EEET2339_Assign1_s3915884

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Royal Melbourne Institute of Technology *

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Electrical Engineering

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Apr 3, 2024

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RMIT University Master of Engineering (Sustainable Energy) Sela Bloomfield (s3915884) EEET2339 Power System Analysis & Control Assignment 1: Transmission Line Modelling and Design Specific to this assignment, each student was generated a unique set of parameters and variables that must be used in the stages of the assignment calculations, listed in Table 1 Table 1: Unique set of parameters/variables Description Symbol Values Units Transmission line rated voltage V rated 730 kV Length of the transmission line l tx-line 276 km Series impedance per-unit length z 0.0168 + j0.301  /km Shunt admittance per-unit length y j0.0000053 S/km Practical limit for phase angle  max 33 degree Receiving end voltage V R 0.9475 pu Rated full load current I FL 1.87 kA Shunt reactive compensation  shunt 72 % Capacitive compensation  series 25 % Part 1 a) Assuming a positive sequence operation, the following sub-parts were calculated i. Propagation constant gamma in units γ = √ zy = √ ( 0.0168 + j 0.301 ) × ( j 0.0000053 ) ¿ √ ¿¿ γ = 3.5234 × 10 − 5 + j 1.2635 × 10 − 3 km − 1 ii. Characteristic impedance Z C Z C = √ z y = √ ( 0.0168 + j 0.301 ) ( j 0.0000053 ) ¿ √ 56792.453 − j 3169.8113 Z C = 238.4044 − j 6.6480 Ω iii. Exact ABCD parameters of the line A = D = cosh ( γ l tx − line ) B = Z C × sinh ( γ l tx − line ) C = 1 Z C × sinh ( γ l tx − line ) First calculate γ l tx − line γ l tx − line = ( 3.5234 × 10 − 5 + j 1.2635 × 10 − 3 ) × 276 ¿ 9.7246 × 10 − 3 + j 0.3487 A = D = cosh ( γ l tx − line ) = cosh ( 9.7245 × 10 − 3 + j 0.3487 ) A = D = 0.9399 + j 3.3227 × 10 − 3 ∨ 0.9399 ∠ 0.2026 B = Z C × sinh ( γ l tx − line ) 1

RMIT University Master of Engineering (Sustainable Energy) Sela Bloomfield (s3915884) ¿ 238.4044 − j 6.6480 × sinh ( 9.7246 × 10 − 3 + j 0.3487 ) ¿ ( 238.4043 + j 6.6479 ) × ( 0.009139 + j 0.3417 ) B = 4.4504 + j 81.4000 ∨ 81.5217 ∠ 86.8705 C = 1 Z C × sinh ( γ l tx − line ) ¿ 1 ( 238.4044 − j 6.6480 ) × sinh ( ( 9.7246 × 10 − 3 + j 0.3487 ) ) ¿ 1 ( 238.4043 + j 6.6479 ) × ( 0.01827 + j 0.6834 ) C =− 1.6296 × 10 − 6 + j 1.4332 × 10 − 3 ∨ 1.4332 × 10 − 3 ∠ 90.0652 iv. Exact PI equivalent circuit model parameters Z ’ and Y ’ Z ' = z l tx − line [ sinh ( γ l tx − line ) γ l tx − line ] Y ' 2 = [ cosh ( γ l tx − line ) − 1 Z ' ] zl tx − line = ( 0.0168 + j 0.301 ) × 276 = 4.6368 + j 83.076 γ l tx − line = 9.7246 × 10 − 3 + j 0.3487 Z ' = ( 4.6368 + j 83.076 ) [ sinh ( 9.7245 × 10 − 3 + j 0.3487 ) 9.7245 × 10 − 3 + j 0.3487 ] ¿ ( 4.6368 + j 83.076 ) × ( 0.9799 + j 1.1169 × 10 − 3 ) Z ' = 4.4507 + j 81.4089 ∨ 81.5305 ∠ 86.8707 Y ' = 2 [ cosh ( γ l tx − line ) − 1 Z ' ] ¿ 2 × [ ( 0.9399 + j 3.3227 × 10 − 3 ) − 1 4.4507 + j 81.4089 ] Y ' = 8.5477 × 10 − 7 + j 1.4775 × 10 − 3 ∨ 1.4775 × 10 − 3 ∠ 89.966 8 b) The Z’ and Y’ for the nominal and exact pi-equivalent models for different transmission line lengths was calculated and graphed from 200km to 1000km - For both the graphs, polar forms were considered, and the magnitude was used for the y-axis values as the angles do not practically change. 2

RMIT University Master of Engineering (Sustainable Energy) Sela Bloomfield (s3915884) 200 300 400 500 600 700 800 900 1000 0 50 100 150 200 250 300 350 Z Z' Transmission line length (km) Magnitude Figure 1: Comparison between Z and Z' 200 276 300 400 500 600 700 800 900 1000 0 0.01 0.01 Y Y' Transmission line length (km) Magnitude Figure 2: Comparison between Y and Y' 3

RMIT University Master of Engineering (Sustainable Energy) Sela Bloomfield (s3915884) Part 2 a) Calculate the following: i. Assuming | ⃗ V R | = | ⃗ V S | = V rated , calculate surge impedance loading (SIL) in MW Surge Impedance Loading = ( V rated ) 2 Z C = ( 730 kV ) 2 238.4044 Ω SIL = 2235.2775 MW ii. Calculate the theoretical maximum real power delivery as a percentage of SIL P R = V R V S B cos ( θ B − δ ) − AV R 2 B cos ( θ B − θ A ) The A and B magnitudes was obtained from Part 1 a(iii) as well as the angles. For theoretical maximum, P R takes a maximum value when cos ( θ B − δ ) = 1 , therefore: V R V S B = ( 730 ) ( 730 ) 81.5217 = 6536.9098 A V R 2 B = ( 0.9399 ) ( 730 ) 2 81.5217 = 6144.0415 P R =( 6536.9098 ) 1 −( 6144.0415 ) cos ( 86.8705 − 0.2026 ) P R = 6179.7976 MW As percentage of SIL results in: P R ( max ) = 6179.7976 MW 2235.278 MW × 100 = 276.4667% of SIL b) Calculate the SIL and theoretical maximum real power if the transmission line operates at 500kV as nominal voltage while the SE and RE voltages are kept at 500kV Surge Impedance Loading = ( V rated ) 2 Z C = ( 500 kV ) 2 238.4044 Ω SIL = 1048.6384 MW V R V S B = ( 500 ) ( 500 ) 81.5217 = 3066.6681 A V R 2 B = ( 0.9399 ) ( 500 ) 2 81.5217 = 2882.3614 P R =( 3066.6681 ) 1 −( 2882.3614 ) cos ( 86.8705 − 0.2026 ) P R = 2899.1356 MW Part 3 4

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