ELEN 30184 Module 1 Elements Of Power System Analysis PDF

ELEN 30184 Module 1 Elements of Power System Analysis Symbols identifying different type of aluminum conductors are as follows: AAC all-aluminum conductors AAAC all-aluminum-alloy conductors...

ELEN 30184 Module 1 Elements of Power System Analysis Symbols identifying different type of aluminum conductors are as follows: AAC all-aluminum conductors AAAC all-aluminum-alloy conductors ACSR all-aluminum conductor, steel reinforced ACAR all conductor alloy reinforced RESISTANCE Symbols identifying different type of aluminum conductors are as follows: AAC all-aluminum conductors AAAC all-aluminum-alloy conductors ACSR all-aluminum conductor, steel reinforced ACAR all conductor alloy reinforced The most significant effect of resistance of transmission line conductors is the generation of copper loss in the line and IR type voltage drop which affects the regulation of the line. The DC resistance is given by the formula, ! 𝑅=𝜌 " 𝑎𝑟𝑒𝑎 𝑅 − 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑜ℎ𝑚𝑠 𝜌 − 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑖𝑡𝑦, 𝑜ℎ𝑚. 𝐿 − 𝑙𝑒𝑛𝑔𝑡ℎ 𝐴 − 𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎 𝑙𝑒𝑛𝑔𝑡ℎ  The temperature dependance of resistance is given by the formula, 𝑅# = 𝑅$ 1 + 𝛼 𝑡# − 𝑡$ 𝑅# 𝑎𝑛𝑑 𝑅$ − 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑡 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑡# 𝑎𝑛𝑑 𝑡$ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 𝛼 − 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡, 𝑐𝑒𝑙𝑠𝑖𝑢𝑠 %$ This is derive from, 𝑅# 𝑇 + 𝑡# = 𝑅$ 𝑇 + 𝑡$ 𝑇 − 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑓𝑜𝑟 𝑎 𝑐𝑒𝑟𝑡𝑎𝑖𝑛 𝑘𝑖𝑛𝑑 𝑜𝑓 𝑐𝑎𝑏𝑙𝑒, 𝑐𝑒𝑙𝑠𝑖𝑢𝑠 INDUCTANCE The inductance of a transmission line is calculated as flux linkages per ampere and if permeability is constant, the sinusoidal current produces sinusoidal varying flux in phase in current as shown by the equation, 𝜆 𝐿= 𝐼 𝑡 𝑤ℎ𝑒𝑟𝑒: 𝐿 − 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒, ℎ𝑒𝑛𝑟𝑦 𝜆 − 𝑓𝑙𝑢𝑥 𝑙𝑖𝑛𝑘𝑎𝑔𝑒𝑠, 𝑊𝑏. 𝐼 − 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝑎𝑚𝑝𝑒𝑟𝑒 𝑚  The inductance of a transmission line is calculated as flux linkages per ampere and if permeability is constant, the sinusoidal current produces sinusoidal varying flux in phase in current as shown by the equation, 𝜆 𝐿= 𝐼 𝑡 𝑤ℎ𝑒𝑟𝑒: 𝐿 − 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒, ℎ𝑒𝑛𝑟𝑦 𝜆 − 𝑓𝑙𝑢𝑥 𝑙𝑖𝑛𝑘𝑎𝑔𝑒𝑠, 𝑊𝑏. 𝐼 − 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝑎𝑚𝑝𝑒𝑟𝑒 𝑚 In order to obtain an accurate value for inductance of a transmission line, consider the flux inside each conductor as well as the external flux as shown in the figure below,  By Ampere’s law, the magnetomotive force (mmf) in ampere-turns around any closed path is equal to the net current in amperes enclosed by the path as shown by equation, 𝑚𝑚𝑓 = H 𝐻 𝑑𝑠 = 𝐼 𝐴𝑡 𝐴𝑡 𝑤ℎ𝑒𝑟𝑒: 𝐻 − 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦, 𝑠 − 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑙𝑜𝑛𝑔 𝑝𝑎𝑡ℎ, 𝑚. 𝐼 − 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑, 𝐴 𝑚 #( #( 𝐼& = H 𝐻& 𝑑𝑠 = 𝐻& H 𝑑𝑠 = 𝐻& 𝑥 2𝜋 − 0 = 2𝜋𝑥𝐻& ' '  By Ampere’s law, the magnetomotive force (mmf) in ampere-turns around any closed path is equal to the net current in amperes enclosed by the path as shown by equation, 𝑚𝑚𝑓 = $ 𝐻 𝑑𝑠 = 𝐼 𝐴𝑡 𝐴𝑡 𝑤ℎ𝑒𝑟𝑒: 𝐻 − 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦, 𝑠 − 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑙𝑜𝑛𝑔 𝑝𝑎𝑡ℎ, 𝑚. 𝐼 − 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑, 𝐴 𝑚 #$ #$ 𝐼! = $ 𝐻! 𝑑𝑠 = 𝐻! $ 𝑑𝑠 = 𝐻! 𝑥 2𝜋 − 0 = 2𝜋𝑥𝐻! " " 𝜋𝑟 # 𝑠𝑖𝑛𝑐𝑒, 𝐼! = # 𝐼 𝜋𝑥 𝜋𝑟 # 𝑥 𝑡ℎ𝑒𝑛, 𝐼 = 2𝜋𝑥𝐻! −−→ 𝐻! = 𝐼 𝜋𝑥 # 2𝜋𝑟 # 𝜇𝑥 𝐵! = 𝜇𝐻! = 𝐼 2𝜋𝑟 # 𝑊𝑏 𝑤ℎ𝑒𝑟𝑒: 𝐵! − 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑚# 𝜇 − 𝑝𝑒𝑟𝑚𝑒𝑎𝑏𝑖𝑙𝑖𝑡𝑦, (𝑤𝑖𝑡ℎ 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 4𝜋 𝑥 10%& 𝑎𝑡 𝑓𝑟𝑒𝑒 𝑠𝑝𝑎𝑐𝑒 𝑎𝑡 𝑆𝐼 𝑢𝑛𝑖𝑡 𝑜𝑓 1)  Considering a tubular element of thickness “x” , the flux is Bx times the cross sectional area of the element normal to the flux lines. The flux per meter length is given by, 𝜇𝑥𝐼 𝑑𝜙 = 𝑑𝑥 2𝜋𝑟 # 𝜋𝑥 # 𝜋𝑥 # 𝜇𝑥𝐼 𝐼𝜇𝑥 ) 𝑑𝜆 = # 𝑑𝜙 = ( # ) 𝑑𝑥 = 𝑑𝑥 𝜋𝑟 𝜋𝑟 2𝜋𝑟 # 2𝜋𝑟 *.. 𝐼𝜇𝑥 ) 𝐼𝜇 ) 𝑑𝑥 = 𝑟 * 0 * 𝐼𝜇 𝐼𝜇 𝜆+,- =H * 𝑑𝑥 = H 𝑥 − = ' 2𝜋𝑟 2𝜋𝑟 * ' 4 4 2𝜋𝑟 * 8𝜋 𝑊𝑏 𝑤ℎ𝑒𝑟𝑒: 𝜙 − 𝑓𝑙𝑢𝑥 𝑝𝑒𝑟 𝑚𝑒𝑡𝑒𝑟 𝑙𝑒𝑛𝑔𝑡ℎ, 𝑚  Considering a tubular element of thickness “x” , the flux is Bx times the cross sectional area of the element normal to the flux lines. The flux per meter length is given by, 𝜇𝑥𝐼 𝑑𝜙 = 𝑑𝑥 2𝜋𝑟 # 𝜋𝑥 # 𝜋𝑥 # 𝜇𝑥𝐼 𝐼𝜇𝑥 ' 𝑑𝜆 = # 𝑑𝜙 = ( # ) 𝑑𝑥 = 𝑑𝑥 𝜋𝑟 𝜋𝑟 2𝜋𝑟 # 2𝜋𝑟 ( , , 𝐼𝜇𝑥 ' 𝐼𝜇 ' 𝑟 ( 0( 𝐼𝜇 𝐼𝜇 𝜆)*+ =$ ( 𝑑𝑥 = $ 𝑥 𝑑𝑥 = − = " 2𝜋𝑟 2𝜋𝑟 ( " 4 4 2𝜋𝑟 ( 8𝜋 𝑊𝑏 𝑤ℎ𝑒𝑟𝑒: 𝜙 − 𝑓𝑙𝑢𝑥 𝑝𝑒𝑟 𝑚𝑒𝑡𝑒𝑟 𝑙𝑒𝑛𝑔𝑡ℎ, 𝑚 For relative permeability of 1, assuming a free space, 𝐼(4𝜋 𝑥 10%& ) 𝐼 𝜆)*+ = = 𝑥10%& 8𝜋 2 𝐼 𝜆)*+ 2 𝑥10%& 1 𝐻 𝐿)*+ = = = 𝑥10%& 𝐼 𝐼 2 𝑚 Consider a conductor with external points P1 and P2 as shown by the figure below, Consider a conductor with external points P1 and P2 as shown by the figure below, Recall, the mmf around the element and the magnetic flux density is given by, 𝐼 𝐼 = 2𝜋𝑥𝐻! −−→ 𝐻! = 2𝜋𝑥 𝐼 𝜇𝐼 𝐵! = 𝜇𝐻! = 𝜇( )= 2𝜋𝑥 2𝜋𝑥 Consider the distance D1 and D2 which is the distance of points P1 and P2 from the center of the conductor respectively, so between these two points the flux linkages is given by, /" 𝜇𝐼 𝜇𝐼 /" 𝑑𝑥 𝜇𝐼 𝐷# 𝜆$# = H 𝑑𝑥 = H = ln /! 2𝜋𝑥 2𝜋 /! 𝑥 2𝜋 𝐷$ For a relative permeability of 1, 𝐷# 𝜆$# = 2𝑥10%0 𝐼 ln 𝐷$ Consider the distance D1 and D2 which is the distance of points P1 and P2 from the center of the conductor respectively, so between these two points the flux linkages is given by, /" 𝜇𝐼 𝜇𝐼 /" 𝑑𝑥 𝜇𝐼 𝐷# 𝜆$# = H 𝑑𝑥 = H = ln /! 2𝜋𝑥 2𝜋 /! 𝑥 2𝜋 𝐷$ For a relative permeability of 1, 𝐷# 𝜆$# = 2𝑥10%0 𝐼 ln 𝐷$ The inductance due to flux only between 1 and 2, 𝐷# 𝐿$# = 2𝑥10%0 ln 𝐷$  Consider a two wire circuit with radius r1 and r2 separated by distance D from each other, one is return circuit for another vice versa, Replacing D2 with D and D1 with r1, to determine the inductance of the circuit due to one conductor only will yield, 1 𝐿+,- = 𝑥10%0 2 %0 𝐷 𝐿# = 2𝑥10 ln 𝑟$ 1 𝐷 𝐿$ = 𝐿+,- + 𝐿# = + 2 ln 𝑥10%0 2 𝑟$  Consider a two wire circuit with radius r1 and r2 separated by distance D from each other, one is return circuit for another vice versa, Replacing D2 with D and D1 with r1, to determine the inductance of the circuit due to one conductor only will yield, 1 𝐿#$% = 𝑥10&' 2 𝐷 𝐿( = 2𝑥10&' ln 𝑟) 1 𝐷 𝐿) = 𝐿#$% + 𝐿( = + 2 ln 𝑥10&' 2 𝑟) u Dividing both sides by 2 will yield, u 𝐿) 1 1 𝐷 = ( ) + 2 ln 𝑥10&' 2 2 2 𝑟)  $ 1 𝑠𝑖𝑛𝑐𝑒 ln 𝑒* = , 𝑡ℎ𝑒𝑛 4 $ 𝐷 𝐿$ = 2𝑥10%0 ln 𝑒* + ln 𝑟$ 𝐷 𝐷 𝐿$ = 2𝑥10%0 ln $ = 2𝑥10%0 ln % 𝑟$ ′ 𝑒 * 𝑟$ 𝐷 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 𝑓𝑜𝑟 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 2, 𝐿# = 2𝑥10%0 ln 𝑟# ′  $ 1 𝑠𝑖𝑛𝑐𝑒 ln 𝑒* = , 𝑡ℎ𝑒𝑛 4 $ 𝐷 𝐿$ = 2𝑥10%0 ln 𝑒* + ln 𝑟$ 𝐷 𝐷 𝐿$ = 2𝑥10%0 ln $ = 2𝑥10%0 ln % 𝑟$ ′ 𝑒 * 𝑟$ 𝐷 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 𝑓𝑜𝑟 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 2, 𝐿# = 2𝑥10%0 ln 𝑟# ′ Thus, for the complete circuit, 𝐷 𝐷 𝐷 𝐿 = 𝐿$ + 𝐿# = 2𝑥10%0 ln + 2𝑥10%0 ln = 4𝑥10%0 ln 𝑟$ ′ 𝑟# ′ 𝑟$ ′𝑟# ′ If the radius is equal for both of the conductor, 𝐷 𝐿 = 4𝑥10%0 ln 𝑟′ 𝑤ℎ𝑒𝑟𝑒: 𝐷 − 𝑚𝑢𝑡𝑢𝑎𝑙 𝐺𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑀𝑒𝑎𝑛 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐺𝑀𝐷 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑟 1 − 𝐺𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑀𝑒𝑎𝑛 𝑅𝑎𝑑𝑖𝑢𝑠 𝐺𝑀𝑅 𝑜𝑟 𝑠𝑒𝑙𝑓 𝐺𝑀𝐷 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 Consider a group of conductors where the sum of the currents is zero as shown by the figure below, The flux linkages of conductor 1 due to I1 is given by, 𝐼$ 𝐷$2 𝐷$2 𝜆$2$ = + 2𝐼$ ln 𝑥10%0 = 2𝑥10%0 𝐼$ ln 1 2 𝑟$ 𝑟$ Consider a group of conductors where the sum of the currents is zero as shown by the figure below, The flux linkages of conductor 1 due to I1 is given by, 𝐼$ 𝐷$2 𝐷$2 𝜆$2$ = + 2𝐼$ ln 𝑥10%0 = 2𝑥10%0 𝐼$ ln 1 2 𝑟$ 𝑟$ The flux linkages with conductor 1 due to I2 but excluding flux beyond point P is given by, 𝐷#2 𝜆$2# = 2𝑥10%0 𝐼# ln 𝐷$# Consider a group of conductors where the sum of the currents is zero as shown by the figure below, The flux linkages of conductor 1 due to I1 is given by, 𝐼- 𝐷-. 𝐷-. 𝜆-.- = + 2𝐼- ln 𝑥10%& = 2𝑥10%& 𝐼- ln / 2 𝑟- 𝑟- The flux linkages with conductor 1 due to I2 but excluding flux beyond point P is given by, 𝐷#. 𝜆-.# = 2𝑥10%& 𝐼# ln 𝐷-# Expanding, 𝐷-. 𝐷#. 𝐷'. 𝐷*. 𝜆-. = 2𝑥10%& (𝐼- ln + 𝐼# ln + 𝐼' ln + ⋯ + 𝐼* ln ) 𝑟-/ 𝐷-# 𝐷-' 𝐷-* 1 1 1 = 2𝑥10%& (𝐼- ln / + 𝐼# ln + 𝐼' ln + ⋯ + 𝐼- ln 𝐷-0 + 𝐼# ln 𝐷#0 + 𝐼' ln 𝐷'0 + ⋯ + 𝐼* ln 𝐷*. ) 𝑟- 𝐷-# 𝐷-' Since the sum of all the currents is zero, 𝐼$ + 𝐼# + 𝐼) + ⋯ + 𝐼, = 0 𝐼, = −(𝐼$ + 𝐼# + 𝐼) + ⋯ + 𝐼,%$ ) Since the sum of all the currents is zero, 𝐼$ + 𝐼# + 𝐼) + ⋯ + 𝐼, = 0 𝐼, = −(𝐼$ + 𝐼# + 𝐼) + ⋯ + 𝐼,%$ ) Substituting In in equation , 1 1 1 1 𝐷$3 𝐷#3 𝐷)3 𝐷(,%$)3 𝜆$2 = 2𝑥10%0 (𝐼$ ln 1 + 𝐼# ln + 𝐼) ln + ⋯ + 𝐼, ln + 𝐼$ ln + 𝐼# ln + 𝐼) ln + ⋯ + 𝐼,%$ ln ) 𝑟$ 𝐷$# 𝐷$) 𝐷$, 𝐷,3 𝐷,3 𝐷,3 𝐷,3 1 1 1 1 𝜆$ = 2𝑥10%0 (𝐼$ ln + 𝐼# ln + 𝐼) ln + ⋯ + 𝐼, ln ) 𝑟$1 𝐷$# 𝐷$) 𝐷$, Consider a single phase line with composite conductors X and Y, Using equation to filament a of conductor X, we obtain, 𝐼 1 1 1 1 𝐼 1 1 1 1 𝜆6 = 2𝑥10%0 ln + ln + ln + ⋯ + 𝐼6 ln − 2𝑥10 %0 ln + ln + ln … + ln 𝑛 𝑟61 𝐷6# 𝐷6) 𝐷6, 𝑚 𝐷661 𝐷671 𝐷681 𝐷69 Using equation to filament a of conductor X, we obtain, 𝐼 1 1 1 1 𝐼 1 1 1 1 𝜆6 = 2𝑥10%0 ln + ln + ln + ⋯ + 𝐼6 ln − 2𝑥10 %0 ln + ln + ln … + ln 𝑛 𝑟61 𝐷6# 𝐷6) 𝐷6, 𝑚 𝐷661 𝐷671 𝐷681 𝐷69 * 𝐷66+ 𝐷67+ 𝐷68 + … 𝐷69 𝜆6 = 2𝑥10%0 𝐼 ln , 𝑟61 𝐷67 𝐷68 … 𝐷6, * 𝜆6 𝐷66+ 𝐷67+ 𝐷68 + … 𝐷69 𝐿6 = = 2𝑛𝑥10%0 𝐼 ln , 1 𝑟61 𝐷67 𝐷68 … 𝐷6, 𝑛 * 𝜆7 𝐷76+ 𝐷77+ 𝐷78 + … 𝐷79 𝐿7 = = 2𝑛𝑥10%0 𝐼 ln 1 , 𝑟71 𝐷76 𝐷78 … 𝐷7, 𝑛 If all the filaments have the same inductance, the inductance of the conductor would be 1/n times the inductance of one filament. The average inductance of a given filaments of conductors is given by is given by, 𝐿6 + 𝐿7 + 𝐿8 + ⋯ + 𝐿, 𝐿6: = 𝑛 If all the filaments have the same inductance, the inductance of the conductor would be 1/n times the inductance of one filament. The average inductance of a given filaments of conductors is given by is given by, 𝐿6 + 𝐿7 + 𝐿8 + ⋯ + 𝐿, 𝐿6: = 𝑛 In this case, all filaments have different inductances, the inductance of conductor X is given by, 𝐿6: 𝐿6 + 𝐿7 + 𝐿8 + ⋯ + 𝐿, 𝐿6: = = 𝑛 𝑛# If all the filaments have the same inductance, the inductance of the conductor would be 1/n times the inductance of one filament. The average inductance of a given filaments of conductors is given by is given by, 𝐿6 + 𝐿7 + 𝐿8 + ⋯ + 𝐿, 𝐿6: = 𝑛 In this case, all filaments have different inductances, the inductance of conductor X is given by, 𝐿6: 𝐿6 + 𝐿7 + 𝐿8 + ⋯ + 𝐿, 𝐿6: = = 𝑛 𝑛# Substituting the logarithmic expression for inductance of each filament, *, (𝐷66+ 𝐷67+ 𝐷68 + … 𝐷69 )(𝐷76+ 𝐷77+ 𝐷78 + … 𝐷79 ) 𝐿& = 2𝑥10%0 ln , 𝑟71 𝐷76 𝐷78 … 𝐷7, Let Dm be the GMD and Ds be the GMR, 𝐷9 𝐿& = 2𝑥10%0 ln 𝐷; Let Dm be the GMD and Ds be the GMR, 𝐷9 𝐿& = 2𝑥10%0 ln 𝐷; Thus, the inductance of the line is, 𝐿 = 𝐿& + 𝐿< Let Dm be the GMD and Ds be the GMR, 𝐷9 𝐿& = 2𝑥10%0 ln 𝐷; Thus, the inductance of the line is, 𝐿 = 𝐿& + 𝐿< Thus, the reactive inductance is given by, 𝐷9 𝐷9 𝑋= = 2𝜋𝑓𝐿& = 2𝜋𝑓 2𝑥10%0 ln = 4𝜋𝑓𝑥10%0 𝑓 ln 𝑜ℎ𝑚/𝑚 𝐷; 𝐷; 𝐷9 1609 𝑚 𝐷9 𝑋= = 4𝜋𝑓𝑥10%0 𝑓 ln 𝑥 = 2.022𝑥10%) 𝑓 ln 𝑜ℎ𝑚/𝑚𝑖 𝐷; 1 𝑚𝑖𝑙𝑒 𝐷;  Assuming an equilateral spaced conductor given by the figure below and a balanced three phase phasor, the inductance is given by,  Assuming an equilateral spaced conductor given by the figure below and a balanced three phase phasor, the inductance is given by, Since Ia=-Ib-Ic, %0 1 1 %0 𝐷 𝜆6 = 2𝑥10 𝐼6 ln − 𝐼6 ln = 2𝑥10 𝐼6 ln 𝐷; 𝐷 𝐷; %0 𝐷 𝐿6 = 2𝑥10 𝐼6 ln 𝐷;  When the conductors of a three phase line is unsymmetrical, the flux linkages of each phase are not the same and thus, causing an unbalanced circuit.  When the conductors of a three phase line is unsymmetrical, the flux linkages of each phase are not the same and thus, causing an unbalanced circuit. Thus, the inductance for this given configuration is given by, 𝐷>? 𝐿6 = 2 𝑥 10%0 ln 𝐷; - 𝑖𝑛 𝑤ℎ𝑖𝑐ℎ 𝐷>? = 𝐷$# 𝐷#) 𝐷)$ 𝜇' 𝐷>? 1 𝐷>? 𝐿= (1 + 4 ln ) = 2 + ln 𝑥10%0 8𝜋 𝐷; 4 𝐷; CAPACITANCE If a long, straight cylindrical conductor lies in a uniform medium such as air and is isolated from other charges so that the charge is uniformly distributed around is periphery, the flux is radial as shown in an isolated conductor below. CAPACITANCE If a long, straight cylindrical conductor lies in a uniform medium such as air and is isolated from other charges so that the charge is uniformly distributed around is periphery, the flux is radial as shown in an isolated conductor below. The electric flux density is given by, 1 𝐶 𝐷@ = 2𝜋𝑥 𝑚# 𝑐𝑜𝑢𝑙𝑢𝑚𝑏𝑠 𝑤ℎ𝑒𝑟𝑒: 𝑞 − 𝑐ℎ𝑎𝑟𝑔𝑒, 𝑥 − 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑚𝑒𝑡𝑒𝑟 𝑚𝑒𝑡𝑒𝑟 The electric field density is given by, 𝑞 𝑉 𝐸= 2𝜋𝑥𝑘 𝑚 𝐹 𝑤ℎ𝑒𝑟𝑒: 𝑘 − 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑖𝑣𝑖𝑡𝑦, 𝑓𝑜𝑟 𝑓𝑟𝑒𝑒 𝑠𝑝𝑎𝑐𝑒, 8.85 𝑥 10%$# 𝑚 The electric field density is given by, 𝑞 𝑉 𝐸= 2𝜋𝑥𝑘 𝑚 𝐹 𝑤ℎ𝑒𝑟𝑒: 𝑘 − 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑖𝑣𝑖𝑡𝑦, 𝑓𝑜𝑟 𝑓𝑟𝑒𝑒 𝑠𝑝𝑎𝑐𝑒, 8.85 𝑥 10%$# 𝑚 Consider a long, straight wire carrying a positive charge of q C/m with points P1 and P2 are located at distances D1 and D2 meters respectively, as shown in the figure below, Thus, the instantaneous voltage drop between P1 and P2 is, /" /" 𝑞 𝑞 𝐷# 𝑣$# = H 𝐸 𝑑𝑥 = H 𝑑𝑥 = ln 𝑉 /! /! 2𝜋𝑥𝑘 2𝜋𝑥𝑘 𝐷$ 𝑤ℎ𝑒𝑟𝑒: 𝑣$# − 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑒𝑜𝑢𝑠 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑑𝑟𝑜𝑝 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑃$ 𝑎𝑛𝑑 𝑃# , 𝑣𝑜𝑙𝑡𝑠 Thus, the instantaneous voltage drop between P1 and P2 is, /" /" 𝑞 𝑞 𝐷# 𝑣$# = H 𝐸 𝑑𝑥 = H 𝑑𝑥 = ln 𝑉 /! /! 2𝜋𝑥𝑘 2𝜋𝑥𝑘 𝐷$ 𝑤ℎ𝑒𝑟𝑒: 𝑣$# − 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑒𝑜𝑢𝑠 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑑𝑟𝑜𝑝 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑃$ 𝑎𝑛𝑑 𝑃# , 𝑣𝑜𝑙𝑡𝑠 The capacitance per unit length of the unit length, 𝑞 𝐹 𝐶= 𝑣 𝑚 Consider a single phase two wire as shown in the figure,  Consider a single phase two wire as shown in the figure, The potential difference between the two conductors is given by, 𝑞6 𝐷 𝑞6 𝑟7 𝑉67 = ln + ln 2𝜋𝑘 𝑟6 2𝜋𝑘 𝐷 Since qa+qb=0 for a two wire line, 𝑞6 𝐷 𝑟7 𝑞6 𝐷# 𝑉67 = ln − ln = (ln ) 2𝜋𝑘 𝑟6 𝐷 2𝜋𝑘 𝑟6 𝑟7 𝑞6 2𝜋𝑘 𝐶67 = = 𝑉67 𝐷# ln( ) 𝑟6 𝑟7  Consider a single phase two wire as shown in the figure, The potential difference between the two conductors is given by, 𝑞. 𝐷 𝑞. 𝑟/ 𝑉./ = ln + ln 2𝜋𝑘 𝑟. 2𝜋𝑘 𝐷 Since qa+qb=0 for a two wire line, 𝑞. 𝐷 𝑟/ 𝑞. 𝐷( 𝑉./ = ln − ln = (ln ) 2𝜋𝑘 𝑟. 𝐷 2𝜋𝑘 𝑟. 𝑟/ 𝑞. 2𝜋𝑘 𝐶./ = = 𝑉./ 𝐷( ln(𝑟 𝑟 ). / If ra=rb=r, 𝜋𝑘 𝐶./ = 𝐷 ln( 𝑟 )  If tap by a transformer having a grounded center tap, the potential difference between each conductor is halved which gives the equation which is shown in the figure, ?0 #(A D 𝐶, = 𝐶6, = 𝐶7, = 102 = 3 𝑡𝑜 𝑛𝑒𝑢𝑡𝑟𝑎𝑙 BC( ) 9 " 4 𝑤ℎ𝑒𝑟𝑒: 𝑟 − 𝑎𝑐𝑡𝑢𝑎𝑙 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑟𝑎𝑑𝑖𝑢𝑠, 𝑚𝑒𝑡𝑒𝑟  If tap by a transformer having a grounded center tap, the potential difference between each conductor is halved which gives the equation which is shown in the figure, ?0 #(A D 𝐶, = 𝐶6, = 𝐶7, = 102 = 3 𝑡𝑜 𝑛𝑒𝑢𝑡𝑟𝑎𝑙 BC( ) 9 " 4 𝑤ℎ𝑒𝑟𝑒: 𝑟 − 𝑎𝑐𝑡𝑢𝑎𝑙 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑟𝑎𝑑𝑖𝑢𝑠, 𝑚𝑒𝑡𝑒𝑟 The capacitive reactance is given by (at free space), 1 2.862 E 𝐷 𝑥8 = = 𝑥 10 ln 𝑜ℎ𝑚. 𝑚𝑒𝑡𝑒𝑟 𝑡𝑜 𝑛𝑒𝑢𝑡𝑟𝑎𝑙 2𝜋𝑓𝐶 𝑓 𝑟 2.862 𝐷 1 𝑚𝑖𝑙𝑒 1.779 𝐷 𝑥8 = 𝑥 10E ln 𝑜ℎ𝑚. 𝑚𝑒𝑡𝑒𝑟 𝑥 = 𝑥10F ln 𝑜ℎ𝑚. 𝑚𝑖𝑙𝑒 𝑡𝑜 𝑛𝑒𝑢𝑡𝑟𝑎𝑙 𝑓 𝑟 1609 𝑚𝑒𝑡𝑒𝑟 𝑓 𝑟