eee 103 2nd exam review

Load Flow Studies

EEE 103 2nd Exam Review

Outline

• Bus Admittance Matrix

• Gauss-Seidel Iterative Method

• Tips for Reliable Solutions

The Bus Admittance Matrix

𝑌𝐵𝑈𝑆 =𝑌11 ⋯ 𝑌1𝑛⋮ ⋱ ⋮𝑌𝑛1 ⋯ 𝑌𝑛𝑛

• YBUS building rules:

– Diagonal elements Yii = sum of all admittances connected to bus i

– Off-diagonal elements Yij = negative of net admittance connected

between bus i and bus j

• Yii is the self-admittance or driving-point admittance

• Yij is the mutual admittance or transfer admittance

Properties of the YBUS

• Complex

• Symmetric

• Sparse (can have zero elements)

• Singular if there are no admittances between

any of the buses and the reference bus

Example 1: 3-bus system


Form the YBUS:

• Y12 = Y21 = -(1 / (0.02 + j0.04 Ω))= -10 + j20

• Y13 = Y31 = -(1 / (0.01 + j0.03 Ω)) = -10 + j30

• Y23 = Y32 = -(1 / (0.0125 + j0.025 Ω)) = -16 + j32

• Y11 = Y12 + Y13 = (10 – j20) + (10 – j30) = 20 – j50

• Y22 = Y21 + Y23 = (10 – j20) + (16 – j32) = 26 – j52

• Y33 = Y31 + Y32 = (10 – j30) + (16 – j32) = 26 – j62


𝑌𝐵𝑈𝑆 =

20 − j50 –10 + j20 −10 + j30−10 + j20 26 − j52 −16 + j32−10 + j30 −16 + j32 26 − j62

Source Transformation

• Is = Vs / Zs

• Ys = 1 / Zs

Injected Current

• The current that flows through a bus

• Current from sources flow into the bus

• Current to admittances flow out of the bus

Injected Current

KCL equation at bus i:

• IS = Iij + Iik + Iil

• IS = Yij(Vi – Vj) + Yik(Vi – Vk) + Yil(Vi – Vl)

• IS = YijVi – YijVj + YikVi – YikVk + YilVi – YilVl

• IS = (Yij + Yik + Yil)Vi – YijVj – YikVk – YilVl

Injected Current

𝐼1⋮𝐼𝑛

=𝑌11 ⋯ 𝑌1𝑛⋮ ⋱ ⋮𝑌𝑛1 ⋯ 𝑌𝑛𝑛

𝑉1⋮𝑉𝑛

𝐼𝐵𝑈𝑆 = 𝑌𝐵𝑈𝑆 𝑉𝐵𝑈𝑆

• IBUS: Bus current vector (i.e. current sources)

• YBUS: Bus admittance matrix

• VBUS: Bus voltage vector

Injected Power

𝑆1⋮𝑆𝑛

=𝑉1 ⋯ 0⋮ ⋱ ⋮0 ⋯ 𝑉𝑛

𝑌11 ⋯ 𝑌1𝑛⋮ ⋱ ⋮𝑌𝑛1 ⋯ 𝑌𝑛𝑛

𝑉1⋮𝑉𝑛

∗

𝑆𝑗 = 𝑉𝑗 𝑌𝑗𝑖𝑉𝑗∗

𝑛

𝑖=1

• The complex power that flows through a bus

The Gauss-Seidel (G/S) Iterative Method

• A solution to a non-linear equation with the form f(x) = 0

• 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) – x(k)|< ε, where ε is the desired accuracy

Example 3: G/S solution

Find a root of f(x) = x3 – 6x2 + 9x – 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



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) – x(1)| = 0.3472

5. Second iteration, x(2) = g(3.8472) = 3.9848


7. Third iteration, x(3) = g(3.9848) = 3.9998




9. Fourth iteration, x(4) = g(3.9998) = 4.0


11. Stop iterations, x = 4.0

The Load Flow Problem

• The ff. information are available

– Source (Generator) • Voltage and real power

delivered

– Line (Transmission) • Impedance

– Load • Complex power

consumed V, P

Z

P, Q

• The ff. are unknown

– Bus voltages

– Bus currents

– Line power flows


• We need the equation f(x) to solve for x

• f(x) is the complex injected power

• x is the bus voltage

• G/S will be extended to a linear system


For a bus i, 𝑓 𝑉𝑖 = 𝑆∗

𝑃𝑖 − j𝑄𝑖 = 𝑉𝑖∗ 𝑌𝑖1𝑉1 +⋯𝑌𝑖𝑖𝑉𝑖 +⋯𝑌𝑖𝑛𝑉𝑛

𝑌𝑖𝑖𝑉𝑖 =𝑃𝑖 − j𝑄𝑖𝑉𝑖∗ − 𝑌𝑖1𝑉1 +⋯𝑌𝑖 𝑖−1 𝑉𝑖 𝑖−1 + 𝑌𝑖(𝑖+1)𝑉𝑖(𝑖+1) +⋯𝑌𝑖𝑛𝑉𝑛

The (k+1)th approximation to the voltage Vi is

𝑉𝑖(𝑘+1)=1

𝑌𝑖𝑖

𝑃𝑖 − j𝑄𝑖

𝑉𝑖(𝑘)∗− 𝑌𝑖1𝑉1

(𝑘)+⋯+ 𝑌𝑖(𝑖−1)𝑉(𝑖−1)

(𝑘)+ 𝑌𝑖(𝑖+1)𝑉(𝑖+1)

(𝑘)+⋯+ 𝑌𝑖𝑛𝑉𝑛

(𝑘)

System Representation

Step-by-step G/S Formulation

1. Build the [YBUS]

2. Initialize voltages (Vi∠δ) and power (Psch,Qsch)

A. PV bus: use given V, set δ = 0.0°, and Psch = Pg – Pl

B. PQ bus: set V = 1.0, δ = 0.0°, Psch = Pg – Pl, and Qsch = Qg – Ql

3. Approximate voltages (except slack bus). Use the latest approximations.

A. PQ bus: 𝑉𝑖(𝑘+1)=1

𝑌𝑖𝑖

𝑃𝑖𝑠𝑐ℎ−j𝑄𝑖

𝑠𝑐ℎ

𝑉𝑖(𝑘)∗ − 𝑌𝑖1𝑉1

(𝑘)+⋯+ 𝑌𝑖(𝑖−1)𝑉(𝑖−1)

(𝑘)+ 𝑌𝑖(𝑖+1)𝑉(𝑖+1)

(𝑘)+⋯+ 𝑌𝑖𝑛𝑉𝑛

(𝑘)

Step-by-step G/S Formulation

3. Approximate voltages (except slack bus). Use the latest approximations.

B. PV bus: i. Approximate Qsch

using 𝑄 = −Im 𝑉𝑖

∗ 𝑌𝑖1𝑉1 +⋯𝑌𝑖𝑖𝑉𝑖 +⋯𝑌𝑖𝑛𝑉𝑛

ii. If there are MVAR limits, Qsch = Qmin if Q ≤ Qmin and Qsch = Qmax if Q ≥ Qmax; otherwise Qsch = Q

iii. Approximate V’ as if it were a PQ bus

iv. V = |V(0)|∠(angle of V’)

4. If max(|V(k+1) – V(k)|) ≤ ε then stop iteration, otherwise repeat steps 3 and 4.

Example 4: 3-bus Load Flow

• Use 100 MVA base and ε = 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) δ (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 - - -


• The [YBUS] was solved earlier in Ex. 1

𝑌𝐵𝑈𝑆 =

20 − j50 –10 + j20 −10 + j30−10 + j20 26 − j52 −16 + j32−10 + j30 −16 + j32 26 − j62

• The initial voltage vector [VBUS] is [1.0500∠0.00° 1.0000∠0.00° 1.0400∠0.00°]

• The scheduled complex powers in per-unit are P2

sch = −4.0, Q2sch = −2.5, and P3

sch = 2.0


• First iteration V2

(1) = [1/(26−j52)+ * ,*(−4.0+j2.5) / (1.0∠0)+ − [(−10+j20) * (1.05∠0) + (−16+j32) * (1.04∠0)]}]

= 0.9755∠−2.49

Q3(1) = − Im{[1.04∠0] * [(−10+j30) * (1.05∠0) +

(−16+j32) * (0.9755∠−2.49) + (26−j62) * (1.04∠0)]}

= 1.16

V3(1) = *1/(26−j62)] * {[(2.0-j 1.16) / (1.04∠0)+ −

[(−10+j30) * (1.05∠0°) + (−16+j32) * (0.9755∠−2.49)]}] = 1.04∠−0.29


• Second iteration V2

(2) = 0.9708∠−2.56

Q3(2) = 1.2911

V3(2) = 1.04∠−0.36

• Third iteration V2

(3) = 0. 9720∠−2.56

Q3(3) = 0.8605

V3(3) = 1.04∠−0.36

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


(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


• Build the bus impedance matrix

• Solve for the bus voltages using G/S. Use 100 MVA base and ε = 0.01


• [YBUS]

8.9852–j44.9260 –3.8156 +j19.0781 –5.1696 +j25.8478 0

–3.8156+j19.0781 8.9852–j44.9260 0 –5.1696+j25.8478

–5.1696+j25.8478 0 8.1933–j40.9663 –3.0237+j15.1185

0 –5.1696+j25.8478 –3.0237+j15.1185 8.1933–j40.9663


• First iteration

V2 = 1.0034∠−1.82°

V3 = 1.0013∠−2.34°

V4 = 1.0200∠−4.28°

Q4 = 1.0468

Max mismatch = 0.0762


• Second iteration

V2 = 1.0032∠−4.26°

V3 = 1.0001∠−3.87°

V4 = 1.0200∠6.45°

Q4 = 1.5222



• Third iteration

V2 = 1.0016∠−5.47°

V3 = 0.9986∠−4.64°

V4 = 1.0200∠7.53°

Q4 = 1.4658



• Fourth iteration

V2 = 1.0006∠−6.07°

V3 = 0.9978∠−5.02°

V4 = 1.0200∠8.07°

Q4 = 1.3588



• Fifth iteration

V2 = 1.0001∠−6.37°

V3 = 0.9973∠−5.21°

V4 = 1.0200∠8.34°

Q4 = 1.2858

Max mismatch = 0.0053, stop iterations


Line R (pu) X (pu)

1-2 0.02 0.10

1-3 0.04 0.15

2-3 0.07 0.25


(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

eee 103 2nd exam review

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