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# ACDC LOAD FLOW For solving the load ow problem of an A

12 ACDC LOAD FLOW For solving the load 64258ow problem of an AC system in which one or more HVDC links are present either of the following two approaches are followed a Simultaneous solution technique b

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## Presentation on theme: "ACDC LOAD FLOW For solving the load ow problem of an A"— Presentation transcript:

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2.12 A.C.-D.C. LOAD FLOW For solving the load ﬂow problem of an A.C. system in which one or more HVDC links are present, either of the following two approaches are followed; a. Simultaneous solution technique b. Sequential solution technique In simultaneous solution technique, the equations pertaining to the A.C. system and the equa- tions pertaining to the DC system are solved together. In the sequential method, the AC and DC systems are solved separately and the coupling between the AC and DC system in accomplished by injecting an equivalent amount of real and

reactive power at the terminal AC buses. In other words, for an HVDC link existing between buses �i� and �j� of an AC system (rectiﬁer at bus �i and inverter at bus �j�), the eﬀect of the DC link in incorporated into the AC system by injections DCi and DCi at the rectiﬁer bus �i� and DCj and DCj at bus �j� (the super scripts �R� and �I� denote the rectiﬁer and inverter respectively). Therefore the net injected power at bus �i� and �j� are: total ACi DCi Total ACi DCi Total ACj DCj Total ACj DCj With these net injected powers the AC system is again solved and

subsequently, the equivalent in- jected powers DCi ,Q DCi ,P DCj ,Q DCi and the total injected powers Total ,Q Total ,P Total ,Q Total are updated. This process of alternately solving AC and DC system quantities is continued till the changes in AC system and DC system quantities between two consecutive iterations become less then a threshold value. Although simultaneous technique gives the solution of the system without any to and fro switching between the AC and DC systems, the sequential solution technique is actually quite easy to implement as we will see later. Now let as look at the

equations of the DC system. 2.12.1 DC system model For deriving a suitable model of a HVDC system for steady state operation, few basic assumptions are adopted as described below; a. The three A.C. voltages at the terminal bus bar are balanced and sinusoidal. b. The converter operation is perfectly balanced. c. The direct current and voltages are smooth. d. The converter transformer is lossless and the magnetizing admittance is ignored. With the above assumptions, the equivalent circuit of the converter (either rectiﬁer or inverter) is shown in Fig. 2.18 . In this ﬁgure, the

notations are as follows; 73
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Figure 2.18: Equivalent circuit of the converter under the steady state operation Magnitude and angle of the terminal bus bar of the converter Converter transformer tap ratio Magnitude and angle of the secondary side of the converter transformer Primary and secondary current of the converter transformer respectively ,I DC voltage and DC current respectively It is to be noted that in Fig. 2.18 , the angles are referred to the common reference of the entire AC-DC system. With the above notations, the basic equation governing the HVDC systems are as

follow: For rectiﬁer dr tr cos dor cos (2.88) dr dor cos cr (2.89) For inverter di ti cos doi cos (2.90) di doi cos ci (2.91) In the above equations, the subscripts �r and �i denote the rectiﬁer and inverter side respectively. The quantity �N denotes the number of six-pulse bridges at any partienlar side and the angle denotes the angular diﬀerence between the terminal voltages and primary current of the transformer, i.e. the power factor of the converter as seen by the AC bus. denotes the commutating reactance of the converter transformer and the angles and denote the

ﬁring angle of the rectiﬁer and the extinction angle of the inverter respectively. 74
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The rectiﬁe and the inverter are interconnected though the following equation: dr di (2.92) In equation (2.92), the quantity denotes the DC link resistance. Equations (2.88)-(2.92) describe the operation of a two-terminal HVDC link. Now, as the basic objective of a HVDC link is to provide complete controllability of power over a transmission corridor, both the rectiﬁes and the inverter stations are suitably controlled and thus, suitable control equations also need

to be incorporated in the above model. We will discuss these control equations shortly. However, to solve the above equations, appropriate solution variables must be chosen. Now, for the reason of simplicity, following set of solution variables is chosen for each converter; cos (2.93) Therefore, for a two terminal HVDC link, the complete set of solution vector is; dr di cos cos (2.94) In equation (2.94), has been taken only once as the DC current is same at both the ends. From equation (2.94) it is observed that there are total 9 unknown variables which need to be solved to completely

determine the HVDC link. However, we have only 5 independent equations as shown in equations (2.88)-(2.92). Therefore, out of 9 unknown variables, any 4 variables need to be speciﬁed and thereafter, remaining 5 variables can be solved using equations (2.88)-(2.92). These 4 variables can be speciﬁed using the control speciﬁcation. There can be several combina- tions of control speciﬁcation and some of their combination are; i) dr di ; ii) dr di iii) dr di ; iv) dr di v) dr ; vi) dr vii) di ; viiii) dr di With any of these four speciﬁed control values, the

remaining 5 variables can be solved from equations (2.88)-(2.92) by using standard Newton-Raphoson technique. However, for the sequential solution techniques, the quantities Dci Dci Dcj and Dcj can be competed in a much easier way by algebraic manipulation of equations (2.88)-(2.92). we will show this procedure by two of the eight combinations listed above. 75
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Combination 1 In this case, dr and di are speciﬁed. With these known quantities, the calculation procedure is as follows: Step 1: We know, dr dr . Or, dr dr dr di (from equation (2.92)). Or, dr dr di dr . Or, dr

di di dr (2.95) From equation (2.95), two values of dr are obtained. Out of these two values, the value of dr which is greater than di is chosen, i.e. dr di di dr (2.96) Step 2: is calculated as, dr dr (2.97) Step 3: Using equation (2.89), dor is calculated as, dor dr cr cos (2.98) Step 4: Using equation (2.88), and cos are calculated as, cos dr dor (2.99) dor tr (2.100) In equation (2.100) tr is known as in the sequential solution method, the terminal voltages are known from the immediate past solution of the AC system equations. Step 5: The quantities DCi and DCi are calculated as; DCi dr

and DCi dr tan (2.101) Step 6: From equation (2.91), doi is calculated as, doi di ci cos (2.102) 76
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Step 7: Using equation (2.90), and cos are calculated as, cos di doi (2.103) πV doi ti (2.104) Step 8: The quantities DCj and DCj are calculated as, DCj di and DCj DCj tan (2.105) With these values of DCi DCi DCj and DCj , the AC system equations are again solved to obtain the updated values of tr and ti and subsequently, steps (1)-(8) are repeated again to update the values of DCi DCi DCj and DCj . This alternate process of solving AC and DC system equations are repeated

till convergence in obtained. Combination 8 In this case, di and dr are known. With these known quantities, the calculation procedure is as follows: Step 1: We know di di di dr di di dr di Or, di di di dr . Or, di dr dr di (2.106) From the two values of di in equation (2.106), the ﬁnal value of di is calculated as, di dr dr di (2.107) Step 2: is calculated as, di di (2.108) With these calculated values of di and , steps (3)-(8) of combination-1 are followed to calculate the Equivalent power injection values, where DCi dr . With these injected power values, the AC and DC systems are

continued to be solved alternately till convergence in achieved. It is to be noted that at the rectiﬁer end, DCi dr and DCi dr as the rectiﬁer draws both real and reactive power from the grid. On the other hand, at the inverter end, DCj di and DCj di as the inverter supplies real power to the AC grid and draws reactive power from the AC grid. In the next lecture, we will look at an example of AC-DC load ﬂow method. 77