EBF 483
Introduction to Electricity Markets

2.5.0 Power Flow Problems: Three-Node Networks

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Now we'll move on to what seems like a much harder problem: power flow in a three-node network with multiple generators. Solving these problems requires a lot more steps, but is not fundamentally different than the two-node problems in the previous section. Following the written examples here are two videos that you can watch to see the steps that are used to calculate power flows on a three node network. On the next page of the lesson is a practice quiz that you can take as many times as you would like, to get practice calculating power flows on the three node network.

Here is our example: Generators G1 and G2 each produce 50 MW of power, which is consumed by the load L at Node 3. All transmission lines have identical resistance R as shown in the picture below.

Enter image and alt text here. No sizes!
Figure 2.15: A triangular three node network with one branch connecting each pair of nodes.
Source: Seth Blumsack

Solving a power flow problem this complicated involves four steps:

  1. Define sign conventions for each branch.
  2. Calculate flows on each branch due to the 50 MW produced at G1
  3. Do the same for G2.
  4. Use superposition to add flows on each branch.

We'll start by defining the following sign conventions, which we'll use repeatedly for these types of problems.

  • Flow from Node 1 to Node 2 is positive (so flow from 2 to 1 is negative)
  • Flow from Node 1 to Node 3 is positive (so flow from 3 to 1 is negative)
  • Flow from Node 2 to Node 3 is positive (so flow from 3 to 2 is negative)

Now we'll calculate flows on each of the three transmission lines due to the 50 MW of power injected at node 1.

We'll start by realizing that there are two paths from G1 to L, as shown in the picture below:

  • Path A: 1 -> 3 with resistance R
  • Path B: 1 -> 2 -> 3 with series resistance 2R
Enter image and alt text here. No sizes!
Figure 2.16: Illustrating the paths that power injected at Node 1 can take to get to the consumer at Node 3. Path A in the three node network uses one branch to connect Nodes 1 and 3. Path B uses two branches in series: The branch from Node 1 to Node 2 and the branch from Node 2 to Node 3.
Source: Seth Blumsack

Next we'll calculate the flows on each path (not each branch, yet - remember that path B consists of two branches) from the 50 MW of power injected at Node 1.

Flow on Path A is:

F A  = 50 × ( 1 R 1 R + 1 2R )        = 50 × ( 1 1+1/2 )       = 50 ×  2 3         = 33 1 3

Flow on Path B is:

F B  = 50 × ( 1 2R 1 R + 1 2R )        = 50 × ( 1/2 1+1/2 )       = 50 ×  1 3         = 16 2 3

Note that since Path B consists of two branches (Node 1 to Node 2 and Node 2 to Node 3) the flow across both of those branches due to the generator at Node 1 will be identical.

So, to summarize, when the generator at Node 1 injects 50 MW of power, this induces the following flows:

  • +33 1/3 MW on Line 1-3
  • + 16 2/3 MW on Line 1-2
  • +16 2/3 MW on Line 2-3

Here, I have included the positive sign explicitly to remind you that all flows are in the direction that we had defined as positive.

The next step is to calculate the flows induced by the 50 MW injected at Node 2. To start this calculation, we realize that there are two parallel paths from Node 2 to Node 3 (the Load), as shown in the figure below. Path A consists of one branch with resistance R. Path B consists of two branches in series with series resistance 2R.

Enter image and alt text here. No sizes!
Figure 2.17: Illustrating the paths that power injected at Node 2 can take to get to the consumer at Node 3. Path A in this example uses one branch to connect Nodes 2 and 3. Path B uses two branches in series: The branch from Node 2 to Node 1 and the branch from Node 1 to Node 3.
Source: Seth Blumsack

The flow on Path A, which consists of one branch, is given by:

F A  = 50 × ( 1 R 1 R + 1 2R )        = 50 × ( 1 1+1/2 )       = 50 ×  2 3         = 33 1 3

As an exercise for yourself, show that flow on Path B from the generator at Node 2 is equal to 16 2/3 MW. You can use the flow equations directly to do this, or you could realize that whatever power does not flow over Path A must flow over Path B. Thus we must have FA, G2 + FB, G2 = 50 MW.

So the magnitude of flow on Path B due to the generator at Node 2 is 16 2/3 MW, which means that there is 16 2/3 MW flowing on the line connecting Node 1 and Node 2, and 16 2/3 MW of flow on the line connection Node 1 and Node 3. We do, however, need to be very careful of the direction in this circumstance. Power injected by the generator at Node 2 and flowing along Path B must flow from Node 2 to Node 1 and then to Node 3. Remember that we had defined flow from Node 2 to Node 1 as being in the negative direction, so we would write that flow as F(2,1), G2 = -16 2/3 MW.

In summary, when the generator at Node 2 injects 50 MW of power, this induces the following flows:

  • +33 1/3 MW on Line 2-3
  • - 16 2/3 MW on Line 1-2
  • +16 2/3 MW on Line 1-3

Now we are ready for the last step - to find the total flow on each branch, we can add the flows on each branch due to each of the two generators. We can do this because of superposition, which we mentioned earlier in this lesson.

What we get in summary is:

  • Line 1-2: + 16 2/3 MW - 16 2/3 MW = 0 MW
  • Line 1-3: + 33 1/3 MW + 16 2/3 MW = 50 MW
  • Line 2-3: + 33 1/3 MW + 16 2/3 MW = 50 MW

As an exercise, repeat the whole problem but assume that each generator produces 100 MW (so demand at Node 3 is 200 MW total). You should get the following flows:

  • Line 1-2: 0 MW
  • Line 1-3: 100 MW
  • Line 2-3: 100 MW

In the second example, we doubled the generator outputs and doubled flows on all the lines (flow on Line 1-2 is still zero because 2 times zero is, well, zero). What this suggests is that we can write more general flow equations that relate line flows to individual generator outputs. For the generator at Node 1, the equations for the three-node network in the last example are:

  • f1,3(G1) = G1 × 2/3
  • f1,2(G1) = G1 × 1/3
  • f2,3(G1) = G1 × 1/3

For the generator at Node 2, the equations are:

  • f1,3(G2) = G2 × 1/3
  • f1,2(G2) = G2 × (-1/3)
  • f2,3(G2) = G2 × 2/3

The coefficients on the flow equations are know as "Power Transfer Distribution Factors," and are defined as follows:

  • The Power Transfer Distribution Factor (or just "distribution factor") of flows on line (j,k) with respect to a generation node i is defined as D( j,k; i ) =  dF j,k / dG i

The distribution factors are used very frequently in electricity markets and in the operation of electric power grids. We will use them extensively, not only to calculate power flows but also to calculate market prices for electricity later in the course.

The table below shows the distribution factors for our three-node network.

Table 2.1: Distribution factors for the three-node network
D. Factors for G1 D. Factors for G2
D(1,3;G1) = 2/3 D(1,3;G2) = 1/3
D(1,2;G1) = 1/3 D(1,2;G2) = -1/3
D(2,3;G1) = 1/3 D(2,3;G2) = 2/3

Just to make sure that you understand the distribution factors, try calculating them yourself. As an example, the distribution factor D(1,3; G1) is calculated as:

D( 1,3; G1 ) =  1 R 1 R + 1 2R  =  1 1+ 1 2  =  2 3

Similarly, where we have placed the negative sign in front to remind us that flow in this case is going in the opposite direction.

Try the other four yourself! You may notice that for each branch along a given path, the distribution factors for each of those branches with respect to the same generator will be equal in magnitude. This is why D(1,2;G1) and D(2,3;G1) are the same, and why D(1,2;G2) and D(1,3;G2) have the same magnitude.

We can use the distribution factors to write the flow equations for the three-node network as follows, for any pair of nodes j,k:

F j,k  = f j,k ( G1 )  + f j,k ( G2 )       = G1 × D( j,k; G1 ) + G2 × D( j,k; G2 )

This suggests another (faster) way to solve power flow problems on the three-node network. First, calculate the six distribution factors (three with respect to each of the two generators), and then calculate flows using the distribution factor flow equation above.

This method is summarized in the following table.

Table 2.2: Calculating power flows in the three node network
G1
Output
D. Factors
for G1
G2
Output
D Factors
for G2
Power
Flows
Lin1 (1,3) 50 MW D(1,3;G1) = 2/3 50 MW D(1,3;G2) = 1/3 (50 x 1/3) + (50 x 1/3) = 50 MW
Line (1,2) 50 MW D(1,2;G1) = 1/3 50 MW D(1,2;G2) = -1/3 (50 x 1/3) - (50 x 1/3) = 0 MW
Line (2,3) 50 MW D(2,3;G1) = 1/3 50 MW D(2,3;G2) = 2/3 (50 x 1/3) + (50 x 1/3) = 50 MW

Here is an exercise for you to try on your own. We'll modify the three-node network to assume that the resistance between node 2 and node 3 is 2R. All other resistances are the same. Each generator produces 50 MW.

First, calculate the distribution factors. Check that you get:

  • f1,3(G1) = G1 × 3/4
  • f1,2(G1) = G1 × 1/4
  • f2,3(G1) = G1 × 1/4
  • f1,3(G2) = G2 × 1/2
  • f1,2(G2) = G2 × (-1/2)
  • f2,3(G2) = G2 × 1/2

Next, check that the power flows you get are:

  • F1,3 = f1,3(G1) + f1,3(G2) = 50 × 3/4 + 50 × 1/2 = 62.5 MW
  • F2,3 = f2,3(G1) + f2,3(G2) = 50 × 1/4 + 50 × 1/2 = 37.5 MW
  • F1,2 = f1,2(G1) + f1,2(G2) = 50 × 1/4 + 50 × (-1/2) = -12.5 MW

Watch the following videos where I work through a pair of example problems.

  • Power Flow for a Three-Node Network Example #1 (14 minutes 53 seconds).
  • Power Flow for a Three-Node Network Example #2 (11 minutes 58 seconds).

Power Flow for a Three-Node Network Example #1
Click for transcript.

so in this video we're going to walk through how to calculate power flows for a three node network. So our network is going to look something like this, so we're gonna have two generators, one at node 1 and one at node 2. And then we're gonna have a load at node , and we're gonna have some lines connecting our generators and our load so that our network looks like a triangle.

So the process of figuring out power flows in this 3 node Network consists of three steps. The first step is that we have to calculate the distribution factors for each of the three lines for the power injected by G1. So we're going to calculate the following things, so we're going to calculate the distribution factor on line 1 to 3 from G1. We're going to calculate the distribution factor on line 1 to 2 from G1. And we're going to calculate the distribution factor on line 2 to 3 from G1. The second thing we're going to do is we're going to calculate the distribution factors for generator 2. Okay, so we're going to calculate the distribution factor on line 1 to 3 from G2, the distribution factor on line 1 to 2 from the power injected by G2, and the distribution factor on line 2 to 3 from G2. Okay, and the third step is we're going to use our flow equations, using the power injected by G1 and G2 and the distribution factors, to calculate the power flows.

So the first step is we're going to have to calculate the distribution factors for the power injected by G1. So the first step is to calculate the distribution factors for G1. So to do this we need a set of resistances. So for this example I'm going to assume that the resistances on all three lines are the same. So line 1 to 3 has resistance R, line 1 to 2 has resistance R and line 1 to 3 has resistance R. So if we think about how the power injected by G1 gets to the load at node 3, there are two possible paths. Some of the power injected at node 1 is going to go straight from node 1 to node 3. Okay, so this is a path with resistance R. The rest of the power injected by G1 is going to go from node 1 to node 2 and then through node 2 all the way to node 3. So the second path is going to have a total resistance of 2R (R plus R). Okay, and so we can use these resistances to calculate the distribution factors. So first we'll do the distribution factor for G1 for line 1 to 3. So we're going to have the distribution factor for line 1 to 3 from G1. Okay, and this is going to be equal to the inverse of the resistance of the path that line 1 3 is a part of. Okay, and so this is this shorter path with resistance R. So we'll take the inverse of that, divided by the sum of the inverse resistances of both paths. So we have one path that has resistance R, that is the direct path, and then we have a second path, the indirect path, which goes from node 1 to 2 and then node 2 to 3 that has total resistance 2R. Okay, so all of the R's are going to cancel here and we will get 2/3. Next we're going to calculate the distribution factor for line 1 to 2. Okay, and again the distribution factor is going to be the resistance of the path on which line 1 to 2 lies. So this path is the indirect path with resistance 2R. so we're going to have 1 over 2R in the numerator and then we're going to have the sum of the inverse resistances of both paths in the denominator. So the same thing as the denominator for the distribution factor on line 1 to 3. So we're going to have one third. And then the distribution factor on 2 to 3 from G1 is going to be exactly the same as the distribution factor for line 1 to 2 from g1 because lines 1 to 2 and 2 to 3 are part of the same path from g1 to the load. So the distribution factor on line 2 to 3 from G1 is going to be equal to 1 over 2R divided by 1 over R plus 1 over 2R and that's going to be equal to 1.

Okay, so the next step is to calculate the distribution factors for G2. so from G2 there are two paths by which the power injected at G2 can flow to the load at node 3. First is the direct path from 2 to 3 which is going to have resistance R. The second path is the indirect path where the power injected at node 2 flows through node 1 and then to node 3. So the indirect path is going to go from node 2 to node 1 all the way to node 3 and this path is going to have total resistance 2R. So the distribution factor on line 1 to 3 from G2 is going to be 1 over 2R (because that's the inverse of the resistance of the longer indirect path of which line 1 3 is apart) divided by 1 over R (the inverse of the resistance of the short path) and 1 over 2R (the resistance of the indirect or longer path) and so we are going to get 1/3. Okay, now the distribution factor for line 1 to 2, we have to be a little bit careful with this one, because remember we often define that flow from node 1 to node 2 is in the positive direction, and flow from node 2 to node 1 is in the negative direction, so the distribution factor on line 1 to 2 from G2 is actually is going to have be a negative number. So there's going to be a negative sign there. And so that's going to be equal to 1 over 2R divided by 1 over R plus 1 over 2R is equal to negative 1/3. So we see that, in the case of G2, the distribution factor on line 1 to 2 is equal in magnitude as the distribution factor on line 1 to 3, but opposite in sign because flow going from node 2 to node 1 is in what we call the negative direction. So the last one, the distribution factor on, oops, line 2 to 3 from G2, is going to be equal to the inverse resistance of the short path. So that's 1 over R divided by the same denominator as in the other distribution factors 1 over R plus 1 over 2R and so we're going to get 2/3.

So our last step is to use the distribution factors that we calculated and the assumed production at G1 and G2 to calculate power flows. So for this example, I'm going to assume that G1 produces 50 megawatts and G2 also produces 50 megawatts. So remember that the flow equation says that the the flow on line j to k is going to be equal to the distribution factor for line j k with respect to G1, times the output at G1, plus the distribution factor on line j to k, with respect to G2 times the output at G2. So we're gonna use our outputs of 50 megawatts at each generator, the distribution factors that we've already calculated, to calculate power flows on all three lines. So first we'll do line one two three. So the flow on line 1 two 3 is going to be equal to the distribution factor on line 1 3 from G1, which was 2/3, times the output of G1, which is 50, plus the distribution factor on line 1 3 from G2, which was 1/3, times the output of G2, which is 50, and so we get 50 megawatts of flow on line 1 3. On line 1 2, we're going to have the distribution factor on line 1 2 from G1, which we had calculated was 1/3, times the output of G1, which was 50, plus the distribution factor on line 1 3 from G2, which we had said was minus 1/3, remember it's minus 1/3 because the flow goes from node 2 to node 1. Okay, so 1/3 times 50, plus minus one third times 50, is equal to zero. So the flow from G1 on line 1 2 and the flow from G2 on line 1 2 exactly cancel each other out. And finally, on line 2 3, we're going to have the distribution factor on line 2 3 from G1, which was 1/3, times 50 (the output from G1) plus the distribution factor on line 2 3 from G2, which we said was 2/3, times 50. And so that's going to give us 50 megawatts. So in sum, we find that flow from node 1 to node 3 is 50 megawatts, flow from node 2 to node 3 is 50 megawatts, and flow from node 1 to node 2 is 0 megawatts.

Power Flow for a Three-Node Network Example #2
Click for transcript.

So we're going to do a second example of calculating power flow on a three node network. This network has two generators, there's a generator at node 1 a generator at node 2 and then there's a load at node 3. In this network, the line connecting node 1 to node 3 has resistance R, the line connecting node 1 to 2 has resistance R and the line connecting node 2 to node 3 has resistance 2R. So the resistance on line 2 to 3 is twice as high as the resistance on either of the other two lines.

So the first step is to calculate the distribution factors for G1. Okay, so first we're going to do the distribution factor on line 1 to 3 from G1, and so the numerator in this is going to be the inverse of the resistance of the path that goes from node 1 to node 3. Okay, so this is one path with resistance R. So that's going to go in the numerator over here, and then in the denominator, we're going to have the sum of the inverse resistances for both paths. So the resistance of the direct path from 1 to 3 is 1 over R, and the resistance of the indirect path, that goes from node 1 to node 2 down to node 3, has total resistance 3R. So the second term in the denominator of the distribution factor equations is going to be 1 over 3R and we're going to get 3/4. Now, we're going to calculate the distribution factor on line 1 2 from G1. And so the numerator is going to be the total resistance of the long path that goes from 1 to node 2 to node 3 which is 3R. So the inverse of that is 1 over 3R, and then in the denominator, we're going to have 1 over R, plus 1 over 3R, and we're gonna get 1/4 there. And last, we're going to do the distribution factor of line 2 to 3 for G1, which because line 2 to 3 is on that same indirect path as line 1 to 2, we're going to have the same distribution factor on line 2 3 for G1 as we did on line 1 2 for G1. So we're going to have 1 over 3R, divided by 1 over R, plus 1 over 3R, and we're going to get 1/4.

So the next step is to calculate the distribution factors for G2, and from G2, some of the power injected at G2 is going to go directly from node 2 to node 3. So this is the direct path which is going to have resistance to R and some of the power injected at G2 is going to go from node 2 through node 1 to node 3 so this indirect path, which is going to have total resistance 2R. Okay, so now we can calculate the distribution factors. So the distribution factor on line 1 to 3 from G2. For that, in the numerator, we're going to have the inverse of the total resistance of the path that contains line 1 3. So this is this indirect path with resistance 2R. And then in the denominator we're going to have the sum of the inverse resistances on both paths. So we have the indirect path with resistance 1 over 2R and we have the direct path which also has resistance 1 over 2R and so when we work this out we're going to get 1/2. The distribution factor on line 1 to 2 from G2. To calculate this, the first thing we need to remember is that power going from node 2 to node 1 is in what we had defined as the negative direction. So there's going to be a negative sign in front of the distribution factor, and remember that on these long indirect paths, both of the lines that make up the path are going to have the same distribution factor in magnitude. So the distribution factor for line 1 2 is going to have the same magnitude as the distribution factor for line 1 3 except there's going to be a negative sign in front of it. So to see how that works out we can write out the whole formula. So in the numerator we're going to have 1 over 2R and in the denominator we're gonna have 1 over 2R plus 1 over 2R and here we're going to get negative 1/2. So the distribution factor on line 1 2 is the same magnitude as line 1 3 just has a negative sign. And then finally, we're going to calculate the distribution factor on line 2 3 from G2. Okay, and so in the numerator here we're going to have the inverse of the resistance of the path that goes directly from 2 to 3. So that's 1 over 2R, and then in the denominator we're going to have the same thing as in the other distribution factors. And again we're gonna get 1/2.

So the last step is we're going to use the distribution factors and the assumed outputs of the two generators to calculate power flows. So in this case, we're going to assume that generator 1 produces 50 megawatts and generator 2 also produces 50 megawatts. So we're going to use our flow equation here, which is that the flow on line jk is equal to the distribution factor on line jk from G1, times the output of G1, plus the distribution factor on line jk from G2, times the output of G2. So we're going to take our outputs for G1 and G2, our distribution factors that we've already calculated, and we're going to calculate the flows for each of the 3 lines. So the flow on line 1 to 3 is going to be equal to the distribution factor on line 1/3 from generator 1, which we had said was 3/4, times 50, the output of G1, plus the distribution factor on line 1 to 3 from G2, which we had said was 1/2 times 50, and so when we add this together we're going to get 62 and 1/2 megawatts. Now for line 1 to 2, we're going to take the distribution factor on line 1 to 2 from G1, which we had said was 1/4, times 50. Then the distribution factor on line 1 to 2 from G2, which remember was equal to minus 1/2, because the flow in this case is going from node 2 to node 1, which we defined as the negative direction. So this is going to be equal to minus 1/2, times 50, and when we multiply this out, we're gonna get minus 12 and 1/2 megawatts. Okay, and remember what that minus sign means, it doesn't mean that we have negative energy. All that minus sign means is that it's telling us that 12 and 1/2 megawatts is flowing on this line but the direction is going from node 2 to node 1. So remember don't get freaked out by the minus sign that just tells us the direction that the power is flowing. So now we'll do the flow on line 2 to 3. So the distribution factor on line 2 to 3 from generator 1, we said was 1/4, times 50, plus the distribution factor on line 2 to 3 from generator 2, which we said was 1/2, times 50. And so when we multiply this out we get 37 and 1/2 megawatts.

so in summary, what we get is the total flow from node 1 to node 3 is 62 and 1/2 megawatts. The total flow from node 2 to node 3 is 37 and 1/2 megawatts. And the total flow on line 1 2 is going from node 2 to node 1, and the magnitude of that flow is 12 and 1/2 megawatts.