1) The document discusses power flow analysis of a three bus power system using circuit theory concepts. It introduces notation for generation (q) and load (x) at each bus and derives an equation relating power flow between two buses to the generation and loads.
2) It models the system as an electric circuit with current sources representing generation (q) and loads (x) at each node. Through techniques like current division and superposition, the document derives the same power flow equation relating the flow between two buses to the differences in generation and loads at those buses.
3) It introduces the concept of "Ohm's law for real power flow" which relates the real power flow on a line to the difference in voltage
1. 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:
( ) ( ) 16
3
1
3
1
≤−−−= BBAAAB xqxqF
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.
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
2. 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
( )AA xq −=
3
1
• 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
( )BB xq −=
3
1
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?
2.0 Conventions and nomenclature
2
3. 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.
F13
x2x1
F12
jx13
jx12
jx23
F23
x3
V2V1
V3
q2
q1
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.
We do note, however, that use of xk for load presents some
problem, because x is universally used to denote line
3
4. 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.
F12
x2x1
jx12
V2V1
q2
q1
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:
( )φcos12112 IVF = (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:
12112 IVF ≈ (2)
A basic fact of power systems is that the voltages usually do not
deviate significantly from their nominal value. Under a system of
4
5. 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:
1212 IF = (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).
Noting that Fig. 2 shows 3 different buses, the circuit drawing will
need the same number of nodes with an additional one for ground.
5
6. 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.
F23
F12
F12
x1q1 x2q2
x3
q3
jx12 jx23
jx13
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
7. F23
F13
F12
P1 P2 P3
jx12 jx23
jx13
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.,
0321 =++ PPP (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:
)( 213 PPP +−= (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
8. F13
F23F12
P1 P2
P3
jx12 jx23
jx13
Node 3
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:
( ) ( )BBAAAB xqxqF −−−=
3
1
3
1
(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
9. 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
F23F12
P1
P3
jx12 jx23
jx13
Fig. 7
Using current division, it is immediately apparent from Fig. 7 that:
++
==
132312
13
1
)1(
23
)1(
12
jxjxjx
jx
PFF (6)
++
+
=
132312
2312
1
)1(
13
jxjxjx
jxjx
PF (7)
where we have used notation )(i
jkF 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
10. F13
F23F12
P2
P3
jx12 jx23
jx13
Fig. 8
Again, using current division, we see that:
++
==−
132312
23
2
)2(
13
)2(
12
jxjxjx
jx
PFF (8)
++
+
=
132312
1312
2
)2(
23
jxjxjx
jxjx
PF (9)
The total flows will then be the sum of the individual flows from
each source. Therefore (and canceling the j’s):
++
−
++
=+=
132312
23
2
132312
13
1
)2(
12
)1(
1212
xxx
x
P
xxx
x
PFFF (10)
++
+
+
++
=+=
132312
1312
2
132312
13
1
)2(
23
)1(
2323
xxx
xx
P
xxx
x
PFFF (11)
++
+
++
+
=+=
132312
23
2
132312
2312
1
)2(
13
)1(
1313
xxx
x
P
xxx
xx
PFFF (12)
When x12=x13=x23=1, then these equations are:
−
=
3
1
3
1
2112 PPF (13)
+
=
3
2
3
1
2123 PPF (14)
+
=
3
1
3
2
2113 PPF (15)
We see that eq. (13) is the same as eq. (1)
( ) ( )BBAAAB xqxqF −−−=
3
1
3
1
(1)
10
11. 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:
F12
x2x1
jx12
V2V1
q2
q1
Fig. 3
Another fundamental relation for power flow F12 (learned
by EE students in EE 303) is:
( )21
12
21
12 sin θθ −=
x
VV
F (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.
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.
11
12. 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,
( ) ( )2121sin θθθθ −≈− (17)
Applying eq. (17) to eq. (16), and using |V1|=|V2|≈1.0, we get:
12
21
12
x
F
θθ −
= (18)
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.
F13
F23F12
P1 P2
P3
jx12 jx23
jx13
Fig. 6
12
13. 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:
Node 1:
13
1
12
21
13121
xx
FFP
θθθ
+
−
=+= (19)
Node 2:
12
21
23
2
12232
xx
FFP
θθθ −
−=−= (20)
Collecting terms with common angles in both (19) and (20), we
get:
12
2
1312
11
111
xxx
P θθ −
+= (21)
++−=
2312
2
12
12
111
xxx
P θθ (22)
Writing eqs. (21) and (22) in matrix form, we get:
+
−
−
+
=
2
1
231212
121312
2
1
111
111
θ
θ
xxx
xxx
P
P
(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).
13
