EEET2339_Assign1_s3915884

docx

School

Royal Melbourne Institute of Technology *

*We aren’t endorsed by this school

Course

2339

Subject

Electrical Engineering

Date

Apr 3, 2024

Type

docx

Pages

Uploaded by CorporalPheasant2100

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

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

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

RMIT University Master of Engineering (Sustainable Energy) Sela Bloomfield (s3915884) a) A transmission line is operating with a sending end voltage (SE) | ⃗ V S | = 1.02 pu where base voltage is equal to the rated voltage. i. Calculate the practical line load ability as a percentage of SIL. ( | ⃗ V R | = 0.98 pu ¿ P R = V R V S B cos ( β − δ ) − AV R 2 B cos ( β − α ) V R V S B = ( 0.98 × 730 )( 1.02 × 730 ) 81.5217 = 6534.2950 A V R 2 B = ( 0.9399 ) ( 0.98 × 730 ) 2 81.5217 = 5900.7375 P R = 6534.2950cos ( 86.8705 − 33 ) − 5900.7375cos ( 86.8705 − 0.2026 ) P R = 3509.7300 MW As percentage of SIL results in: P R ( max ) = 3509.7300 MW 2235.2775 MW × 100 = 157.0154% of SIL b) Practical load-ability if nominal voltage is 500 kV P R = V R V S B cos ( β − δ ) − AV R 2 B cos ( β − α ) The A and B magnitudes was obtained from Part 1 a(iii) as well as the angles relative to A and B V R V S B = ( 0.98 × 500 )( 1.02 × 500 ) 81.5217 = 3065.4415 A V R 2 B = ( 0.9398 ) ( 0.98 × 500 ) 2 81.5217 = 2767.9254 P R = 3065.4415cos ( 86.8705 − 33 ) − 2767.9254 cos ( 86.8705 − 0.2026 ) P R = 1646.5410 MW Part 4 Calculate the following: i. Sending end line to line voltage and power angle when it undergoes full-load operation at pf = 1 V S = ( A×V R ) +( B × I FL ) The A and B magnitudes was obtained from Part 1 a(iii) & V R and I FL from Table 1 5

RMIT University Master of Engineering (Sustainable Energy) Sela Bloomfield (s3915884) V S = [ ( 0.9399 ∠ 0.2026 ) × ( 0.9475 × 730 √ 3 ) ] + [ ( 81.5217 ∠ 86.8705 ) × 1.87 ] ¿ ( 375.3385 ∠ 0.2026 ) + ( 152.4456 ∠ 86.8705 ) V S = 413.2434 ∠ 21.8120 kV The voltage sending end line to line is V S ( L − L ) = 413.2434 ∠ 21.8120 × √ 3 V S ( L − L ) = 715.7586 ∠ 21.8120 kV ii. Calculate the no-load receiving end voltage V R ( NL ) = V S A V R ( NL ) = 715.7586 ∠ 21.8120 0.9399 ∠ 0.2026 V R ( NL ) = 761.5263 ∠ 21.6094 kV iii. Calculate the voltage regulation for the transmission line Voltageregulation = | V NL | − | V FL | | V FL | × 100% Voltageregulation = 892.218 − ( 0.9475 × 730 ) ( 0.9475 × 730 ) × 100% Voltageregulation = 10.0989% Part 5 a) Calculate the following: i. The new A and B for the exact-pi equivalent transmission line model Z' new = Z ' = 81.5305 ∠ 86.8707 Y ' new = G ' + j ( 1 − ε shunt 100 ) B ' Y ' new = ( 8.5477 × 10 − 7 ) + j ( 1 − 72 100 ) 1.4775 × 10 − 3 Y ' new = 8.5477 × 10 − 7 + j 4.1370 × 10 − 4 ∨ 4.137 × 10 − 4 ∠ 89.8816 S A eq = 1 + Y ' new ×Z ' new 2 ¿ 1 + ( 4.137 × 10 − 4 ∠ 89.8816 ) × ( 81.5305 ∠ 86.8707 ) 2 A eq = 0.9832 ∠ 0.0557 pu B eq = Z ' new 6

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

RMIT University Master of Engineering (Sustainable Energy) Sela Bloomfield (s3915884) B eq = 81.5305 ∠ 86.8707 Ω ii. The sending end line to line voltage and the power angle during the FL operation at pf = 1 V S = ( A eq ×V R ) +( B eq ×I FL ) V S = [ ( 0.9832 ∠ 0.0557 ) × ( 0.9475 × 730 √ 3 ) ] + [ ( 81.5305 ∠ 86.8707 ) × 1.87 ] ¿ ( 392.6299 ∠ 0.0557 ) + ( 152.4620 ∠ 86.8707 ) V S = 429.0159 ∠ 20.8386 kV The voltage sending end line to line is V S ( L − L ) = 429.0159 ∠ 20.8386 × √ 3 V S ( L − L ) = 743.0773 ∠ 20.8386 kV iii. Calculate the no-load RE voltage V R ( NL ) = V S A eq V R ( NL ) = 743.0773 ∠ 20.8386 0.9832 ∠ 0.0557 V R ( NL ) = 755.7744 ∠ 20.7829 kV iv. Voltage regulation for transmission line as % Voltageregulation = | V NL | − | V FL | | V FL | × 100% Voltageregulation = 755.7744 − ( 0.9475 × 730 ) ( 0.9475 × 730 ) × 100% Voltageregulation = 9.2673% v. RE voltage under no-load operation V R ( NL ) = V S A eq V R ( NL ) = 715.7586 ∠ 21.8120 0.9832 ∠ 0.0557 V R ( NL ) = 727.9888 ∠ 21.7563 kV vi. Voltage regulation for transmission line as % operating with switched shunt reactive compensation 7

RMIT University Master of Engineering (Sustainable Energy) Sela Bloomfield (s3915884) Voltageregulation = | V NL | − | V FL | | V FL | × 100% Voltageregulation = 727.9888 − ( 0.9475 × 730 ) ( 0.9475 × 730 ) × 100% Voltageregulation = 5.2501% b) Repeat 5a vi with transmission line operating at 500kV nominal voltage Voltageregulation = | V NL | − | V FL | | V FL | × 100% Voltageregulation = 727.9888 − ( 0.9475 × 500 ) ( 0.9475 × 500 ) × 100% Voltageregulation = 53.6652% It can be seen that the voltage regulation increases considerably with a smaller nominal voltage. Part 6 a) Calculate the following: i. Capacitive element related impedance Z cap Z cap =− j 1 2 X ' × ( ε series 100 ) ¿ − j 1 2 ( 81.4089 ) × 25 100 Z cap =− j 10.1761 ii. The equivalent A and B for the series compensated exact-pi model for transmission line The values of A, B, C and D are obtained from Part 1 a iii and Z cap from Part 6 a i A eq = A +( C×Z cap ) A eq = ( 0.9399 ∠ 0.2026 ) + [ ( 1.4332 × 10 − 3 ∠ 90.0652 ) × − j 10.1761 ] A eq = 0.9545 ∠ 0.2005 pu B eq = ( C×Z cap 2 ) + [ ( A + D ) Z cap ] + B B eq =[ ( 1.4332 × 10 − 3 ∠ 90.0652 ¿ × ( − j 10.1761 ) 2 ) ]+ [ ( 0.9399 ∠ 0.2026 + 0.9399 ∠ 0.2026 ) × − j 10.1 ¿ ( 0.1484 ∠ − 89.9348 ) + ( 19.1290 ∠ − 89.7974 ) + ( 81.5217 ∠ 86.8705 ) B eq = 62.2869 ∠ 85.8400 Ω iii. Theoretical maximum real power delivery 8

RMIT University Master of Engineering (Sustainable Energy) Sela Bloomfield (s3915884) P R = V R ,com V S, com B eq cos ( θ B, eq − δ ) − A eq V R ,com 2 B eq cos ( θ B ,eq − θ A, eq ) ¿ ( 730 ) ( 730 ) 62.2869 ( 1 ) − ( 0.9545 )( 730 ) 2 62.2869 cos ( 85.8400 − 0.2005 ) P R = 7934.6745 MW iv. Difference between the theoretical maximum real power delivery values of compensated and uncompensated transmission line as percentage of SIL Δ P Rmax = 7934.6745 − 6179.7976 = 1754.8769 MW As percentage of SIL results in: P R ( max ) = 1754.8769 MW 2235.2775 MW × 100 = 78.5082% of SIL v. Practical line load-ability P R = V R V S B cos ( β − δ ) − AV R 2 B cos ( β − α ) V R V S B = ( 0.98 × 730 )( 1.02 × 730 ) 62.2869 = 8552.1488 A V R 2 B = ( 0.9545 ) ( 0.98 × 730 ) 2 62.2869 = 7842.9074 P R = 8552.1488cos ( 85.8400 − 33 ) − 7842.9074 cos ( 85.8400 − 0.2005 ) P R = 4569.5560 MW vi. Difference between practical maximum real power delivery on compensated and uncompensated transmission line Δ P Rmax = 4569.5560 − 3509.7300 = 1059.8260 MW As percentage of SIL results in: P R ( max ) = 1059.8260 MW 2235.278 MW × 100 = 47.4136 % b. Repeat Part 6 a (v) and practical line load-ability. Explain if same practical load-ability can be obtained by the addition of capacitive compensation series P R = V R V S B cos ( β − δ ) − AV R 2 B cos ( β − α ) V R V S B = ( 0.98 × 500 )( 1.02 × 500 ) 62.2869 = 4012.0795 9

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

RMIT University Master of Engineering (Sustainable Energy) Sela Bloomfield (s3915884) A V R 2 B = ( 0.9545 ) ( 0.98 × 500 ) 2 62.2869 = 3679.3523 P R = 4012.0795cos ( 85.8400 − 33 ) − 3679.3523cos ( 85.8400 − 0.2005 ) P R = 2143.7209 MW The addition of series capacitive compensation can aid to the reduction of Z cap magnitude, in relation the B eq magnitude will also decrease. This then will overall increase the practical line load-ability. 10