2.12.2 Example for A.C-D.

C load flow
To illustrate the application of the above procedure, let us first consider the 5-bus system. In this
system, it is now assumed that one bipolar HVDC link is connected between bus 4 and 5 (rectifier
at bus 4 and inverter at bus 5). Other relevant data for this link are as follows; Rd = 10.0 Ω;
3 3
Nr = Ni = 2; Xcr = Xci = 6.0 Ω. Further, let us also assume that the specified values have been
π π
taken according to combination 1 and the values are as follows:

Combination-1

α = 5o , Pdr = 100 MW; γ = 18o , Vdi = 250 kV

With the above specification, the calculation procedure for the DC system is as follows. Initially,
the flat start is assumed for all the buses in the system. Therefore, ∣V4 ∣ = ∣V5 ∣ = 1.0 p.u. Let us also
assume that the base voltage of the AC system is 132 kV. Now, before commencing the AC load flow,
the equivalent power injections (both real and reactive) at buses 4 and 5 need to be calculated. For
this purpose, equations (2.95) - (2.105) are used to calculate the values of different DC variables as
follows: Vdr = 253.938 kV; id = 393.8 Amp.; Qdr = 16.276 MVAR; Pdi = 98.45 MW; Qdi = 35.024
MVAR.
Now, let us look at equations (2.95) - (2.105) more closely. Equations (2.96) - (2.99) show that
the quantities Vdr , id , Vdor and cos Φr depend only on the DC system data. As the DC system data
are constant, the calculatd values of these four quantities would also be constant (i.e. their values
would not change from iteration to iteration). Similarly, from equations (2.102) and (2.103) it can
be seen that the quantities Vdoi and cos Φi are also constant. As cos Φr and cos Φi are constant,
(R) (I)
from equations (2.101) and (2.105), QDCi and QDCj are also constant. Thus, the equivalent real
and reactive power injections at buses 4 and 5 are constant (they need not be updated at every
iteration) and hence these values can be pre-calculated and suitably adjusted into the injected real
and reactive powers at buses 4 and 5 before solving the AC system equations. Thus, for the example
at hand, the net injected real and reactive powers at bus 4 and bus 5 can be calculated as follows;
P4 = −1.15 − 1.0 = −2.15 p.u., Q4 = −0.6 − 0.16276 = −0.76276 p.u., P5 = −0.85 + 0.9845 = 0.1345
p.u. and Q5 = −0.4 − 0.35024 = −0.75024 p.u. With these net injected real and reactive powers,
the load flow solution of the AC system is computed and the final solution is shown in Table 2.36.
It is to be noted that in Table 2.36, it has been assumed that no violation of reactive power limit
has taken place for any of the generators. After the final solution of voltage magnitudes is obtained,
the quantities ar and ai can be calculated from equations (2.100) and (2.104) as ar = 0.8714 and
ai = 0.8149. Please note that in these two equations, the quantities Etr and Eti should be taken in
actual values (i.e. in kV), not in per unit.
Let us now turn our attention to combination 8. Following the same reasonong as described
above, from equations (2.107) - (2.108) it can be observed that the quantities Vdi and Id are con-
stant. Moreover, as steps (3)-(8) of combination 8 are same as in combination 1, it immediately
follows that for combination 8 also, the equivalent real and reactive power injections (representing

78
Table 2.36: Final Results of AC-DC load flow of 5 bus system without any generator Q limit violation

Without generator Q limit

Bus no. ∣V ∣ θ Pinj Qinj
(p.u) (deg) (p.u) (p.u)
1 1.0 0 0.68984 0.46301
2 1.0 -0.63995 0.5 -0.17235
3 1.0 -4.91128 1.0 1.54134
4 0.82813 -17.48682 -2.15 -0.76277
5 0.91332 -3.89028 0.13449 -0.75025
Total iteration = 5

the DC system) are constant. Therefore, by pre-calculating these equivalent power injections and
subsequently incorporating these calculated values into net bus power injections, standard AC load
flow solution can be computed to obtain the solution of the composite AC-DC system.
From the above discussion regarding combination 1 and 8, it may appear that the equivalent real
and reactive power injections (representing the DC system) are always constant for any combination
of the specified control variables. However, this is not true. Depending on the specified control
variables, the equivalent real and reactive power injections may vary from iteration to iteration and
therefore, they need to be calculated in every iteration. As an example, let us consider combination
3. For this combination, the various steps are as follows:
Step 1: Initialise all the bus voltages with flat start. Hence, Etr and Eti are known.
Step 2: From the specified values of Pdr and Vdi , calculate Vdr and Id using equations (2.96)
and (2.97) respectively.
Step 3: Calculate Vdor from equation (2.100).
Step 4: Calculate cos α and cos Φr using equations (2.98) and (2.99) respectively.
(R)
Step 5: From the knowledge of Pdr and cos Φr , calculate QDCi using equation (2.101).
Step 6: Calculate Vdoi from equation (2.104).
Step 7: Calculate cos γ and cos Φi using equations (2.102) and (2.103) respectively.
(I) (I)
Step 8: Calculate PDCj and QDCj from equation (2.105).

(R)
Please note that in steps 3-5, the quantities Vdor , cos Φr and QDCi are all dependent on the
(I)
rectifier side AC bus voltage, Etr . Similarly, in steps 6-8, the quantities Vdoi , cos Φi and QDCj are all
(I)
dependent on the inverter side AC bus voltage, Eti . The quantity PDCj however, depends only on
the DC system quantities and hence remain constant. Thus, the equivalent reactive power injections
at both rectifier and inverter side depend on the AC bus voltage magnitudes (although the equivalent
real power injections at both the sides are independent of AC bus voltage magnitudes). Hence, the
equivalent reactive power injections need to be updated at each iteration and with these updated
power injection values, another iteration of AC load flow is carried out. This process is continued till
convergence is achieved. To illustrate this procedure further, let us assume that the specified values
corresponding to combination 3 are as follows:

79
 ar = 1.0; Pdr = 100 MW; ai = 1.0; Vdi = 250 kV

From the information of Pdr and Vdi , the quantities Pdi and Id are calculated as; Pdi = 98.45
MW and Id = 393.8 Amp. The calculated values for different other DC quantities corresponding to
first 3 iterations are shown in Table 2.37. In this table, the symbols ‘In’ and ‘MM’ denote ‘iteration
number’ and ‘mismatch’ respectively. Proceeding in this fashion, the algorithm finally converges in
70 iterations with a convergence threshold value of 1.0e−12 . The final converged values of different
(R)
DC quantities are as follows: Vdr = 253.938 kV; α = 19.82o ; γ = 34.84o ; QDCi = 41.52 MVAR;
Q(I)
DCj = 72.44 MVAR. The final converged values of the AC system quantities are shown in Table
2.38. Note that, as in the case of Table 2.36, in this case also, no generator reactive power violation
has been assumed.

Table 2.37: Calculated DC quantities for first three iterations in 5 bus system

V4 V5 Vdor φr Q(R)
DCi α Vdoi φi Q(I)
DCj γ
In MM
(p.u.) (p.u.) (kV) (rad.) (MVAR) (deg.) (kV) (rad.) (MVAR) (deg.)
0 1.0 1.0 356.52 0.778 98.54 43.48 356.52 0.7937 100.09 44.4 2.15
1 0.7843 0.8731 279.63 0.432 46.11 22.33 311.29 0.638 73.04 35.08 0.5578
2 0.7682 0.8703 273.91 0.3842 40.43 19.20 310.29 0.634 72.37 34.82 0.038

Table 2.38: Final Results of AC-DC load flow of 5 bus system for combination 3 without any
generator Q limit violation

Without generator Q limit

Bus no. ∣V ∣ θ Pinj Qinj
(p.u) (deg) (p.u) (p.u)
1 1.0 0 0.75382 0.78420
2 1.0 -0.99374 0.5 -0.16893
3 1.0 -5.52481 1.0 2.17338
4 0.82813 -18.69153 -2.15 -1.01522
5 0.91332 -3.65137 0.13449 -1.12440
Total iteration = 70

For further illustration, let us now consider the 30-bus system. In this system, it is now assumed
that one bipolar HVDC link is connected between bus 9 and 28 (rectifier at bus 9 and inverter at bus
3 3
28). Other relevant data for this link are as follows; Rd = 10.0 Ω; Nr = Ni = 2; Xcr = Xci = 6.0 Ω.
π π
The load flow has been solved for combination-1, combination-3 and combination-8 (of specified quan-
tities). The specified values which have been considered are as follows;

Combination-1

80
 α = 5o ; Pdr = 100 MW; γ = 18o ; Vdi = 250 kV

Combination-3

ar = 0.75; Pdr = 100 MW; ai = 0.75; Vdi = 250 kV

Combination-8

α = 5o ; Pdi = 100 MW; γ = 18o ; Vdr = 250 kV

The results of the 30-bus system for combination 1 and 8 are shown in Table 2.39 for a tolerance of
10−12 p.u. Furthermore, the results corresponding to combination 3 are shown in Table 2.40. It is to
be noted that for these results, no reactive power limit on the generators have been considered. The
final solutions of corresponding DC system quantities are also shown in these tables for these three
cases. Comparison of Tables 2.18, 2.39 and 2.40 shows that because of the reactive power absorption
at both bus 9 and 28, the overall voltage profile of the system is lower in the presence of HVDC
link. Moreover, when the equivalent injected real and reactive powers are constant (i.e. do not vary
from iteration to iteration), the number of iterations taken by the algorithm is quiet comparable
with that taken by the normal NRLF (polar) method (without any HVDC link). However, when
these equivalent injected powers vary from iteration to iteration, the number of iterations taken by
the sequential algorithm is appreciably more as compared that taken by the normal NRLF (polar)
method (without any HVDC link).
In the above, the detail calculation procedures for three combinations (1, 3 and 8) have been
shown. For the remaining combinations, the DC quantities can be calculated following the procedure
of either combination 1 or combination 3 and thus, these are not detailed here.
Let us now turn our attention to simultaneous techniques. As discussed earlier, in the simultane-
ous technique, the AC and DC system equations are solved together. Now, in a N-bus, M-generator
power system having a HVDC link between lens ‘k’ and ‘l’ (bus ‘k’ being the rectifier and bus ‘l’ being
the inverter), the total number of unknown are (N − 1) + (N − M ) + 5 = 2N − M + 4. To solve these
unknowns, we also have (N − 1) + (N − M ) + 5 = 2N − M + 4 equations. Therefore the size of Jaco-
brian matrix would be (2N −M +4)×(2N −M +4), as compared to the (2N −M −1)×(2N −M −1)
Jacobrian matrix of the AC system.
The additional 5 rows and 5 columns pertain to the DC equations which need to be evaluated
in each iteration. Also, the Jacobian matrix also needs to be inverted in each iteration, thereby
increasing the computation burden appreciably. Apart from that, depending upon the combination
of specified quantities, the DC equations [(2.88)-(2.92)] need to be recasted appropriately before
starting the solution procedure. Therefore, the simultaneous solution technique does not give any
computational advantage vis-à-vis the sequential method and thus, this method is not further dis-
cussed here.

81
 Table 2.39: Results of the 30 bus system with a bipolar HVDC link between bus 9 and 28

With combination 1 With combination 8

Bus no. ∣V ∣ θ Pinj Qinj ∣V ∣ θ Pinj Qinj
(p.u) (deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u)
1 1.05 0 2.42039 -0.26099 1.05 0 2.42229 -0.26023
2 1.0338 -5.0387 0.3586 0.06523 1.0338 -5.04273 0.3586 0.06857
3 1.02345 -8.08233 -0.024 -0.012 1.02325 -8.08809 -0.024 -0.012
4 1.01637 -9.72937 -0.076 -0.016 1.01613 -9.73654 -0.076 -0.016
5 1.0058 -13.69708 -0.6964 0.10883 1.0058 -13.70427 -0.6964 0.11036
6 1.01055 -11.35395 0 0 1.01026 -11.35703 0 0
7 0.99437 -13.96765 -0.628 -0.109 0.99419 -13.97288 -0.628 -0.109
8 1.023 -12.18127 -0.45 0.47454 1.023 -12.18265 -0.45 0.48393
9 1.01796 -20.00269 -1 -0.16277 1.01711 -20.13217 -1.01653 -0.16832
10 1.00713 -19.52942 -0.058 -0.02 1.00633 -19.6175 -0.058 -0.02
11 1.0913 -18.07884 0.1793 0.38778 1.0913 -18.20671 0.1793 0.39224
12 1.04132 -16.16702 -0.112 -0.075 1.04109 -16.21744 -0.112 -0.075
13 1.0883 -14.97002 0.1691 0.36696 1.0883 -15.02017 0.1691 0.36879
14 1.02408 -17.34016 -0.062 -0.016 1.02379 -17.39439 -0.062 -0.016
15 1.01562 -17.64022 -0.082 -0.025 1.01524 -17.69657 -0.082 -0.025
16 1.01841 -17.83191 -0.035 -0.018 1.01792 -17.89732 -0.035 -0.018
17 1.00471 -19.23196 -0.09 -0.058 1.00399 -19.31323 -0.09 -0.058
18 0.99979 -19.02914 -0.032 -0.009 0.99925 -19.09708 -0.032 -0.009
19 0.9938 -19.66267 -0.095 -0.034 0.99317 -19.73758 -0.095 -0.034
20 0.99623 -19.69028 -0.022 -0.007 0.99555 -19.76851 -0.022 -0.007
21 0.99497 -19.72658 -0.175 -0.112 0.99419 -19.81012 -0.175 -0.112
22 0.99578 -19.62149 0 0 0.99501 -19.7033 0 0
23 1.00064 -18.30125 -0.032 -0.016 1.00016 -18.3595 -0.032 -0.016
24 0.98942 -18.8339 -0.087 -0.067 0.98881 -18.89458 -0.087 -0.067
25 0.98869 -16.70952 0 0 0.98823 -16.73336 0 0
26 0.97048 -17.15432 -0.035 -0.023 0.97001 -17.17859 -0.035 -0.023
27 0.99711 -15.12179 0 0 0.99676 -15.12284 0 0
28 1.00723 -9.39718 0.98449 -0.35025 1.00683 -9.35718 1 -0.35723
29 0.97669 -16.41847 -0.024 -0.009 0.97633 -16.42047 -0.024 -0.009
30 0.96489 -17.35084 -0.106 -0.019 0.96452 -17.35353 -0.106 -0.019
Total iteration = 4 Total iteration = 4

DC system solutions DC system solutions

Vdr = 253.938 kV; idr = 393.8 Amp.; Vdi = 245.93 kV; idr = 406.613 Amp.;
ar = 0.7088; Qdr = 16.276 MVAR; ar = 0.6988; Pdr = 101.65 MW;
ai = 0.7389; Pdi = 98.45 MW; ai = 0.7275; Qdr = 16.831 MVAR;
Qdi = 35.024 MVAR; Qdi = 35.723 MVAR;

We are now at the end of our theoretical study of various load flow techniques. However, in
production grade implementatin of these techniques, the sparsity of the linear equations (connecting
the mismatch and solution vectors) is exploited to reduce the computation time as well as mem-
ory requirement. From the next lecture, we will study some methods for solution of sparse linear

82
Table 2.40: Further results of the 30 bus system with a bipolar HVDC link between bus 9 and 28

With combination 3
Bus no. ∣V ∣ θ Pinj Qinj
(p.u) (deg) (p.u) (p.u)
1 1.05000 0.00000 2.42241 -0.25233
2 1.03380 -5.04581 0.35860 0.09175
3 1.02180 -8.07176 -0.02400 -0.01200
4 1.01437 -9.71785 -0.07600 -0.01600
5 1.00580 -13.71987 -0.69640 0.12191
6 1.00804 -11.32909 0.00000 0.00000
7 0.99286 -13.96633 -0.62800 -0.10900
8 1.02300 -12.20102 -0.45000 0.55554
9 1.00248 -20.07763 -1.00000 -0.33808
10 0.99624 -19.56171 -0.05800 -0.02000
11 1.09130 -18.12405 0.17930 0.46906
12 1.03787 -16.26936 -0.11200 -0.07500
13 1.08830 -15.06838 0.16910 0.39377
14 1.01972 -17.44774 -0.06200 -0.01600
15 1.01042 -17.72802 -0.08200 -0.02500
16 1.01181 -17.89806 -0.03500 -0.01800
17 0.99507 -19.27534 -0.09000 -0.05800
18 0.99250 -19.10546 -0.03200 -0.00900
19 0.98530 -19.73134 -0.09500 -0.03400
20 0.98713 -19.74984 -0.02200 -0.00700
21 0.98445 -19.76760 -0.17500 -0.11200
22 0.98541 -19.66188 0.00000 -0.00000
23 0.99404 -18.37832 -0.03200 -0.01600
24 0.98099 -18.89414 -0.08700 -0.06700
25 0.98192 -16.75594 0.00000 0.00000
26 0.96358 -17.20702 -0.03500 -0.02300
27 0.99144 -15.15766 -0.00000 -0.00000
28 1.00345 -9.34449 0.98449 -0.38369
29 0.97089 -16.46954 -0.02400 -0.00900
30 0.95901 -17.41321 -0.10600 -0.01900
Total iteration = 23

DC system solutions
Vdr = 253.938 kV; idr = 393.8 Amp.;
α = 15.21o ; Qdr = 33.808 MVAR;
γ = 18.31o ; Pdi = 98.45 MW;
Qdi = 38.37 MVAR;

equations.

83