Three Bus Power Flow Analysis
1.0
Introduction
In discussions that you will have in the next few weeks, you will
make reference to the power system diagram of Fig. 1, where G
denotes generators and C denotes consumers.
Fig. 1
You will use the notation
 qA, qB, and qC to denote MW delivery (generation) into buses
A, B, C by the generators and
 xA, xB, and xC to denote MW consumption (demand, or load)
from buses A, B, C by the consumers.
Finally, you will write down the following equation:
FAB 
1
q A  x A   1 qB  xB   16
3
3
where the left-hand-side is the flow on the line from region A to
region B. Our goal for this presentation is to clarify for you the
source of this kind of equation and to enable you to write them
down for more general networks configurations.
1
First note that what is inside the parentheses of each term in the
above equation is the power injection at nodes A and B,
respectively. Let’s make the following assumptions:
1. The impedance of each line is j1 pu.
2. Node C has no active injection but rather is short circuited to
ground.
3. We may treat power just like we treat current.
With these assumptions, one can apply current division to the
network of Fig. 1 to obtain the flow on line A-B.

Flow from A to C along line A-B:
Path A-C Impedance
-------------------------------------------------------Node A Injection
Path A-C Impedance+Path A-B-C Impedance


1
q A  x A 
3
Flow from B to C along line A-B:
Path B-C Impedance
-------------------------------------------------------Node B Injection
Path B-C Impedance+Path B-A-C Impedance

1
qB  xB 
3
To obtain the composite flow, we subtract Flow computed in (2)
from flow computed in (1).
Question:
On what basis can we make our assumptions 2 and 3?
2
2.0
Conventions and nomenclature
We mention at the outset that there are conventions and
nomenclature used in the previous page are generally
unfamiliar to the power engineer.
Convention: The one-line diagram is similar to that drawn
in Fig. 1, but the conventional diagram uses lines for nodes,
circles for generators, and arrows for loads. Fig. 2 is the
one-line diagram that is equivalent to Fig. 1.
V1
V2
F12
q2
jx12
q1
jx13
x1
jx23
x2
F13
F23
V3
x3
q3
Fig. 2
Nomenclature: Economists like to use q for supply and x
for demand, of anything, including electric power. Power
engineers like to use P for any real power quantity: supply,
demand (load), or flow, with the only distinguishing
difference being (sometimes) the subscripts, e.g., PD1, PG1,
P12. In these notes, we will
 Use q and x for supply and demand, but we will use
numerical subscripts instead of A,B,C, e.g., qk,xk, k=1,3.
 Use Pk to denote the injection into bus, i.e., Pk=qk-xk.
 We will also use Fkj to denote the real power flow across
the circuit connecting buses k and j.
3
We do note, however, that use of xk for load presents some
problem, because x is universally used to denote line
reactance. We will deal with this by using xjk to denote
reactance of the line between buses j and k, and xk to denote
the consumption at bus k.
Finally, we will use (only briefly), Vk to denote the voltage
(a phasor) at bus k, | Vk| the magnitude, and θk the angle.
This nomenclature is used in Fig. 2.
3.0
A basic power relation
Consider a single transmission line connecting two buses,
as shown in Fig. 3.
V1
V2
F12
q2
jx12
q1
x1
x2
Fig. 3
A very basic relation for power system engineers, which
EEs learn in EE 201, expresses the real power flow across a
transmission circuit as:
F12  V1 I 12 cos 
(1)
Here, φ is the angle by which the voltage leads the current and is
called the power factor angle.
If we assume that electric loads are purely resistive, so that only
real power flows in the network, then φ≈0 (φ will not be exactly
zero because of line reactance). In this case, eq. (1) is:
F12  V1 I 12
(2)
4
A basic fact of power systems is that the voltages usually do not
deviate significantly from their nominal value. Under a system of
normalization (called per-unit), where all voltages are normalized
with respect to this nominal voltage, it will be the case that
|Vk|≈1.0. As a result, eq. (2) becomes:
F12  I 12
(3)
In other words, the numerical value of the real power flowing on
the circuit is the same as the numerical value of the current
magnitude flowing on that circuit (under the system of
normalization).
If, again, the electric load is purely resistive, then all currents will
have almost the same angle, and one can treat the current
magnitude as if it were the current phasor (and in phase with
voltages, so that if we assume any one voltage or current is at 0
degrees, then all voltage and currents will be at zero degrees).
Useful conclusion: If we assume voltage magnitudes are all unity,
and all loads are purely resistive, then whatever rules we have of
dealing with currents also work with real power flows!
4.0
Drawing a circuit from a one-line diagram
Our goal here is to draw the one-line diagram of Fig. 2 as an
electric circuit. To do that, we will model the generation q1, q2, q3,
and the loads, x1, x2, and x3, as power injections (a similar notion as
a current injection).
We will use the circuit symbol for current source, which is
, to
model these power injections (which makes sense because, as
concluded in Section 3.0, we can treat real power just as we treat
currents).
5
Noting that Fig. 2 shows 3 different buses, the circuit drawing will
need the same number of nodes with an additional one for ground.
Each generator appears as a power source from ground, like
this:
Each load appears as a power “source” into ground, like this:
The circuit drawing for the one-line diagram of Fig. 2 is shown in
Fig. 4.
F12
jx13
F12
F23
jx12
q1
x1
jx23
q2
x2
q3
x3
Fig. 4
5.0
Netting the power sources
Since the two power “sources” at each node are in parallel, we can
“net” them to get a single power source corresponding to what we
have previously defined as the power injection: Pk=qk-xk. Note
that the power injection is


Positive if generation qk is larger than load xk
Negative if generation qk is smaller than load xk.
The resulting circuit is shown in Fig. 5.
6
F13
jx13
F12
F23
jx12
P1
jx23
P2
P3
Fig. 5
6.0
Accounting for power balance
Power balance is required, meaning, for a lossless system,
the total net power injection must be zero, i.e.,
P1  P2  P3  0
(4)
This means that power injections may only be specified at 2
nodes, and then power injection at 3rd node is determined.
Let’s assume that the node where the power injection is
determined is node 3. Therefore:
P3  ( P1  P2 )
(5)
It is common in power system engineering to refer to node
3 as the “swing bus” or “slack bus.”
The proper way to model the branch which includes P3, in
order to account for eq. (5), is to make it a short circuit.
One can easily see that this is the case by writing a KCL
equation at the ground node of our circuit, as in Fig. 6.
7
F13
jx13
F12
F23
jx12
P1
Node 3
jx23
P3
P2
P1  P2  P3  0
Fig. 6
Note, however, that P3 is now a short circuit, with current
determined by the network, and not a source (with current
specified). Two comments are in order here:

Our assumption that all voltage magnitudes (including
the “ground”) are 1.0 means that the node 3 voltage
magnitude is not zero.

The fact that voltages are phasors and thus have a
magnitude and an angle, means that the voltages may
differ at the various nodes, even though their magnitudes
are the same (see Section 8 below for “Ohm’s Law for
real power flow for more on interpretation of angles).
7.0
Obtain currents in branches
We desire to obtain the branch flows to see if the above
circuit results in what was used in previous notes, which
was eq. (1), repeated here for convenience:
FAB 
1
q A  x A   1 qB  x B 
3
3
(1)
There are several ways we could analyze the circuit of Fig. 6. For
example, we can write 2 KCL equations at nodes 1 and 2 as a
8
function of voltage variables at those nodes (and we will do so
later). For now, let’s take a simpler way: superposition, where we
compute flows from each source one at a time, and then add the
results for a given circuit from each calculation.
We begin with P1, according to the circuit of Fig. 7.
F13
jx13
F12
F23
jx12
jx23
P3
P1
Fig. 7
Using current division, it is immediately apparent from Fig. 7 that:


jx13
F12(1)  F23(1)  P1 

jx

jx

jx
23
13 
 12
(6)

jx12  jx 23 
F13(1)  P1 

(7)
 jx12  jx 23  jx13 
(i )
where we have used notation F jk to denote flow from node j to
node k due to power source at node i.
Now we analyze source at node 2, P2, using the circuit of
Fig. 8.
9
F13
jx13
F12
F23
jx12
jx23
P3
P2
Fig. 8
Again, using current division, we see that:


jx 23
 F12( 2 )  F13( 2 )  P2 

 jx12  jx 23  jx13 

jx12  jx13 
F23( 2 )  P2 

 jx12  jx 23  jx13 
(8)
(9)
The total flows will then be the sum of the individual flows from
each source. Therefore (and canceling the j’s):




x13
x23
F12  F12(1)  F12( 2 )  P1 

P
 2
 (10)
x

x

x
x

x

x
23
13 
23
13 
 12
 12


 x12  x13 
x13
F23  F23(1)  F23( 2 )  P1 

P

2
 (11)
 x12  x23  x13 
 x12  x23  x13 
 x12  x23 


x23
F13  F13(1)  F13( 2 )  P1 

P

2
 (12)
x

x

x
x

x

x
12
23
13
12
23
13




When x12=x13=x23=1, then these equations are:
1
1
F12  P1    P2  
 3
 3
10
(13)
1
2
F23  P1    P2  
3
3
2
1
F13  P1    P2  
3
3
(14)
(15)
We see that eq. (13) is the same as eq. (1)
FAB 
1
q A  x A   1 qB  x B 
3
3
(1)
with P1=qA-xA and P2=qB-xB.
8.0
Ohm’s Law For Real Power Flow
Let’s return to Fig. 3, repeated here for convenience:
V1
V2
F12
q2
jx12
q1
x1
x2
Fig. 3
Another fundamental relation for power flow F12 (learned
by EE students in EE 303) is:
F12 
V1 V2
sin1   2 
x12
(16)
We will derive this equation later in the course. Here, θ1
and θ2 are the angles of the voltage phasors at buses 1 and 2
respectively.
Recall that we applied some approximations to eq. (1). One
of these was that |V1|=|V2|≈1.0 under a special system of
normalization, i.e., all voltage magnitudes are unity. We
will apply this approximation here as well.
11
We will also apply another approximation here, and that is
that θ1-θ2, the angular separation across the transmission
circuit, is relatively small. This means that the sin function
of eq. (16) has a small argument. The situation is illustrated
in Fig. 9.
Fig. 9
One observes from Fig. 9 that the vertical distance denoted
by the dark, small line segment, which is the sin of the
corresponding unit circle, is almost exactly the same as the
radial distance around the corresponding circumference of
the circle. This radial distance is exactly the denoted angle,
when we measure the angle in radians. In other words,
sin1   2   1   2 
(17)
Applying eq. (17) to eq. (16), and using |V1|=|V2|≈1.0, we get:
1   2
F12 
(18)
x12
Consider thinking of the left-hand side of eq. (18) as current
(which we have already been doing). This means that, if we think
of the angles (when measured in radians) as voltages, then eq. (18)
is “Ohm’s Law for Real Power Flow” !!!
With this, let’s go back to Fig. 6, repeated here for convenience.
12
F13
jx13
F12
F23
jx12
P1
jx23
P3
P2
P1  P2  P3  0
Fig. 6
9.0
Relation between real power injections and
angles
As promised in Section 7.0, let’s write 2 KCL equations at nodes 1
and 2 as a function of voltage variables at those nodes, using
“Ohm’s Law for Real Power Flow.” It will be:
1   2 1
P

F

F


1
12
13
Node 1:
(19)
x12
x13
 2 1   2
Node 2: P2  F23  F12  x  x
(20)
23
12
Collecting terms with common angles in both (19) and (20), we
get:
 1
1 
1
   2
P1   1 

(21)
x12
 x12 x13 
 1
1
1 

  2 

x12
x
x
23 
 12
Writing eqs. (21) and (22) in matrix form, we get:
P2  1
13
(22)
1
1 
 1

 P1   x12 x13
x12   1 
 
P     1
1
1
(23)
  2 
 2 

 x12
x12 x 23 
The matrix of eq. (23) has a special name. It is called the B’ matrix
and represents the relation of power injections to the angles of the
bus voltage phasors in our 3-bus network.
We can now specify a procedure to get the real power flows based
on eq. (23). This is called the DC Power Flow Procedure.
1. Given power injections, solve (23) for angles.
2. Use angles to compute power flows on branches (see eqs. (19)
and (20).
14