Load Flow Studies
EEE 103 2nd Exam Review
Outline
โข Bus Admittance Matrix
โข Gauss-Seidel Iterative Method
โข Tips for Reliable Solutions
The Bus Admittance Matrix
๐๐ต๐๐ =๐11 โฏ ๐1๐โฎ โฑ โฎ๐๐1 โฏ ๐๐๐
โข YBUS building rules:
โ Diagonal elements Yii = sum of all admittances connected to bus i
โ Off-diagonal elements Yij = negative of net admittance connected
between bus i and bus j
โข Yii is the self-admittance or driving-point admittance
โข Yij is the mutual admittance or transfer admittance
Properties of the YBUS
โข Complex
โข Symmetric
โข Sparse (can have zero elements)
โข Singular if there are no admittances between
any of the buses and the reference bus
Example 1: 3-bus system
Example 1: 3-bus system
Form the YBUS:
โข Y12 = Y21 = -(1 / (0.02 + j0.04 ฮฉ))= -10 + j20
โข Y13 = Y31 = -(1 / (0.01 + j0.03 ฮฉ)) = -10 + j30
โข Y23 = Y32 = -(1 / (0.0125 + j0.025 ฮฉ)) = -16 + j32
โข Y11 = Y12 + Y13 = (10 โ j20) + (10 โ j30) = 20 โ j50
โข Y22 = Y21 + Y23 = (10 โ j20) + (16 โ j32) = 26 โ j52
โข Y33 = Y31 + Y32 = (10 โ j30) + (16 โ j32) = 26 โ j62
Example 1: 3-bus system
๐๐ต๐๐ =
20 โ j50 โ10 + j20 โ10 + j30โ10 + j20 26 โ j52 โ16 + j32โ10 + j30 โ16 + j32 26 โ j62
Source Transformation
โข Is = Vs / Zs
โข Ys = 1 / Zs
Injected Current
โข The current that flows through a bus
โข Current from sources flow into the bus
โข Current to admittances flow out of the bus
Injected Current
KCL equation at bus i:
โข IS = Iij + Iik + Iil
โข IS = Yij(Vi โ Vj) + Yik(Vi โ Vk) + Yil(Vi โ Vl)
โข IS = YijVi โ YijVj + YikVi โ YikVk + YilVi โ YilVl
โข IS = (Yij + Yik + Yil)Vi โ YijVj โ YikVk โ YilVl
Injected Current
๐ผ1โฎ๐ผ๐
=๐11 โฏ ๐1๐โฎ โฑ โฎ๐๐1 โฏ ๐๐๐
๐1โฎ๐๐
๐ผ๐ต๐๐ = ๐๐ต๐๐ ๐๐ต๐๐
โข IBUS: Bus current vector (i.e. current sources)
โข YBUS: Bus admittance matrix
โข VBUS: Bus voltage vector
Injected Power
๐1โฎ๐๐
=๐1 โฏ 0โฎ โฑ โฎ0 โฏ ๐๐
๐11 โฏ ๐1๐โฎ โฑ โฎ๐๐1 โฏ ๐๐๐
๐1โฎ๐๐
โ
๐๐ = ๐๐ ๐๐๐๐๐โ
๐
๐=1
โข The complex power that flows through a bus
Example 2: 4-bus system
Example 2: 4-bus system
Example 2: 4-bus system
The Gauss-Seidel (G/S) Iterative Method
โข A solution to a non-linear equation with the form f(x) = 0
โข Steps:
1. Rewrite function into x = g(x)
2. Initialize x(k), k = 0
3. Approximate x(k+1) with g(x(k))
4. Stop approximations when |x(k+1) โ x(k)|< ฮต, where ฮต is the desired accuracy
Example 3: G/S solution
Find a root of f(x) = x3 โ 6x2 + 9x โ 4 = 0
Desired accuracy is 0.01
1. x = -(1/9)x3 + (6/9)x2 + (4/9) = g(x)
2. Let x(0) = 3.5
Example 3: G/S solution
Find a root of f(x) = x3 โ 6x2 + 9x โ 4 = 0
3. First iteration, x(1) = g(3.5) = -(1/9)(3.5)3 + (6/9)(3.5)2 + (4/9) = 3.8472
4. Check convergence |x(0) โ x(1)| = 0.3472
5. Second iteration, x(2) = g(3.8472) = 3.9848
6. Check convergence |x(1) โ x(2)| = 0.1376
7. Third iteration, x(3) = g(3.9848) = 3.9998
Example 3: G/S solution
Find a root of f(x) = x3 โ 6x2 + 9x โ 4 = 0
8. Check convergence |x(3) โ x(2)| = 0.015
9. Fourth iteration, x(4) = g(3.9998) = 4.0
10. Check convergence |x(4) โ x(3)| = 0.0002
11. Stop iterations, x = 4.0
The Load Flow Problem
โข The ff. information are available
โ Source (Generator) โข Voltage and real power
delivered
โ Line (Transmission) โข Impedance
โ Load โข Complex power
consumed V, P
Z
P, Q
โข The ff. are unknown
โ Bus voltages
โ Bus currents
โ Line power flows
The Load Flow Problem
โข We need the equation f(x) to solve for x
โข f(x) is the complex injected power
โข x is the bus voltage
โข G/S will be extended to a linear system
The Load Flow Problem
For a bus i, ๐ ๐๐ = ๐โ
๐๐ โ j๐๐ = ๐๐โ ๐๐1๐1 +โฏ๐๐๐๐๐ +โฏ๐๐๐๐๐
๐๐๐๐๐ =๐๐ โ j๐๐๐๐โ โ ๐๐1๐1 +โฏ๐๐ ๐โ1 ๐๐ ๐โ1 + ๐๐(๐+1)๐๐(๐+1) +โฏ๐๐๐๐๐
The (k+1)th approximation to the voltage Vi is
๐๐(๐+1)=1
๐๐๐
๐๐ โ j๐๐
๐๐(๐)โโ ๐๐1๐1
(๐)+โฏ+ ๐๐(๐โ1)๐(๐โ1)
(๐)+ ๐๐(๐+1)๐(๐+1)
(๐)+โฏ+ ๐๐๐๐๐
(๐)
System Representation
Step-by-step G/S Formulation
1. Build the [YBUS]
2. Initialize voltages (Viโ ฮด) and power (Psch,Qsch)
A. PV bus: use given V, set ฮด = 0.0ยฐ, and Psch = Pg โ Pl
B. PQ bus: set V = 1.0, ฮด = 0.0ยฐ, Psch = Pg โ Pl, and Qsch = Qg โ Ql
3. Approximate voltages (except slack bus). Use the latest approximations.
A. PQ bus: ๐๐(๐+1)=1
๐๐๐
๐๐๐ ๐โโj๐๐
๐ ๐โ
๐๐(๐)โ โ ๐๐1๐1
(๐)+โฏ+ ๐๐(๐โ1)๐(๐โ1)
(๐)+ ๐๐(๐+1)๐(๐+1)
(๐)+โฏ+ ๐๐๐๐๐
(๐)
Step-by-step G/S Formulation
3. Approximate voltages (except slack bus). Use the latest approximations.
B. PV bus: i. Approximate Qsch
using ๐ = โIm ๐๐
โ ๐๐1๐1 +โฏ๐๐๐๐๐ +โฏ๐๐๐๐๐
ii. If there are MVAR limits, Qsch = Qmin if Q โค Qmin and Qsch = Qmax if Q โฅ Qmax; otherwise Qsch = Q
iii. Approximate Vโ as if it were a PQ bus
iv. V = |V(0)|โ (angle of Vโ)
4. If max(|V(k+1) โ V(k)|) โค ฮต then stop iteration, otherwise repeat steps 3 and 4.
Example 4: 3-bus Load Flow
โข Use 100 MVA base and ฮต = 0.01
Line R (pu) X (pu)
1-2 0.02 0.04
1-3 0.01 0.03
2-3 0.0125 0.025
Bus Type V (pu) ฮด (deg) PGEN
(MW)
QGEN
(MVAR) PLOAD
(MW)
QLOAD
(MVAR)
1 SLACK 1.05 0 - - - -
2 LOAD - - 0 0 400 250
3 GEN 1.04 - 200 - - -
Example 4: 3-bus Load Flow
โข The [YBUS] was solved earlier in Ex. 1
๐๐ต๐๐ =
20 โ j50 โ10 + j20 โ10 + j30โ10 + j20 26 โ j52 โ16 + j32โ10 + j30 โ16 + j32 26 โ j62
โข The initial voltage vector [VBUS] is [1.0500โ 0.00ยฐ 1.0000โ 0.00ยฐ 1.0400โ 0.00ยฐ]
โข The scheduled complex powers in per-unit are P2
sch = โ4.0, Q2sch = โ2.5, and P3
sch = 2.0
Example 4: 3-bus Load Flow
โข First iteration V2
(1) = [1/(26โj52)+ * ,*(โ4.0+j2.5) / (1.0โ 0)+ โ [(โ10+j20) * (1.05โ 0) + (โ16+j32) * (1.04โ 0)]}]
= 0.9755โ โ2.49
Q3(1) = โ Im{[1.04โ 0] * [(โ10+j30) * (1.05โ 0) +
(โ16+j32) * (0.9755โ โ2.49) + (26โj62) * (1.04โ 0)]}
= 1.16
V3(1) = *1/(26โj62)] * {[(2.0-j 1.16) / (1.04โ 0)+ โ
[(โ10+j30) * (1.05โ 0ยฐ) + (โ16+j32) * (0.9755โ โ2.49)]}] = 1.04โ โ0.29
Example 4: 3-bus Load Flow
โข Second iteration V2
(2) = 0.9708โ โ2.56
Q3(2) = 1.2911
V3(2) = 1.04โ โ0.36
โข Third iteration V2
(3) = 0. 9720โ โ2.56
Q3(3) = 0.8605
V3(3) = 1.04โ โ0.36
Max. voltage mismatch: 0.0037, stop iterations.
Example 5: 4-bus Load Flow Line R (pu) X (pu)
1-2 0.01008 0.05040
1-3 0.00744 0.03720
2-4 0.00744 0.03720
3-4 0.01272 0.06360
Bus Type V (pu) ฮด (deg) PGEN
(MW)
QGEN
(MVAR) PLOAD
(MW)
QLOAD
(MVAR)
1 SLACK 1.0 0 - - - -
2 LOAD - - - - 170 105.35
3 LOAD - - - - 200 123.94
4 GEN 1.02 - 318 - 80 49.58
Example 5: 4-bus Load Flow
โข Build the bus impedance matrix
โข Solve for the bus voltages using G/S. Use 100 MVA base and ฮต = 0.01
Example 5: 4-bus Load Flow
โข [YBUS]
8.9852โj44.9260 โ3.8156 +j19.0781 โ5.1696 +j25.8478 0
โ3.8156+j19.0781 8.9852โj44.9260 0 โ5.1696+j25.8478
โ5.1696+j25.8478 0 8.1933โj40.9663 โ3.0237+j15.1185
0 โ5.1696+j25.8478 โ3.0237+j15.1185 8.1933โj40.9663
Example 5: 4-bus Load Flow
โข First iteration
V2 = 1.0034โ โ1.82ยฐ
V3 = 1.0013โ โ2.34ยฐ
V4 = 1.0200โ โ4.28ยฐ
Q4 = 1.0468
Max mismatch = 0.0762
Example 5: 4-bus Load Flow
โข Second iteration
V2 = 1.0032โ โ4.26ยฐ
V3 = 1.0001โ โ3.87ยฐ
V4 = 1.0200โ 6.45ยฐ
Q4 = 1.5222
Max mismatch = 0.0428
Example 5: 4-bus Load Flow
โข Third iteration
V2 = 1.0016โ โ5.47ยฐ
V3 = 0.9986โ โ4.64ยฐ
V4 = 1.0200โ 7.53ยฐ
Q4 = 1.4658
Max mismatch = 0.0211
Example 5: 4-bus Load Flow
โข Fourth iteration
V2 = 1.0006โ โ6.07ยฐ
V3 = 0.9978โ โ5.02ยฐ
V4 = 1.0200โ 8.07ยฐ
Q4 = 1.3588
Max mismatch = 0.0105
Example 5: 4-bus Load Flow
โข Fifth iteration
V2 = 1.0001โ โ6.37ยฐ
V3 = 0.9973โ โ5.21ยฐ
V4 = 1.0200โ 8.34ยฐ
Q4 = 1.2858
Max mismatch = 0.0053, stop iterations
Example 6: 3-bus Load Flow
Line R (pu) X (pu)
1-2 0.02 0.10
1-3 0.04 0.15
2-3 0.07 0.25
Bus Type V (pu) ฮด (deg) PGEN
(MW)
QGEN
(MVAR) PLOAD
(MW)
QLOAD
(MVAR)
1 SLACK 1.0 0 - - 0 0
2 GEN 1.0 - 44.1 - 0 0
3 LOAD - - - - 100 66.67
Tips for Reliable Solutions
1. Remember these equations: [I] = [YBUS][V] and [P-jQ] = diag([V*])[YBUS][V]
2. Report per-unit quantities in four (4.0000) decimal places and angles in two (2.00) decimal places
3. YBUS
A. Complete the upper-diagonal only. Remember that the matrix is symmetric
B. Start with off-diagonal elems.
C. The diagonal elems. have positive real terms
Tips for Reliable Solutions
4. System Modelling
A. Positive power flows from generators
B. Negative power flows into load
5. G/S Iterative Method
A. Write the voltages in tabular form
B. Setup all equations first
C. Use the latest information available
D. All quantities are in per-unit
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