect to reference, as shown in Fig. 4.27.
Figure 4.27: Calculation of
¯
Z
qi
Calculation of
¯
Z
qi
As all other bus currents are zero, bus voltages can be written as,
129
¯
V
1
=
¯
Z
1i
¯
I
i
¯
V
2
=
¯
Z
2i
¯
I
i
?
¯
V
p
=
¯
Z
pi
¯
I
i
?
¯
V
m
=
¯
Z
mi
I
i
¯
V
q
=
¯
Z
qi
¯
I
i
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
(4.35)
Also from Fig. 4.27,
¯
V
p
and
¯
V
q
can be related as,
¯
V
q
=
¯
V
p
- ¯ v
pq
(4.36)
Where ¯ v
pq
is the voltage across the added element ‘p-q’. Also, the currents in the elements of
the network can be related to the voltages across the elements as,

¯
i
pq
¯
i
?s
= 
¯ y
pq,pq
¯ y
pq,?s
¯ y
?s,pq
¯ y
?s,?s

¯ v
pq
¯ v
?s
(4.37)
Where,
¯
i
pq
= the current through the added element ‘p-q’.
¯
i
?s
= (m× 1) current vector of the elements of the partial network.
¯ v
?s
= (m× 1) voltage vector of the elements of the partial network.
¯ y
pq,pq
¯ y
pq,?s
= (m× 1) vector of mutual admittances between the added element ‘p-q’ and the
elements ‘?-s’ of the partial network.
¯ y
?s,?s
= (m×m) primitive admittance matrix of the partial network.
¯ y
?s,pq
= ¯ y
pq,?s

T
The diagonal elements of primitive impedance matrix [¯ z] are the self impedance of the individual
elements while the o?-diagonal elements are mutual impedances between the elements. The inverse
of primitive impedance matrix is the primitive admittance matrix [¯ y]. This can be explained with
the help of an illustrative example.
A single line diagram of a power system is shown in Fig. 4.28. The self-impedances of lines are
written by the side of the line and are in p.u.. The two lines between nodes 1 and 3 are mutually
coupled with a mutual impedance ofj0.10. The primitive impedance matrix for the network can be
written as,
¯ z=
?
?
?
?
?
?
?
?
?
?
?
?
?
j0.60 0 0 0 0
0 j0.50 0 0 0
0 0 j0.50 0 0
0 0 0 j0.25 j1.0
0 0 0 j1.0 j0.20
?
?
?
?
?
?
?
?
?
?
?
?
?
130
Figure 4.28: Sample Power System
The inverse of ¯ z is ¯ y, the primitive admittance matrix.
¯ y=
?
?
?
?
?
?
?
?
?
?
?
?
?
-j1.67 0 0 0 0
0 -j2.0 0 0 0
0 0 -j2.0 0 0
0 0 0 -j5.0 j2.5
0 0 0 j2.5 -j6.25
?
?
?
?
?
?
?
?
?
?
?
?
?
The current
¯
i
pq
in the added branch ‘p-q’ equal to zero as node ‘q’ is open.
¯
i
pq
= 0 (4.38)
Thevoltage ¯ v
pq
of the partial network. Thus, the voltage across other elements of the network can be expressed as,
¯ v
?s
=
¯
V
?
-
¯
V
s
(4.39)
where
¯
V
?
and
¯
V
s
are the voltages of the nodes of the partial network. With
¯
i
pq
= 0 from equation
(4.37) one can write,
¯ y
pq,pq
¯ v
pq
+ ¯ y
pq,?s
¯ v
?s
= 0
Hence,
¯ v
pq
=-
¯ y
pq,?s
¯ v
?s
¯ y
pq,pq
(4.40)
131
Page 5

4.3
¯
Z
BUS
formation considering mutual coupling between
elements
Assume that the bus impedance matrix 
¯
Z
m
BUS
 is known for a partial network of ‘m’ nodes and
a reference node. The bus voltage and bus current relation for the partial network, shown in Fig.
4.26, can be expressed as:
Figure 4.26: Partial Network with m-buses

¯
V
BUS
= 
¯
Z
m
BUS

¯
I
BUS
 (4.33)
In equation (4.33),
¯
V
BUS
is (m× 1) bus voltage vector
¯
I
BUS
is (m× 1) bus current vector
¯
Z
m
BUS
is (m×m) bus impedance matrix
The new added element ‘p-q’ may be a branch or may be a link as discussed in the previous
algorithm.
128
4.3.1 Addition of a branch to this partial network:
The performance equation of the network with an added branch ‘p-q’ is
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
V
1
¯
V
2
?
¯
V
p
?
¯
V
m
¯
V
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
Z
11
¯
Z
12
?
¯
Z
1p
?
¯
Z
1m
¯
Z
1q
¯
Z
21
¯
Z
22
?
¯
Z
2p
?
¯
Z
2m
¯
Z
2q
?
¯
Z
p1
¯
Z
p2
?
¯
Z
pp
?
¯
Z
pm
¯
Z
pq
?
¯
Z
m1
¯
Z
m2
?
¯
Z
mp
?
¯
Z
mm
¯
Z
mq
¯
Z
q1
¯
Z
q2
?
¯
Z
qp
?
¯
Z
qm
¯
Z
qq
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
I
1
¯
I
2
?
¯
I
p
?
¯
I
m
¯
I
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
(4.34)
The network is assumed to contain bilateral passive elements and hence,
¯
Z
qi
=
¯
Z
iq
for i =
1, 2,?m, i ? q. The added branch ‘p-q’ is assumed to be mutually coupled with one or more
elements of the partial network.
To determine element
¯
Z
qi
, inject a current ati
th
node and calculate the voltage atq
th
node with
respect to reference, as shown in Fig. 4.27.
Figure 4.27: Calculation of
¯
Z
qi
Calculation of
¯
Z
qi
As all other bus currents are zero, bus voltages can be written as,
129
¯
V
1
=
¯
Z
1i
¯
I
i
¯
V
2
=
¯
Z
2i
¯
I
i
?
¯
V
p
=
¯
Z
pi
¯
I
i
?
¯
V
m
=
¯
Z
mi
I
i
¯
V
q
=
¯
Z
qi
¯
I
i
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
(4.35)
Also from Fig. 4.27,
¯
V
p
and
¯
V
q
can be related as,
¯
V
q
=
¯
V
p
- ¯ v
pq
(4.36)
Where ¯ v
pq
is the voltage across the added element ‘p-q’. Also, the currents in the elements of
the network can be related to the voltages across the elements as,

¯
i
pq
¯
i
?s
= 
¯ y
pq,pq
¯ y
pq,?s
¯ y
?s,pq
¯ y
?s,?s

¯ v
pq
¯ v
?s
(4.37)
Where,
¯
i
pq
= the current through the added element ‘p-q’.
¯
i
?s
= (m× 1) current vector of the elements of the partial network.
¯ v
?s
= (m× 1) voltage vector of the elements of the partial network.
¯ y
pq,pq
¯ y
pq,?s
= (m× 1) vector of mutual admittances between the added element ‘p-q’ and the
elements ‘?-s’ of the partial network.
¯ y
?s,?s
= (m×m) primitive admittance matrix of the partial network.
¯ y
?s,pq
= ¯ y
pq,?s

T
The diagonal elements of primitive impedance matrix [¯ z] are the self impedance of the individual
elements while the o?-diagonal elements are mutual impedances between the elements. The inverse
of primitive impedance matrix is the primitive admittance matrix [¯ y]. This can be explained with
the help of an illustrative example.
A single line diagram of a power system is shown in Fig. 4.28. The self-impedances of lines are
written by the side of the line and are in p.u.. The two lines between nodes 1 and 3 are mutually
coupled with a mutual impedance ofj0.10. The primitive impedance matrix for the network can be
written as,
¯ z=
?
?
?
?
?
?
?
?
?
?
?
?
?
j0.60 0 0 0 0
0 j0.50 0 0 0
0 0 j0.50 0 0
0 0 0 j0.25 j1.0
0 0 0 j1.0 j0.20
?
?
?
?
?
?
?
?
?
?
?
?
?
130
Figure 4.28: Sample Power System
The inverse of ¯ z is ¯ y, the primitive admittance matrix.
¯ y=
?
?
?
?
?
?
?
?
?
?
?
?
?
-j1.67 0 0 0 0
0 -j2.0 0 0 0
0 0 -j2.0 0 0
0 0 0 -j5.0 j2.5
0 0 0 j2.5 -j6.25
?
?
?
?
?
?
?
?
?
?
?
?
?
The current
¯
i
pq
in the added branch ‘p-q’ equal to zero as node ‘q’ is open.
¯
i
pq
= 0 (4.38)
Thevoltage ¯ v
pq
of the partial network. Thus, the voltage across other elements of the network can be expressed as,
¯ v
?s
=
¯
V
?
-
¯
V
s
(4.39)
where
¯
V
?
and
¯
V
s
are the voltages of the nodes of the partial network. With
¯
i
pq
= 0 from equation
(4.37) one can write,
¯ y
pq,pq
¯ v
pq
+ ¯ y
pq,?s
¯ v
?s
= 0
Hence,
¯ v
pq
=-
¯ y
pq,?s
¯ v
?s
¯ y
pq,pq
(4.40)
131
Substituting ¯ v
?s
from equation (4.39) and ¯ v
pq
equation (4.36) in equation (4.40) one gets,
¯
V
q
=
¯
V
p
+
¯ y
pq,?s
(
¯
V
?
-
¯
V
s
)
¯ y
pq,pq
Substitution of
¯
I
i
= 1 pu in equation (4.35) results in
¯
V
p
,
¯
V
q
,
¯
V
?
and
¯
V
s
being replaced by their
corresponding impedances and hence,
¯
Z
qi
=
¯
Z
pi
+
¯ y
pq,?s
(
¯
Z
?i
-
¯
Z
si
)
¯ y
pq,pq
(4.41)
?i= 1, 2,?,m, i?q
For calculating the self impedance
¯
Z
qq
, a current
¯
I
q
= 1 p.u. is injected into q
th
node with all
other currents equal to zero as shown in Fig. 4.29. Then the voltages of the nodes are calculated
from equation (4.35) , as
¯
V
1
=
¯
Z
1q
¯
I
q
¯
V
2
=
¯
Z
2q
¯
I
q
?
¯
V
p
=
¯
Z
pq
¯
I
q
?
¯
V
m
=
¯
Z
mq
I
q
¯
V
q
=
¯
Z
qq
¯
I
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
(4.42)
With
¯
I
q
= 1 p.u.,
¯
Z
qq
Can be calculated directly by calculating
¯
V
q
.
also,
¯
V
q
=
¯
V
p
- ¯ v
pq
(4.43)
and
¯
i
pq
=-
¯
I
q
=-1 (4.44)
Hence, from equation (4.37) one gets
¯
i
pq
=-1= ¯ y
pq,pq
¯ v
pq
+ ¯ y
pq,?s
¯ v
?s
(4.45)
And thus ¯ v
pq
can be written as,
¯ v
pq
=-
1+ ¯ y
pq,?s
¯ v
?s
¯ y
pq,pq
(4.46)
Substituting ¯ v
pq
and ¯ v
?s
, the above equation can be rewritten as,
¯
V
q
=
¯
V
p
+
1+ ¯ y
pq,?s
(
¯
V
?
-
¯
V
s
)
y
pq,pq
(4.47)
132
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!