EBF 483
Introduction to Electricity Markets

2.4.0 Power Flow Problems: Two-Node Networks

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We'll now start working through some problems where we use our knowledge of electric power flow and apply it to simple networks. We will walk through some theory and simple examples first, then at the bottom of the page are a couple of videos that you can watch to see additional examples. The next page of the lesson also has a self-evaluation quiz so you can go through as many examples as you would like to get practice.

In this section we will start by calculating power flows for a two-node network with parallel lines, like the one shown below.

Enter image and alt text here. No sizes!
Figure 2.14: A two node network with two parallel branches connecting nodes G and L.
Source: Seth Blumsack

In this network, R1 and R2 are the resistances of the transmission lines connecting G and L.

 
Let's suppose that W units of power are injected into the transmission network by the generator at node G. If we ignore transmission losses, we know that two things will happen:

  • The customer at node L will consume exactly W units of power, because supply and demand have to balance at all times.
  • The power will divide between the parallel lines based on the ratio of the (inverse) resistance of each line to the equivalent parallel resistance.

Flow on each of the two transmission lines can be written using the following equations:

Flow on Line 1:

F 1  = W × ( 1 R1 1 R1 + 1 R2 )

Flow on Line 2:

F 2  = W × ( 1 R2 1 R1 + 1 R2 )

We get these equations from what we know about equivalent resistances along parallel lines. The terms in parentheses are called "distribution factors," and those describe the proportion of the total power injected at node G that will flow along each of the two lines.

As an example with actual numbers, let's say that R1 = 0.01 Ω and R2 = 0.05 Ω. 100 MW is injected at node G and consumed at node L. Flows on the two lines would be:

Flow on Line 1:

F 1  = 100 × ( 1 0.01 1 0.01 + 1 0.05 )  83

Flow on Line 2:

F 2  = 100 × ( 1 0.05 1 0.01 + 1 0.05 )  17

Note that F1 + F2 = 100 MW, the total amount of power injected into the network. If you add up the flows on the lines connected to the load node in our toy power networks, you will always get the total amount of power injected into the network (and consumed at the load). Note that if you add up the two distribution factors, you will get 1. This is a good way to check your work on two-node networks. If the distribution factors do not add up to 1, and if the flows on the two lines don't add up to the total amount of power injected into the network at node G, then you have done something wrong in your calculations.

Here is an intuitive example that you can use to test yourself. If R1=R2, then (as you might expect) the flow on each line will be identical. If W = 100 MW, and if R1=R2=0.05 Ω, use the formulas for the two-node network to convince yourself that this is actually the case. If you don't get flow of 50 MW on each line, then you have made a mistake!

An important thing to realize is that only relative resistance matters. The actual resistances, in Ohms, do not matter when calculating the distribution factors or the power flows. Let's suppose that R2 = αR1 in the two-node network. We can then write our power flow equations as:

Flow on Line 1:

F 1  = W × ( 1 R1 1 R1 + 1 αR1 )       = W × ( 1 1+ 1 α )

Flow on Line 2:

F 2  = W × ( 1 αR1 1 R1 + 1 αR1 )       = W × ( 1 α 1+ 1 α )

Note that both R1 and R2 have totally disappeared, and all we are left with is α, which describes the size of R1 relative to R2.

Calculating Distribution Factors for a Two-Node Network
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So in this video we're going to walk through the steps of calculating the the distribution factors and power flow for a simple two node electricity network. So our network is going to look something like this. We're going to have a generator at one node and we're going to have a load at the other node. These two nodes are going to be connected by two lines, one of which has resistance R1 and the other one has resistance R2. Now, when the power goes from the generator to the load, it's going to divide itself among these two lines based on the relative resistances R1 versus R2. The absolute resistances don't matter, it's just the relative resistance. So to make our lives a little easier, what we're going to do is we're going to write R2 is equal to alpha times R1. Okay, so the proportion of the power flow that goes on line one and on line two, those are described by the distribution factors. So the distribution factor for this line right here with resistance R1 which we call D1 is going to be equal to the inverse of R1 (one over R1) divided by 1 over R1 plus one over R2. Okay, and then we're going to make our substitution alpha R 1 for R2. Okay, and all of the R1s are gonna cancel. Okay, and so what we're left with is that the distribution factor on this line, which is the proportion of the power that goes from G to L that flows over this line, is going to be equal to one divided by one plus one over alpha. Okay, and we can go through the same thing for D2 which is the proportion of the power that flows over this line. Okay, and what we're going to get is that D2 is going to be equal to one over alpha divided by one plus one over alpha. Alright, so now we're going to do an example and this example is going to have alpha equal to one. And alpha is equal 1, means that the two resistances are exactly the same so R1 is equal to R2. So to calculate D1, the proportion of power that flows on on this line right here, we're going to plug alpha equals one into the formula for D1. Okay, and so what we're going to get is we're going to get one over one plus one over one which is just equal to 1/2. So D2 is going to be equal to 1 over 1 divided by 1 plus 1 over 1 which is 1/2. So as a second example let's take alpha is equal to 2 and so what this means is that the resistance of R2 is twice as big as the resistance of R1. So when we calculate D1, in this case, we're going to get D1 is equal to 1 over 1 plus 1/2, okay, or D1 is equal to 2/3. And, so D2 is going to be equal to 1 over 2 divided by 1 plus 1 over 2 which is 1/3.

Power Flow for a Two-Node Network
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So to use the distribution factors to calculate power flows, the thing that we need the thing that we realize is that the distribution factors simply tell us the proportion of power that's going to flow on each of these two lines. And so the total amount that's going to flow on each of these two lines is the amount that's produced at the generator times the distribution factor on each of the two lines. So the flow on line one is going to be equal to the amount of power produced at the generator times the distribution factor on line one. Which remember is 1 over 1 plus 1 over alpha. So in this equation remember that W is the the power produced at the generator. Okay, and this term right here is just the the distribution factor for line 1. So the flow on line 2, okay, it's going to be the amount of power that's produced at the generator times the distribution factor on line 2. Right, which you remember was 1 over alpha divided by 1 plus 1 over alpha. As an example, let's take W is equal to 100 megawatts, and we're going to take alpha is going to be equal to 2. So, the flow on line 1, is going to be equal to W times 1 over 1 plus 1/2. So this is going to be equal to a hundred times, and remember our distribution factor here in this case was going to be 2/3, okay so what we're gonna get is 66 and 2/3 megawatts flowing on line 1. So for the flow on line 2, okay, that's going to be equal to W or 100, times the distribution factor for line 2, which is going to be 1/2 times 1 ,sorry, 1/2 divided by 1 plus 1/2. So this is going to be equal to 100 times 1/3 which is 33 and 1/3 megawatts.