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SvM Lec Power Flow Solutions — Transcript

by Prof. Zack Subin · 5,406 words · 858 segments · language en · Watch on YouTube

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0:04Hello everyone.
0:06This is the second uh part of the
0:08lecture on power flow analysis. And here
0:12we'll get a little bit more into the
0:14meat of how do you actually perform this
0:17analysis? Uh what are the solution
0:19methods?
0:21Um we had looked at this overall picture
0:27of the power flow equations
0:31which specify real power P and reactive
0:35power Q injected at each bus throughout
0:38a network. Um and
0:43we can appreciate that these are
0:45functions of the characteristics the
0:49properties of the transmission system
0:51the transmission network which will be
0:54described in terms of G's and B's
0:57conductances or susceptances those are
0:59the real and imaginary parts of the
1:01admittance
1:03um of each branch and they are functions
1:07of the voltage magnitudes and angles at
1:12each of the buses. Uh in these
1:16equations, we use lowercase delta. We're
1:18going to shift over to using uh theta
1:21for the angle. It means exactly the same
1:23thing. Different references use uh
1:26different notations.
1:28Um so our problem can be fundamentally
1:33stated as
1:35given P and Q for each bus find V and
1:40theta voltage magnitude and angle for
1:42each bus that will be describing the
1:47state of the system. Those are the state
1:49variables
1:51um that will drive the physical power
1:55flow and define the operating state of
1:58the system at a particular time.
2:01There is a twist however which is that
2:05the scheduled P's and Q's at each bus
2:08which is all the economic transactions
2:10that we would like to um conduct at each
2:14location
2:16don't actually add up to the entire
2:21sum total real and reactive power demand
2:24for the system. Why not? Because there
2:28are losses in the system. Uh there are
2:30real and reactive line losses I squared
2:33R and I squared X where the real line
2:37losses I squared R represent actual
2:39physical energy being dissipated. The
2:41transmission lines get hot and the I^ 2
2:44X is more of an accounting difference uh
2:48between reactive power uh injected
2:52um in quotation marks by the uh all the
2:56generation resources or capacitors.
2:59um and the reactive power quote consumed
3:02by the load. There will be some reactive
3:04power quote consumed by transmission
3:07lines and transformers. Um it's actually
3:10possible that I squared X would be
3:13negative that they would in some behave
3:15more capacitively.
3:17Uh typically that's not the case. Um so
3:20I squared X is usually systemwide a
3:23positive number. But the important
3:26observation here is we don't know what
3:28the number is. So we are not in a
3:31position to write down what is the total
3:34power demand of the network at any
3:36instant because that total will depend
3:40on how the power routes itself
3:44throughout the network. How the power
3:46actually flows as a result of these
3:49different voltage magnitudes and angles.
3:53So this is like saying until we've
3:55solved the problem, we can't even really
3:58state the problem. We don't know the
4:01total power yet.
4:03Um so [snorts]
4:05that is one way to understand why this
4:08will require an iterative solution
4:11method that's essentially
4:13um a fancy refined version of trial and
4:16error.
4:19So let's with that in mind um let's talk
4:23about the variables again
4:26um there are different types of buses in
4:28power flow analysis and we generally
4:31distinguish uh generator and load buses.
4:35The load buses actually could also be
4:38generator buses but they're um really
4:41differentiated by the types of control
4:43options that we have. So for a load bus
4:47we are just given the load demand in
4:49terms of realum reactive power
4:52and from that we want to find voltage
4:55magnitude and angle and it is quite
4:58possible as an option to specify a
5:00generator bus in that way. Um it is a
5:04generator that does not regulate
5:05voltage. It is just set to
5:09inject some amount of P and Q. But as
5:12we've seen with uh synchronous
5:15generators, they're often controlled in
5:18such a way as to maintain some constant
5:20bus voltage magnitude
5:23uh by adjusting the reactive power
5:27internally with the excitation current
5:30and uh the rotor field.
5:33Um and in that case the power flow
5:37analysis we would already know what the
5:40bus voltage magnitude is at that
5:42location and we would have to do what's
5:46in a way the simpler exercise which is
5:48to find okay so now what's the reactive
5:50power that's injected in order to
5:52maintain that bus voltage magnitude. Um
5:56so in a sense having a lot of generator
5:58buses simplifies the problem a little
6:00bit. it's it's more specification uh in
6:04place. So for a generator bus that means
6:09we are given the voltage magnitude and
6:11we're given the real power
6:14uh injection. We have to find the
6:17voltage phase angle and um and the
6:20reactive power.
6:23So now
6:25um have we forgotten anything?
6:28Yes.
6:31We have that problem that we don't
6:33really know the total power
6:38demand of the system because we have not
6:41accounted for losses.
6:43Um so
6:46if we initially specify all the P's and
6:49Q's
6:51uh for the generation to exactly add up
6:55to what all the load is going to demand
6:58whether we've forecast that or we have
7:00some contract for it um or whatever
7:05we try to equate all the P's and Q's
7:09ahead of time and we can never do that
7:12exactly
7:13because we don't know how the power will
7:16actually flow. So we don't know what the
7:18losses are going to be. There is some
7:22uncertainty and for that we have to
7:24build in some slack
7:26and
7:28mathematically
7:30we delegate one location to serve as the
7:34slack bus. And that slack bus also
7:37called a swing bus is a bus where we
7:42have to find the real power that it
7:46needs to inject in order to make
7:48everything add up. So all the
7:51uncertainty from the line losses that we
7:54didn't know ahead of time before
7:56performing the power flow analysis, all
7:58that uncertainty will be shoved into
8:00that slack bus that then is going to
8:03produce exactly as much real power as it
8:06takes to make up for the difference and
8:08exactly as much reactive power as it
8:11takes to make up for all the losses.
8:16That slack bus is also the place where
8:20by convention, this is not a
8:22mathematical necessity although it would
8:24get very messy if we didn't do it this
8:26way. Uh that is where we set the voltage
8:30angle to be zero and we can just choose
8:33our coordinate system. Um so that's not
8:38really saying anything physical. It's
8:40just saying this is what we call time t
8:43equal to zero.
8:46So and the voltage magnitude um
8:50there we will also set and you know that
8:53may be exactly one per unit or it may be
8:56something different but basically we
8:58have two variables specified for every
9:01bus and we are to find through our power
9:05flow analysis
9:07uh two variables for every bus in the
9:10work.
9:16So we slightly revise the problem
9:19statement um to allow for generator or
9:23load buses. So given P and either Q or
9:27voltage magnitude for each bus except
9:29the slack bus, find the missing
9:31information uh voltage phase angle theta
9:34and either Q or V for each bus and P for
9:37the slack bus.
9:39Now we start by writing down power flow
9:42equations for each bus
9:45and uh you may not want to drill into
9:49this for the sake of some semblance of
9:53completeness. I will go through the
9:55steps of how we expand the power flow
9:58equations in the complex format here. So
10:01complex power s would be the product of
10:04voltage times current phaser at each bus
10:07where we take the complex conjugate or
10:10the negative imaginary part um for the
10:13current and that gives us um the voltage
10:17phase angle difference between uh
10:19voltage and current angle
10:22uh in the complex power. So
10:26we expand complex power into its real
10:29and imaginary part and we will expand um
10:34the voltages into phasers.
10:37Um
10:40so again here we use theta instead of
10:43delta. Don't worry about that.
10:49And then
10:52we write those down for every bus I.
10:56And for every bus I the power injection
10:59is going to be its voltage. And these
11:02are now uh complex phasers
11:06um multiplied by the admmitance Y and
11:12the voltage complex voltage at its
11:15neighboring bus K. summing up over all
11:19the K neighboring buses.
11:23So the admittance Y
11:26um
11:28is again that that's where the G's and
11:30B's came from the real and imaginary
11:32parts uh conductance and susceptance.
11:36So we can uh expand that for the E bus
11:40and we get this equation that we um had
11:45earlier. And now mind you this includes
11:50real as well as uh reactive power
11:54because s is complex.
11:57Um so here we separate them out for
11:59convenience.
12:01And um
12:04just to drive you crazy different people
12:07in the literature, different textbooks
12:09uh use different conventions of what
12:12letters they use for the counting index.
12:15Um and then there will be another
12:17counting number which is when we do
12:20trial and error and we count iterations.
12:23Um so they might use new instead of I.
12:26Um you don't have to worry about that
12:28because you don't really have to perform
12:30power flow analysis. We're just
12:32spectators here. Um
12:35so but this is cool. Uh we this is the
12:40heart of the process. We're going to
12:42start with a guess for all the voltage
12:45magnitudes and angles at each bus.
12:49We're going to compute by plugging them
12:51into the P and Q equations the the P's
12:54and Q's that would result.
12:57Then we have a thing that's called a
13:00mismatch function that looks at how
13:04wrong those P's and Q's are compared to
13:08what we wanted. Right? Remember, we
13:11started out with great economic
13:12ambitions. We're going to sell power
13:15from here to there and so much power P
13:17and Q is going to go in at one point and
13:19come out at the other point. So from our
13:22guess of V's and thetas, we're going to
13:25calculate physically what the P's and
13:27Q's are for that state and then see,
13:30well, is that what we wanted? And of
13:34course, they'll not be exactly what we
13:36wanted. And the whole heart of the
13:41clever power flow analysis technique is
13:45making a correction to the initial guess
13:49of what the v's and thetas would be. So
13:53the heart of iterative computation is
13:57learning from the error is maximizing
14:00the information that you get out of the
14:02error and then plug that into a revised
14:08guess and compute the P's and Q's again
14:12and repeat however many times you need
14:15to. And if you make clever corrections
14:17each time you'll get closer. your
14:21mismatch will shrink and you get closer
14:23to the answer that you wanted for the
14:26P's and Q's and you set some uh
14:30threshold for the error and once you get
14:34below that threshold you say okay our
14:36work here is done close enough right
14:40that's the heart of the methodology
14:44now along the way what will happen is to
14:48get a consistent solution
14:50we will have to adjust the real power at
14:54the slack bus.
14:56Um,
14:58and let me say that in practical reality
15:03more than one generator would serve to
15:06pick up the slack of uncertainty. Right?
15:10In fact, we talked about how this is
15:11done. there's there's load following,
15:14there's frequency regulation. Um, and
15:17all that of course accounts for the
15:19losses that vary in real time.
15:22So this is kind of a mathematical
15:26representation of what if only a single
15:29generator uh were to take up the slack.
15:36Um
15:39so again given the state of the system
15:43given the state variables it will be
15:45easy to compute currents and power flows
15:48along all the lines. That part isn't
15:50complicated. It's getting to the v and
15:53thetas.
15:55So
15:57in a way I could I could stop here.
16:01I want to tell you a little bit more
16:04about um
16:07how we do this. Um and the first step is
16:11going to be writing
16:13essentially Kilch's laws for every
16:16branch of the network. And that means
16:19having an admittance for every branch of
16:21the network.
16:23And that information is captured in a
16:27tidy linear algebra format in a matrix
16:29called the bus admittance matrix.
16:32There's actually a whole separate way of
16:36doing um doing this with an impedance
16:41matrix um that's more complicated and
16:45less often done.
16:47Basically, the nice thing about the bus
16:49admittance matrix is that there's zero
16:52admittance connecting many pairs of
16:54buses, right? So, it's a sparse matrix.
16:59Um, so that bus admittance matrix
17:03captures everything that's known about
17:07the physical network across which things
17:10uh connect. And from that
17:15we can given the currents calculate the
17:17voltages or given the voltages calculate
17:20the currents.
17:22Uh that will be totally deterministic
17:25based on Kchov's laws.
17:30Um or actually I should say Ohm's law
17:32it's not it's this is this that's really
17:36a statement of Ohm's law. I stand
17:38corrected.
17:40Um, but we also use Kushchoff's laws to
17:44make sure that all the currents add up
17:47at each node.
17:49Um, don't worry about the comment here
17:51about the Norton equivalent circuit. Um,
17:54that's if you had to actually create bus
17:58admittance matrices,
18:00uh, which you don't have to do. Um, so
18:05here's a little bit of a um an
18:08explanation of iterative computation.
18:10Again, you won't have to do this. Um,
18:14but I want you to understand
18:17um how it works and it's actually I
18:21think very cool. This is really the
18:24foundation of numerical analysis which
18:27is nowadays how a lot of math problems
18:32in re in the real world are answered not
18:35by solving elegantly closed form um you
18:39know solutions uh solving equations.
18:42It's just you plug in numbers and you
18:44see what you get and you make an
18:46adjustment and you repeat. And um with
18:51computers there's not much penalty for
18:54doing a lot of that and you can do it
18:56very fast.
18:58So this is just uh really about
19:00notation.
19:02Uh and this this slide repeats what I
19:05told you a minute ago um about seeing
19:09the mismatch or the error uh when you
19:13calculate your target function uh which
19:16is basically all the P's and Q's that
19:19you want and you repeat that until your
19:22error is small enough.
19:24Um it's best illustrated graphically
19:28where um in this example uh we have some
19:33function and um we want to know for what
19:38value uh of x is this function zero. So
19:43we're going to format the mismatch or
19:46the error function such that when it's
19:50zero we have our correct result. And
19:54this is the sort of example where we
19:56can't just go backwards and turn the
19:58function inside out. We have um some x
20:02and we can calculate what f ofx is. Um
20:06and then we can also look at the slope
20:09and using the slope at some point x
20:13we've evaluated we can say okay what
20:15adjustment should we make for x? In this
20:19case the slope is positive. So we should
20:22and we have a positive f ofx. So if we
20:24want f ofx to be zero, we should choose
20:26a smaller x. And depending on how
20:29steeply this function uh is rising here,
20:34we may want to make a bigger or a
20:37smaller adjustment delta x. then try
20:40evaluating the function again and
20:45um and again and again until we find the
20:49x for which the function is zero.
20:53Um so in this case the x would represent
20:57the v's and the thetas the voltage
20:59magnitudes and angles and the f ofx
21:03would be the p's and the q's the real
21:05and reactive power.
21:07So, what's going to happen is we're
21:10going to plug in an initial guess.
21:13We're going to look at the slope of the
21:16power flow equations
21:19and we're going to evaluate the P's and
21:21Q's by by just plugging into the power
21:24flow equations. At that point from that
21:27information we make an a clever educated
21:31guess for adjusting v's and thetas and
21:34repeat until
21:38um
21:39the error is small enough.
21:42Now this illustration shows that
21:47you know for a curvy function that is a
21:49nonlinear function there may be more
21:52than one place that it crosses zero. So
21:56we might have to be a little careful uh
21:58to be sure that the solution that we're
22:00getting is actually the one we wanted or
22:03that it's actually physically plausible.
22:06Um
22:09the slope
22:11the inverse uh of the slope
22:15is
22:16has a name for this analysis is called a
22:20Jacobian matrix.
22:22And why is it a matrix? Because there's
22:25a lot of value v variables. There's a
22:29lot of V's and X's. There may be 10,000
22:32V's and X's and a lot of
22:36um P's and Q's. And this Jacobian matrix
22:40J basically
22:44is something that shows us the
22:47sensitivity or the rate of change of
22:51power real and reactive power with
22:55respect to
22:57the angles and voltage magnitudes at
23:00each bus.
23:03So this Jacobian matrix um
23:07basically says how much would power
23:12injection at
23:14the E bus, in this case bus 3,
23:19change if we made a change to the
23:21voltage magnitude or angle at any one of
23:25the other buses, in this case bus two.
23:29And so there's a Jacobian matrix for the
23:33whole network.
23:35And this matrix has to be inverted
23:40because when we um
23:44you can [snorts] see here that if we're
23:46trying to use the slope to see what
23:49adjustment we make for our guess x, we
23:53have to divide
23:55by the derivative frime of x. we have to
23:58divide by the slope. So here we have the
24:02inverse of the matrix. We're dividing by
24:04the matrix.
24:06Um and that taking that inverse that's
24:10the really computational uh
24:13computationally expensive step in here.
24:16Now there's different algorithms for
24:18doing that. They differ in how easy it
24:21is to compute that
24:24the the next guess for the next
24:25iteration. how many iterations it takes
24:28to converge and how certain you are of
24:31converging at all and getting a solution
24:34out of it at all.
24:36Um,
24:38so
24:43we're this just explains the process.
24:47I'm going to skip over it. You can read
24:50about it in the book uh in more detail
24:52if you want to.
24:54Let me illustrate the simplest possible
24:57example uh of how this method would
25:00work. Um and in this simple example,
25:05we have two buses. One bus with a
25:09generator, another bus with a load.
25:14And we are told that we want exactly one
25:18per unit voltage magnitude
25:21at our slack bus, which will also have
25:24zero voltage angle. And so the question
25:27is in order to deliver 200 megawws of
25:32real power and 100 megavars of reactive
25:34power from bus one to bus 2,
25:38what do the actual voltage magnitudes
25:42and angles have to be? And in this case,
25:45we're just solving for a single voltage
25:48magnitude and angle at bus two.
25:51And even that which you know seems like
25:56a
25:58really elementary uh question is
26:01actually not trivial to calculate. If I
26:04told you what the voltages are, it would
26:07be pretty trivial once you know what
26:09you're doing um to calculate the power
26:12flow. But to go from the power flow to
26:15the voltages is actually tricky. So this
26:19example which um is discussed in the in
26:23the text
26:25um kind of walks through applying the
26:28power flow equations. Uh it walks
26:31through how you construct the bus
26:33admittance matrix which for a single
26:36branch is already a 2x two matrix.
26:41um and then walks through
26:44the iterations,
26:46how you write down the Jacobian matrix,
26:50first of all, how you write the the
26:52power flow equations,
26:54um how you take the derivatives of them,
26:57and then um how step by step you arrive
27:03at a solution. And in this case the
27:08solution turns out to be that
27:11bus 2 will have a 0855 per unit voltage.
27:17So the voltage magnitude is a little
27:19lower and it will have an angle of
27:22minus3 12° uh where we start by
27:26definition with the slack bus at 0
27:28degrees. So with the power flowing from
27:31bus one to bus two, we go to a more
27:34negative
27:35uh voltage angle
27:38and we also
27:41calculate that to make all this happen
27:44the generator has to supply
27:48168.3 megavars
27:51because even though the load is only
27:54consuming a consuming in quotation marks
27:56a 100 megavars of reactive power. This
28:00line
28:02is acting in an inductive manner. This
28:04line has a an impedance of 0.1 J meaning
28:09only an imaginary component. It has no
28:11resistance
28:13because the line has no resistance. 200
28:16megawws injected equal 200 megawws
28:20withdrawn. No I squared R losses. No
28:23resistance here. But there are I squared
28:25X losses. And so those amount to the
28:2968.3.
28:31That's the difference.
28:33And
28:35um
28:37this just shows that three iterations of
28:40this method would get you to an error
28:43that's on the order of uh one part in
28:4710,000, which is close enough. Um so the
28:51solution then comes out um to be
28:56this angle. And so here there's another
28:59little you don't really need to worry
29:01about this. Uh this angle is given in
29:03radians not degrees.
29:06Um 08554
29:08uh is the magnitude and this is how we
29:11would write it uh in the notation as an
29:13angle. Um so we have calculated with an
29:18iterative method
29:20um the
29:22complex bus voltage or the the bus
29:26voltage phaser for bus 2 that will make
29:31it so that will allow for
29:35200 megawatts and 100 megavars to be
29:38delivered at this bus. And the whole
29:41takeaway is like, wow, that was more
29:44complicated than you would have
29:46expected.
29:47And it gets worse because
29:51that is actually not the only
29:53mathematically possible solution. It
29:55turns out to be the solution that we
29:57want.
29:58But
30:02from a mathematical standpoint,
30:06there is a perfectly plausible solution
30:10that yields the same answer. 200 megawws
30:14and 100 mega bars delivered.
30:17But the bus voltage is now.26
30:20per unit or only 26% of what it was
30:23nominally supposed to be. and the
30:25voltage phase angle is almost 50 degrees
30:28behind.
30:30So physically that's not realistic.
30:32There will have been circuit breakers
30:34popping open, relays tripping open. This
30:37is not okay. Um you aren't supposed to
30:40run that lower voltage or 50 degrees of
30:44phase angle uh difference is really
30:46stretching it. So this low voltage
30:49solution is not physically plausible.
30:51And look at the reactive power
30:56loss on this transmission line. So
30:59what's happening here is we are
31:02supplying this power demand with a huge
31:06amount of current through the
31:08transmission line where actually the
31:10transmission line is the dominant
31:12reactive load. Right? The transmission
31:15line can't be a real load because it has
31:17no resistance. But the transmission line
31:20is the dominant load here. And so this
31:24example illustrates the fact that just
31:26based on the megawatt and the megavar
31:29injections,
31:31you can't actually pin down with
31:34mathematical certainty
31:36what the state variables are.
31:39Um there's generally a low voltage and a
31:44high voltage solution that's
31:45mathematically possible and almost
31:48always it's the higher one the higher
31:50voltage that's physically
31:54uh the plausible one. What's really
31:57interesting is that
32:01which solution your algorithm arrives at
32:05depends on the algorithm and it may
32:07depend on your starting guess. So this
32:10is actually a graphic that illustrates
32:14where you land
32:17whether you converge to the plausible
32:20solution which is the red one um or the
32:24implausible solution uh which is the
32:26yellow one for all the different
32:29possible starting guesses where the
32:32vertical uh axis is voltage magnitude
32:35centered on one here and the horizontal
32:38axis is the starting guess for the angle
32:41the voltage phase angle. Um and
32:46generally speaking because we would like
32:50to um maintain a pretty easy [snorts]
32:56smooth voltage profile across the
32:58network. So, we want to keep voltages
33:01close to one per unit generally. And we
33:04want phase angle differences to be
33:07pretty moderate, not too great. Uh it's
33:10typical to use as a starting guess uh a
33:15one per unit magnitude and a 0 degree
33:18angle. And that would be this starting
33:22guess uh right here.
33:25Um but now what's interesting is that
33:30there's not a simple boundary. This is
33:32actually a fractal boundary between the
33:35regions of attraction uh for different
33:38starting guesses. And so, you know, if
33:42you're into um dynamic systems,
33:45nonlinear dynamic systems,
33:48um this is a fabulous uh thing for
33:52further study because it's it's really
33:55cool and you understand sort of what
33:57structures give rise to chaotic
33:59behavior. And there's many applications
34:01of that uh in many other
34:04areas of physics and science and life in
34:08general. Um but so it's worth knowing
34:13that there is this sort of latent
34:16instability or possibility
34:19um for an AC network to just really
34:23behave out of the um out outside the
34:28bounds of um of stable and desired
34:32behavior.
34:34Um
34:40I won't go into detail about different
34:43uh iteration methods. What I I just want
34:46you to know that what I've presented is
34:48the new Newton Rafson method. That's the
34:51most common. Um you know power
34:55engineering courses will al also teach
34:58Gaus Zidal method. Um they have
35:01different pros and cons. Um [snorts] and
35:03so you should just know that there's not
35:06just one single method of doing this and
35:08different algorithms may be more uh
35:10efficient uh in certain cases than
35:14others. Um I want to show you just
35:17briefly
35:19uh the derivatives written out here of
35:22the power flow equations
35:25uh with respect to voltage magnitudes
35:27and angles. So these are the
35:29sensitivities
35:31and the point I want to make is why it
35:36is I want to justify the rule of thumb
35:39that I told you in the very beginning um
35:42that you know real power flow is mostly
35:45associated with the timing or voltage
35:47phase angle and reactive power mostly
35:50associated uh with voltage magnitude.
35:53Um, and we've justified that with some
35:57physical intuition having to do with
35:59generator controls. But we can also see
36:02it from these sensitivities or partial
36:06derivatives here. Um, and the way to see
36:10that is that if the transmission lines
36:12are mostly inductive in their behavior,
36:16in other words, the conductances, the
36:18real parts, the G's are small numbers.
36:21So these terms that are multiplied by
36:24the G's are small and kind of are
36:28irrelevant go away.
36:31Um and if we have relatively small angle
36:37voltage phase angle differences not some
36:39pathological
36:41phase angle separations the signs of
36:44those angles will also be small. So that
36:47means that these sensitivities of real
36:50power with respect to voltage magnitude
36:52and reactive power with respect to angle
36:56those derivatives will tend to almost
36:59vanish. They will not be dominant and
37:03those will not be major sensitivities.
37:07If we look at the the dependence of real
37:10power on angle and reactive power on
37:14voltage magnitude we find okay so yeah
37:18this I'm just following the animation
37:20steps here those are often negligible
37:24um
37:27once again if the transmission lines are
37:30mostly inductive reactants
37:32um the G's here are also small and the
37:36signs of these angles are small.
37:40Whereas these terms, the B's and the
37:44cosiness of the angle differences, those
37:46are really the meat in in the equations.
37:50That's that's where the the meat is
37:52here. And in fact, if the angles are
37:57not very different from each other, then
37:58the coine of the angle difference should
38:00be close to one. So from that is where
38:05we get
38:07um
38:09approximations
38:11that um
38:15these the the sensitivity of real power
38:19with respect to angle can be roughly
38:21approximated as
38:23the product of voltage magnitudes
38:25between adjacent nodes times that um
38:30uh susceptance B and then uh the
38:35reactive power flow between so it's it's
38:38I've written it here between nodes two
38:40and three um or
38:44power injection at node two with respect
38:47to the voltages at node three um and
38:52that would be approximated
38:55by the voltage um at node two and um
39:00this acceptance uh B3 of the path
39:03between them. Um so to show that another
39:08way is that in the Jacobian matrix which
39:12has four partitions for the four types
39:15of uh sensitivities
39:17they're basically two partitions that I
39:19can often ignore. And when I do that
39:23it's called decoupled power flow. In
39:26other words, I've decoupled
39:29um real power and uh and voltage angle
39:34as a pair of va v variables from the
39:37other pair of variables reactive power
39:40and voltage magnitude. So when you hear
39:43decoupled power flow, that's what
39:45they're talking about.
39:47And then finally, DC power flow. So
39:51that's a really important thing to know
39:54because when you hear that terminology,
39:56oh we ran a DC power flow analysis,
39:59that doesn't mean that they actually
40:02assumed the whole system is running on
40:04direct current. Uh what it means is that
40:08they just ignored the variable pairs of
40:12uh Q and V and turned the
40:16two-dimensional problem into a
40:17one-dimensional problem. Uh so it's just
40:21saying let's assume all the voltages are
40:24one
40:26and never mind reactive power. We're
40:29trying to get a first cut at roughly how
40:32is this power going to distribute across
40:33the network. So we're going to use we're
40:37going to write down a different matrix a
40:39B matrix. It's like the admittance
40:40matrix but with only with not resistive
40:44parts accounted for at all just the B's.
40:48um and you know invert that or divide
40:54the power injections by it and that
40:56gives you the set of thetas and this is
40:59a linear algebra problem that you can
41:01just solve exactly. It's a closed form
41:04solution. So what that does is it gives
41:08you exactly the right answer to what you
41:12know is the wrong problem because you
41:15know that what you've specified
41:17is not the situation that exists in the
41:20real world but you've made an
41:22approximation
41:24that you can sort of understand how
41:27wrong it is. Right?
41:31the voltages are not exactly one but
41:33let's suppose they are and based on that
41:38how will I change the angles if I change
41:42the PS or vice versa so that is a DC
41:47power flow analysis with a you know
41:49single variables that often is used for
41:54first cut at uh contingency analysis
41:57because if I've got to run a hundred
42:00different possible contingency
42:01scenarios,
42:03then you know it's going to be really
42:05computation intensive to do power flows
42:07for each one. So I might start with the
42:11DC power flow and that'll run very
42:14quickly and then I can set some flags
42:17and say okay if I get some results that
42:20are borderline maybe not okay I'm going
42:24to go back and do a full power flow to
42:29make sure I'm not missing maybe a
42:31violation or some uh condition that
42:35doesn't work out. Um so
42:40um
42:42this is a summary of shortcuts that I
42:46can do in Powerflow analysis. not
42:49terribly relevant for you because you
42:52won't be sitting there laboring for
42:54hours on doing power flow problems which
42:58you know honestly even for like three
43:02a three bus system doing it by hand
43:06um is very time consuming and you know
43:10for like four five six buses you already
43:13have to do some coding and computer
43:16programs to do it. Um, and it I mean
43:20it's sort of a character building
43:21exercise, but I won't put you through
43:23it. Um, we won't put you through it, but
43:27it it's really to appreciate. I mean, I
43:30hope that what you got out of this
43:31lecture is that there's a lot more to it
43:35than meets the eye. And um, so there's
43:39shortcuts. When you hear about these
43:40decoupled power flow, there's a a fast
43:44version of it um where again you just
43:48make an assumption about let's say the
43:49voltages are close to one per unit. Uh
43:53and the most important one being the DC
43:55power flow that will you know will not
43:59give a correct solution uh because it
44:02ignores reactive power flow completely.
44:05Um,
44:07that's
44:09what I wanted to share about the math
44:12and I'll see you next time.

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