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PowerFlow2
1.0 Introduction
We derive two forms of the power flow
equations in these notes. We also provide an
example of using these equations.
2.0 Derivation of power flow equations
Our goal here is to express the complex
power injected into a bus as a function of the
network voltage phasors (magnitude and
angle) and the network impedances.
We work entirely in per-unit.
To do this, we consider a general node, any
node, in the network. Let’s assume, for
example, that we have a 4 node network,
and we want to express the complex power
for node 3. This is just going to be:
S3  V3 I 3*
(1)
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The complex power S3 is defined positive
when power is injected into the bus (as a
generator) and negative when power is
required from the bus (as a load). Defining
SG3 and SD3 as the generation and load
components of S3, respectfully, then, we
have:
S3  SG 3  S D 3
(2)
Recall the Ybus relation.
 I 1  Y11
 I  Y
 2    21
 I 3  Y31
  
 I 4  Y41
Y12 Y13 Y14  V1 
Y22 Y23 Y24  V2 
Y32 Y33 Y34  V3 
 
Y42 Y43 Y44  V4 
(3)
Then, I3 is given by the third equation in (3):
I 3  Y31V1  Y32V2  Y33V3  Y34V4
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  Y3 kVk
(4)
k 1
The conjugate of I3 will be then
2
4
I  Y V
*
3
*
3k
k 1
*
k
(5)
Substitution of (5) into (1) results in
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S3  V3  Y3*kVk*
(6)
k 1
Now let’s bring V3 into the summation:
4
S3   Y3*kV3Vk*
(7)
k 1
Expressing the voltages in their polar form:
4
S 3   Y3*k V3 e j 3 Vk e  j k
k 1
(8)
Note that the negative sign on the Vk angle
comes from the conjugation of Vk in (7).
Combining the exponentials in (8), we get:
4
S 3   Y V3 Vk e
k 1
*
3k
3
j ( 3  k )
(9)
Now consider Y3k in eq. (9). It is a complex
number and can therefore be written in
either rectangular or polar form, i.e.,
Y3 k  G3k  jB3 k  Y3 k e
j 3 k
(10)
where
 3k
B3 k
 tan
G3 k
(11)
Remember that Y3k is the element in
position row 3, column k, of the Y bus. It is
NOT the admittance of the circuit
connecting buses 3 and k, and therefore G3k
and B3k are NOT the conductance and
susceptance, respectively, of the circuit
connecting buses 3 and k (contrary to what
the book says on page 326).
But when k≠3, Y3k is the NEGATIVE of the
admittance connecting buses 3 and k, and
therefore G3k and B3k are the NEGATIVE of
the
conductance
and
susceptance,
respectively, of the circuit connecting buses
3 and k.
4
Since transmission lines and transformers
always have
- positive resistance and positive reactance,
- positive conductance and negative
susceptance,
the numerical value of off-diagonal Ybus
elements will have negative real part and
positive imaginary part, whereas the
numerical value of diagonal Ybus elements
will be just the opposite.
Now we will develop two different forms of
the power flow equations.
2.1 Polar Form of Ybus elements
Substituting the polar form of (10) into (9),
we get:
4
S3   Y3 k e
k 1
 j 3 k
V3 Vk e
j ( 3  k )
(12)
where the negative angle results from the
conjugation of Y3k in eq. (9).
5
Combining exponentials, we obtain:
4
S3   Y3 k V3 Vk e j ( 3  k  3 k )
k 1
(13)
Now apply Euler to the exponential in (13):
4
S3   Y3 k V3 Vk cos( 3   k   3 k )  j sin( 3   k   3 k )
k 1
(14)
Distributing the product of magnitudes:
S3   Y3 k V3 Vk cos( 3   k   3 k )
4
k 1
 j Y3 k V3 Vk sin( 3   k   3 k )
(15)
Now recalling that
S3  P3  jQ3
(16)
it must be the case that
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P3   Y3 k V3 Vk cos( 3   k   3 k )
(17a)
Q3   Y3 k V3 Vk sin( 3   k   3 k )
(17b)
k 1
4
k 1
The above development could be done for
any bus i, and for any network size n.
Therefore,
6
n
Pi   Y3 i Vi Vk cos( i   k   ik )
(18a)
Qi   Y3 i V3 Vi sin( i   k   ik )
(18b)
k 1
n
k 1
Equations (18a) and (18b) are one form in
which you may see the power flow
equations expressed in some books. For
example, reference [1] on page 330
expresses them this way, and so does
reference [2], pp 231. (In both of these
references, there is one small difference in
that the argument of the trigonometric
functions is negated, which has no effect on
the real power equation but requires a
negative sign out front of the reactive power
equation).
However, eqs. (18a) and (18b) are not the
form that your text uses. We develop that
form in the next subsection.
7
2.2 Rectangular Form of Ybus elements
Returning to our efforts in regards to bus 3,
we substitute the rectangular form of (10)
into (9) to obtain:
4
S 3   G3 k  jB3 k V3 Vk e j ( 3  k )
k 1
(19)
where the negative sign in front of jB3k
results from the conjugation of Y3k in eq.
(9).
Shifting the first term in the summation to
the right,
4
S 3   V3 Vk G3 k  jB3 k e j ( 3  k )
k 1
(20)
Now apply Euler to the exponential in (20):
4
S3   V3 Vk G3 k  jB3 k cos( 3   k )  j sin( 3   k )
k 1
(21)
Now perform the complex multiplication
indicated in eq. (21):
8
4
S3   V3 Vk G3 k cos( 3   k )  B3 k sin( 3   k )
k 1
 j G3 k sin( 3   k )  B3 k cos( 3   k )
(22)
Again recalling that
S3  P3  jQ3
(16)
it must be the case that
4
P3   V3 Vk G3 k cos( 3   k )  B3 k sin( 3   k )
k 1
(23a)
4
Q3   V3 Vk G3 k sin( 3   k )  B3 k cos( 3   k )
k 1
(23b)
The above development could be done for
any bus i, and for any network size n.
Therefore,
n
Pi   Vi Vk Gik cos( i   k )  Bik sin( i   k )
k 1
(24a)
n
Qi   Vi Vk Gik sin( i   k )  Bik cos( i   k )
k 1
(24b)
9
The text does one more thing to eqs. (24a)
and (24b). It defines the angular difference
in those equations as θik, i.e., θik=θi-θk. With
this change, eqs. (24a) and (24b) become:
n
Pi   Vi Vk Gik cos  ik  Bik sin  ik 
(25a)
Qi   Vi Vk Gik sin  ik  Bik cos  ik 
(25b)
k 1
n
k 1
Equations (25a) and (25b) are the same as
eqs. (10.5) in the text.
You should feel comfortable working with
either the angular notation of eqs. (24a) and
(24b) or the angular notation of eqs. (25a)
and (25b).
Example: Consider the test system of
Example 10.8 in the text. It is redrawn
below in Fig. 1. All circuit admittances are
1-j10 pu. Write down the power flow
equations for all 5 buses. Then show that the
solution is given by
10
 2    5 
  


10

 3  
  V4  1.0
 4    10  ,     
  
  V5  1.0
 5    15 
Pg1
Pg2=0.8830
Pg3=0.2076
|V2| =1
|V3| =1
V1=1∟0°
SD3=0.2+j0.1
V5
V4
SD3=1.7137+j0.5983
SD3=1.7355+j0.5496
Fig. 1
First we need to obtain the Y-bus. It is:
0
 1  j10
0
 2  j 20  1  j10

  1  j10 3  j 30  1  j10  1  j10

0


Y 
0
 1  j10 2  j 20
0
 1  j10



1

j
10

1

j
10

1

j
10
3

j
30

1

j
10



0
0
 1  j10  1  j10 2  j 20 
11
Writing the power flow equations for bus 1:
5
P1   V1 Vk G1k cos(1   k )  B1k sin(1   k )
k 1
5
Q1   V1 Vk G1k sin(1   k )  B1k cos(1   k )
k 1
Recognizing that θ1=0, the real power is:
5
P1   V1 Vk G1k cos( k )  B1k sin(  k )
k 1
 V1 V1 G11 cos(1 )  B11 sin( 1 )
 V1 V2 G12 cos( 2 )  B12 sin(  2 )
 V1 V3 G13 cos( 3 )  B13 sin(  3 )
 V1 V4 G14 cos( 4 )  B14 sin(  4 )
 V1 V5 G15 cos( 5 )  B12 sin(  5 )
Filling in the values from the given solution,
(note all voltage magnitudes are 1.0) we get:
P1  2  1 cos(5)  10 sin( 5)  1 cos(10)  10 sin(10)
 2.6270
12
Likewise, for the reactive power,
5
Q1   V1 Vk G1k sin(  k )  B1k cos( k )
k 1
 V1 V1 G11 sin( 1 )  B11 cos(1 )
 V1 V2 G12 sin(  2 )  B12 cos( 2 )
 V1 V3 G13 sin(  3 )  B13 cos( 3 )
 V1 V4 G14 sin(  4 )  B14 cos( 4 )
 V1 V5 G15 sin(  5 )  B15 cos( 5 )
Filling in the values from the given solution,
(note all voltage magnitudes are 1.0) we get:
Q1  20  1 sin( 5)  10 cos(5)  1 sin(10)  10 cos(10)
 0.0708
Now does this show that the solution is
right? NO!!!
Why not?
Because we have nothing to which we might
compare these values of P1 and Q1. These
values were not specified in the input data,
and so we cannot compare our calculated
values to anything.
13
The reason for this is that this bus is the
swing bus. Remember, we do NOT specify
the swing bus power injection (real or
reactive). The swing bus real and reactive
power injection is obtained only as a
function of the solution to the power flow
problem (the “solution” to the power flow
problem is all voltage magnitudes and
angles at all buses) and therefore offers us
no basis to check the solution.
So let’s compute Pi and Qi for another bus.
The text (page 358) computes for bus 4.
That is a good choice because we know both
P4 and Q4. Look at Fig. 1, and tell me what
they are. So it is easy to compare the
computed values of P4 and Q4 to the
specified values of P4 and Q4. Doing so
validates that the “solution” (the bus voltage
magnitudes and angles) is correct.
14
Bus 5 would be another good choice to
make, because, like bus 4, it has P4 and Q4
specified.
But let’s live our lives on the edge. Let’s
select bus 3 instead.
Writing the power flow equations for bus 3:
5
P3   V3 Vk G3 k cos( 3   k )  B3 k sin( 3   k )
k 1
5
Q3   V3 Vk G3 k sin( 3   k )  B3 k cos( 3   k )
k 1
With θ3=-10, the real power is:
5
P3   V3 Vk G3 k cos(10   k )  B3 k sin( 10   k )
k 1
 V3 V1 G31 cos(10  1 )  B31 sin( 10  1 )
 V3 V2 G32 cos(10   2 )  B32 sin( 10   2 )
 V3 V3 G33 cos(10   3 )  B33 sin( 10   3 )
 V3 V4 G34 cos(10   4 )  B34 sin( 10   4 )
 V3 V5 G35 cos(10   5 )  B32 sin( 10   5 )
 1 cos(5)  10 sin( 5)  2  1 cos(5)  10 sin( 5)
 0.0076
15
Likewise, for the reactive power,
5
Q3   V3 Vk G3 k sin( 10   k )  B3 k cos(10   k )
k 1
 V3 V1 G31 sin( 10  1 )  B11 cos(10  1 )
 V3 V2 G32 sin( 10   2 )  B32 cos(10   2 )
 V3 V3 G33 sin( 10   3 )  B33 cos(10   3 )
 V3 V4 G34 sin( 10   4 )  B34 cos(10   4 )
 V3 V5 G35 sin( 10   5 )  B35 cos(10   5 )
 1 sin( 5)  10 cos(5)  20  1 sin( 5)  10 cos(5)
 0.0761
Now do these bus 3 calculations show that
the solution is right? Yes, and no.
Why do I say “yes and no”?
The P3 calculation is 0.0076. Referring back
to Fig. 1, we see that Pg3=0.2076 and
PD3=0.2, therefore
P3=PG3-PD3=0.2076-0.2=0.0076.
So the computed P3 agreed with the
specified P3. Solution is right.
16
But what about Q3. Referring back to Fig. 1,
we see that QD3=0.1, but QG3 is not
specified. Therefore we do not have a
“specified” Q3. So we cannot compare our
computed Q3 to anything. Again, we are in
the same situation that we were in for the
swing bus, i.e., Q3 is something that is
obtained only as a function of the solution
and therefore offers us no basis to check the
solution.
[1], J. Grainger and W. Stevenson, “Power system analysis,”
McGraw-Hill, 1994.
[2] O. Elgerd, “Electric Energy Systems Theory,” McGraw-Hill,
1982.
17