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- Figure 1 shows the one-line diagram of a four-bus system. Table 1 gives the line
Figure 1: Sample system for 1Q
impedances identified by the buses on which these terminate. The shunt admittance at all the buses is assumed to be negligible.
Table 1: Line, Bus to bus R (p.u) X (p.u) 1–2 0.05 0. 1–3 0.10 0. 2–3 0.15 0. 2–4 0.10 0. 3–4 0.05 0.
(a) Find YBus, assuming that the line shown dotted is not connected. (b) What modifications need to be carried out in YBus if the line shown dotted is connected.
Table 2: Line G (p.u) B (p.u) 1–2 2 - 1–3 1 - 2–3 0.666 - 2–4 1 - 3–4 2 -
Solution: (a) From Table 1, Table 2 is obtained from which YBus for the system can be written as
YBus =
Y 11 Y 12 Y 13 Y 14 Y 21 Y 22 Y 23 Y 24 Y 31 Y 32 Y 33 Y 34 Y 41 Y 42 Y 43 Y 44
YBus =
y 13 0 −y 13 0 0 (y 23 + y 24 ) −y 23 −y 24 −y 13 −y 23 (y 31 + y 32 + y 34 ) −y 34 0 −y 24 −y 34 (y 43 + y 42 )
YBus =
1 − j 3 0 −1 + j 3 0 0 1. 666 − j 5 − 0 .666 + j 2 −1 + j 3 −1 + j 3 − 0 .666 + j 2 3. 666 − j 11 −2 + j 6 0 −1 + j 3 −2 + j 6 3 − j 9
(b) The following elements of YBus of part (i) are modified when a line is added between buses 1 and 2.
Y 11 ,new = Y 11 ,old + (2 − j6) = 3 − j 9
Y 12 ,new = Y 12 ,old − (2 − j6) = −2 + j6 = Y 21 ,new Y 22 ,new = Y 22 ,old + (2 − j6) = 3. 666 − j 11 Modified YBus is written as,
YBus =
3 − j 9 −2 + j 6 −1 + j 3 0 −2 + j 6 3. 666 − j 11 − 0 .666 + j 2 −1 + j 3 −1 + j 3 − 0 .666 + j 2 3. 666 − j 11 −2 + j 6 0 −1 + j 3 −2 + j 6 3 − j 9
- The following is the system data for a load flow solution. The line admittances are given in Table 3. The scheduled active and reactive powers are given in Table 4. Determine the voltages
Table 3: The line admittances Line Admittance 1–2 2-j8. 1–3 1-j4. 2–3 0.666-j2. 2–4 1-j4. 3–4 2-j8.
−(−2 + j8)(1 + j 0 .0)] = 0. 994119 − j 0. 029248 V (^31) ,acc = (1.0 + j 0 .0) + 1.6[0. 994119 − j 0. 029248 − 1 − j 0 .0] = 0. 99059 − j 0. 0467968
V 41 =
Y 44
[
P 4 − jQ 4 V 4 ∗
− Y 42 V 21 − Y 43 V 31
]
V 41 =
3 − j 12
[
− 0 .3 + j 0. 1 1 − j 0. 0
− (−1 + j 4 .0)(1. 01899 − j 0 .046208)
−(−2 + j8)(0. 99059 − j 0 .0467968)] = 0. 9716032 − j 0. 064684 V (^41) ,acc = (1.0 + j 0 .0) + 1.6[0. 9716032 − j 0. 064684 − 1 − j 0 .0] = 0. 954565 − j 0. 1034944
- If in above problem, bus 2 is taken as a generator bus with |V 2 |=1.04 and reactive power constraint is 0. 1 ≤ Q 2 ≤ 1. 0 Determine the voltages starting with a flat voltage profile and assuming acceleration factor as 1.0. Assume that bus 2 has generator connected on it and injects P 2 = 0.5 p.u.
Solution: Since bus 2 is taken as a generator bus Q 2 is not specified and P 2 = 0.5. To find V 21 we first find Q 2 with V 2 = 1.04 + j 0 .0 as the phase angle of the voltage is 0.0 to begin with.
P 2 − jQ 2 = V 2 ∗
q = 14Y 2 qVq = V 2 ∗ [Y 21 V 1 + Y 22 V 2 + Y 23 V 3 + Y 24 V 4 ]
Q 2 = −Imag[V 20 ∗(Y 21 V 1 + Y 22 V 2 + Y 23 V 3 + Y 24 V 4 )] Q 2 = −Imag[(1. 04 − j 0 .0)(−2 + j 8 .0)(1.06) + (3. 666 − j 14 .664)(1.04) +(− 0 .666 + j 2 .664)(1 + j 0 .0) + (−1 + j 4 .0)(1.0)] = 0. 1108 (1) Since Q 2 lies within the limits, therefore V 2 will be taken as |V 2 |psec and phase angle as in this iteration,
V 2 =
Y 22
[
P 2 − jQ 2 V 2 ∗
− Y 21 V 1 − Y 23 V 30 − Y 24 V 40
]
Bus 2 being a generator bus, P 2 and Q 2 are to be taken as positive and value of P 2 as the specified and Q 2 as the one calculated above i.e. Q 2 = 0.1108.
V 21 =
(3. 666 − j 4 .664)
[
- 5 − j 0. 1108
- 04 − j 0. 0
− (−2 + j8)(1.06) − (− 0 .666 + j 2 .664)(1.0) − (−1 + j4)(
V 21 = 1.0472846 + j 0. 0291476
δ = 1. 770 V 21 = 1. 04 ̸ 1 .77 = 1.0395985 + j 0. 02891158
V 31 =
Y 33
[
P 3 − jQ 3 V 3 ∗
− Y 31 V 1 − Y 32 V 21 − Y 34 V 40
]
V 31 =
- 666 − j 14. 664
[
− 0 .4 + j 0. 3 1 − j 0. 0
−(−1+j4)(1.06)−(− 0 .666+j 2 .664)(1.0395985+j 0 .02891158)
−(−2 + j8)(1 + j 0 .0)] V 31 = 0. 9978866 − j 0. 015607057
Similarly V 41 can be obtained and it will be found to be
V 41 = 0. 998065 − j 0. 022336
- For the sample system of Figure 2, the generator is connected at bus number 1, while the loads are at buses 2, 3 and 4. The values of real and reactive powers are listed in
Figure 2: Sample system
Table 5.
Table 5: Bus P Gi (p.u.) QGi (p.u.) P Di (p.u.) QDi (p.u.) Vi (p.u.) Type of bus 1 — — — — 1.05 ̸ 0 Slack 2 — — 0.45 0.15 — PQ 3 — — 0.51 0.25 — PQ 4 — — 0.60 0.30 — PQ
All buses other than slack are of PQ-type. Line data are given in Table 6. Find the voltages and the bus angles at the three buses using the Newton-Raphson method. [Ans: δ 2 = -0.09696 rad; δ 3 = -0.11217 rad; δ 4 = -0.06928 rad; V 2 = 0. p.u.; V 3 = 0.92451 p.u.; V 4 = 0.95715 p.u.]
The real powers (P 2 , P 3 , and P 4 ) are computed as follows:
Pi =
∑^ n
k=
ViVk[Gikcos(δi − δk) + Biksin(δi − δk)]
The above equation can be rewritten as
Pi = GiiV (^) i^2 +
∑^ n
k̸ =i
ViVk[Gikcos(δi − δk) + Biksin(δi − δk)]
P 2 =
∑^4
k=
V 2 Vk[G 2 kcos(δ 2 − δk) + B 2 ksin(δ 2 − δk)] = − 8. 620691 × 10 −^2 p.u.
P 3 =
∑^4
k=
V 3 Vk[G 3 kcos(δ 3 − δk) + B 3 ksin(δ 3 − δk)] = − 2. 384186 × 10 −^7 p.u.
P 4 =
∑^4
k=
V 4 Vk[G 4 kcos(δ 4 − δk) + B 4 ksin(δ 4 − δk)] = − 1. 999998 × 10 −^1 p.u.
Real power residuals are calculated as
∆P 2 = P 2 sp − P 2 = − 0. 3638 p.u.
∆P 3 = P 3 sp − P 3 = − 0. 51 p.u. ∆P 4 = P 4 sp − P 4 = − 0. 4 p.u.
The reactive powers (Q 2 , Q 3 , and Q 4 ) are computed as follows:
Qi =
∑^ n
k=
ViVk[Giksin(δi − δk) − Bikcos(δi − δk)]
The above equation can be rewritten as
Qi = −BiiV (^) i^2 +
∑^ n
k̸ =i
ViVk[Giksin(δi − δk) − Biksin(δi − δk)]
Q 2 =
∑^4
k=
V 2 Vk[G 2 ksin(δ 2 − δk) − B 2 kcos(δ 2 − δk)] = − 2. 15517 × 10 −^1 p.u.
Q 3 =
∑^4
k=
V 3 Vk[G 3 ksin(δ 3 − δk) − B 3 kcos(δ 3 − δk)] = − 4. 768372 × 10 −^7 p.u.
Q 4 =
∑^4
k=
V 4 Vk[G 4 ksin(δ 4 − δk) − B 4 kcos(δ 4 − δk)] = − 3. 999996 × 10 −^1 p.u.
Reactive power residuals are calculated as
∆Q 2 = Qsp 2 − Q 2 = 0. 0655 p.u.
∆Q 3 = Qsp 3 − Q 3 = − 0. 25 p.u. ∆Q 4 = Qsp 4 − Q 4 = 0. 1 p.u.
The convergence criterion is checked to stop the iterations, i.e., maximum [∆Pi (i=2,3,4) and ∆Qi (i=2,3,4)] ≤ ϵ (0.001) maximum [0.3636,0.51,0.40,0.0655,0.25,0.1] =0.51 > 0.
The convergence criteria is not satisfied, so the changes in variables at the end of the first iteration are obtained as follows:
H 22 H 23 H 24 N 22 N 23 N 24 H 32 H 33 H 34 N 32 N 33 N 34 H 42 H 43 H 44 N 42 N 43 N 44 J 22 J 23 J 24 L 22 L 23 L 24 J 32 J 33 J 34 L 32 L 33 L 34 J 42 J 43 J 44 L 42 L 43 L 44
∆δ 2 ∆δ 3 ∆δ 4 ∆V 2 V 2 ∆V 3 V 3 ∆V 3 V 3
=
∆P 2 ∆P 3 ∆P 4 ∆Q 2 ∆Q 3 ∆Q 4
Where,
H 22 =
∂P 2
∂δ 2
= −Q 2 − B 22 V 22 = 12. 0259
H 23 =
∂P 2
∂δ 3
= V 2 V 3 [G 23 sin(δ 2 − δ 3 ) − B 23 cos(δ 2 − δ 3 )] = − 7. 5
H 24 =
∂P 2
∂δ 4
= V 2 V 4 [G 24 sin(δ 2 − δ 4 ) − B 24 cos(δ 2 − δ 4 )] = 0. 0
H 32 =
∂P 3
∂δ 2
= V 3 V 2 [G 32 sin(δ 3 − δ 2 ) − B 32 cos(δ 3 − δ 2 )] = − 7. 5
H 33 =
∂P 3
∂δ 3
= −Q 3 − B 33 V 32 = 14. 1038
H 34 =
∂P 3
∂δ 4
= V 3 V 4 [G 34 sin(δ 3 − δ 4 ) − B 34 cos(δ 3 − δ 4 )] = − 6. 6038
H 42 =
∂P 4
∂δ 2
= V 4 V 2 [G 42 sin(δ 4 − δ 2 ) − B 42 cos(δ 4 − δ 2 )] = 0. 0
Using the guass elimination method, in which triangularization and back substitution
Modified values of δi and Vi are computed as
δ 2 = δ 2 + ∆δ 2 = 0. 0 − 0 .09696 = − 0. 09696 rad
δ 3 = δ 3 + ∆δ 3 = 0. 0 − 0 .11217 = − 0. 11217 rad δ 2 = δ 2 + ∆δ 2 = 0. 0 − 0 .06928 = − 0. 06928 rad
V 2 = 1. 0 −
V 3 = V 3 +
∆V 3
V 3
V 3 = 1. 0 −
× 1 .0 = 0. 92451 p.u.
V 4 = V 4 +
∆V 4
V 4
V 4 = 1. 0 −
× 1 .0 = 0. 95715 p.u.
- For the sample system of Figure 2, the generators are connected at buses 1 and 2, while the loads are at buses 3 and 4. The values of real and reactive powers are listed in Table 7 along with the type of buses.
Table 7: Bus P Gi (p.u.) QGi (p.u.) P Di (p.u.) QDi (p.u.) Vi (p.u.) Type of bus 1 — — — — 1.05 ̸ 0 Slack 2 0.45 — — — 1.00 ̸ 0 PV 3 — — 0.51 0.25 — PQ 4 — — 0.60 0.30 — PQ
Line data are given in Table 6. Find the voltages and the bus angles at the three buses using the Newton-Raphson method. [Ans: δ 2 = 0.01166 rad; δ 3 = -0.0404 rad; δ 4 = -0.0406 rad; V 3 = 0.9626 p.u.; V 4 = 0.9776 p.u.]
Solution: Calculating the Ybus:
Ybus =
- 724138 − j 12. 310340 − 1 .724138 + j 4. 310345 0 + j 0 − 4 .0 + j 8. 0 − 1 .724138 + j 4. 310345 4. 224138 − j 11. 810340 − 2 .5 + j 7. 5 0 + j 0 0 + j 0 − 2 .5 + j 7. 5 4. 386792 − j 14. 103770 − 1 .886792 + j 6. 603774 − 4 .0 + j 8. 0 0 + j 0 − 1 .886792 + j 6. 603774 5. 886792 − j 14. 603770
The real and imaginary parts are separated to obtain the G and B matrices.
G =
- 724138 − 1. 724138 0 − 4. 0 − 1. 724138 4. 224138 − 2. 5 0 0 − 2. 5 4. 386792 − 1. 886792 − 4. 0 0 − 1. 886792 5. 886792
The reactive powers (Q 3 , and Q 4 ) are computed as follows:
Qi =
∑^ n
k=
ViVk[Giksin(δi − δk) − Bikcos(δi − δk)]
The above equation can be rewritten as
Qi = −BiiV (^) i^2 +
∑^ n
k̸ =i
ViVk[Giksin(δi − δk) − Biksin(δi − δk)]
Q 3 =
∑^4
k=
V 3 Vk[G 3 ksin(δ 3 − δk) − B 3 kcos(δ 3 − δk)] = − 4. 768372 × 10 −^7 p.u.
Q 4 =
∑^4
k=
V 4 Vk[G 4 ksin(δ 4 − δk) − B 4 kcos(δ 4 − δk)] = − 3. 999996 × 10 −^1 p.u.
Reactive power residuals are calculated as
∆Q 3 = Qsp 3 − Q 3 = − 0. 25 p.u.
∆Q 4 = Qsp 4 − Q 4 = 0. 1 p.u.
The convergence criterion is checked to stop the iterations, i.e., maximum [∆Pi (i=2,3,4) and ∆Qi (i=3,4)] ≤ ϵ (0.001) maximum [0.5362,0.51,0.40,0.25,0.1] =0.51 > 0.
The convergence criteria is not satisfied, so the changes in variables at the end of the first iteration are obtained as follows:
H 22 H 23 H 24 N 23 N 24 H 32 H 33 H 34 N 33 N 34 H 42 H 43 H 44 N 43 N 44 J 32 J 33 J 34 L 33 L 34 J 42 J 43 J 44 L 43 L 44
∆δ 2 ∆δ 3 ∆δ 4 ∆V 3 V 3 ∆V 3 V 3
=
∆P 2 ∆P 3 ∆P 4 ∆Q 3 ∆Q 4
Where,
H 22 =
∂P 2
∂δ 2
= −Q 2 − B 22 V 22 = 12. 0259
H 23 =
∂P 2
∂δ 3
= V 2 V 3 [G 23 sin(δ 2 − δ 3 ) − B 23 cos(δ 2 − δ 3 )] = − 7. 5
H 24 =
∂P 2
∂δ 4
= V 2 V 4 [G 24 sin(δ 2 − δ 4 ) − B 24 cos(δ 2 − δ 4 )] = 0. 0
H 32 =
- ∂P - ∂δ - = V 4 V 3 [G 43 sin(δ 4 − δ 3 ) − B 43 cos(δ 4 − δ 3 )] = − - ∂P H 44 = - ∂δ - = −Q 4 − B 44 V 42 = 15. - N 22 = V - ∂P - ∂V - = P 2 + G 22 V 22 = 4.
- N 23 = V - ∂P - ∂V - = V 2 V 3 [G 23 cos(δ 2 − δ 3 ) + B 23 sin(δ 2 − δ 3 )] = −
- N 24 = V - ∂P - ∂V - = V 2 V 4 [G 24 cos(δ 2 − δ 4 ) + B 24 sin(δ 2 − δ 4 )] = 0.
- N 32 = V - ∂P - ∂V - = V 3 V 2 [G 32 cos(δ 3 − δ 2 ) + B 32 sin(δ 3 − δ 2 )] = − - N 33 = V - ∂P - ∂V - = P 3 + G 33 V 32 = 4.
- N 34 = V - ∂P - ∂V - = V 3 V 4 [G 34 cos(δ 3 − δ 4 ) + B 34 sin(δ 3 − δ 4 )] = − - N 42 = V - ∂P - ∂V - = V 4 V 2 [G 42 cos(δ 4 − δ 2 ) + B 42 sin(δ 4 − δ 2 )] = 0.
- N 43 = V - ∂P - ∂V - = V 4 V 3 [G 43 cos(δ 4 − δ 3 ) + B 43 sin(δ 4 − δ 3 )] = − - N 44 = V - ∂P - ∂V - = P 4 + G 44 V 42 = 5. - ∂Q J 22 = - ∂δ - = P 2 − G 22 V 22 = − - ∂Q J 23 = - ∂δ - = V 2 V 3 [−G 23 cos(δ 2 − δ 3 ) − B 23 sin(δ 2 − δ 3 )] = 2. - ∂Q J 24 = - ∂δ - = V 2 V 4 [−G 24 cos(δ 2 − δ 4 ) − B 24 sin(δ 2 − δ 4 )] = 0. - ∂Q J 32 = - ∂δ - = V 3 V 2 [−G 32 cos(δ 3 − δ 2 ) − B 32 sin(δ 3 − δ 2 )] = 2. - ∂Q J 33 = - ∂δ - = P 3 − G 33 V 32 = − - ∂Q J 34 = - ∂δ - = V 3 V 4 [−G 34 cos(δ 3 − δ 4 ) − B 34 sin(δ 3 − δ 4 )] = 1. - ∂Q J 42 = - ∂δ - = V 4 V 2 [−G 42 cos(δ 4 − δ 2 ) − B 42 sin(δ 4 − δ 2 )] = 0. - ∂Q J 43 = - ∂δ - = V 4 V 3 [−G 43 cos(δ 4 − δ 3 ) − B 43 sin(δ 4 − δ 3 )] = 1. - ∂Q J 44 = - ∂δ - = P 4 − G 44 V 42 = − - L 22 = V - ∂Q - ∂V - = Q 2 − B 22 V 22 = 11. - L 23 = V - ∂Q - ∂V - = V 2 V 3 [G 23 sin(δ 2 − δ 3 ) − B 23 cos(δ 2 − δ 3 )] = − - L 24 = V - ∂Q - ∂V - = V 2 V 4 [G 24 sin(δ 2 − δ 4 ) − B 24 cos(δ 2 − δ 4 )] = 0. - L 32 = V - ∂Q - ∂V - = V 3 V 2 [G 32 sin(δ 3 − δ 2 ) − B 32 cos(δ 3 − δ 2 )] = − - L 33 = V - ∂Q - ∂V - = Q 3 − B 33 V 32 = 14.
- L 34 = V - ∂Q - ∂V - = V 3 V 4 [G 34 sin(δ 3 − δ 4 ) − B 34 cos(δ 3 − δ 4 )] = − - L 42 = V - ∂Q - ∂V - = V 4 V 2 [G 42 sin(δ 4 − δ 2 ) − B 42 cos(δ 4 − δ 2 )] = 0.
- L 43 = V - ∂Q - ∂V - = V 4 V 3 [G 43 sin(δ 4 − δ 3 ) − B 43 cos(δ 4 − δ 3 )] = − - L 44 = V - ∂Q - ∂V - = Q 4 − B 44 V 42 = 14.
- ∆δ 2 = − 0 09696 , ∆δ 3 = − 0 11217 , ∆δ 4 = − process is performed to compute the changes in δ and V, we have
- ∆V - V - ∆V = − 0 05504 , - V - ∆V = − 0 07549 , - V - = − - ∆V V 2 = V 2 + - V - ∂P - ∂δ - = V 3 V 2 [G 32 sin(δ 3 − δ 2 ) − B 32 cos(δ 3 − δ 2 )] = − - ∂P H 33 = - ∂δ - = −Q 3 − B 33 V 32 = 14. - ∂P H 34 = - ∂δ - = V 3 V 4 [G 34 sin(δ 3 − δ 4 ) − B 34 cos(δ 3 − δ 4 )] = − - ∂P H 42 = - ∂δ - = V 4 V 2 [G 42 sin(δ 4 − δ 2 ) − B 42 cos(δ 4 − δ 2 )] = 0. - ∂P H 43 = - ∂δ - = V 4 V 3 [G 43 sin(δ 4 − δ 3 ) − B 43 cos(δ 4 − δ 3 )] = − - ∂P H 44 = - ∂δ - = −Q 4 − B 44 V 42 = 15.
- N 23 = V - ∂P - ∂V - = V 2 V 3 [G 23 cos(δ 2 − δ 3 ) + B 23 sin(δ 2 − δ 3 )] = −
- N 24 = V - ∂P - ∂V - = V 2 V 4 [G 24 cos(δ 2 − δ 4 ) + B 24 sin(δ 2 − δ 4 )] = 0. - N 33 = V - ∂P - ∂V - = P 3 + G 33 V 32 = 4.
- N 34 = V - ∂P - ∂V - = V 3 V 4 [G 34 cos(δ 3 − δ 4 ) + B 34 sin(δ 3 − δ 4 )] = −
- N 43 = V - ∂P - ∂V - = V 4 V 3 [G 43 cos(δ 4 − δ 3 ) + B 43 sin(δ 4 − δ 3 )] = − - N 44 = V - ∂P - ∂V - = P 4 + G 44 V 42 = 5. - ∂Q J 32 = - ∂δ - = V 3 V 2 [−G 32 cos(δ 3 − δ 2 ) − B 32 sin(δ 3 − δ 2 )] = 2. - ∂Q J 33 = - ∂δ - = P 3 − G 33 V 32 = − - ∂Q J 34 = - ∂δ - = V 3 V 4 [−G 34 cos(δ 3 − δ 4 ) − B 34 sin(δ 3 − δ 4 )] = 1. - ∂Q J 42 = - ∂δ - = V 4 V 2 [−G 42 cos(δ 4 − δ 2 ) − B 42 sin(δ 4 − δ 2 )] = 0. - ∂Q J 43 = - ∂δ - = V 4 V 3 [−G 43 cos(δ 4 − δ 3 ) − B 43 sin(δ 4 − δ 3 )] = 1. - ∂Q J 44 = - ∂δ - = P 4 − G 44 V 42 = −
