Lec 43: Three-Phase Load Flow (Classification of Known Quantities)

Hello, welcome to this lecture on Modeling Analysis and Estimation of Three-Phase Unbalanced Power Network. So, in the last lecture we have been talking about the unknown and the known quantities. Unknown quantities we said that basically first of all these are the voltage magnitudes at all original buses corresponding to all three phases. So, if I have got n bus system, so then I have got altogether total 3n unknown quantities. So, and these unknown quantities we are trying to put it in a vector V1a, V1b, V1c up to Vna, Vnb, Vnc. And then we also have the unknown angles except the angle at bus 1 at phase a, which is taken to be the reference angle except this angle all the other angles are unknown. So, then therefore total number of angles which are unknown is 3n minus 1. And these angles also we have put into one vector and that vector is theta 1b, theta 1c and then dot, dot, dot, theta na, theta nb, theta nc. And then apart from these two, we also have our voltage magnitudes at the internal buses. So, then therefore if there are m number of generators in the system, so then there are also m number of internal buses in our formulation, in the formulation. And as we are assuming that our internal buses are actually balanced buses, so then therefore at each and every of this internal bus the voltage magnitudes at all phases are same and the angles are only differing by an angle 120 degree. So, then therefore theta b will be equal to theta a minus 120 degree and theta c will be theta a plus 120 degree. So, then therefore at each and every internal bus, we have to only know the voltage magnitude of phase a as well as the voltage angle of phase a because if we know this voltage magnitude and voltage angle of phase a, we can immediately calculate the voltage magnitude and voltage angle at phase b and phase c as well. So, then therefore going by this logic, we have got a total m number of voltage magnitude unknowns and what we are doing? We have said that if there is a generator bus which is connected at original bus number i, so in that case the corresponding internal bus would be numbered as n plus i and that I mean that is actually the convention we have been following. So, then therefore we said that my voltage magnitudes are nothing but Vn plus 1, Vn plus 1 to Vn plus 3 a dot dot dot Vn plus ma and then similarly we also have unknown quantities are all the voltage angles corresponding to phase a for all the internal buses. So, then therefore these are theta n plus 1 a, theta n plus 2 a dot dot dot theta n plus ma. So, with all these, so we said we have counted that our total number of unknown quantities are 6n plus 2m minus 1. So, then therefore we need 6n plus 2m minus 1 known quantities also and after that we have talked about this known quantities. We first said that for every original bus, we know the quantities P1, Pia, Pib, Pic, Qia, Qib, Qic, these are all known and we also said that if there is actually any load at this any buses. So, then therefore this injected buses injected powers will be nothing but the negative of the load powers and if there is at all nothing connected at these buses. So, then therefore this injected power will be 0, but then again 0 is also a known value and so then therefore total number of known quantities here we are essentially getting it as a 6n and then we also know the generated power at each of these m minus 1 generators. Because as we already said that if I have got total m number of generators, then of course we really cannot pre-specify the real power generated by each of these generator. We have to keep at least I mean we have to keep the real power generated by one generator to be free, so that it can act as a slack bus. So, then therefore for the remaining m minus 1 generators, we will be able to pre-specify the total real power generation. So, then therefore we will know that, so then therefore from here we know total m minus 1 number of real power generation. And then we also know what is the pre-specified voltage at each of this generator terminal. So, then therefore if I have got total m number of generators and so then therefore these m number of generators would be able to control the voltage magnitude at its terminal. So, then therefore we also know that what is the pre-specified voltage at the terminal of each of these generators. So, then therefore from there also we would be knowing that what are the, so then therefore from this also we will be knowing the total m number of known quantities. So, then again here total number of known quantities are also 6n plus 2m minus 1. So, as a result, so the total number of known quantities is exactly equal to the total number of unknown quantities. So, then therefore the system will be solvable. Now, as we have classified the unknown quantities, let us first classify the known quantities. So, the classification, so our discussion is now classification known quantities. So, this classification of known quantities as we have done. So, all three phases. So, we again term it as pre-specified vector and this vector is nothing but would be P1a, P1b, P1c and P2a and then so on and so forth dot dot dot Pna, Pnb, Pnc transpose. So, here it would be 3n cross 1 vector, because for each of this you have got 3n cross 1 vector. So, for each bus there are three quantities Pia, Pib, Pic. So, for n buses there are 3n quantities, so I have got 3n cross 1. Then next classification is injected reactive power at all original buses corresponding to all three phases. So, we again denote it as Q sp, Q specified vector. So, it is again Q1a, Q1b, Q1c, Q2a dot dot dot Qna, Qnb transpose 3n cross 1. So, this also would be 3n cross 1. Third classification is as we said this is generated power by M-1 generators. So, we denote it as a P gen vector and this we say that it is now here we assume that my generator at bus 1. Now, here we have got one convention, again we are taking one convention that all the generators that is all M generators are connected at first M buses of the system. So, then therefore if there are let us say 5 generators in the system, so then therefore these 5 generators are connected at bus 1, bus 2, bus 3, bus 4 and bus 5. And out of these first M buses we are also assuming that the generator at bus 1 is acting as a slack bus. So, then therefore we really cannot pre-specify the power of the generator at bus 1. So, we can only specify the power of generator 2, generator 3 up to generator M. So, then therefore we say that my generators are let us say we say that P2 gen, P3 gen dot dot dot P M gen. So, this is total is M-1 cross 1. So, this is also vector. Now, here our convention Pig that is we say that the total real power generation, total we should not use this convention, it will cause confusion, it is better. Pg2, Pg3, PgM. So, my Gi is the total real power generation And then specified voltages, then the last classification is pre-specified voltage magnitude at every generator terminal. So, we denote it as say V specified and we say that this is V1 specified, V2 specified dot dot dot V M specified transpose. So, it will be an M cross 1 vector. What is Vi specified? It is the specified voltage magnitude at the terminal of generator connected at bus i, where i is 1. So, we have got specified voltage magnitude for all the generators, but we have only pre-specified generator power, not for all generators, for all the generators except the generator which is treated as a slack bus. So, I mean these are the, so then therefore, I mean this is actually the classification of the known quantities. So, we have got the details of all unknown quantities as well as the known quantities. So, now our task is to find out the mathematical relation between these known quantities and the unknown quantities and essentially these mathematical relations will give us the required equations and we have to solve those equations, calculate the unknown quantities. So, our next topic is the equations. It is basically the equations relating known quantities to these are the equations. So, now to do that first we let us understand that we will again recapitulate the expression for Iia. What is Iia? This is an complex quantity Iia and this Iia is nothing but we have already seen or rather we have not seen Iia, we should have said Iip. So, this Iip is l is equal to a b c and j is equal to 1 to n plus m. I mean why it should be n plus m that we have already discussed. So, then therefore, we know that it is b ip v j l y i j p l cosine theta ip minus theta j l minus alpha i j p l. So, these we have already derived several already derived. So, we are not really explaining. We have already derived these expressions of p ip as well as q ip. So, let us use that. So, this is p ip and q ip. These we have already derived and this is v ip v j l y i p l sin theta ip minus theta j l minus alpha i j p l. So, these power flow equations we have already derived. So, these are nothing but power flow equations. These we have already derived in some earlier lectures. So, then therefore, this equation now of course here we know that. So, for this case, so this is i is essentially 1 to n and p is a. So, then therefore, by this p ip, so that means by this, so all the injected power at all the original buses, they are being expressed in terms of the unknown quantities v ip v j l theta ip and theta j l. Similarly, by this equation of p ip, all the known injected reactive power at all the original buses, they are also being expressed in terms of the voltage magnitudes as well as the voltage angles at all the buses. Now, here we need to understand one thing. Here in this expression, j is actually going from n to n, j is actually going from n to n. So, then therefore, if I have a bus, let us say n plus ith bus, n plus ith bus or i is 1. So, this n plus ith bus, it is actually the internal bus of the generator connected at bus i because it is an internal generator. So, then therefore, v n plus i a is equal to v n plus i b is equal to v n plus i c and we have to know that theta i n plus i b is equal to theta n plus i a minus 120 degree and theta n plus i c is equal to theta n plus i a plus 120 degree. So, then therefore, when j is varying from n plus 1 to n plus m, we have to replace this v j b and v j c by these expressions. So, then therefore, we should say that here that v j b is equal to v j a and similarly v j c is also equal to v j a, where for j is n plus m to n plus 1, right. So, this we must take care of and similarly theta j b is equal to theta j a minus 120 degree. So, these relations we must use and theta j c is equal to theta j a plus 120 degree again for j n plus n and n plus 1, right. We said that our internal buses is numbered as n plus 1, n plus 2 up to n plus n. So, when j is going from n plus m to n plus m, right, so these relations I must use, I must remember this. So, in my equations we also have to use, well. So, these are the equations for injected real and reactive power. So, these cover the first two categories. So, first two known classes of known quantities are covered, correct. So, now we have to talk about the equations corresponding to total real power generation by the generators as well as the pre-specified voltage magnitudes at every generator terminal, which we will do in the next lecture. Thank you.