IMPEDANCE AND REACTANCE DIAGRAM OF THE PS.ppt

B4.2 Power System One-Line Diagrams
A convenient way to represent power systems
uses “one-line” diagrams. The one-line
diagram can be obtained from a per-unitized
circuit by:
1. Omitting the neutral.
2. Representing each component by simple,
standardized symbols.
1

Two - winding power
transformer
Machine or rotating
armature (basic)
Three - winding power
transformer
Power circuit breaker
oil or other liquid
Air circuit breaker
Three- phase,
delta connection
Three - phase wye
neutral ungrounded
Three- phase wye
neutral grounded
Standard symbols for one-line diagrams.
2

Another important symbol is the “bus”
or node, which typically represents a
substation or a generation plant switchyard.
Loads are always represented at a bus
with an arrow. Generators and motors are
also always represented at a bus.
or G
If a circle does not have
a “G” or “M” inside,
assume it is a generator.
3

G
   
T1 Transmission Line T2
M1 M2
Z
Load
Consider this per-unitized circuit:
G
M1
M2
T1 T2
Load
Transmission Line
The one-line diagram is:
4

Often, for an even simpler representation,
we omit breakers and connection types, and
we use a single “squiggle” for transformers.
G
M1
M2
T1 T2
Load
Transmission Line
G
M
M
The “simpler” representation becomes:
5

Brief comments on transformers (xfmrs):
Xfmrs convert voltages from one level to another.
Power systems use 3 phase xfmrs, which can be
thought of as 3 single phase xfmrs with primary and
secondary windings connected either in Wye or Delta.
Per-phase equivalent circuits represent xfmrs as single
phase.
6

Brief comments on transformers (xfmrs):
The simplest xfmr model is the ideal xfmr.
The next simplest is the ideal xfmr with a reactance.
We called this “approximate equivalent circuit #3” in our
previous treatment of transformers.
The transformation ratio of ideal xfmrs in per-unitized
circuits is 1:1. Why is this?
7
Therefore, in per-unit, we only need to represent the
reactance (when using the “approximate model #3”).
Nominal voltage = 69 kV
Choose
Vbase1=69 kV
for this side.
Xfmr turns
ratio=V1/V2=
69/161
Compute
Vbase2=69 kV*161/69=161 kV
for this side.
1 1 1
2 2 2
/
/
pu base
pu base
V V V
V V V
 2
1
2 1
69 161
1.0
161 69
base
base
V
V
V V
  

Definition:
A “section” of a system is a set of interconnected
components not separated by a transformer. If you
can trace the one-line diagram from one component to
another without crossing a transformer, these two
components are in the same section.
Different sections are connected by transformers.
A single section may be connected to 1 or more other
sections.
8

6.9 kV/115 kV 115 kV/13.8 kV
13.8 kV/115 kV 115 kV/13.8 kV
13.8 kV/115 kV 115 kV/13.8 kV
Voltages
specified in
xfmr ratios
are assumed
to be line-to-
line unless
otherwise
indicated.
How many
sections are
in each of
these
systems?
9

Choosing bases for per unitizing systems with xfmrs
1. Select the system power base.
2. Select the voltage base for one section of the system.
3. Compute the voltage bases for all other sections. This
computation is guided by the following rule:
The ratio of voltage bases (line to line voltage bases and
phase to neutral voltage bases) in two connected
sections must be the same as the line-to-line voltage ratio
across the transformer connecting the two sections.
Therefore, we can use the transformer ratios
indicated on the one-line to get the voltage bases.
10
4. Compute the current and impedance base
for each section using the system power
base and the appropriate voltage base.

Example: Select voltage base for one section &
compute voltage bases for other sections. Then
compute current & impedance bases for all sections of
this system. Assume 3 phase power base is 100 MVA.
6.9 kV/115 kV 115 kV/13.8 kV
Section 1
Section 2
Section 3
11

6.9 kV/115 kV 115 kV/13.8 kV
Section 1
Section 2
Section 3
Select section 3 voltage base as 13.8 kV. Then
kV
9
.
6
kV
115
115
9
.
6
base
voltage
base
voltage
kV
115
kV
8
.
13
8
.
13
115
base
voltage
base
voltage
2
,
2
,
1
1
3
,
3
,
2
2








V
V
V
V
V
V
V
V
xfmr
xfmr
xfmr
xfmr
Now we can compute current and impedance bases.
Note: All voltage bases are line to line, so we will use relations
with line to line voltages (we could also compute line-to-neutral
voltage bases, then use the relations with line-to-neutral voltages).
12



3
2
3
power
base
)
voltage
(base
imped.
base
,
)
voltage
base
(
3
power
base
current
base
S
V
Z
V
S
I LL
LL





























9044
.
1
10
100
)
10
8
.
13
(
Z
imped.
base
amps
7
.
4183
)
10
8
.
13
(
3
10
100
base
current
25
.
132
10
100
)
10
115
(
Z
imped.
base
amps
04
.
502
)
10
115
(
3
10
100
base
current
4761
.
0
10
100
)
10
9
.
6
(
Z
imped.
base
amps
4
.
8367
)
10
9
.
6
(
3
10
100
base
current
6
2
3
3
3
6
3
6
2
3
2
3
6
2
6
2
3
1
3
6
1
I
I
I
13

Example: Repeat the calculations from the
previous example. 3 phase power base is 100 MVA
6.9 kV/115 kV 114 kV/13.8 kV
Section 1
Section 2
Section 3
Watch out! Xfmr voltage ratios characterize the base
voltage ratios but not necessarily the base voltages
themselves. You must be careful if a section is
connected by multiple xfmrs.
14

Select section 3 voltage base as 13.8 kV. Then
kV
84
.
6
kV
114
115
9
.
6
base
voltage
base
voltage
kV
114
kV
8
.
13
8
.
13
114
base
voltage
base
voltage
2
,
2
,
1
1
3
,
3
,
2
2








V
V
V
V
V
V
V
V
xfmr
xfmr
xfmr
xfmr
Now we can compute current and impedance bases. Note
that all voltage bases are line to line, so we will again use
the relations with line to line voltages (we could also
compute line-to-neutral voltage bases, then use the
relations with line-to-neutral voltages).
15



3
2
3
power
base
)
voltage
(base
imped.
base
,
)
voltage
base
(
3
power
base
current
base
S
V
Z
V
S
I LL
LL





























9044
.
1
10
100
)
10
8
.
13
(
Z
imped.
base
amps
7
.
4183
)
10
8
.
13
(
3
10
100
base
current
96
.
129
10
100
)
10
114
(
Z
imped.
base
amps
45
.
506
)
10
114
(
3
10
100
base
current
4679
.
0
10
100
)
10
84
.
6
(
Z
imped.
base
amps
2
.
8440
)
10
84
.
6
(
3
10
100
base
current
6
2
3
3
3
6
3
6
2
3
2
3
6
2
6
2
3
1
3
6
1
I
I
I
16

Converting pu impedances from one base to another
Often impedances given in pu on one base must be converted
to a pu value given on another base. The most common
application of this is when
•component impedances are given on the component base
•and you want to represent them within a system on another
base.
You can have
•a change of power base or
•a change of voltage base or
•a change of both.
In any of these three situations, the impedance base will
change, and so the pu value of the impedance will change.
17

The module B4 text develops the following relation.
)
base
power
base
power
(
)
base
base
( 2
,
,
old
new
new
old
old
pu
new
pu
S
S
V
V
Z
Z 
• base voltages can either be line to line or phase to neutral,
but not both.
• Base powers can either be 3 phase or per phase,
but not both.
18
The above relation is obtained by expressing
, ,
, ,
,
pu new pu old
base new base old
Z Z
Z Z
Z Z
 
 
and then substituting appropriate expressions for Zbase,new and
Zbase,old, solving for ZΩ in both cases, and equating the two results.

Example: system power base is 100 MVA
4.1 kV/116 kV
10%, 100 MVA
120 kV/13.8 kV, 8%, 10 MVA
Section 1
Section 2
Section 3
115 kV/13.8 kV, 7%,20 MVA
Section 4
T1
T2
T3
Transformer reactances are as shown.
Transmission line has impedance of 0.01+j0.1 pu
on base of 50 MVA, 115 kV.
Convert 3 transformer reactances and line impedance
to system base.
Line
19
These pu values
are given on
component bases,
not system bases.

4.1 kV/116 kV
10%, 100 MVA
120 kV/13.8 kV, 8%, 10 MVA
Section 1
Section 2
Section 3
115 kV/13.8 kV, 7%,20 MVA
Section 4
base voltage V1=4.1 kV
base voltage V2=116 kV
base voltage V3=116x13.8/120=13.34 kV
base voltage V4=116x13.8/115=13.92 kV
T1
T2
T3
Line
20

Assume xfmr impedances are given on rated voltage
bases. Note that per cent is 100 times pu value.
Xfmr 1: No change of base is necessary.
Xfmr 2:
8561
.
0
10
100
)
116
120
(
08
.
0
base
base
)
base
base
( 2
2
,
,
2
,
,
2 


old
new
new
old
old
pu
T
new
pu
T
S
S
V
V
X
X
Xfmr 3:
344
.
0
20
100
)
116
115
(
07
.
0
base
base
)
base
base
( 2
2
,
,
3
,
,
3 


old
new
new
old
old
pu
T
new
pu
T
S
S
V
V
X
X
21

Transmission line:
197
.
0
0197
.
0
50
100
)
116
115
)(
1
.
0
01
.
0
( 2
,
, j
j
Z new
pu
L 



22

These impedances would be used in the following
per unit circuit.
Line
j0.1 0.0197+j0.197
Load
T1
T2 T3
j0.8561 j0.344
23

A comment on loads:
Loads can be represented as constant impedance.
Assume we are given line-to-line load voltage magnitude
VLL,load and three phase power consumption P3ϕ+jQ3ϕ.
There are two different equivalent representations.
1. R and jX in parallel:
Rp jXp
VLN,Load
-
+
24
2 2
, ,
1 3
LN Load LL Load
p
V V
R
P P
 
 
2 2
, ,
1 3
LN Load LL Load
p
V V
X
Q Q
 
 
See example B3.1, Module 3.
Also see problem 1a, Module 4.
2. R and jX in series:
Rs
jXs
VLN,Load
-
+
2 2
, ,
1 1 3 3
LN Load LL Load
s
V V
Z
P jQ P jQ
   
 
 
   
Re ; X =Im
s s s s
R Z Z

Then, in both cases, you
can obtain the pu values
as usual:
Rpu=R/Zbase
Xpu=X/Zbase

IMPEDANCE AND REACTANCE DIAGRAM OF THE PS.ppt

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IMPEDANCE AND REACTANCE DIAGRAM OF THE PS.ppt