Lec 56: Three-Phase Load Flow (Derivation of J13, J14, J15 and J16, Algorithm and Example)

Okay, welcome to this lecture on modeling analysis and estimation of three-phase unbalanced power network. So, we are now at the last leg of deriving the Jacobian matrix. So, now what we have to now only do that we have to now basically only derive the Jacobian matrices J 13, J 14, J 15, J 16. So, these Jacobian matrices are nothing but del V specified by del theta, del V specified by del V, del V specified by del theta internal and del V specified by del V. Now, what is, now we have already looked into that what are these elements of these vectors theta V, theta internal, V internal many times. So, we are not going to repeat them but let us look at these array V specified. So, what is this array V specified? It would be basically V specified 1, V specified 2, dot dot dot V specified and V specified I, we already said that it is 1 by 3 V I A plus V plus V I C that we have already denoted somewhere. So, the essentially what is V specified I that is basically the specified voltage magnitude to be maintained at the generator terminal. Now, as we have said that in the case of an unbalanced network, the voltage magnitudes at the terminal of any particular generator is not going to be equal for all the phases and because every generator has got only one field winding. So, it is not possible to control all these three voltage magnitudes separately by using only one single field winding right. I mean by using one single field winding current, it can only control only one particular value of voltage by essentially changing the amount of field current. So, then therefore, there can be many functions of the three phase voltage magnitudes. The most common function of the three phase voltage magnitudes is the average. So, then therefore, V specified I is one third of V I A plus V I B plus V I C. So, what it means that this generator field winding or rather this particular generator excitation system is required to maintain the average voltage of all the three phase voltages average of all these three phase voltages at a particular specified value. So, I mean that is essentially it means that the field winding of any particular I H generator is actually required to maintain the average of all these three phase voltage magnitudes at any particular value. So, this is the function. So, obviously, so then therefore, for this so then therefore, I have to say that so my quantity. So, my general del V specified I by del theta I p would be the I mean general. So, then for this let me write it in a little. Let us first look at this matrix del V specified by del theta. Let us first look at this matrix. So, for this matrix general term is del V specified I by del theta J p, right, where I is equal to 1 to N, J is equal to 1, sorry, I is equal to 1 to N, J is equal to 1 to N and p is equal to A, B, C. This is wrong, this is N. Obviously, as we can see that this expression V specified I is not function of any angle. So, therefore, very happily 0 for all I, J, p, right. So, it is 0 for all I, J, p. Then let us look at the expression del V specified, it will be done very quickly, del theta internal. So, the general term del V specified I by del theta internal J, where I, J is equal to 1 to N. As you can see that this expression of V specified I is not also the function of any theta internal or rather any internal angles. So, then therefore, del V specified I by del theta internal J is also 0 for all I, J, correct. So, this is also, so this matrix is also null matrix. Then let us look at del V specified, this matrix del V specified by del V internal. The general term is del V specified I by del V internal J, I, J is equal to 1 to N. As you can see again from this expression that the expression of V specified I is not also dependent on any internal voltage, it is only dependent on the terminal voltage. So, then therefore, again very happily is equal to 0, this is also. So, out of this 4, we have already covered 3 and we have just now seen. So, this is 0, this is 0, this is 0. So, these 3 are 0, only this one is to be discussed. So, let us discuss this. What is del V specified by del V? So, the last one what is remaining is del V specified by del V. So, the general term would be, there is actually general term. General term is del V specified I by del V J P, I is equal to 1 to N, J is equal to 1 to N, P equal to A, B, C. Now, let us again write down the expression of V specified I, again write down, so that V iA plus V iB. So, if that is the case, so we can see that this expression of V specified I is only dependent on V iA, V iB and V iC. So, that means, the partial derivative of V specified I with respect to any other bus terminal voltage would be 0 because this is not, so because this is not dependent on any other V J P. So, then therefore, I can say that del V specified I del V J P is 0 for J not equal to I and for all P. If J is not equal to I, for any P it is 0, always 0 because V specified I is not the function of any V J P, correct. Only it will be V del V specified I by del V iA would be equal to 1 by third, del V specified I by del V iB would be one third and del V specified I del V is equal to. So, in the last row of the Jacobian sub matrices, last row of the Jacobian sub matrix says only these 3 terms are non-zero, all the other terms are 0. So, this completes our Jacobian matrix formulation. So, then therefore, what is the algorithm? Algorithm is as same as that. So, the main algorithm, I will not go into much detail because this algorithm is now very, anybody who is conversant with the course on computer aided power system analysis, they would know that what is the main, I mean what is the basic algorithm of any Newton Raphson method. So, essentially what are the main algorithm is that first we form the Weibull's matrix, then we initialize voltage magnitudes and angle. Now, because it is a 3 phase case, all voltage magnitudes V iP is all 1, 1 angle 0, right, for all I and all P. So, for all bus and all phase, this voltage magnitude is 1 angle 0. And for theta IA, we will take that any theta IA is equal to 0, theta IB we will take minus 2 pi by 3, theta IC is equal to 2 pi by 3. So, these are the initialized conditions, right. So, after that what we will do is that we calculate the mismatch vector m and if then we vector n and if then we calculate if max of absolute value of m, max of absolute of m is less than epsilon, that is basically m is basically vector quantity. So, it has got let us say 100 terms. So, what we do first is that we take the absolute values of all these 100 terms and then we take the, so then therefore, after we take this absolute value, so then therefore, this particular vector becomes an positive non-negative vector. And then after that we take the absolute value of all these 100 terms. And if that absolute value is less than some pre-specified tolerance, then we say that it is converge. Yes, if it is no, then go to step 5. What do we do in 5 that we calculate Jacobian matrix, calculate matrix J and then we go to step 6 is calculate vector x, right, calculate vector delta x, calculate the vector delta x or other vector x, x is the correction vector. Let me write it down x is the calculate the correction vector and then 7 is update the voltage magnitudes and angles and go to step 3. This is the standard, very standard algorithm, very standard algorithm. Please note that in this algorithm we have not considered the issue of generator reactive power limit etcetera, etcetera. So, that we can do by some more modification in this algorithm, but we are not doing this in this course. So, this is the very basic algorithm. So, this algorithm is actually known once anybody, I mean basically this algorithm is very much known to anybody who is reasonably conversant with the course of computed power system analysis. So, this is the very basic algorithm. So, now let us look at an example. Of course, in this example we cannot show the detailed calculations, but just an example. So, let us look at a 5 bus single line diagram. So, this is, so essentially there are 5 buses, bus 1, bus 2, bus 3, bus 4 and bus 5. Out of these, please note all these buses are 3 phase buses and all these lines are 3 phase lines, all these lines everything are 3 phase lines. At bus 1, bus 2 and bus 3 there are 3 generators connected. You see there are 3 generator symbols. Bus 4 and bus 5, these are PQ bus that have got some loads and I have got essentially 4 lines. One line 1 is between bus 2 and bus 5, line 2 is between bus 3 and bus 5, line 3 is between bus 4 and bus 5 and line 4 is between bus 1 and bus 4. So, these are the 4 lines and I have got 3 and I have also got 2 transformers. One transformer is between bus 1 and bus 2, say this transformer is YG, YG and another is between bus 2 and bus 3, it is also YG, YG. Just as an example we have taken these and the data are also given like this. So, then basically these are the line data between bus 1 and 2, 2 and 3, 3 and 4, 4 and 1. So, essentially between 1 and 4 that is line 4, line 3 is between 4 and 5, line 2 is between 3 and 5, line 2 is between 3 and 5 and line 1 is between 2 and 5, 2 and 5 and their length is 65, 64, 76 and 26 kilometer and the impedance of all these lines per kilometer in ohm is given by this and these values have been calculated by following the method we have already discussed in this course and all these lines also has got shunt admittance data matrix and those are given as in actually per unit beyond per unit per kilometer. So, this is also given and we have got our base MVA is 100 MVA and base KVA is 230 KVA and out of these 3 generators bus 1 we have taken a slack bus, bus 2 and bus 3 are generator bus and these 3 generators have got 0 sequence, positive sequence and negative sequence impedances which are given. So, from this 0 sequence, positive sequence and negative sequence impedances we can calculate back essentially the reactance matrix in terms of ABC matrix that we can calculate easily, we have just by simple reverse transformation very easily we can calculate. So, essentially for these 3 generators there will be 3 more internal buses. So, for generator 1 internal bus will be bus number 6, generator 2 internal bus will be bus number 7, generator 3 internal bus will be bus number 8 right and these 3 internal buses will be behind this internal impedance and these transformers between bus 1 and 2 and bus 2 and 3. So, they have got their per unit impedances given and their connections are given, we calculate their admittance matrix as compared to as given by the procedure which also we have discussed in this course. So, these are the all data which are necessary for us and these data are also sufficient. So, with these data and of course, we have got this bus data. So, this bus data I have also got. So, we said that bus 1 is basically slack bus and bus 2 is essentially basically a generator bus. So, it does not have any load there, at bus 3 although it is a generator bus, but it has got some load at its terminal, bus 1 and other generator 1 and 2 they do not have any load at their terminal, but generator 3 has got also load in both phase in all the phases at its terminal. So, it is and phase A 150 megawatt, phase B 100 megawatt, phase C 200 megawatt and 30 MVR in phase A, 30 MVR in phase B, 30 MVR in phase C and bus 4 and bus 5 these are also loads. So, 150, 250, 350 in phase A, phase B, phase C in bus 4 in megawatt and in MVR 75, 75, 75 and in bus 5 the loads are 125, 125 and 225 in phase A, B, C respectively in megawatt and the reactive power loads are all 50 MVR in all the phases. So, essentially this is the loading data, essentially the load data. So, bus data is nothing but the load data and this generator bus data is for example, the basically the specified voltage for bus 1 is 1.0 as we have said this particular specified voltage is nothing but the average of these 3 voltages, 3 phase voltages phase A, phase B, phase C. Similarly, for bus or rather generator 2 and 3 the specified voltage is 1.045 per unit and because bus 1 or rather the generator 1 is the slag bus therefore, we did not. So, then therefore, its generator reactive power cannot be pre-specified, but for generator 2 it is generating an active power of 450 megawatt and bus and generator 3 is generating an active power of 225 megawatt. So, these are the essentially specifications. With these specifications we have formed, we have followed this particular algorithm what we have just now said and here this epsilon has been taken as 10 to the power minus 12 and these are the results. So, this is phase A voltage magnitude, phase B voltage magnitude, phase C voltage magnitudes and these are the angles. So, you see all the voltage magnitudes are close to 1.0 per unit of course, in phase A little higher, phase B also it is relatively higher, but phase C it is almost equal to 1.0. So, these are highly unbalanced, in phase A it is quite high 1.08, 1.07, 08, 08, 08, phase B it is 1.0504, 048, 0504 and if we look at the angles that the angles in phase A they are closer to 0 degree, angles in phase B these are closer to minus 120 degree and angles in phase C for all the buses they are closer to plus 120 degree. But the angular differences between the phases is not equal to 120 degree for any bus, neither the voltage magnitudes is all equal for any bus. So, then therefore this system is in unbalanced mode. So, then therefore this particular system is an unbalanced system which you have already given and this is a kind of unbalance. So, I mean basically that is how the result essentially looks like. So, here we have not given the internal bus voltage angles and magnitudes because of the simple reason that these internal bus voltage and magnitude they do not exist physically, they are just our mathematical construction for the convenience of our calculation or rather for the convenience of our formulation. Physically only these voltage magnitudes and angles will exist in the system, internal bus voltage magnitudes and angles they would not be there physically, we cannot really measure them but you can calculate them but those are only in the sphere of calculation, we cannot measure them, we can only measure them. So, then therefore we have only listed these particular voltage magnitudes and angles. So, this completes this example and with this example of three-phase network load flow we also complete this long session or rather this long series of lectures on three-phase load flow analysis of the transmission system. So, from the next lecture onwards we shall be talking about the static state estimation of a three-phase network but again many of the things are already known if actually anybody is actually conversant in the course on computer aided power system analysis. So, then in that case we will not go deep into the details of the basic state estimation algorithms because we shall assume in this course that the basic state estimation algorithm is already known to the audience, we will only look into the case that how that basic state estimation algorithm can be utilized for solving the static state estimation problem for the three-phase unbalanced power network, right. So, with that we will finish this course. Thank you so much.