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POWER SUPPLY SYSTEMS& DISTRIBUTION SYSTEMS

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Duddu Rajesh
Duddu Rajesh

Transmission and distribution systems- D.C 2-wire and 3- wire systems, A.C single phase, three phase and 4-wire systems, comparison of copper efficiency. Primary and secondary distribution systems, concentrated & uniformly distributed loads on distributors fed at both ends, ring distributor, voltage drop and power loss calculation, Kelvin’slaw.

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By D.Rajesh Asst. Prof. EEE Department 6 October 2020 D.Rajesh Asst.Prof. EEE Department 1
 Unit-i Electric Power Generation& Economic Considerations  Unit-ii Power Supply Systems& Distribution Systems  Unit-iii Inductance & Capacitance Calculations  Unit-iv Performance Of Transmission Lines  Unit-v Overhead Line Insulators 6 October 2020 D.Rajesh Asst.Prof. EEE Department 2
POWER SUPPLY SYSTEMS& DISTRIBUTION SYSTEMS:  Transmission and distribution systems- D.C 2-wire and 3- wire systems, A.C single phase , three phase and 4-wire systems, comparison of copper efficiency.  Primary and secondary distribution systems, concentrated & uniformly distributed loads on distributors fed at both ends, ring distributor, voltage drop and power loss calculation,  Kelvin’slaw. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 3
Power Transmission D.C D.C. two-wire D.C. two-wire with mid-point earthed D.C. three-wire A.C Single-phase A.C. system Single-phase two- wire Single-phase two- wire with mid- point earthed Single-phase three-wire Two-phase A.C. system Two-phase four- wire Two-phase three wire. Three-phase A.C. system Three-phase three-wire Three-phase four- wire 6 October 2020 D.Rajesh Asst.Prof. EEE Department 4
6 October 2020 D.Rajesh Asst.Prof. EEE Department 5 Overhead Under ground
(i) Same power (P watts) transmitted by each system. (ii) The distance (l metres) over which power is transmitted remains the same. (iii) The line losses (W watts) are the same in each case. (iv) The maximum voltage between any conductor and earth (𝑉 π‘š) is the same in each case. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 6
Max. voltage between conductors = 𝑉 π‘š , Power to be transmitted = P= 𝑉 π‘šπΌ1 1 ∴ Load current, 𝐼1 = P/ 𝑉 π‘š If 𝑅1 is the resistance of each line conductor, then, 𝑅1 = ρ l/ π‘Ž1 where π‘Ž1 is the area of X-section of the conductor. copper losses in line W = 𝐼1 2 𝑅1 + 𝐼1 2 𝑅1 =2𝐼1 2 𝑅1 6 October 2020 D.Rajesh Asst.Prof. EEE Department 7 1 1 2
The maximum voltage between any conductor and earth is 𝑉 π‘š so that maximum voltage between conductors is 2 𝑉 π‘š. 𝑅2 is the resistance of each line conductor Power to be transmitted P= 𝑉 π‘šπΌ2+ 𝑉 π‘šπΌ2=2 𝑉 π‘šπΌ2 Load current, 𝐼2 = P/2 𝑉 π‘š Since from assumption 1 the power transmitted is same for all systems so from eq 1& 3 2𝑉 π‘šπΌ2= 𝑉 π‘šπΌ1 𝐼2 = 𝐼1 2 copper losses in line W = 𝐼2 2 𝑅2+ 𝐼2 2 𝑅2 = 2 𝐼2 2 𝑅2 From eq 4 W=2( 𝐼1 2 )2 𝑅2 = 𝐼1 2 2 𝑅2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 8 3 4 5
 Compare eq 2 & 5 2𝐼1 2 𝑅1 = ( 𝐼1 2 2 ) 𝑅2 𝑅1 𝑅2 = 1 4 π‘Ž1 π‘Ž2 =4 (where R = ρ l/ π‘Ž) π‘Ž1=X-area of conductor in 2 wire dc system π‘Ž2= X-area of conductor in 2 wire dc system with mid-point earthed Then volume of copper wire Twoβˆ’wire D.C system with midβˆ’point earthed volume of copper wire Twoβˆ’wire D.C system = π‘Ž2βˆ—π‘™2 π‘Ž1βˆ—π‘™1 = 1 4 π‘Ž1βˆ—2𝑙 π‘Ž1βˆ—2𝑙 = 1 4 where 𝑙1 = 𝑙2=l+l Hence, the volume of conductor material required in Twoβˆ’wire D.C system with midβˆ’point earthed is one-fourth of that required in a two-wire D.C system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 9
If the load is balanced, the current in the neutral wire is zero. Assuming balanced loads, Power to be transmitted P= 𝑉 π‘šπΌ3+ 𝑉 π‘šπΌ3=2 𝑉 π‘šπΌ3 6 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 6 2𝑉 π‘šπΌ3= 𝑉 π‘šπΌ1 𝐼3 = 𝐼1 2 7 Let 𝑅3 is the resistance of each line conductor, the area of X-section of neutral wire to be half that of the outer wire, copper losses in line W = 𝐼3 2 𝑅3+ 𝐼3 2 𝑅3 = 2𝐼3 2 𝑅3 8 From eq 7 W=2( 𝐼1 2 )2 𝑅3 = 𝐼1 2 2 𝑅3 9 6 October 2020 D.Rajesh Asst.Prof. EEE Department 10
 Compare eq 2 & 8 2𝐼1 2 𝑅1 = 𝐼1 2 2 𝑅3 𝑅1 𝑅3 = 1 4 𝑅3 = ρ l/ π‘Ž3 where π‘Ž3 is the area of X-section of the conductor. π‘Ž1 π‘Ž3 =4 Then volume of copper wire Threeβˆ’wire D.C volume of copper wire Twoβˆ’wire D.C system = π‘Ž3βˆ—π‘™3 π‘Ž1βˆ—π‘™1 = 1 4 π‘Ž1βˆ—2.5𝑙 π‘Ž1βˆ—2𝑙 = 2.5 8 where 𝑙3=l+l+0.5l =0.3125 Hence, the volume of conductor material required in Threeβˆ’wire D.C system is 0.3125 times of that required in a two-wire D.C system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 11
The maximum voltage between conductors is 𝑉 π‘š, so that r.m.s. value of voltage between them is π‘‰π‘š 2 . Assuming the load power factor to be cos Ο†, load current 𝐼4 Power supplied P= π‘‰π‘š 2 𝐼4 cos Ο† 10 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 10 π‘‰π‘š 2 𝐼4 cos Ο†= 𝑉 π‘šπΌ1 𝐼4= 2𝐼1 cos Ο† 11 Let 𝑅4 is the resistance of each line conductor copper losses in line W = 𝐼4 2 𝑅4+ 𝐼4 2 𝑅4 = 2𝐼4 2 𝑅4 =2( 2𝐼1 cos Ο† )2 𝑅4 =4 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅4 12 6 October 2020 D.Rajesh Asst.Prof. EEE Department 12
 Comparing copper losses with 2 wire d.c system 4 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅4= 2𝐼1 2 𝑅1 𝑅4 𝑅1 = π‘π‘œπ‘ 2Ο† 2 π‘Ž4 π‘Ž1 = 2 π‘π‘œπ‘ 2Ο† 13 volume of copper wire 1βˆ’phase 2βˆ’ wire volume of copper wire Twoβˆ’wire D.C system = π‘Ž4𝑙4 π‘Ž1𝑙1 = 2 π‘π‘œπ‘ 2Ο†π‘Ž12𝑙 π‘Ž12𝑙 = 2 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required in this system is 2 π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system with the one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 13
 The two wires possess equal and opposite voltages to earth (i.e., 𝑉 π‘š).  Therefore, the maximum voltage between the two wires is π‘‰π‘š 2 . The r.m.s. value of voltage between conductors is =2 π‘‰π‘š 2 = 2 𝑉 π‘š. Assuming the power factor of the load to be cos Ο†, load current 𝐼5  Power supplied P= 2 𝑉 π‘šπΌ5 cos Ο† 14 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 14 2 𝑉 π‘šπΌ5 cos Ο†= 𝑉 π‘šπΌ1 𝐼5= 𝐼1 2 cos Ο† 15 6 October 2020 D.Rajesh Asst.Prof. EEE Department 14
Let 𝑅5 is the resistance of each line conductor copper losses in line W = 𝐼5 2 𝑅5+ 𝐼5 2 𝑅5 = 2𝐼5 2 𝑅5 =2( 𝐼1 2 cos Ο† )2 𝑅5 = 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅5 Comparing copper losses with 2 wire d.c system 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅5 =2𝐼1 2 𝑅1 𝑅5 𝑅1 = 2π‘π‘œπ‘ 2Ο† π‘Ž5 π‘Ž1 = 1 2π‘π‘œπ‘ 2Ο† volume of copper wire 1βˆ’phase 2βˆ’ wire π‘€π‘–π‘‘β„Ž π‘šπ‘–π‘‘ βˆ’ π‘π‘œπ‘–π‘›π‘‘ π‘’π‘Žπ‘Ÿπ‘‘β„Žπ‘’π‘‘ volume of copper wire Twoβˆ’wire D.C system π‘Ž5𝑙5 π‘Ž1𝑙1 = 1 2π‘π‘œπ‘ 2Ο† π‘Ž12𝑙 π‘Ž12𝑙 = 1 2π‘π‘œπ‘ 2Ο† Hence the volume of conductor material required in this system is 1 2π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 15
 If the load is balanced, the current through the neutral wire is zero. Assuming balanced load,  The two wires possess equal and opposite voltages to earth (i.e., 𝑉 π‘š).  Therefore, the maximum voltage between the two wires is π‘‰π‘š 2 . The r.m.s. value of voltage between conductors is 2 π‘‰π‘š 2 = 2 𝑉 π‘š. Assuming the power factor of the load to be cos Ο†, load current 𝐼6  Power supplied P= 2 𝑉 π‘šπΌ6 cos Ο† 16 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 16 2 𝑉 π‘šπΌ6 cos Ο†= 𝑉 π‘šπΌ1 𝐼6= 𝐼1 2 cos Ο† 17 6 October 2020 D.Rajesh Asst.Prof. EEE Department 16
Let 𝑅6 is the resistance of each line conductor copper losses in line W = 𝐼6 2 𝑅6+ 𝐼6 2 𝑅6 = 2𝐼6 2 𝑅6 =2( 𝐼1 2 cos Ο† )2 𝑅6 = 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅6 Comparing copper losses with 2 wire d.c system 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅6 =2𝐼1 2 𝑅1 𝑅6 𝑅1 = 2π‘π‘œπ‘ 2Ο† π‘Ž6 π‘Ž1 = 1 2π‘π‘œπ‘ 2Ο† volume of copper wire 1βˆ’phase3βˆ’wire system volume of copper wire Twoβˆ’wire D.C system π‘Ž6𝑙6 π‘Ž1𝑙1 = 1 2π‘π‘œπ‘ 2Ο† π‘Ž12.5𝑙 π‘Ž12𝑙 = 0.625 π‘π‘œπ‘ 2Ο† Hence the volume of conductr material required in this system is 0.625 π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 17
 This system can be considered as two independent single- phase systems, each transmitting one half of the total power.  As the voltage is same (𝑉 π‘š) as in a Single phase 2-wire system with mid-point earthed, so the current will be half of the Single phase system  i.e. 𝐼7= 𝐼5 2 = 𝐼1 2 cos Ο† 2 = 𝐼1 2 2 cos Ο†  If the current density remains constant the X-area will be half of the single-phase system  i.e. π‘Ž7= π‘Ž5 2 = 1 2π‘π‘œπ‘ 2Ο† π‘Ž1 2 = π‘Ž1 4π‘π‘œπ‘ 2Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 18
 Since wires are four  Then  volume of copper wireTwo phase, 4βˆ’wire a.c. system volume of copper wire Twoβˆ’wire D.C system π‘Ž7𝑙7 π‘Ž1𝑙1 = 1 4π‘π‘œπ‘ 2Ο† π‘Ž14𝑙 π‘Ž12𝑙 = 1 2π‘π‘œπ‘ 2Ο† Hence the volume of conductor material required in this system is 1 2π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 19
The third or neutral wire is taken from the junction of two-phase windings whose voltages are in quadrature with each other. Obviously, each phase transmits one half of the total power. The R.M.S. voltage between outgoing conductor and neutral = π‘‰π‘š 2 , Let 𝑅8 is the resistance of each line conductor Total power transmitted=2 π‘‰π‘š 2 𝐼8 cos Ο† Comparing the power with 2-wire D.C 2 π‘‰π‘š 2 𝐼8 cos Ο†= 𝑉 π‘šπΌ1 𝐼8 = 𝐼1 2cos Ο† The middle wire will act as a common return and the current through it will be 𝐼8 2 + 𝐼8 2 = 2 𝐼8 6 October 2020 D.Rajesh Asst.Prof. EEE Department 20
 Assuming the current density to be constant, the area of X-section of the neutral wire will be 2 times that of either of the outers.  ∴ Resistance of neutral wire= 𝑅8 2  Copper losses in this system W= 𝐼8 2 𝑅8+ 𝐼8 2 𝑅8 + ( 2 𝐼8)2𝑅8 2 = 2𝐼8 2 𝑅8+ 2𝐼8 2 𝑅8=(2+ 2) 𝐼8 2 𝑅8 Where 𝐼8 = 𝐼1 2cos Ο† W=(2+ 2) ( 𝐼1 2cos Ο†)2 𝑅8=(2+ 2) 𝐼1 2𝑅8 2π‘π‘œπ‘ 2Ο† Comparing copper losses with 2 wire d.c system (2+ 2) 𝐼1 2𝑅8 2π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅8 𝑅1 = 4π‘π‘œπ‘ 2Ο† (2+ 2) π‘Ž8 π‘Ž1 = (2+ 2) 4π‘π‘œπ‘ 2Ο† volume of copper wireTwo phase, 3βˆ’wire a.c. system volume of copper wire Twoβˆ’wire D.C system π‘Ž8𝑙8 π‘Ž1𝑙1 = (2+ 2) 4π‘π‘œπ‘ 2Ο† π‘Ž1(2+ 2)𝑙 π‘Ž12𝑙 = 1.457 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 1.457 π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 21
This system is almost universally adopted for transmission of electric power. The 3-phase, 3- wire system may be star connected or delta connected. Maximum voltage between each conductor and neutral =𝑉 π‘š Let 𝑅9 is the resistance of each line conductor R.M.S. voltage per phase = π‘‰π‘š 2 Power supplied=3( π‘‰π‘š 2 I9cos Ο† )  Comparing the power with 2-wire D.C 3 π‘‰π‘š 2 𝐼9 cos Ο†= 𝑉 π‘šπΌ1 𝐼9 = 2𝐼1 3cos Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 22
 Total copper losses= 3𝐼9 2 𝑅9 =3 ( 2𝐼1 3cos Ο† )2 𝑅9= 2𝐼1 2 𝑅9 3π‘π‘œπ‘ 2Ο†  Comparing copper losses with 2 wire d.c system 2𝐼1 2 𝑅9 3π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅9 𝑅1 = 3π‘π‘œπ‘ 2Ο† π‘Ž9 π‘Ž1 = 1 3π‘π‘œπ‘ 2Ο† volume of copper wire3βˆ’Phase, 3βˆ’wire system volume of copper wire Twoβˆ’wire D.C system = π‘Ž9𝑙9 π‘Ž1𝑙1 = 1 3π‘π‘œπ‘ 2Ο†π‘Ž13𝑙 π‘Ž12𝑙 = 0.5 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 0.5 π‘π‘œπ‘ 2Ο† times that required for 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 23
In this case, 4th or neutral wire is taken from the neutral point. The area of X- section of the neutral wire is generally one-half that of the line conductor. If the loads are balanced, then current through the neutral wire is zero. Assuming balanced loads and p.f. of the load as cos Ο†, Let 𝑅10 is the resistance of each line conductor R.M.S. voltage per phase = π‘‰π‘š 2 Power supplied=3( π‘‰π‘š 2 I10cos Ο† )  Comparing the power with 2-wire D.C 3 π‘‰π‘š 2 𝐼10 cos Ο†= 𝑉 π‘šπΌ1 𝐼10 = 2𝐼1 3cos Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 24
 Total copper losses= 3𝐼10 2 𝑅10 =3 ( 2𝐼1 3cos Ο† )2 𝑅10= 2𝐼1 2 𝑅10 3π‘π‘œπ‘ 2Ο†  Comparing copper losses with 2 wire d.c system 2𝐼1 2 𝑅10 3π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅10 𝑅1 = 3π‘π‘œπ‘ 2Ο† π‘Ž10 π‘Ž1 = 1 3π‘π‘œπ‘ 2Ο† volume of copper wire3βˆ’Phase, 3βˆ’wire system volume of copper wire Twoβˆ’wire D.C system = π‘Ž10𝑙10 π‘Ž1𝑙1 = 1 3π‘π‘œπ‘ 2Ο†π‘Ž13.5𝑙 π‘Ž12𝑙 = 0.583 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 0.583 π‘π‘œπ‘ 2Ο† times that required for 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 25
In the under-ground system the maximum stress exists on the insulation between the conductors. (i) Same power (P watts) transmitted by each system. (ii) The distance (l meters) over which power is transmitted remains the same. (iii) The line losses (W watts) are the same in each case. (iv) The maximum voltage between conductors (𝑉 π‘š) is the same in each case. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 26
Max. voltage between conductors = 𝑉 π‘š , Power to be transmitted = P= 𝑉 π‘šπΌ1 1 ∴ Load current, 𝐼1 = P/ 𝑉 π‘š If 𝑅1 is the resistance of each line conductor, then, 𝑅1 = ρ l/ π‘Ž1 where π‘Ž1 is the area of X-section of the conductor. copper losses in line W = 𝐼1 2 𝑅1 + 𝐼1 2 𝑅1 =2𝐼1 2 𝑅1 6 October 2020 D.Rajesh Asst.Prof. EEE Department 27 1 1 2
Maximum voltage between conductors is 𝑉 π‘š. 𝑅2 is the resistance of each line conductor Power to be transmitted P= 𝑉 π‘šπΌ2 Load current, 𝐼2 = P/ 𝑉 π‘š Since from assumption 1 the power transmitted is same for all systems so from eq 1& 3 𝑉 π‘šπΌ2= 𝑉 π‘šπΌ1 𝐼2 = 𝐼1 copper losses in line W = 𝐼2 2 𝑅2+ 𝐼2 2 𝑅2 = 2 𝐼2 2 𝑅2 From eq 4 W=2( 𝐼1)2 𝑅2 =2𝐼1 2 𝑅2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 28 3 4 5
 Compare eq 2 & 5 2𝐼1 2 𝑅1 = (𝐼1 2 ) 𝑅2 𝑅1 𝑅2 =1 π‘Ž1 π‘Ž2 =1 (where R = ρ l/ π‘Ž) π‘Ž1=X-area of conductor in 2 wire dc system π‘Ž2= X-area of conductor in 2 wire dc system with mid-point earthed Then volume of copper wire Twoβˆ’wire D.C system with midβˆ’point earthed volume of copper wire Twoβˆ’wire D.C system = π‘Ž2βˆ—π‘™2 π‘Ž1βˆ—π‘™1 = π‘Ž1βˆ—2𝑙 π‘Ž1βˆ—2𝑙 =1 where 𝑙1 = 𝑙2=l+l Hence, the volume of conductor material required in this system is the same as that for 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 29
If the load is balanced, the current in the neutral wire is zero. Assuming balanced loads, Power to be transmitted P= 𝑉 π‘šπΌ3 6 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 6 𝑉 π‘šπΌ3= 𝑉 π‘šπΌ1 𝐼3 = 𝐼1 7 Let 𝑅3 is the resistance of each line conductor, the area of X-section of neutral wire to be half that of the outer wire, copper losses in line W = 𝐼3 2 𝑅3+ 𝐼3 2 𝑅3 = 2𝐼3 2 𝑅3 8 From eq 7 W=2( 𝐼1)2 𝑅3 =2𝐼1 2 𝑅3 9 6 October 2020 D.Rajesh Asst.Prof. EEE Department 30
 Compare eq 2 & 8 2𝐼1 2 𝑅1 = 2(𝐼1 2 ) 𝑅3 𝑅1 𝑅3 =1 π‘Ž1 π‘Ž3 =1 𝑅3 = ρ l/ π‘Ž3 where π‘Ž3 is the area of X-section of the conductor. Then volume of copper wire Threeβˆ’wire D.C volume of copper wire Twoβˆ’wire D.C system = π‘Ž3βˆ—π‘™3 π‘Ž1βˆ—π‘™1 = π‘Ž1βˆ—2.5𝑙 π‘Ž1βˆ—2𝑙 = 2.5 2 where 𝑙3=l+l+0.5l =1.25 Hence, the volume of conductor material required in the system is 1Β·25 times that required for 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 31
The maximum voltage between conductors is 𝑉 π‘š, so that r.m.s. value of voltage between them is π‘‰π‘š 2 . Assuming the load power factor to be cos Ο†, load current 𝐼4 Power supplied P= π‘‰π‘š 2 𝐼4 cos Ο† 10 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 10 π‘‰π‘š 2 𝐼4 cos Ο†= 𝑉 π‘šπΌ1 𝐼4= 2𝐼1 cos Ο† 11 Let 𝑅4 is the resistance of each line conductor copper losses in line W = 𝐼4 2 𝑅4+ 𝐼4 2 𝑅4 = 2𝐼4 2 𝑅4 =2( 2𝐼1 cos Ο† )2 𝑅4 =4 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅4 12 6 October 2020 D.Rajesh Asst.Prof. EEE Department 32
 Comparing copper losses with 2 wire d.c system 4 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅4= 2𝐼1 2 𝑅1 𝑅4 𝑅1 = π‘π‘œπ‘ 2Ο† 2 π‘Ž4 π‘Ž1 = 2 π‘π‘œπ‘ 2Ο† 13 volume of copper wire 1βˆ’phase 2βˆ’ wire volume of copper wire Twoβˆ’wire D.C system = π‘Ž4𝑙4 π‘Ž1𝑙1 = 2 π‘π‘œπ‘ 2Ο†π‘Ž12𝑙 π‘Ž12𝑙 = 2 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required in this system is 2 π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 33
The maximum value of voltage between the outers is 𝑉 π‘š. Therefore, the r.m.s. value of voltage is π‘‰π‘š 2 Assuming the power factor of the load to be cos Ο†, load current 𝐼5 Power supplied P= π‘‰π‘š 2 𝐼5 cos Ο† 14 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 14 π‘‰π‘š 2 𝐼5 cos Ο†= 𝑉 π‘šπΌ1 𝐼5= 2𝐼1 cos Ο† 15 6 October 2020 D.Rajesh Asst.Prof. EEE Department 34
Let 𝑅5 is the resistance of each line conductor copper losses in line W = 𝐼5 2 𝑅5+ 𝐼5 2 𝑅5 = 2𝐼5 2 𝑅5 =2( 2𝐼1 cos Ο† )2 𝑅5 = 4𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅5 Comparing copper losses with 2 wire d.c system 4𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅5 =2𝐼1 2 𝑅1 𝑅5 𝑅1 = π‘π‘œπ‘ 2Ο† 2 π‘Ž5 π‘Ž1 = 2 π‘π‘œπ‘ 2Ο† volume of copper wire 1βˆ’phase 2βˆ’ wire π‘€π‘–π‘‘β„Ž π‘šπ‘–π‘‘ βˆ’ π‘π‘œπ‘–π‘›π‘‘ π‘’π‘Žπ‘Ÿπ‘‘β„Žπ‘’π‘‘ volume of copper wire Twoβˆ’wire D.C system π‘Ž5𝑙5 π‘Ž1𝑙1 = 2 π‘π‘œπ‘ 2Ο† π‘Ž12𝑙 π‘Ž12𝑙 = 2 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required in this system is 2 π‘π‘œπ‘ 2Ο† times that required in 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 35
 If the load is balanced, the current through the neutral wire is zero. Assuming balanced load,  The maximum value of voltage between the outers is 𝑉 π‘š. Therefore, the r.m.s. value of voltage is π‘‰π‘š 2 .  Assuming the power factor of the load to be cos Ο†, load current 𝐼6  Power supplied P= π‘‰π‘š 2 𝐼6 cos Ο† 16 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 16 π‘‰π‘š 2 𝐼6 cos Ο†= 𝑉 π‘šπΌ1 𝐼6= 2𝐼1 cos Ο† 17 6 October 2020 D.Rajesh Asst.Prof. EEE Department 36
Let 𝑅6 is the resistance of each line conductor copper losses in line W = 𝐼6 2 𝑅6+ 𝐼6 2 𝑅6 = 2𝐼6 2 𝑅6 =2( 2𝐼1 cos Ο† )2 𝑅6 = 4𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅6 Comparing copper losses with 2 wire d.c system 4𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅6 =2𝐼1 2 𝑅1 𝑅6 𝑅1 = π‘π‘œπ‘ 2Ο† 2 π‘Ž6 π‘Ž1 = 2 π‘π‘œπ‘ 2Ο† volume of copper wire 1βˆ’phase 2βˆ’ wire π‘€π‘–π‘‘β„Ž π‘šπ‘–π‘‘ βˆ’ π‘π‘œπ‘–π‘›π‘‘ π‘’π‘Žπ‘Ÿπ‘‘β„Žπ‘’π‘‘ volume of copper wire Twoβˆ’wire D.C system π‘Ž6𝑙6 π‘Ž1𝑙1 = = 2 π‘π‘œπ‘ 2Ο† π‘Ž12.5𝑙 π‘Ž12𝑙 = 2.5 π‘π‘œπ‘ 2Ο†  Hence, the volume of conductor material required in this system is 2.5 π‘π‘œπ‘ 2Ο† times that required in 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 37
 This system can be considered as two independent single-phase systems, each transmitting one half of the total power.  Current (𝐼7) in each conductor will be half that in single-phase 2- wire system. Consequently, area of X-section of each conductor is also half but as there are four wires, so volume of conductor material used is the same as in a single phase, 2-wire system  i.e. 𝐼7= 𝐼5 2 = 2𝐼1 cos Ο† 2 = 𝐼1 2 cos Ο†  If the current density remains constant the X-area will be half of the single-phase system  i.e. π‘Ž7= π‘Ž5 2 = 2 π‘π‘œπ‘ 2Ο† 2 = 1 π‘π‘œπ‘ 2Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 38
 Since wires are four  Then  volume of copper wireTwo phase, 4βˆ’wire a.c. system volume of copper wire Twoβˆ’wire D.C system π‘Ž7𝑙7 π‘Ž1𝑙1 = 1 π‘π‘œπ‘ 2Ο† π‘Ž14𝑙 π‘Ž12𝑙 = 2 π‘π‘œπ‘ 2Ο† Hence, volume of conductor material required in this system is 2 π‘π‘œπ‘ 2Ο† times that required in 2- wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 39
 Let us suppose that maximum voltage between the outers is 𝑉 π‘š. Then maximum voltage between either outer and neutral wire is π‘‰π‘š 2  R.M.S. voltage between outer and neutral wire= π‘‰π‘š 2 2 = π‘‰π‘š 2  Total power transmitted=2 π‘‰π‘š 2 𝐼8 cos Ο† = 𝑉 π‘š 𝐼8 cos Ο†  Comparing the power with 2-wire D.C 𝑉 π‘š 𝐼8 cos Ο†= 𝑉 π‘šπΌ1 𝐼8 = 𝐼1 cos Ο† The middle wire will act as a common return and the current through it will be 𝐼8 2 + 𝐼8 2 = 2 𝐼8 6 October 2020 D.Rajesh Asst.Prof. EEE Department 40
 Assuming the current density to be constant, the area of X-section of the neutral wire will be 2 times that of either of the outers.  ∴ Resistance of neutral wire= 𝑅8 2  Copper losses in this system W= 𝐼8 2 𝑅8+ 𝐼8 2 𝑅8 + ( 2 𝐼8)2𝑅8 2 = 2𝐼8 2 𝑅8+ 2𝐼8 2 𝑅8=(2+ 2) 𝐼8 2 𝑅8 Where 𝐼8 = 𝐼1 cos Ο† W=(2+ 2) ( 𝐼1 cos Ο†)2 𝑅8=(2+ 2) 𝐼1 2𝑅8 π‘π‘œπ‘ 2Ο† Comparing copper losses with 2 wire d.c system (2+ 2) 𝐼1 2𝑅8 π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅8 𝑅1 = 2π‘π‘œπ‘ 2Ο† (2+ 2) π‘Ž8 π‘Ž1 = (2+ 2) 2π‘π‘œπ‘ 2Ο† volume of copper wireTwo phase, 3βˆ’wire a.c. system volume of copper wire Twoβˆ’wire D.C system π‘Ž8𝑙8 π‘Ž1𝑙1 = (2+ 2) 2π‘π‘œπ‘ 2Ο† π‘Ž1(2+ 2)𝑙 π‘Ž12𝑙 = 2.914 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 2.914 π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 41
This system is almost universally adopted for transmission of electric power. The 3-phase, 3- wire system may be star connected or delta connected. Maximum voltage between conductors =𝑉 π‘š, Let 𝑅9 is the resistance of each line conductor Then maximum voltage between each phase and neutral is π‘‰π‘š 3 R.M.S. voltage per phase = π‘‰π‘š 3 2 = π‘‰π‘š 6 Power supplied=3( π‘‰π‘š 6 I9cos Ο† )  Comparing the power with 2-wire D.C 3 π‘‰π‘š 6 𝐼9 cos Ο†= 𝑉 π‘šπΌ1 𝐼9 = 6𝐼1 3cos Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 42
 Total copper losses= 3𝐼9 2 𝑅9 =3 ( 6𝐼1 3cos Ο† )2 𝑅9= 2𝐼1 2 𝑅9 π‘π‘œπ‘ 2Ο†  Comparing copper losses with 2 wire d.c system 2𝐼1 2 𝑅9 π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅9 𝑅1 = π‘π‘œπ‘ 2Ο† π‘Ž9 π‘Ž1 = 1 π‘π‘œπ‘ 2Ο† volume of copper wire3βˆ’Phase, 3βˆ’wire system volume of copper wire Twoβˆ’wire D.C system = π‘Ž9𝑙9 π‘Ž1𝑙1 = 1 π‘π‘œπ‘ 2Ο†π‘Ž13𝑙 π‘Ž12𝑙 = 1.5 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 1.5 π‘π‘œπ‘ 2Ο† times that required for 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 43
In this case, 4th or neutral wire is taken from the neutral point. The area of X- section of the neutral wire is generally one-half that of the line conductor. If the loads are balanced, then current through the neutral wire is zero. Assuming balanced loads and p.f. of the load as cos Ο†, Let 𝑅10 is the resistance of each line conductor R.M.S. voltage per phase = π‘‰π‘š 6 Power supplied=3( π‘‰π‘š 6 I10cos Ο† )  Comparing the power with 2-wire D.C 3 π‘‰π‘š 6 𝐼10 cos Ο†= 𝑉 π‘šπΌ1 𝐼10 = 6𝐼1 3cos Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 44
 Total copper losses= 3𝐼10 2 𝑅10 =3 ( 6𝐼1 3cos Ο† )2 𝑅10= 2𝐼1 2 𝑅10 π‘π‘œπ‘ 2Ο†  Comparing copper losses with 2 wire d.c system 2𝐼1 2 𝑅10 π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅10 𝑅1 = π‘π‘œπ‘ 2Ο† π‘Ž10 π‘Ž1 = 1 π‘π‘œπ‘ 2Ο† volume of copper wire3βˆ’Phase, 3βˆ’wire system volume of copper wire Twoβˆ’wire D.C system = π‘Ž10𝑙10 π‘Ž1𝑙1 = 1 π‘π‘œπ‘ 2Ο†π‘Ž13.5𝑙 π‘Ž12𝑙 = 1.75 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 1.75 π‘π‘œπ‘ 2Ο†times that required for 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 45
System Same maximum voltage to earth Same maximum voltage between conductors 1. D.C. system (i) Two-wire (ii) Two-wire mid-point earthed (iii) 3-wire 1 0.25 0.3125 1 1 1.25 2. Single phase system (i) Two-wire (ii) Two-wire mid-point earthed (iii) 3-wire 2 π‘π‘œπ‘ 2Ο† 0.5 π‘π‘œπ‘ 2Ο† 1.625 π‘π‘œπ‘ 2Ο† 2 π‘π‘œπ‘ 2Ο† 0.5 π‘π‘œπ‘ 2Ο† 2.5 π‘π‘œπ‘ 2Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 46
System Same maximum voltage to earth Same maximum voltage between conductors 3. Two-phase system (i) 2–phase 4-wire (ii)2-phase 3-wire 0.5 π‘π‘œπ‘ 2Ο† 1.457 π‘π‘œπ‘ 2Ο† 2 π‘π‘œπ‘ 2Ο† 2.914 π‘π‘œπ‘ 2Ο† 4. Three-phase system (i) 3-phase 3-wire (ii)3-phase 4-wire , 0.5 π‘π‘œπ‘ 2Ο† 0.583 π‘π‘œπ‘ 2Ο† 1.5 π‘π‘œπ‘ 2Ο† 1.75 π‘π‘œπ‘ 2Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 47
What is the percentage saving in feeder copper if the line voltage in a 2-wire d.c. system is raised from 200 volts to 400 volts for the same power transmitted over the same distance and having the same power loss ? 6 October 2020 D.Rajesh Asst.Prof. EEE Department 48
 (i) Conductors  (ii) Step-up and step-down transformers  (iii) Line insulators  (iv) Support  (v) Protective devices  (vi) Voltage regulating devices 6 October 2020 D.Rajesh Asst.Prof. EEE Department 49
The following two fundamental economic principles which closely influence the electrical design of a transmission line (i) Economic choice of transmission voltage (ii) Economic choice of conductor size 6 October 2020 D.Rajesh Asst.Prof. EEE Department 50
 The effect of increased voltage level of transmission on 1. Volume of conductor material used for transmission 2. Line Drop 6 October 2020 D.Rajesh Asst.Prof. EEE Department 51
Let a 3-phase A.C. system is used for the transmission, The resistance per conductor is given by R= ρ l/A Power transmitted P= 3𝑉𝐼cos Ο† I= 𝑃 3𝑉cos Ο† Total copper losses 𝑃𝑐𝑒= 3𝐼2R =3( 𝑃 3𝑉cos Ο† )2ρ l 𝐴 A= 𝑃2ρ l 𝑃𝑐𝑒𝑉2π‘π‘œπ‘ 2Ο† The volume of copper used is V=3(A*l) Volume of copper= 3𝑃2ρ 𝑙2 𝑃𝑐𝑒𝑉2π‘π‘œπ‘ 2Ο† Volume of copper 𝛼 1 𝑉2π‘π‘œπ‘ 2Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 52 Thus greater is the transmission voltage level, lesser is the volume of the copper required.
Line Drop=IR=I ρ l/A Current density(J)= 𝐼 𝐴 Then Line Drop= 𝐼ρ l 𝐼 𝐽 = Jρ l %Line Drop= Jρ l 𝑉 *100 Higher the transmission voltage level, then lesser the %Line Drop 6 October 2020 D.Rajesh Asst.Prof. EEE Department 53
 The most economical area of conductor is that for which the total annual cost of transmission line is minimum . This is known as Kelvin’s Law  The total annual cost of transmission line can be divided broadly into two parts viz., ο‚‘ Annual charge on capital outlay ο‚‘ Annual cost of energy wasted in the conductor. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 54
 This is on account of interest and depreciation on the capital cost of complete installation of transmission line. In case of overhead system, it will be the annual interest and depreciation on the capital cost of conductors, supports and insulators and the cost of their erection.  Now, for an overhead line, insulator cost is constant, the conductor cost is proportional to the area of X-section and the cost of supports and their erection is partly constant annual charge.  Transmission line can be expressed as : π΄π‘›π‘›π‘’π‘Žπ‘™ π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’ = 𝑃1 + 𝑃2π‘Ž 𝑃1 constant depending on the copper conductor 𝑃2 constant depending on the X-area of the conductor a is the area of X-section of the conductor 6 October 2020 D.Rajesh Asst.Prof. EEE Department 55
 This is on account of energy lost mainly in the conductor due to 𝐼2R losses.  Assuming a constant current in the conductor throughout the year,  The energy lost in the conductor is proportional to resistance. where R = ρ l/ π‘Ž  The energy lost in the conductor is inversely proportional to area of X-section. energy lost Ξ± 1 π‘Ž  Thus, the annual cost of energy wasted in an overhead transmission line can be expressed as : Annual cost of energy wasted = 𝑃3 π‘Ž where P3 is a constant. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 56
 Total annual cost, C=𝑃1 + 𝑃2π‘Ž+ 𝑃3 π‘Ž only area of X-section a is variable. Therefore, the total annual cost of transmission line will be minimum if differentiation of C w.r.t. a is zero i.e. ⅆ𝐢 β…†π‘Ž =0 β…† β…†π‘Ž (𝑃1 + 𝑃2π‘Ž+ 𝑃3 π‘Ž )=0 𝑃2π‘Ž βˆ’ 𝑃3 π‘Ž2=0 𝑃2π‘Ž = 𝑃3 π‘Ž2 i.e. Variable part of annual charge = Annual cost of energy wasted Therefore Kelvin’s Law can also be stated The most economical area of conductor is that for which the variable part of annual charge is equal to the cost of energy losses per year. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 57
 The straight line shows the relation between the annual charge (i.e., 𝑃1 + 𝑃2π‘Ž) and the area of X- section a of the conductor  The rectangular hyperbola gives the relation between annual cost of energy wasted ( 𝑃3 π‘Ž ) and X-sectional area a.  Curve 3 shows the relation between total annual cost of transmission line and area of X-section a. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 58
 (i) It is not easy to estimate the energy loss in the line without actual load curves, which are not available at the time of estimation.  (ii) The assumption that annual cost on account of interest and depreciation on the capital outlay is in the form 𝑃1 + 𝑃2π‘Ž is strictly speaking not true. For instance, in cables neither the cost of cable dielectric and sheath nor the cost of laying vary in this manner.  (iii) This law does not take into account several physical factors like safe current density, mechanical strength, corona loss etc.  (iv) The conductor size determined by this law may not always be practicable one because it may be too small for the safe carrying of necessary current.  (v) Interest and depreciation on the capital outlay cannot be determined accurately. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 59
A 2-conductor cable 1 km long is required to supply a constant current of 200 A throughout the year. The cost of cable including installation is Rs. (20 a + 20) per meter where β€˜a’ is the area of X-section of the conductor in π‘π‘š2 . The cost of energy is 5P per kWh and interest and depreciation charges amount to 10%. Calculate the most economical conductor size. Assume resistivity of conductor material to be 1Β·73 ΞΌ Ξ© cm. economical conductor size a =? Energy lost per annum=2𝐼2 𝑅𝑑 Resistance of one conductor=ρ l/ π‘Ž = 1.73βˆ—10βˆ’6βˆ—105 π‘Ž = 0.173 π‘Ž Ξ© Energy lost per annum= 2βˆ— 2002 βˆ— 0.173 π‘Ž βˆ— 8760 = 121238.4 π‘Ž kWh 6 October 2020 D.Rajesh Asst.Prof. EEE Department 60
 Annual cost of energy lost = Cost per kWh Γ— Annual energy loss =Rs 0.05* 121238.4 π‘Ž kWh=Rs 6062 π‘Ž The capital cost (variable) of the cable is given to be Rs 20 a per metre. Therefore, for 1 km length of the cable, the capital cost (variable) is Rs. 20 a Γ— 1000 = Rs. 20,000 a. Variable annual charge = Annual interest and depreciation on capital cost (variable) of cable = Rs 0Β·1 Γ— 20,000 a = Rs 2000 a According to Kelvin’s law, for most economical X-section of the conductor, Variable annual charge = Annual cost of energy lost 2000 a = 6062 π‘Ž a= 6062 2000 =1.74 π‘π‘š2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 61
A 2-wire feeder carries a constant current of 250 A throughout the year. The portion of capital cost which is proportional to area of X-section is Rs 5 per kg of copper conductor. The interest and depreciation total 10% per annum and the cost of energy is 5P per kWh. Find the most economical area of X-section of the conductor. Given that the density of copper is 8Β·93gm/ π‘π‘š3 and its specific resistance is 1Β·73 Γ— 10βˆ’8 Ξ© m. Consider 1 metre length of the feeder. Let a be the most economical area of X-section of each conductor in π‘š2 Energy lost per annum=2𝐼2 𝑅𝑑 Resistance of one conductor=ρ l/ π‘Ž = 1.73βˆ—10βˆ’8βˆ—1 π‘Ž = 1.73βˆ—10βˆ’8 π‘Ž Ξ© Energy lost per annum= 2βˆ— 2502 βˆ— 1.73βˆ—10βˆ’8 π‘Ž βˆ— 8760 = 18,94,350 π‘Ž βˆ— 10βˆ’8 kWh 6 October 2020 D.Rajesh Asst.Prof. EEE Department 62
 Annual cost of energy lost = Cost per kWh Γ— Annual energy loss =Rs 0.05* 18,94,350 π‘Ž βˆ— 10βˆ’8 kWh=Rs 94,717.5 π‘Ž βˆ— 10βˆ’8 Mass of 1 metre feeder = 2 (Volume Γ— density) = 2 Γ— a Γ— 1 Γ— 8Β·93 Γ— 103kg = 17Β·86 Γ— 103 a kg Capital cost (variable) = Rs 5 Γ— 17Β·86 Γ— 103 a = Rs 89Β·3 Γ— 103 a Variable annual charge = 10% of capital cost (variable) = 0Β·1 Γ— 89Β·3 Γ— 103 a = Rs 8930 a According to Kelvin’s law, for most economical X-section of the conductor, Variable annual charge = Annual cost of energy lost 8930 a = 94,717.5 π‘Ž βˆ— 10βˆ’8 a= 94,717.5 8930 βˆ— 10βˆ’8 =3.25βˆ— 10βˆ’4 π‘š2 = 3.25 π‘π‘š2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 63
Determine the most economical cross-section for a 3-phase transmission line, 1 km long to supply at a constant voltage of 110 kV for the following daily load cycle : 6 hours 20 MW at p.f. 0Β·8 lagging 12 hours 5 MW at p.f. 0Β·8 lagging 6 hours 6 MW at p.f. 0Β·8 lagging The line is used for 365 days yearly. The cost per km of line including erection is Rs (9000 +6000 a) where β€˜a’ is the area of X- section of conductor in cm2. The annual rate of interest and depreciation is 10% and the energy costs 6P per kWh. The resistance per km of each conductor is0Β·176/a. Power transmitted P= 3𝑉𝐼cos Ο† Energy lost per annum=3R(𝐼2 𝑑) 6 October 2020 D.Rajesh Asst.Prof. EEE Department 64
 The load currents at various loads are :  At 20 MW, 𝐼1= 𝑃 3𝑉cos Ο† = 20βˆ—106 3βˆ—110βˆ—103βˆ—0.8 = 131.2𝐴  At 5 MW, 𝐼1= 𝑃 3𝑉cos Ο† = 5βˆ—106 3βˆ—110βˆ—103βˆ—0.8 = 32.8𝐴  At 6 MW, 𝐼1= 𝑃 3𝑉cos Ο† = 6βˆ—106 3βˆ—110βˆ—103βˆ—0.8 = 39.36𝐴  Energy loss per day in 3-phase line =3* 0.176 π‘Ž (131.22 βˆ— 6+ 32.82 βˆ— 12+ 36.362 βˆ— 6) = 66.26 π‘Ž kWh Energy lost per annum = 66.26 π‘Ž *365= = 24,184.9 π‘Ž kWh Annual cost of energy=Rs 6 100 βˆ— 24,184.9 π‘Ž = = 1451.09 π‘Ž Variable annual charge = 10% of capital cost (variable) of line = Rs 0Β·1 Γ— 6000 a = Rs 600 a According to Kelvin’s law, for most economical X-section of the conductor, Variable annual charge = Annual cost of energy 600 a= 1451.09 π‘Ž a= 1451.09 600 =1.56 π‘π‘š2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 65
The conveyance of electric power from a power station to consumers’ premises is known as electric supply system.  Electric supply system consist of three principle components: 1) Power station 2) Transmission lines 3) Distribution system  The electric supply system can be broadly classified into: 1) DC or AC system 2)Overhead or underground system 6 October 2020 D.Rajesh Asst.Prof. EEE Department 66
Different blocks of typical acpowersupply scheme are asper following:: 1)Generating Station: Ingenerating station power is producedby three phase alternators operating inparallel. The usual generation voltage is 11kV. Foreconomy in the transmission of electric power, the generation voltage is stepped upto 132 kVor more at generating station with the help of three phase transformer. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 67
2)Primarytransmission: Theelectric power at 132 kVis transmitted by 3-phase,3 wire overhead system to the out 3)Secondary transmission: Atthe receiving station the voltage is reduced to33kV by step down transformer. 4)Primary distribution: Atthe substation voltage is reduced 33kV to 11kV.The11kV line run along important roadsides to city. This forms the primary distribution.
5)Secondarydistribution: Theelectric power form primary distribution line is delivered to distribution substation. The substation is located nearthe consumers location and step down the voltage to 400V, 3- phase ,4-wire for secondary distribution. The voltage between any two phases is 400V and between any phase and neutral is 230V.
Basis for Comparison AC Transmission Line DC Transmission Line Definition The ac transmission line transmit the alternating current. The dc transmission line is used for transmitting the direct current. Number of Conductors Three Two Inductance & surges Have Don’t Have Voltage drop High Low Skin Effect Occurs Absent Need of Insulation More Less Interference Have Don’t Have Corona Loss Occur Don’t occur Dielectric Loss Have Don’t have Synchronizing and Stability Problem No difficulties Difficulties Cost Expensive Cheap Length of conductors Small More Repairing and Maintenance Easy and Inexpensive Difficult and Expensive Transformer Requires Not Requires 6 October 2020 D.Rajesh Asst.Prof. EEE Department 70
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Which distributes electric power for local use is known as distribution system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 73
 (i) Feeders. A feeder is a conductor which connects the sub-station to the area where power is to be distributed.  (ii) Distributor. A distributor is a conductor from which tappings are taken for supply to the consumers.  (iii) Service mains. A service mains is generally a small cable which connects the distributor to the consumers’ terminals. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 74
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Primary distribution system 6 October 2020 D.Rajesh Asst.Prof. EEE Department 76 Secondary distribution system
2-wire d.c. system 6 October 2020 D.Rajesh Asst.Prof. EEE Department 77 3 -wire d.c. system
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 (i) Radial System. In this system, separate feeders radiate from a single substation and feed the distributors at one end only. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 80
 Limitations  (a) The end of the distributor nearest to the feeding point will be heavily loaded.  (b) The consumers are dependent on a single feeder and single distributor. Therefore, any fault on the feeder or distributor cuts off supply to the consumers who are on the side of the fault away from the substation.  (c) The consumers at the distant end of the distributor would be subjected to serious voltage fluctuations when the load on the distributor changes. Due to these limitations, this system is used for short distances only. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 81
 (ii) Ring main system. In this system, the primaries of distribution transformers form a loop. The loop circuit starts from the substation bus-bars, makes a loop through the area to be served, and returns to the substation. βœ“ substation supplies to the closed feeder LMNOPQRS. βœ“ The distributors are tapped from different points M, O and Q of the feeder through distribution transformers. ➒ There are less voltage fluctuations at consumer’s terminals. ➒ The system is very reliable as each distributor is fed via two feeders. ➒ In the event of fault on any section of the feeder, the continuity of supply is maintained. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 82
 (iii) Interconnected system. When the feeder ring is energized by two or more than two generating stations or substations, it is called inter-connected system. (a) It increases the service reliability. (b) Any area fed from one generating station during peak load hours can be fed from the other generating station. This reduces reserve power capacity and increase efficiency of the system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 83
 (i) Proper voltage.  One important requirement of a distribution system is that voltage variations at consumer’s terminals should be as low as possible.  The changes in voltage are generally caused due to the variation of load on the system.  Low voltage causes loss of revenue, inefficient lighting and possible burning out of motors.  High voltage causes lamps to burn out permanently and may cause failure of other appliances.  Thus, if the declared voltage is 230 V, then the highest voltage of the consumer should not exceed 244 V while the lowest voltage of the consumer should not be less than 216 V. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 84
 (ii) Availability of power on demand. Power must be available to the consumers in any amount that they may require from time to time.  (iii) Reliability. Modern industry is almost dependent on electric power for its operation.  The reliability can be improved to a considerable extent by (a) interconnected system (b) reliable automatic control system (c) providing additional reserve facilities. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 85
(i) Distributor fed at one end (ii) Distributor fed at both ends (iii) Distributor fed at the centre (iv) Ring distributor. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 86
 (i) Distributor fed at one end 6 October 2020 D.Rajesh Asst.Prof. EEE Department 87  (ii) Distributor fed at both ends
 (iii) Distributor fed at the centre 6 October 2020 D.Rajesh Asst.Prof. EEE Department 88  (iv) Ring distributor
(i) Concentrated loading (ii) Uniform loading (iii) Both concentrated and uniform loading. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 89
 2-wire d.c. distributor AB fed at one end A and having concentrated loads 𝐼1, 𝐼2, 𝐼3 and 𝐼4 tapped off at points C, D, E and F respectively.  Let π‘Ÿ1 , π‘Ÿ2, π‘Ÿ3 π‘Žπ‘›π‘‘ π‘Ÿ4 be the resistances of both wires (go and return) of the sections AC, CD, DE and EF of the distributor respectively.  Current fed from point A = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4 Current in section AC = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4  Current in section CD = 𝐼2 + 𝐼3 + 𝐼4 Current in section DE = 𝐼3 + 𝐼4  Current in section EF = 𝐼4  Voltage drop in section AC = π‘Ÿ1 (𝐼1 + 𝐼2 + 𝐼3 + 𝐼4)  Voltage drop in section CD = π‘Ÿ2 (𝐼2 + 𝐼3 + 𝐼4) Voltage drop in section DE = π‘Ÿ3 (𝐼3 + 𝐼4)  Voltage drop in section EF = π‘Ÿ4 𝐼4  ∴ Total voltage drop in the distributor =π‘Ÿ1 (𝐼1 + 𝐼2 + 𝐼3 + 𝐼4)+π‘Ÿ2 (𝐼2 + 𝐼3 + 𝐼4)+π‘Ÿ3 (𝐼3 + 𝐼4)+ π‘Ÿ4 𝐼4 6 October 2020 D.Rajesh Asst.Prof. EEE Department 90
 2-wire d.c. distributor AB fed at one end A and loaded uniformly with i amperes per metre length.  The load tapped is i amperes. Let 𝑙 metres be the length of the distributor and r ohm be the resistance per metre run. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 91
 Consider a point C on the distributor at a distance x metres from the feeding point A  Then current at point C is = i π‘™βˆ’ i x amperes = i (π‘™βˆ’ x) amperes Now, consider a small length dx near point C. Its resistance is r dx and the voltage drop over length dx is dv = i (𝑙 βˆ’ x) r dx = i r (𝑙 βˆ’ x) dx Total voltage drop in the distributor upto point C is v =‫׬‬0 π‘₯ ir (l βˆ’ x)dx =ir 𝑙π‘₯ βˆ’ π‘₯2 2 The voltage drop upto point B (i.e. over the whole distributor) can be obtained by putting x = 𝑙 in the above expression. ∴ Voltage drop over the distributor AB = ir 𝑙 βˆ— 𝑙 βˆ’ 𝑙2 2 = 1 2 π‘–π‘Ÿπ‘™2 = 1 2 𝑖𝑙 π‘Ÿπ‘™ = 1 2 𝐼𝑅 Thus, in a uniformly loaded distributor fed at one end, the total voltage drop is equal to that produced by the whole of the load assumed to be concentrated at the middle point. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 92
 The two ends of the distributor may be supplied with (i) equal voltages (ii) unequal voltages. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 93
 Consider a distributor AB fed at both ends with equal voltages V volts and having concentrated loads 𝐼1, 𝐼2, 𝐼3, 𝐼4 and 𝐼5 at points C, D, E, F and G respectively  As we move away from one of the feeding points, say A, p.d. goes on decreasing till it reaches the minimum value at some load point, say E, and then again starts rising and becomes V volts as we reach the other feeding point B.  All the currents tapped off between points A and E (minimum p.d. point) will be supplied from the feeding point A while those tapped off between B and E will be supplied from the feeding point B.  The current tapped off at point E itself will be partly supplied from A and partly from B. If these currents are x and y respectively, then, x+y= 𝐼3  Therefore, we arrive at a very important conclusion that at the point of minimum potential, current comes from both ends of the distributor.  Voltage drop from feeding point A to the point of minimum potential difference is equal to the voltage drop from feeding point B to the point of minimum potential difference 6 October 2020 D.Rajesh Asst.Prof. EEE Department 94
 Distributor AB fed with unequal voltages ; end A being fed at 𝑉1volts and end B at 𝑉2 volts.  In order to determine the point of minimum potential, assume of 𝐼𝐴 amperes is supplied from feeding point A  Let r be the resistance of each conductor per unit length(l=1) for both go & return  Resistance of portions AC,CD, DE,EF,FB are π‘Ÿ1 , π‘Ÿ2, π‘Ÿ3 , π‘Ÿ4, π‘Žπ‘›π‘‘ π‘Ÿ5 respectively  Voltage drop between A and B = Voltage drop over AB 𝑉1 βˆ’ 𝑉2 = πΌπ΄π‘Ÿ1+(𝐼𝐴- 𝐼1) π‘Ÿ2 + (𝐼𝐴- 𝐼1- 𝐼2) π‘Ÿ3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) π‘Ÿ4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 βˆ’ 𝐼4) π‘Ÿ5 The load point at which currents are coming from both sides of the distributor will be the point of minimum potential 6 October 2020 D.Rajesh Asst.Prof. EEE Department 95
 A 2-wire d.c. street mains AB, 600 m long is fed from both ends at 220 V. Loads of 20 A, 40 A, 50 A and 30 A are tapped at distances of 100m, 250m, 400m and 500 m from the end A respectively. If the area of X-section of distributor conductor is 1π‘π‘š2 , find the minimum consumer voltage. Take ρ = 1Β·7 Γ— 10βˆ’6 Ξ© cm.  Voltage drop between A and B = Voltage drop over AB R= ρ l/A  Resistance of section AC of distributor=2Γ— 1Β·7 Γ— 10βˆ’6βˆ—100βˆ—102 1 = 3Β·4 Γ— 10βˆ’2 Ξ©  Resistance of section CD of distributor =0.051 Ξ©  Resistance of section DE of distributor =0.051 Ξ©  Resistance of section EF of distributor =0.034 Ξ©  Resistance of section FB of distributor =0.034 Ξ©  Voltage drop between A and B = Voltage drop over AB  𝑉𝐴 βˆ’ 𝑉𝐡 = πΌπ΄π‘Ÿ1+(𝐼𝐴- 𝐼1) π‘Ÿ2 + (𝐼𝐴- 𝐼1- 𝐼2) π‘Ÿ3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) π‘Ÿ4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 βˆ’ 𝐼4) π‘Ÿ5  220-200=0.034 𝐼𝐴+(𝐼𝐴-20)0.051+ (𝐼𝐴-60)0.051+(𝐼𝐴-110)0.034+(𝐼𝐴-140)0.034 𝐼𝐴=61.7A 6 October 2020 D.Rajesh Asst.Prof. EEE Department 96
 Actual distribution of currents in the various sections of the distributor is  Current in section EF, 𝐼𝐸𝐹 = πΌπ΄βˆ’110 =61Β·7βˆ’110 =βˆ’48Β·3A from E to F = 48Β·3 A from F to E  Current in section DE, 𝐼𝐷𝐸 = 𝐼𝐴 βˆ’60=61.7-60=1.7A from D to E 6 October 2020 D.Rajesh Asst.Prof. EEE Department 97 It is clear that currents are coming to load point E from both sides i.e. from point D and point F. Hence, E is the point of minimum potential. ∴ Minimum consumer voltage, 𝑉𝐸 = 𝑉𝐴-(𝐼𝐴𝐢 π‘Ÿ1+ 𝐼𝐢𝐷 π‘Ÿ2+ 𝐼𝐷𝐸 π‘Ÿ3) =220-(61Β·7 Γ— 0Β·034 + 41Β·7 Γ— 0Β·051 + 1Β·7 Γ— 0Β·051] =215.69V
A two-wire d.c. distributor AB, 600 metres long is loaded as under : Distance from A (metres) : 150 300 350 450 Loads in Amperes : 100 200 250 300 The feeding point A is maintained at 440 V and that of B at 430 V. If each conductor has a resistance of 0Β·01 Ξ© per 100 metres, calculate : (i) the currents supplied from A to B, (ii) the power dissipated in the distributor. Let 𝐼𝐴amperes be the current supplied from the feeding point A. Then currents in the various sections of the distributor are Resistance of 100 m length of distributor (both wires) = 2 Γ— 0Β·01 = 0Β·02 Ξ© Resistance of section AC, π‘Ÿ1 = 0Β·02 Γ— 150/100 = 0Β·03 Ξ© Resistance of section CD, π‘Ÿ2 = 0Β·02 Γ— 150/100 = 0Β·03 Ξ© Resistance of section DE, π‘Ÿ3= 0Β·02 Γ— 50/100 = 0Β·01 Ξ© Resistance of section EF, π‘Ÿ4= 0Β·02 Γ— 100/100 = 0Β·02 Ξ© Resistance of section FB, π‘Ÿ5= 0Β·02 Γ— 150/100 = 0Β·03 Ξ© 6 October 2020 D.Rajesh Asst.Prof. EEE Department 98
 Voltage drop between A and B = Voltage drop over AB  𝑉𝐴 βˆ’ 𝑉𝐡 = πΌπ΄π‘Ÿ1+(𝐼𝐴- 𝐼1) π‘Ÿ2 + (𝐼𝐴- 𝐼1- 𝐼2) π‘Ÿ3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) π‘Ÿ4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 βˆ’ 𝐼4) π‘Ÿ5  440-430=0.03 𝐼𝐴+(𝐼𝐴-100)0.03+ (𝐼𝐴-300)0.01+(𝐼𝐴-550)0.02+(𝐼𝐴-850)0.03 𝐼𝐴=437.5A  The actual distribution of currents in the various sections of the distributor are  Current supplied from end A, 𝐼𝐴 = 𝐼𝐴𝐢 = 437Β·5 A  Current supplied from end B, 𝐼𝐡 = 𝐼𝐹𝐡 = 𝐼𝐴 -850= - 412Β·5 A  𝐼𝐢𝐷 = 337.5𝐴 𝐼𝐷𝐸 = 137.5𝐴 𝐼𝐸𝐹 = βˆ’112.5𝐴  Therefore E is the point of minimum potential.  Power loss in the distributor =𝐼𝐴𝐢 2 π‘Ÿ1+ 𝐼𝐢𝐷 2 π‘Ÿ2 + 𝐼𝐷𝐸 2 π‘Ÿ3+ 𝐼𝐸𝐹 2 π‘Ÿ4+ 𝐼𝐹𝐡 2 π‘Ÿ5 =(437.5)2 βˆ— 0.03 + 337.5 2 βˆ— 0.03 + 137.5 2 βˆ— 0.01 + 112.5 2 βˆ— 0.02 + (412.5)2 βˆ— 0.03 =14.705kW 6 October 2020 D.Rajesh Asst.Prof. EEE Department 99
 (i) Distributor fed at both ends with equal voltages.  Consider a distributor AB of length 𝑙 metres, having resistance r ohms per metre run and with uniform loading of 𝑖 amperes per metre run.  Let the distributor be fed at the feeding points A and B at equal voltages, say V volts.  The total current supplied to the distributor is 𝑖𝑙.  As the two end voltages are equal, therefore, current supplied from each feeding point is 𝑖𝑙 2 i.e. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 100
 Consider a point C at a distance x metres from the feeding point A. Then current at point C is = 𝑖𝑙 2 - 𝑖π‘₯= 𝑖( 𝑙 2 -π‘₯) Now, consider a small length dx near point C. Its resistance is r dx and the voltage drop over length dx is dv = i ( 𝑙 2 βˆ’ x) r dx = i r ( 𝑙 2 βˆ’ x) dx Total voltage drop in the distributor upto point C is v =‫׬‬ 0 π‘₯ ir ( 𝑙 2 βˆ’ x)dx =ir 𝑙π‘₯ 2 βˆ’ π‘₯2 2 = π‘–π‘Ÿ 2 (𝑙π‘₯- π‘₯2) Obviously, the point of minimum potential will be the mid-point. Therefore, maximum voltage drop will occur at mid-point i.e. where x = 𝑙 2 Max. voltage drop= π‘–π‘Ÿ 2 (𝑙π‘₯- π‘₯2)= π‘–π‘Ÿ 2 (𝑙 𝑙 2 - ( 𝑙 2 )2)= 1 8 π‘–π‘Ÿπ‘™2= 1 8 𝑖𝑙 π‘Ÿπ‘™ = 1 8 IR Minimum voltage=V- 1 8 IR volts 6 October 2020 D.Rajesh Asst.Prof. EEE Department 101
 (ii) Distributor fed at both ends with unequal voltages.  Consider a distributor AB of length l metres having resistance r ohms per metre run and with a uniform loading of i amperes permetre run. Let the distributor be fed from feeding points A and B at voltages 𝑉𝐴 and 𝑉𝐡 respectively.  Suppose that the point of minimum potential C is situated at a distance x metres from the feeding point A. Then current supplied by the feeding point A will be i x.  ∴ Voltage drop in section AC= 1 2 π‘–π‘Ÿπ‘₯2 volts  As the distance of C from feeding point B is (l βˆ’ x), therefore, current fed from B is i (l βˆ’ x).  ∴ Voltage drop in section BC = 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 volts 6 October 2020 D.Rajesh Asst.Prof. EEE Department 102
 Voltage at point C, 𝑉𝐢 = 𝑉𝐴 βˆ’ Drop over AC = 𝑉𝐴 - 1 2 π‘–π‘Ÿπ‘₯2 Voltage at point C, 𝑉𝐢 = 𝑉𝐡 βˆ’ Drop over BC = 𝑉𝐡 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 From above equations we get, 𝑉𝐴 - 1 2 π‘–π‘Ÿπ‘₯2 = 𝑉𝐡 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 𝑉𝐴- 𝑉𝐡 = 1 2 π‘–π‘Ÿπ‘₯2 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 = 1 2 π‘–π‘Ÿπ‘₯2 - 1 2 π‘–π‘Ÿπ‘™2 - 1 2 π‘–π‘Ÿπ‘₯2 + 1 2 π‘–π‘Ÿ2𝑙π‘₯ = π‘–π‘Ÿπ‘™ {π‘₯- 1 2 𝑙 } π‘₯= π‘‰π΄βˆ’ 𝑉𝐡 π‘–π‘Ÿπ‘™ + 𝑙 2 The point on the distributor where minimum potential occurs can be calculated 6 October 2020 D.Rajesh Asst.Prof. EEE Department 103
 Expression for the power loss in a uniformly loaded distributor fed at both ends  current supplied from each feeding point is 𝑖𝑙 2  Consider a small length dx of the distributor at point P which is at a distance x from the feeding end A.  Resistance of length dx = r dx  Current in length dx = 𝑖𝑙 2 - 𝑖π‘₯= 𝑖( 𝑙 2 -π‘₯)  Power loss in length dx = (current in dx)2 Γ— Resistance of dx = 𝑖( 𝑙 2 βˆ’π‘₯) 2 Γ—rdx  Total power loss in the distributor is=‫׬‬ 0 𝑙 𝑖( 𝑙 2 βˆ’π‘₯) 2 Γ—rdx =𝑖2 π‘Ÿ ‫׬‬ 0 𝑙 𝑙2 4 βˆ’ 𝑙π‘₯ + π‘₯2 𝑑π‘₯ = 𝑖2π‘Ÿ 𝑙2π‘₯ 4 βˆ’ 𝑙π‘₯2 2 + π‘₯3 3 𝑙 0 P= 𝑖2π‘Ÿπ‘™3 12 6 October 2020 D.Rajesh Asst.Prof. EEE Department 104
A 2-wire d.c. distributor AB 500 metres long is fed from both ends and is loaded uniformly at the rate of 1Β·0 A/metre. At feeding point A, the voltage is maintained at 255 V and at B at 250 V. If the resistance of each conductor is 0Β·1 Ξ© per kilometre, determine : (i) the minimum voltage and the point where it occurs (ii) the currents supplied from feeding points A and B Let the minimum potential occur at a point C distant x metres from the feeding point A. π‘₯= π‘‰π΄βˆ’ 𝑉𝐡 π‘–π‘Ÿπ‘™ + 𝑙 2 Resistance of distributor/m, r = 2 Γ— 0Β·1/1000 = 0Β·0002 Ξ© π‘₯= 255βˆ’ 250 1βˆ—0.0002βˆ—500 + 500 2 =300m i.e. minimum potential occurs at 300 m from point A. Minimum voltage, 𝑉𝐢 = 𝑉𝐴 - 1 2 π‘–π‘Ÿπ‘₯2 = 𝑉𝐡 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 255- 1βˆ—0.0002βˆ— (300)2 2 =246V Current supplied from A = i x = 1 Γ— 300 = 300 A Current supplied from B = i (𝑙 βˆ’ x) = 1 (500 βˆ’ 300) = 200 A 6 October 2020 D.Rajesh Asst.Prof. EEE Department 105
A 800 metres 2-wire d.c. distributor AB fed from both ends is uniformly loaded at the rate of 1Β·25 A/metre run. Calculate the voltage at the feeding points A and B if the minimum potential of 220 V occurs at point C at a distance of 450 metres from the end A. Resistance of each conductor is 0Β·05 Ξ©/km.  Minimum voltage, 𝑉𝐢 = 𝑉𝐴 - 1 2 π‘–π‘Ÿπ‘₯2 = 𝑉𝐡 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2  For 𝑉𝐴 𝑉𝐢 = 𝑉𝐴 - 1 2 π‘–π‘Ÿπ‘₯2 𝑉𝐴=220+ 1.25βˆ—0.0001βˆ— (450)2 2 =232.65V  For 𝑉𝐡 𝑉𝐢 = 𝑉𝐡 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 𝑉𝐡 =220+ 1.25βˆ—0.0001βˆ— (800βˆ’450)2 2 =227.65V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 106
 (i) A uniformly loaded distributor is fed at the centre. Show that maximum voltage drop = I R/8 where I is the total current fed to the distributor and R is the total resistance of the distributor.  (ii) A 2-wire d.c. distributor 1000 metres long is fed at the centre and is loaded uniformly at the rate of 1Β·25 A/metre. If the resistance of each conductor is 0Β·05 Ξ©/km, find the maximum voltage drop in the distributor.  ∴ Max. voltage drop = Voltage drop in half distributor = 1 2 ( 𝑖𝑙 2 )( π‘Ÿπ‘™ 2 ) = 1 8 𝑖𝑙 π‘Ÿπ‘™ = 1 8 IR where i l = I, the total current fed to the distributor r l = R, the total resistance of the distributor  (ii) Total current fed to the distributor is  I = i l = 1Β·25 Γ— 1000 = 1250 A  Total resistance of the distributor is  R = r l = 2 Γ— 0Β·05 Γ— 1 = 0Β·1 Ξ©  Max. voltage drop = 1 8 IR = 1 8 *1250*0.1 =15.62V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 107
 A two-wire d.c. distributor cable 1000 metres long is loaded with 0Β·5 A/metre. Resistance of each conductor is 0Β·05 Ξ©/km. Calculate the maximum voltage drop if the distributor is fed from both ends with equal voltages of 220 V. What is the minimum voltage and where it occurs ?  Current loading, i = 0Β·5 A/m  Resistance of distributor/m, r = 2 Γ— 0Β·05/1000 = 0Β·1 Γ— 10βˆ’3 Ξ©  Length of distributor, l = 1000 m  Total current supplied by distributor, I = i l = 0Β·5 Γ— 1000 = 500 A  Total resistance of the distributor, R = r l = 0Β·1 Γ— 10βˆ’3 Γ— 1000 = 0Β·1 Ξ©  ∴ Max. voltage drop = 1 8 IR = 6Β·25 V  Minimum voltage will occur at the mid-point of the distributor and its value is  = 220 βˆ’ 6Β·25 = 213Β·75 V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 108
The total drop over any section of the distributor is equal to the sum of drops due to concentrated and uniform loading in that section. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 109
Two conductors of a d.c. distributor cable AB 1000 m long have a total resistance of 0Β·1 Ξ©. The ends A and B are fed at 240 V. The cable is uniformly loaded at 0Β·5 A per metre length and has concentrated loads of 120 A, 60 A, 100 A and 40 A at points distant 200 m, 400 m, 700 m and 900 m respectively from the end A. Calculate (i) the point of minimum potential (ii) currents supplied from ends A and B (iii) the value of minimum potential. Distributor resistance per metre length, r = 0Β·1/1000 = 10βˆ’4 Ξ© (i) Point of minimum potential. The point of minimum potential is not affected by the uniform loading of the distributor. Therefore, let us consider the concentrated loads. Suppose the current supplied by end A is 𝐼𝐴. Then currents in the various sections will be 6 October 2020 D.Rajesh Asst.Prof. EEE Department 110
 Voltage drop between A and B = Voltage drop over AB  𝑉𝐴 βˆ’ 𝑉𝐡 = πΌπ΄π‘Ÿ1+(𝐼𝐴- 𝐼1) π‘Ÿ2 + (𝐼𝐴- 𝐼1- 𝐼2) π‘Ÿ3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) π‘Ÿ4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 βˆ’ 𝐼4) π‘Ÿ5  0= 10βˆ’4( ) 200 𝐼𝐴+(πΌπ΄βˆ’120)200+ (πΌπ΄βˆ’180)300+(πΌπ΄βˆ’280)200+(πΌπ΄βˆ’320)100 𝐼𝐴=166A The actual distribution of currents in the various sections of the distributor due to concentrated loading. It is clear from this figure that D is the point of minimum potential. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 111
(ii) The feeding point A will supply 166 A due to concentrated loading plus 0Β·5 Γ— 400 = 200 A due to uniform loading. ∴ Current supplied byA, 𝐼𝐴 = 166 + 200 = 366 A The feeding point B will supply a current of 154 A due to concentrated loading plus 0Β·5 Γ— 600 = 300 A due to uniform loading. ∴ Current supplied by B, 𝐼𝐡 = 154 + 300 = 454 A (iii) ∴ Minimum potential, 𝑉𝐷 = 𝑉𝐴 βˆ’ Drop in AD due to conc. loading – Drop in AD due to uniform loading 𝑉𝐷 Now, Drop in AD due to conc. loading = 𝐼𝐴𝐢 𝑅𝐴𝐢 + 𝐼𝐢𝐷 𝑅𝐢𝐷 = 166 Γ— 10βˆ’4 Γ— 200 + 46 Γ— 10βˆ’4 Γ— 200 = 3Β·32 + 0Β·92 = 4Β·24 V Drop in AD due to uniform loading = 1 2 π‘–π‘Ÿπ‘™2 = 1 2 0.5 βˆ— 10βˆ’4 βˆ— (400)2 =4V ∴ 𝑉𝐷 = 240 βˆ’ 4Β·24 βˆ’ 4 = 231Β·76 V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 112
A d.c. 2-wire distributor AB is 500m long and is fed at both ends at 240 V. The distributor is loaded as shown in Fig. The resistance of the distributor (go and return) is 0Β·001Ξ© per metre. Calculate (i) the point of minimum voltage and (ii) the value of this voltage. Let D be the point of minimum potential and let x be the current flowing in section CD. Then current supplied by end B will be (60 βˆ’ x). 6 October 2020 D.Rajesh Asst.Prof. EEE Department 113
 (i) If r is the resistance of the distributor (go and return) per metre length, then, Voltage drop in length AD = 𝐼𝐴𝐢 𝑅𝐴𝐢 + 𝐼𝐢𝐷 𝑅𝐢𝐷 =(100+x)*100r+x*150r Voltage drop in length BD= 1 2 π‘–π‘Ÿπ‘™2 +(60-x)*250r = 1 2 1 βˆ— π‘Ÿ βˆ— (200)2 +(60-x)*250r Since the feeding points A and B are at the same potential (100+x)*100r+x*150r= 1 2 1 βˆ— π‘Ÿ βˆ— (200)2 +(60-x)*250r X=50A The actual directions of currents in the various sections of the distributor are currents supplied by A and B meet at D. Hence point D is the point of minimum 6 October 2020 D.Rajesh Asst.Prof. EEE Department 114
 (ii) Total current = 160 + 1 Γ— 200 = 360 A  Current supplied by A, 𝐼𝐴 = 100 + x = 100 + 50 = 150 A  Current supplied by B, 𝐼𝐡 = 360 βˆ’ 150 = 210 A  Minimum potential, 𝑉𝐷 = 𝑉𝐴 βˆ’ (𝐼𝐴𝐢 𝑅𝐴𝐢 + 𝐼𝐢𝐷 𝑅𝐢𝐷 ) = 240 βˆ’ 150 Γ— (100 Γ— 0Β·001) βˆ’ 50 Γ— (150 Γ— 0Β·001) = 240 βˆ’ 15 βˆ’ 7Β·5 = 217Β·5 V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 115
 A distributor arranged to form a closed loop and fed at one or more points is called a ring distributor.  For the purpose of calculating voltage distribution, the distributor can be considered as consisting of a series of open distributors fed at both ends.  The principal advantage of ring distributor is that by proper choice in the number of feeding points, great economy in copper can be affected. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 116
 Sometimes a ring distributor has to serve a large area. In such a case, voltage drops in the various sections of the distributor may become excessive.  In order to reduce voltage drops in various sections, distant points of the distributor are joined through a conductor called interconnector. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 117
 The ring distributor ABCDEA. The points B and D of the ring distributor are joined through an interconnector BD.  Network can be obtained by applying Thevenin’s theorem.  (i) Consider the interconnector BD to be disconnected and find the potential difference between B and D. This gives Thevenin’s equivalent circuit voltage 𝐸0.  (ii) Next, calculate the resistance viewed from points B and D of the network composed of distribution lines only. This gives Thevenin’s equivalent circuit series resistance 𝑅0.  (iii) If 𝑅𝐡𝐷 is the resistance of the interconnector BD, then Thevenin’s equivalent circuit will be ∴ Current in interconnector BD = 𝐸0 𝑅0+𝑅𝐡𝐷  Therefore, current distribution in each section and the voltage of load points can be calculated. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 118
A 2-wire d.c. distributor ABCDEA in the form of a ring main is fed at point A at 220 V and is loaded as under : 10A at B ; 20A at C ; 30A at D and 10 A at E. The resistances of various sections (go and return) are : AB = 0Β·1 Ξ© ; BC = 0Β·05 Ξ© ; CD = 0Β·01 Ξ© ; DE = 0Β·025 Ξ© and EA = 0Β·075 Ξ©. Determine :  (i) the point of minimum potential  (ii) current in each section of distributor Let us suppose that current I flows in section AB of the distributor. Then currents in the various sections of the distributor are According to Kirchhoff’s voltage law, the voltage drop in the closed loop ABCDEA is zero i.e. 𝐼𝐴𝐡𝑅𝐴𝐡+𝐼𝐡𝐢𝑅𝐡𝐢+ 𝐼𝐢𝐷𝑅𝐢𝐷+ 𝐼𝐷𝐸𝑅𝐷𝐸 + 𝐼𝐸𝐴𝑅𝐸𝐴 =0 0.1𝐼+(𝐼-10)0.05+ (𝐼-30)0.01+(𝐼-60)0.025+(𝐼-70)0.075=0 0.26 𝐼 = 7.55 𝐼 = 29.04A 6 October 2020 D.Rajesh Asst.Prof. EEE Department 119
 (ii) Current in section AB = I = 29Β·04 A from A to B  Current in section BC = I βˆ’ 10 = 29Β·04 βˆ’ 10 = 19Β·04 A from B to C  Current in section CD = I βˆ’ 30 = 29Β·04 βˆ’ 30 = βˆ’ 0Β·96 A = 0Β·96 A from D to C  Current in section DE = I βˆ’ 60 = 29Β·04 βˆ’ 60 = βˆ’ 30Β·96 A = 30Β·96 A from E to D  Current in section EA = I βˆ’ 70 = 29Β·04 βˆ’ 70 = βˆ’ 40Β·96 A = 40Β·96 A from A to E 6 October 2020 D.Rajesh Asst.Prof. EEE Department 120 ∴ C is the point of minimum potential.
 A d.c. ring main ABCDA is fed from point A from a 250 V supply and the resistances (including both lead and return) of various sections are as follows : AB = 0Β·02 Ξ© ; BC = 0Β·018 Ξ© ; CD = 0Β·025 Ξ© and DA = 0Β·02 Ξ©. The main supplies loads of 150 A at B ; 300 A at C and 250 A at D. Determine the voltage at each load point. If the points A and C are linked through an interconnector of resistance 0Β·02 Ξ©, determine the new voltage at each load point.  Without Interconnector.  Let us suppose that a current I flows in section AB of the distributor. Then currents in various sections of the distributor will be  According to Kirchhoff’s voltage law, the voltage drop in the closed loop ABCDA is zero i.e.  𝐼𝐴𝐡𝑅𝐴𝐡+𝐼𝐡𝐢𝑅𝐡𝐢+ 𝐼𝐢𝐷𝑅𝐢𝐷+ 𝐼𝐷𝐴𝑅𝐷𝐴 =0  0.02𝐼+(𝐼-150)0.018+ (𝐼-450)0.025+(𝐼-700)0.02=0 0Β·083 I = 27Β·95 ∴ I = 27Β·95/0Β·083 = 336Β·75 A 6 October 2020 D.Rajesh Asst.Prof. EEE Department 121
The actual distribution of currents are ..  Current in section AB = I = 336.75 A from A to B  Current in section BC = I βˆ’ 150 = 336.75 βˆ’ 150 = 186.75 A from B to C  Current in section CD = I βˆ’ 450 = 336.75 βˆ’ 450 = βˆ’ 113.25 A = 113.25 A from D to C  Current in section DA = I βˆ’ 700 = 336.75 βˆ’ 700 = βˆ’ 363.25A = 363.25 A from A to D Voltage drop in AB = 336Β·75 Γ— 0Β·02 = 6Β·735 V Voltage drop in BC = 186Β·75 Γ— 0Β·018 = 3Β·361 V Voltage drop in CD = 113Β·25 Γ— 0Β·025 = 2Β·831 V Voltage drop in DA = 363Β·25 Γ— 0Β·02 = 7Β·265 V ∴ Voltage at point B = 250 βˆ’ 6Β·735 = 243Β·265 V Voltage at point C = 243Β·265 βˆ’ 3Β·361 = 239Β·904 V Voltage at point D = 239Β·904 + 2Β·831 = 242Β·735 V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 122
 With Interconnector.  The ring distributor with interconnector AC. The current in the interconnector can be found by applying Thevenin’s theorem. 𝐸0 = Voltage between points A and C = 250 βˆ’ 239Β·904 = 10Β·096 V 𝑅0 = Resistance viewed from points A and C = 0.02+0.018 0.02+0.025 0.02+0.018 + 0.02+0.025 = 0.02Ξ© 𝑅𝐴𝐢 = Resistance of interconnector = 0Β·02 Ξ©  From Thevenin’s equivalent circuit Current in interconnector AC is 𝐸0 𝑅0 + 𝑅𝐴𝐢 = 10.096 0.02 + 0.02 = 252.4𝐴 π‘“π‘Ÿπ‘œπ‘š 𝐴 π‘‘π‘œ 𝐢 Let us suppose that current in section AB is 𝐼1 . Then current in section BC will be 𝐼1 βˆ’ 150. As the voltage drop round the closed mesh ABCA is zero, 0.02𝐼1+(𝐼1-150)0.018-0.02*252.4=0 0.038 𝐼1=7.748 𝐼1=203.15A 6 October 2020 D.Rajesh Asst.Prof. EEE Department 123
The actual distribution of currents are ..  Current in section AB = 𝐼1 = 203.15A from A to B  Current in section BC = 𝐼1 βˆ’ 150 = 203.15 βˆ’ 150 = 53.15 A from B to C  Current in section CD = 𝐼1 βˆ’ 450+252.4 = 203.15βˆ’ 450 +252.4= 5.55A C to D  Current in section DA = 𝐼1 βˆ’ 700 = 203.15+252.5 βˆ’ 700 = βˆ’ 244.45A = 244.45A from A to D  Drop in AB = 203Β·15 Γ— 0Β·02 = 4Β·063 V  Drop in BC = 53Β·15 Γ— 0Β·018= 0Β·960 V  Drop in AD = 244Β·45 Γ— 0Β·02 = 4Β·9 V  ∴ Potential of B = 250 βˆ’ 4Β·063= 245Β·93 V  Potential of C = 245Β·93 βˆ’ 0Β·96 = 244Β·97 V  Potential of D = 250 βˆ’ 4Β·9 = 245Β·1 V It may be seen that with the use of interconnector, the voltage drops in the various sections of the distributor are reduced. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 124
 A.C. distribution calculations differ from those of d.c. distribution in the following respects :  (i) In case of d.c. system, the voltage drop is due to resistance alone. However, in a.c. system, the voltage drops are due to the combined effects of resistance, inductance and capacitance.  (ii) In a d.c. system, additions and subtractions of currents or voltages are done arithmetically but in case of a.c. system, these operations are done vectorially.  (iii) In an a.c. system, power factor (p.f.) has to be taken into account. Loads tapped off form the distributor are generally at different power factors. There are two ways of referring power factor viz (a) It may be referred to supply or receiving end voltage which is regarded as the reference vector. (b) It may be referred to the voltage at the load point itself. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 125
 The power factors of load currents may be given (i) w.r.t. receiving or sending end voltage or (ii) w.r.t. to load voltage itself. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 126
 (i) Power factors referred to receiving end voltage. Consider an a.c. distributor AB with concentrated loads of 𝐼1and 𝐼2 tapped off at points C and B. Taking the receiving end voltage 𝑉𝐡 as the reference vector, let lagging power factors at C and B be cos Ο†1 and cos Ο†2 w.r.t. 𝑉𝐡 . Let 𝑅1, 𝑋1 and 𝑅2, 𝑋2 be the resistance and reactance of sections AC and CB of the distributor. Impedance of section AC, 𝑍𝐴𝐢= 𝑅1 + j 𝑋1 Impedance of section CB,𝑍𝐢𝐡 = 𝑅2 + j 𝑋2 Load current at point C, 𝐼1 = 𝐼1 (cos Ο†1 βˆ’ j sin Ο†1) Load current at point B, 𝐼2 = 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) Current in section CB, 𝐼𝐢𝐡= 𝐼2 = 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) Current in section AC, 𝐼𝐴𝐢 = 𝐼1+ 𝐼2 =𝐼1 (cos Ο†1 βˆ’ j sin Ο†1)+𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) 6 October 2020 D.Rajesh Asst.Prof. EEE Department 127
Ο†1  Voltage drop in section CB,  𝑉𝐢𝐡= 𝐼𝐢𝐡 𝑍𝐢𝐡= 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) (𝑅2 + j 𝑋2 )  Voltage drop in section AC,  𝑉𝐴𝐢= 𝐼𝐴𝐢 𝑍𝐴𝐢=(𝐼1+ 𝐼2) 𝑍𝐴𝐢  =𝐼1 (cos Ο†1 βˆ’ j sin Ο†1)+𝐼2 (cos Ο†2 βˆ’ j sin Ο†2)(𝑅1 + j 𝑋1 )  Sending end voltage, 𝑉𝐴= 𝑉𝐡 + 𝑉𝐢𝐡 + 𝑉𝐴𝐢  Sending end current, 𝐼𝐴= 𝐼1+ 𝐼2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 128 𝑉𝑩 𝐼𝟐 𝐼𝟏 𝐼𝑨π‘ͺ Ο†2 ΞΈ
Ο†1  (ii) Power factors referred to respective load voltages.  Suppose the power factors of loads are referred to their respective load voltages. Then Ο†1 is the phase angle between V𝐢 and I1 and Ο†2 is the phase angle between V𝐡 and I2. The vector diagram under these conditions is  Voltage drop in section CB,  𝑉𝐢𝐡= 𝐼2 𝑍𝐢𝐡= 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) (𝑅2 + j 𝑋2 )  Voltage at point C = 𝑉𝐡 + Drop in section CB = V𝐢 ∠ Ξ± (say)  Now 𝐼1 = 𝐼1 ∠ βˆ’ Ο†1 w.r.t. voltage V𝐢  ∴ . 𝐼1 = 𝐼1 ∠ βˆ’ (Ο†1 βˆ’ Ξ±) w.r.t. voltage V𝐡  i.e. 𝐼1 = 𝐼1 [cos (Ο†1 βˆ’ Ξ±) βˆ’ j sin (Ο†1 βˆ’ Ξ±)]  Now 𝐼𝐴𝐢 = 𝐼1+ 𝐼2 =𝐼1 [cos (Ο†1 βˆ’ Ξ±) βˆ’ j sin (Ο†1 βˆ’ Ξ±)]+ 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2)  Voltage drop in section AC, 𝑉𝐴𝐢= 𝐼𝐴𝐢 𝑍𝐴𝐢  ∴ Voltage at point A = V𝐡 + Drop in CB + Drop in AC 6 October 2020 D.Rajesh Asst.Prof. EEE Department 129 𝑉𝑩 𝐼𝟐 𝐼𝟏 𝐼𝑨π‘ͺ Ο†2 Ξ±
A single phase a.c. distributor AB 300 metres long is fed from end A and is loaded as under : (i) 100 A at 0Β·707 p.f. lagging 200 m from point A (ii) 200 A at 0Β·8 p.f. lagging 300 m from point A The load resistance and reactance of the distributor is 0Β·2 Ξ© and 0Β·1 Ξ© per kilometre. Calculate the total voltage drop in the distributor. The load power factors refer to the voltage at the far end. Impedance of distributor/km = (0Β·2 + j 0Β·1) Ξ© Impedance of section AC, 𝑍𝐴𝐢 = (0Β·2 + j 0Β·1) Γ— 200/1000 = (0Β·04 + j 0Β·02) Ξ© Impedance of section CB, 𝑍𝐢𝐡 = (0Β·2 + j 0Β·1) Γ— 100/1000 = (0Β·02 + j 0Β·01) Ξ© Taking voltage at the far end B as the reference vector, we have, Load current at point B, 𝐼2 = 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) = 200 (0Β·8 βˆ’ j 0Β·6) = (160 βˆ’ j 120) A Load current at point C, 𝐼1 = 𝐼1 (cos Ο†1 βˆ’ j sin Ο†1) = 100 (0Β·707 βˆ’ j 0Β·707) = (70Β·7βˆ’ j 70Β·7) A Current in section CB,𝐼𝐢𝐡= 𝐼2 = (160 βˆ’ j 120) A Current in section AC,𝐼𝐴𝐢= 𝐼1+ 𝐼2 = (70Β·7 βˆ’ j 70Β·7) + (160 βˆ’ j 120)= (230Β·7 βˆ’ j 190Β·7) A Voltage drop in section CB, 𝑉𝐢𝐡= 𝐼𝐢𝐡 𝑍𝐢𝐡 = (160 βˆ’ j 120) (0Β·02 + j 0Β·01) = (4Β·4 βˆ’ j 0Β·8) volts Voltage drop in section AC, 𝑉𝐴𝐢= 𝐼𝐴𝐢 𝑍𝐴𝐢 = (230Β·7 βˆ’ j 190Β·7) (0Β·04 + j 0Β·02) = (13Β·04 βˆ’ j 3Β·01) volts Voltage drop in the distributor = 𝑉𝐴𝐢 + 𝑉𝐢𝐡 = (13Β·04 βˆ’ j 3Β·01) + (4Β·4 βˆ’ j 0Β·8) = (17Β·44 βˆ’ j 3Β·81) volts Magnitude of drop = 17.44 2 + 3.81 2 = 17Β·85 V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 130
 A single phase ring distributor ABC is fed at A. The loads at B and C are 20 A at 0.8 p.f. lagging and 15 A at 0.6 p.f. lagging respectively ; both expressed with reference to the voltage at A. The total impedance of the three sections AB, BC and CA are (1 + j 1), (1+ j2) and (1 + j3) ohms respectively. Find the total current fed at A and the current in each section. Use Thevenin’s theorem to obtain the results. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 131

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POWER SUPPLY SYSTEMS& DISTRIBUTION SYSTEMS

  • 1. By D.Rajesh Asst. Prof. EEE Department 6 October 2020 D.Rajesh Asst.Prof. EEE Department 1
  • 2.  Unit-i Electric Power Generation& Economic Considerations  Unit-ii Power Supply Systems& Distribution Systems  Unit-iii Inductance & Capacitance Calculations  Unit-iv Performance Of Transmission Lines  Unit-v Overhead Line Insulators 6 October 2020 D.Rajesh Asst.Prof. EEE Department 2
  • 3. POWER SUPPLY SYSTEMS& DISTRIBUTION SYSTEMS:  Transmission and distribution systems- D.C 2-wire and 3- wire systems, A.C single phase , three phase and 4-wire systems, comparison of copper efficiency.  Primary and secondary distribution systems, concentrated & uniformly distributed loads on distributors fed at both ends, ring distributor, voltage drop and power loss calculation,  Kelvin’slaw. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 3
  • 4. Power Transmission D.C D.C. two-wire D.C. two-wire with mid-point earthed D.C. three-wire A.C Single-phase A.C. system Single-phase two- wire Single-phase two- wire with mid- point earthed Single-phase three-wire Two-phase A.C. system Two-phase four- wire Two-phase three wire. Three-phase A.C. system Three-phase three-wire Three-phase four- wire 6 October 2020 D.Rajesh Asst.Prof. EEE Department 4
  • 5. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 5 Overhead Under ground
  • 6. (i) Same power (P watts) transmitted by each system. (ii) The distance (l metres) over which power is transmitted remains the same. (iii) The line losses (W watts) are the same in each case. (iv) The maximum voltage between any conductor and earth (𝑉 π‘š) is the same in each case. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 6
  • 7. Max. voltage between conductors = 𝑉 π‘š , Power to be transmitted = P= 𝑉 π‘šπΌ1 1 ∴ Load current, 𝐼1 = P/ 𝑉 π‘š If 𝑅1 is the resistance of each line conductor, then, 𝑅1 = ρ l/ π‘Ž1 where π‘Ž1 is the area of X-section of the conductor. copper losses in line W = 𝐼1 2 𝑅1 + 𝐼1 2 𝑅1 =2𝐼1 2 𝑅1 6 October 2020 D.Rajesh Asst.Prof. EEE Department 7 1 1 2
  • 8. The maximum voltage between any conductor and earth is 𝑉 π‘š so that maximum voltage between conductors is 2 𝑉 π‘š. 𝑅2 is the resistance of each line conductor Power to be transmitted P= 𝑉 π‘šπΌ2+ 𝑉 π‘šπΌ2=2 𝑉 π‘šπΌ2 Load current, 𝐼2 = P/2 𝑉 π‘š Since from assumption 1 the power transmitted is same for all systems so from eq 1& 3 2𝑉 π‘šπΌ2= 𝑉 π‘šπΌ1 𝐼2 = 𝐼1 2 copper losses in line W = 𝐼2 2 𝑅2+ 𝐼2 2 𝑅2 = 2 𝐼2 2 𝑅2 From eq 4 W=2( 𝐼1 2 )2 𝑅2 = 𝐼1 2 2 𝑅2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 8 3 4 5
  • 9.  Compare eq 2 & 5 2𝐼1 2 𝑅1 = ( 𝐼1 2 2 ) 𝑅2 𝑅1 𝑅2 = 1 4 π‘Ž1 π‘Ž2 =4 (where R = ρ l/ π‘Ž) π‘Ž1=X-area of conductor in 2 wire dc system π‘Ž2= X-area of conductor in 2 wire dc system with mid-point earthed Then volume of copper wire Twoβˆ’wire D.C system with midβˆ’point earthed volume of copper wire Twoβˆ’wire D.C system = π‘Ž2βˆ—π‘™2 π‘Ž1βˆ—π‘™1 = 1 4 π‘Ž1βˆ—2𝑙 π‘Ž1βˆ—2𝑙 = 1 4 where 𝑙1 = 𝑙2=l+l Hence, the volume of conductor material required in Twoβˆ’wire D.C system with midβˆ’point earthed is one-fourth of that required in a two-wire D.C system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 9
  • 10. If the load is balanced, the current in the neutral wire is zero. Assuming balanced loads, Power to be transmitted P= 𝑉 π‘šπΌ3+ 𝑉 π‘šπΌ3=2 𝑉 π‘šπΌ3 6 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 6 2𝑉 π‘šπΌ3= 𝑉 π‘šπΌ1 𝐼3 = 𝐼1 2 7 Let 𝑅3 is the resistance of each line conductor, the area of X-section of neutral wire to be half that of the outer wire, copper losses in line W = 𝐼3 2 𝑅3+ 𝐼3 2 𝑅3 = 2𝐼3 2 𝑅3 8 From eq 7 W=2( 𝐼1 2 )2 𝑅3 = 𝐼1 2 2 𝑅3 9 6 October 2020 D.Rajesh Asst.Prof. EEE Department 10
  • 11.  Compare eq 2 & 8 2𝐼1 2 𝑅1 = 𝐼1 2 2 𝑅3 𝑅1 𝑅3 = 1 4 𝑅3 = ρ l/ π‘Ž3 where π‘Ž3 is the area of X-section of the conductor. π‘Ž1 π‘Ž3 =4 Then volume of copper wire Threeβˆ’wire D.C volume of copper wire Twoβˆ’wire D.C system = π‘Ž3βˆ—π‘™3 π‘Ž1βˆ—π‘™1 = 1 4 π‘Ž1βˆ—2.5𝑙 π‘Ž1βˆ—2𝑙 = 2.5 8 where 𝑙3=l+l+0.5l =0.3125 Hence, the volume of conductor material required in Threeβˆ’wire D.C system is 0.3125 times of that required in a two-wire D.C system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 11
  • 12. The maximum voltage between conductors is 𝑉 π‘š, so that r.m.s. value of voltage between them is π‘‰π‘š 2 . Assuming the load power factor to be cos Ο†, load current 𝐼4 Power supplied P= π‘‰π‘š 2 𝐼4 cos Ο† 10 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 10 π‘‰π‘š 2 𝐼4 cos Ο†= 𝑉 π‘šπΌ1 𝐼4= 2𝐼1 cos Ο† 11 Let 𝑅4 is the resistance of each line conductor copper losses in line W = 𝐼4 2 𝑅4+ 𝐼4 2 𝑅4 = 2𝐼4 2 𝑅4 =2( 2𝐼1 cos Ο† )2 𝑅4 =4 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅4 12 6 October 2020 D.Rajesh Asst.Prof. EEE Department 12
  • 13.  Comparing copper losses with 2 wire d.c system 4 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅4= 2𝐼1 2 𝑅1 𝑅4 𝑅1 = π‘π‘œπ‘ 2Ο† 2 π‘Ž4 π‘Ž1 = 2 π‘π‘œπ‘ 2Ο† 13 volume of copper wire 1βˆ’phase 2βˆ’ wire volume of copper wire Twoβˆ’wire D.C system = π‘Ž4𝑙4 π‘Ž1𝑙1 = 2 π‘π‘œπ‘ 2Ο†π‘Ž12𝑙 π‘Ž12𝑙 = 2 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required in this system is 2 π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system with the one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 13
  • 14.  The two wires possess equal and opposite voltages to earth (i.e., 𝑉 π‘š).  Therefore, the maximum voltage between the two wires is π‘‰π‘š 2 . The r.m.s. value of voltage between conductors is =2 π‘‰π‘š 2 = 2 𝑉 π‘š. Assuming the power factor of the load to be cos Ο†, load current 𝐼5  Power supplied P= 2 𝑉 π‘šπΌ5 cos Ο† 14 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 14 2 𝑉 π‘šπΌ5 cos Ο†= 𝑉 π‘šπΌ1 𝐼5= 𝐼1 2 cos Ο† 15 6 October 2020 D.Rajesh Asst.Prof. EEE Department 14
  • 15. Let 𝑅5 is the resistance of each line conductor copper losses in line W = 𝐼5 2 𝑅5+ 𝐼5 2 𝑅5 = 2𝐼5 2 𝑅5 =2( 𝐼1 2 cos Ο† )2 𝑅5 = 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅5 Comparing copper losses with 2 wire d.c system 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅5 =2𝐼1 2 𝑅1 𝑅5 𝑅1 = 2π‘π‘œπ‘ 2Ο† π‘Ž5 π‘Ž1 = 1 2π‘π‘œπ‘ 2Ο† volume of copper wire 1βˆ’phase 2βˆ’ wire π‘€π‘–π‘‘β„Ž π‘šπ‘–π‘‘ βˆ’ π‘π‘œπ‘–π‘›π‘‘ π‘’π‘Žπ‘Ÿπ‘‘β„Žπ‘’π‘‘ volume of copper wire Twoβˆ’wire D.C system π‘Ž5𝑙5 π‘Ž1𝑙1 = 1 2π‘π‘œπ‘ 2Ο† π‘Ž12𝑙 π‘Ž12𝑙 = 1 2π‘π‘œπ‘ 2Ο† Hence the volume of conductor material required in this system is 1 2π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 15
  • 16.  If the load is balanced, the current through the neutral wire is zero. Assuming balanced load,  The two wires possess equal and opposite voltages to earth (i.e., 𝑉 π‘š).  Therefore, the maximum voltage between the two wires is π‘‰π‘š 2 . The r.m.s. value of voltage between conductors is 2 π‘‰π‘š 2 = 2 𝑉 π‘š. Assuming the power factor of the load to be cos Ο†, load current 𝐼6  Power supplied P= 2 𝑉 π‘šπΌ6 cos Ο† 16 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 16 2 𝑉 π‘šπΌ6 cos Ο†= 𝑉 π‘šπΌ1 𝐼6= 𝐼1 2 cos Ο† 17 6 October 2020 D.Rajesh Asst.Prof. EEE Department 16
  • 17. Let 𝑅6 is the resistance of each line conductor copper losses in line W = 𝐼6 2 𝑅6+ 𝐼6 2 𝑅6 = 2𝐼6 2 𝑅6 =2( 𝐼1 2 cos Ο† )2 𝑅6 = 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅6 Comparing copper losses with 2 wire d.c system 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅6 =2𝐼1 2 𝑅1 𝑅6 𝑅1 = 2π‘π‘œπ‘ 2Ο† π‘Ž6 π‘Ž1 = 1 2π‘π‘œπ‘ 2Ο† volume of copper wire 1βˆ’phase3βˆ’wire system volume of copper wire Twoβˆ’wire D.C system π‘Ž6𝑙6 π‘Ž1𝑙1 = 1 2π‘π‘œπ‘ 2Ο† π‘Ž12.5𝑙 π‘Ž12𝑙 = 0.625 π‘π‘œπ‘ 2Ο† Hence the volume of conductr material required in this system is 0.625 π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 17
  • 18.  This system can be considered as two independent single- phase systems, each transmitting one half of the total power.  As the voltage is same (𝑉 π‘š) as in a Single phase 2-wire system with mid-point earthed, so the current will be half of the Single phase system  i.e. 𝐼7= 𝐼5 2 = 𝐼1 2 cos Ο† 2 = 𝐼1 2 2 cos Ο†  If the current density remains constant the X-area will be half of the single-phase system  i.e. π‘Ž7= π‘Ž5 2 = 1 2π‘π‘œπ‘ 2Ο† π‘Ž1 2 = π‘Ž1 4π‘π‘œπ‘ 2Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 18
  • 19.  Since wires are four  Then  volume of copper wireTwo phase, 4βˆ’wire a.c. system volume of copper wire Twoβˆ’wire D.C system π‘Ž7𝑙7 π‘Ž1𝑙1 = 1 4π‘π‘œπ‘ 2Ο† π‘Ž14𝑙 π‘Ž12𝑙 = 1 2π‘π‘œπ‘ 2Ο† Hence the volume of conductor material required in this system is 1 2π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 19
  • 20. The third or neutral wire is taken from the junction of two-phase windings whose voltages are in quadrature with each other. Obviously, each phase transmits one half of the total power. The R.M.S. voltage between outgoing conductor and neutral = π‘‰π‘š 2 , Let 𝑅8 is the resistance of each line conductor Total power transmitted=2 π‘‰π‘š 2 𝐼8 cos Ο† Comparing the power with 2-wire D.C 2 π‘‰π‘š 2 𝐼8 cos Ο†= 𝑉 π‘šπΌ1 𝐼8 = 𝐼1 2cos Ο† The middle wire will act as a common return and the current through it will be 𝐼8 2 + 𝐼8 2 = 2 𝐼8 6 October 2020 D.Rajesh Asst.Prof. EEE Department 20
  • 21.  Assuming the current density to be constant, the area of X-section of the neutral wire will be 2 times that of either of the outers.  ∴ Resistance of neutral wire= 𝑅8 2  Copper losses in this system W= 𝐼8 2 𝑅8+ 𝐼8 2 𝑅8 + ( 2 𝐼8)2𝑅8 2 = 2𝐼8 2 𝑅8+ 2𝐼8 2 𝑅8=(2+ 2) 𝐼8 2 𝑅8 Where 𝐼8 = 𝐼1 2cos Ο† W=(2+ 2) ( 𝐼1 2cos Ο†)2 𝑅8=(2+ 2) 𝐼1 2𝑅8 2π‘π‘œπ‘ 2Ο† Comparing copper losses with 2 wire d.c system (2+ 2) 𝐼1 2𝑅8 2π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅8 𝑅1 = 4π‘π‘œπ‘ 2Ο† (2+ 2) π‘Ž8 π‘Ž1 = (2+ 2) 4π‘π‘œπ‘ 2Ο† volume of copper wireTwo phase, 3βˆ’wire a.c. system volume of copper wire Twoβˆ’wire D.C system π‘Ž8𝑙8 π‘Ž1𝑙1 = (2+ 2) 4π‘π‘œπ‘ 2Ο† π‘Ž1(2+ 2)𝑙 π‘Ž12𝑙 = 1.457 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 1.457 π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 21
  • 22. This system is almost universally adopted for transmission of electric power. The 3-phase, 3- wire system may be star connected or delta connected. Maximum voltage between each conductor and neutral =𝑉 π‘š Let 𝑅9 is the resistance of each line conductor R.M.S. voltage per phase = π‘‰π‘š 2 Power supplied=3( π‘‰π‘š 2 I9cos Ο† )  Comparing the power with 2-wire D.C 3 π‘‰π‘š 2 𝐼9 cos Ο†= 𝑉 π‘šπΌ1 𝐼9 = 2𝐼1 3cos Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 22
  • 23.  Total copper losses= 3𝐼9 2 𝑅9 =3 ( 2𝐼1 3cos Ο† )2 𝑅9= 2𝐼1 2 𝑅9 3π‘π‘œπ‘ 2Ο†  Comparing copper losses with 2 wire d.c system 2𝐼1 2 𝑅9 3π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅9 𝑅1 = 3π‘π‘œπ‘ 2Ο† π‘Ž9 π‘Ž1 = 1 3π‘π‘œπ‘ 2Ο† volume of copper wire3βˆ’Phase, 3βˆ’wire system volume of copper wire Twoβˆ’wire D.C system = π‘Ž9𝑙9 π‘Ž1𝑙1 = 1 3π‘π‘œπ‘ 2Ο†π‘Ž13𝑙 π‘Ž12𝑙 = 0.5 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 0.5 π‘π‘œπ‘ 2Ο† times that required for 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 23
  • 24. In this case, 4th or neutral wire is taken from the neutral point. The area of X- section of the neutral wire is generally one-half that of the line conductor. If the loads are balanced, then current through the neutral wire is zero. Assuming balanced loads and p.f. of the load as cos Ο†, Let 𝑅10 is the resistance of each line conductor R.M.S. voltage per phase = π‘‰π‘š 2 Power supplied=3( π‘‰π‘š 2 I10cos Ο† )  Comparing the power with 2-wire D.C 3 π‘‰π‘š 2 𝐼10 cos Ο†= 𝑉 π‘šπΌ1 𝐼10 = 2𝐼1 3cos Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 24
  • 25.  Total copper losses= 3𝐼10 2 𝑅10 =3 ( 2𝐼1 3cos Ο† )2 𝑅10= 2𝐼1 2 𝑅10 3π‘π‘œπ‘ 2Ο†  Comparing copper losses with 2 wire d.c system 2𝐼1 2 𝑅10 3π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅10 𝑅1 = 3π‘π‘œπ‘ 2Ο† π‘Ž10 π‘Ž1 = 1 3π‘π‘œπ‘ 2Ο† volume of copper wire3βˆ’Phase, 3βˆ’wire system volume of copper wire Twoβˆ’wire D.C system = π‘Ž10𝑙10 π‘Ž1𝑙1 = 1 3π‘π‘œπ‘ 2Ο†π‘Ž13.5𝑙 π‘Ž12𝑙 = 0.583 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 0.583 π‘π‘œπ‘ 2Ο† times that required for 2-wire d.c. system with one conductor earthed. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 25
  • 26. In the under-ground system the maximum stress exists on the insulation between the conductors. (i) Same power (P watts) transmitted by each system. (ii) The distance (l meters) over which power is transmitted remains the same. (iii) The line losses (W watts) are the same in each case. (iv) The maximum voltage between conductors (𝑉 π‘š) is the same in each case. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 26
  • 27. Max. voltage between conductors = 𝑉 π‘š , Power to be transmitted = P= 𝑉 π‘šπΌ1 1 ∴ Load current, 𝐼1 = P/ 𝑉 π‘š If 𝑅1 is the resistance of each line conductor, then, 𝑅1 = ρ l/ π‘Ž1 where π‘Ž1 is the area of X-section of the conductor. copper losses in line W = 𝐼1 2 𝑅1 + 𝐼1 2 𝑅1 =2𝐼1 2 𝑅1 6 October 2020 D.Rajesh Asst.Prof. EEE Department 27 1 1 2
  • 28. Maximum voltage between conductors is 𝑉 π‘š. 𝑅2 is the resistance of each line conductor Power to be transmitted P= 𝑉 π‘šπΌ2 Load current, 𝐼2 = P/ 𝑉 π‘š Since from assumption 1 the power transmitted is same for all systems so from eq 1& 3 𝑉 π‘šπΌ2= 𝑉 π‘šπΌ1 𝐼2 = 𝐼1 copper losses in line W = 𝐼2 2 𝑅2+ 𝐼2 2 𝑅2 = 2 𝐼2 2 𝑅2 From eq 4 W=2( 𝐼1)2 𝑅2 =2𝐼1 2 𝑅2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 28 3 4 5
  • 29.  Compare eq 2 & 5 2𝐼1 2 𝑅1 = (𝐼1 2 ) 𝑅2 𝑅1 𝑅2 =1 π‘Ž1 π‘Ž2 =1 (where R = ρ l/ π‘Ž) π‘Ž1=X-area of conductor in 2 wire dc system π‘Ž2= X-area of conductor in 2 wire dc system with mid-point earthed Then volume of copper wire Twoβˆ’wire D.C system with midβˆ’point earthed volume of copper wire Twoβˆ’wire D.C system = π‘Ž2βˆ—π‘™2 π‘Ž1βˆ—π‘™1 = π‘Ž1βˆ—2𝑙 π‘Ž1βˆ—2𝑙 =1 where 𝑙1 = 𝑙2=l+l Hence, the volume of conductor material required in this system is the same as that for 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 29
  • 30. If the load is balanced, the current in the neutral wire is zero. Assuming balanced loads, Power to be transmitted P= 𝑉 π‘šπΌ3 6 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 6 𝑉 π‘šπΌ3= 𝑉 π‘šπΌ1 𝐼3 = 𝐼1 7 Let 𝑅3 is the resistance of each line conductor, the area of X-section of neutral wire to be half that of the outer wire, copper losses in line W = 𝐼3 2 𝑅3+ 𝐼3 2 𝑅3 = 2𝐼3 2 𝑅3 8 From eq 7 W=2( 𝐼1)2 𝑅3 =2𝐼1 2 𝑅3 9 6 October 2020 D.Rajesh Asst.Prof. EEE Department 30
  • 31.  Compare eq 2 & 8 2𝐼1 2 𝑅1 = 2(𝐼1 2 ) 𝑅3 𝑅1 𝑅3 =1 π‘Ž1 π‘Ž3 =1 𝑅3 = ρ l/ π‘Ž3 where π‘Ž3 is the area of X-section of the conductor. Then volume of copper wire Threeβˆ’wire D.C volume of copper wire Twoβˆ’wire D.C system = π‘Ž3βˆ—π‘™3 π‘Ž1βˆ—π‘™1 = π‘Ž1βˆ—2.5𝑙 π‘Ž1βˆ—2𝑙 = 2.5 2 where 𝑙3=l+l+0.5l =1.25 Hence, the volume of conductor material required in the system is 1Β·25 times that required for 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 31
  • 32. The maximum voltage between conductors is 𝑉 π‘š, so that r.m.s. value of voltage between them is π‘‰π‘š 2 . Assuming the load power factor to be cos Ο†, load current 𝐼4 Power supplied P= π‘‰π‘š 2 𝐼4 cos Ο† 10 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 10 π‘‰π‘š 2 𝐼4 cos Ο†= 𝑉 π‘šπΌ1 𝐼4= 2𝐼1 cos Ο† 11 Let 𝑅4 is the resistance of each line conductor copper losses in line W = 𝐼4 2 𝑅4+ 𝐼4 2 𝑅4 = 2𝐼4 2 𝑅4 =2( 2𝐼1 cos Ο† )2 𝑅4 =4 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅4 12 6 October 2020 D.Rajesh Asst.Prof. EEE Department 32
  • 33.  Comparing copper losses with 2 wire d.c system 4 𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅4= 2𝐼1 2 𝑅1 𝑅4 𝑅1 = π‘π‘œπ‘ 2Ο† 2 π‘Ž4 π‘Ž1 = 2 π‘π‘œπ‘ 2Ο† 13 volume of copper wire 1βˆ’phase 2βˆ’ wire volume of copper wire Twoβˆ’wire D.C system = π‘Ž4𝑙4 π‘Ž1𝑙1 = 2 π‘π‘œπ‘ 2Ο†π‘Ž12𝑙 π‘Ž12𝑙 = 2 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required in this system is 2 π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 33
  • 34. The maximum value of voltage between the outers is 𝑉 π‘š. Therefore, the r.m.s. value of voltage is π‘‰π‘š 2 Assuming the power factor of the load to be cos Ο†, load current 𝐼5 Power supplied P= π‘‰π‘š 2 𝐼5 cos Ο† 14 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 14 π‘‰π‘š 2 𝐼5 cos Ο†= 𝑉 π‘šπΌ1 𝐼5= 2𝐼1 cos Ο† 15 6 October 2020 D.Rajesh Asst.Prof. EEE Department 34
  • 35. Let 𝑅5 is the resistance of each line conductor copper losses in line W = 𝐼5 2 𝑅5+ 𝐼5 2 𝑅5 = 2𝐼5 2 𝑅5 =2( 2𝐼1 cos Ο† )2 𝑅5 = 4𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅5 Comparing copper losses with 2 wire d.c system 4𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅5 =2𝐼1 2 𝑅1 𝑅5 𝑅1 = π‘π‘œπ‘ 2Ο† 2 π‘Ž5 π‘Ž1 = 2 π‘π‘œπ‘ 2Ο† volume of copper wire 1βˆ’phase 2βˆ’ wire π‘€π‘–π‘‘β„Ž π‘šπ‘–π‘‘ βˆ’ π‘π‘œπ‘–π‘›π‘‘ π‘’π‘Žπ‘Ÿπ‘‘β„Žπ‘’π‘‘ volume of copper wire Twoβˆ’wire D.C system π‘Ž5𝑙5 π‘Ž1𝑙1 = 2 π‘π‘œπ‘ 2Ο† π‘Ž12𝑙 π‘Ž12𝑙 = 2 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required in this system is 2 π‘π‘œπ‘ 2Ο† times that required in 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 35
  • 36.  If the load is balanced, the current through the neutral wire is zero. Assuming balanced load,  The maximum value of voltage between the outers is 𝑉 π‘š. Therefore, the r.m.s. value of voltage is π‘‰π‘š 2 .  Assuming the power factor of the load to be cos Ο†, load current 𝐼6  Power supplied P= π‘‰π‘š 2 𝐼6 cos Ο† 16 Since from assumption 1 the power transmitted is same for all systems so from eq 1& 16 π‘‰π‘š 2 𝐼6 cos Ο†= 𝑉 π‘šπΌ1 𝐼6= 2𝐼1 cos Ο† 17 6 October 2020 D.Rajesh Asst.Prof. EEE Department 36
  • 37. Let 𝑅6 is the resistance of each line conductor copper losses in line W = 𝐼6 2 𝑅6+ 𝐼6 2 𝑅6 = 2𝐼6 2 𝑅6 =2( 2𝐼1 cos Ο† )2 𝑅6 = 4𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅6 Comparing copper losses with 2 wire d.c system 4𝐼1 2 π‘π‘œπ‘ 2Ο† 𝑅6 =2𝐼1 2 𝑅1 𝑅6 𝑅1 = π‘π‘œπ‘ 2Ο† 2 π‘Ž6 π‘Ž1 = 2 π‘π‘œπ‘ 2Ο† volume of copper wire 1βˆ’phase 2βˆ’ wire π‘€π‘–π‘‘β„Ž π‘šπ‘–π‘‘ βˆ’ π‘π‘œπ‘–π‘›π‘‘ π‘’π‘Žπ‘Ÿπ‘‘β„Žπ‘’π‘‘ volume of copper wire Twoβˆ’wire D.C system π‘Ž6𝑙6 π‘Ž1𝑙1 = = 2 π‘π‘œπ‘ 2Ο† π‘Ž12.5𝑙 π‘Ž12𝑙 = 2.5 π‘π‘œπ‘ 2Ο†  Hence, the volume of conductor material required in this system is 2.5 π‘π‘œπ‘ 2Ο† times that required in 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 37
  • 38.  This system can be considered as two independent single-phase systems, each transmitting one half of the total power.  Current (𝐼7) in each conductor will be half that in single-phase 2- wire system. Consequently, area of X-section of each conductor is also half but as there are four wires, so volume of conductor material used is the same as in a single phase, 2-wire system  i.e. 𝐼7= 𝐼5 2 = 2𝐼1 cos Ο† 2 = 𝐼1 2 cos Ο†  If the current density remains constant the X-area will be half of the single-phase system  i.e. π‘Ž7= π‘Ž5 2 = 2 π‘π‘œπ‘ 2Ο† 2 = 1 π‘π‘œπ‘ 2Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 38
  • 39.  Since wires are four  Then  volume of copper wireTwo phase, 4βˆ’wire a.c. system volume of copper wire Twoβˆ’wire D.C system π‘Ž7𝑙7 π‘Ž1𝑙1 = 1 π‘π‘œπ‘ 2Ο† π‘Ž14𝑙 π‘Ž12𝑙 = 2 π‘π‘œπ‘ 2Ο† Hence, volume of conductor material required in this system is 2 π‘π‘œπ‘ 2Ο† times that required in 2- wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 39
  • 40.  Let us suppose that maximum voltage between the outers is 𝑉 π‘š. Then maximum voltage between either outer and neutral wire is π‘‰π‘š 2  R.M.S. voltage between outer and neutral wire= π‘‰π‘š 2 2 = π‘‰π‘š 2  Total power transmitted=2 π‘‰π‘š 2 𝐼8 cos Ο† = 𝑉 π‘š 𝐼8 cos Ο†  Comparing the power with 2-wire D.C 𝑉 π‘š 𝐼8 cos Ο†= 𝑉 π‘šπΌ1 𝐼8 = 𝐼1 cos Ο† The middle wire will act as a common return and the current through it will be 𝐼8 2 + 𝐼8 2 = 2 𝐼8 6 October 2020 D.Rajesh Asst.Prof. EEE Department 40
  • 41.  Assuming the current density to be constant, the area of X-section of the neutral wire will be 2 times that of either of the outers.  ∴ Resistance of neutral wire= 𝑅8 2  Copper losses in this system W= 𝐼8 2 𝑅8+ 𝐼8 2 𝑅8 + ( 2 𝐼8)2𝑅8 2 = 2𝐼8 2 𝑅8+ 2𝐼8 2 𝑅8=(2+ 2) 𝐼8 2 𝑅8 Where 𝐼8 = 𝐼1 cos Ο† W=(2+ 2) ( 𝐼1 cos Ο†)2 𝑅8=(2+ 2) 𝐼1 2𝑅8 π‘π‘œπ‘ 2Ο† Comparing copper losses with 2 wire d.c system (2+ 2) 𝐼1 2𝑅8 π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅8 𝑅1 = 2π‘π‘œπ‘ 2Ο† (2+ 2) π‘Ž8 π‘Ž1 = (2+ 2) 2π‘π‘œπ‘ 2Ο† volume of copper wireTwo phase, 3βˆ’wire a.c. system volume of copper wire Twoβˆ’wire D.C system π‘Ž8𝑙8 π‘Ž1𝑙1 = (2+ 2) 2π‘π‘œπ‘ 2Ο† π‘Ž1(2+ 2)𝑙 π‘Ž12𝑙 = 2.914 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 2.914 π‘π‘œπ‘ 2Ο† times that of 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 41
  • 42. This system is almost universally adopted for transmission of electric power. The 3-phase, 3- wire system may be star connected or delta connected. Maximum voltage between conductors =𝑉 π‘š, Let 𝑅9 is the resistance of each line conductor Then maximum voltage between each phase and neutral is π‘‰π‘š 3 R.M.S. voltage per phase = π‘‰π‘š 3 2 = π‘‰π‘š 6 Power supplied=3( π‘‰π‘š 6 I9cos Ο† )  Comparing the power with 2-wire D.C 3 π‘‰π‘š 6 𝐼9 cos Ο†= 𝑉 π‘šπΌ1 𝐼9 = 6𝐼1 3cos Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 42
  • 43.  Total copper losses= 3𝐼9 2 𝑅9 =3 ( 6𝐼1 3cos Ο† )2 𝑅9= 2𝐼1 2 𝑅9 π‘π‘œπ‘ 2Ο†  Comparing copper losses with 2 wire d.c system 2𝐼1 2 𝑅9 π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅9 𝑅1 = π‘π‘œπ‘ 2Ο† π‘Ž9 π‘Ž1 = 1 π‘π‘œπ‘ 2Ο† volume of copper wire3βˆ’Phase, 3βˆ’wire system volume of copper wire Twoβˆ’wire D.C system = π‘Ž9𝑙9 π‘Ž1𝑙1 = 1 π‘π‘œπ‘ 2Ο†π‘Ž13𝑙 π‘Ž12𝑙 = 1.5 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 1.5 π‘π‘œπ‘ 2Ο† times that required for 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 43
  • 44. In this case, 4th or neutral wire is taken from the neutral point. The area of X- section of the neutral wire is generally one-half that of the line conductor. If the loads are balanced, then current through the neutral wire is zero. Assuming balanced loads and p.f. of the load as cos Ο†, Let 𝑅10 is the resistance of each line conductor R.M.S. voltage per phase = π‘‰π‘š 6 Power supplied=3( π‘‰π‘š 6 I10cos Ο† )  Comparing the power with 2-wire D.C 3 π‘‰π‘š 6 𝐼10 cos Ο†= 𝑉 π‘šπΌ1 𝐼10 = 6𝐼1 3cos Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 44
  • 45.  Total copper losses= 3𝐼10 2 𝑅10 =3 ( 6𝐼1 3cos Ο† )2 𝑅10= 2𝐼1 2 𝑅10 π‘π‘œπ‘ 2Ο†  Comparing copper losses with 2 wire d.c system 2𝐼1 2 𝑅10 π‘π‘œπ‘ 2Ο† = 2𝐼1 2 𝑅1 𝑅10 𝑅1 = π‘π‘œπ‘ 2Ο† π‘Ž10 π‘Ž1 = 1 π‘π‘œπ‘ 2Ο† volume of copper wire3βˆ’Phase, 3βˆ’wire system volume of copper wire Twoβˆ’wire D.C system = π‘Ž10𝑙10 π‘Ž1𝑙1 = 1 π‘π‘œπ‘ 2Ο†π‘Ž13.5𝑙 π‘Ž12𝑙 = 1.75 π‘π‘œπ‘ 2Ο† Hence, the volume of conductor material required for this system is 1.75 π‘π‘œπ‘ 2Ο†times that required for 2-wire d.c. system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 45
  • 46. System Same maximum voltage to earth Same maximum voltage between conductors 1. D.C. system (i) Two-wire (ii) Two-wire mid-point earthed (iii) 3-wire 1 0.25 0.3125 1 1 1.25 2. Single phase system (i) Two-wire (ii) Two-wire mid-point earthed (iii) 3-wire 2 π‘π‘œπ‘ 2Ο† 0.5 π‘π‘œπ‘ 2Ο† 1.625 π‘π‘œπ‘ 2Ο† 2 π‘π‘œπ‘ 2Ο† 0.5 π‘π‘œπ‘ 2Ο† 2.5 π‘π‘œπ‘ 2Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 46
  • 47. System Same maximum voltage to earth Same maximum voltage between conductors 3. Two-phase system (i) 2–phase 4-wire (ii)2-phase 3-wire 0.5 π‘π‘œπ‘ 2Ο† 1.457 π‘π‘œπ‘ 2Ο† 2 π‘π‘œπ‘ 2Ο† 2.914 π‘π‘œπ‘ 2Ο† 4. Three-phase system (i) 3-phase 3-wire (ii)3-phase 4-wire , 0.5 π‘π‘œπ‘ 2Ο† 0.583 π‘π‘œπ‘ 2Ο† 1.5 π‘π‘œπ‘ 2Ο† 1.75 π‘π‘œπ‘ 2Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 47
  • 48. What is the percentage saving in feeder copper if the line voltage in a 2-wire d.c. system is raised from 200 volts to 400 volts for the same power transmitted over the same distance and having the same power loss ? 6 October 2020 D.Rajesh Asst.Prof. EEE Department 48
  • 49.  (i) Conductors  (ii) Step-up and step-down transformers  (iii) Line insulators  (iv) Support  (v) Protective devices  (vi) Voltage regulating devices 6 October 2020 D.Rajesh Asst.Prof. EEE Department 49
  • 50. The following two fundamental economic principles which closely influence the electrical design of a transmission line (i) Economic choice of transmission voltage (ii) Economic choice of conductor size 6 October 2020 D.Rajesh Asst.Prof. EEE Department 50
  • 51.  The effect of increased voltage level of transmission on 1. Volume of conductor material used for transmission 2. Line Drop 6 October 2020 D.Rajesh Asst.Prof. EEE Department 51
  • 52. Let a 3-phase A.C. system is used for the transmission, The resistance per conductor is given by R= ρ l/A Power transmitted P= 3𝑉𝐼cos Ο† I= 𝑃 3𝑉cos Ο† Total copper losses 𝑃𝑐𝑒= 3𝐼2R =3( 𝑃 3𝑉cos Ο† )2ρ l 𝐴 A= 𝑃2ρ l 𝑃𝑐𝑒𝑉2π‘π‘œπ‘ 2Ο† The volume of copper used is V=3(A*l) Volume of copper= 3𝑃2ρ 𝑙2 𝑃𝑐𝑒𝑉2π‘π‘œπ‘ 2Ο† Volume of copper 𝛼 1 𝑉2π‘π‘œπ‘ 2Ο† 6 October 2020 D.Rajesh Asst.Prof. EEE Department 52 Thus greater is the transmission voltage level, lesser is the volume of the copper required.
  • 53. Line Drop=IR=I ρ l/A Current density(J)= 𝐼 𝐴 Then Line Drop= 𝐼ρ l 𝐼 𝐽 = Jρ l %Line Drop= Jρ l 𝑉 *100 Higher the transmission voltage level, then lesser the %Line Drop 6 October 2020 D.Rajesh Asst.Prof. EEE Department 53
  • 54.  The most economical area of conductor is that for which the total annual cost of transmission line is minimum . This is known as Kelvin’s Law  The total annual cost of transmission line can be divided broadly into two parts viz., ο‚‘ Annual charge on capital outlay ο‚‘ Annual cost of energy wasted in the conductor. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 54
  • 55.  This is on account of interest and depreciation on the capital cost of complete installation of transmission line. In case of overhead system, it will be the annual interest and depreciation on the capital cost of conductors, supports and insulators and the cost of their erection.  Now, for an overhead line, insulator cost is constant, the conductor cost is proportional to the area of X-section and the cost of supports and their erection is partly constant annual charge.  Transmission line can be expressed as : π΄π‘›π‘›π‘’π‘Žπ‘™ π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’ = 𝑃1 + 𝑃2π‘Ž 𝑃1 constant depending on the copper conductor 𝑃2 constant depending on the X-area of the conductor a is the area of X-section of the conductor 6 October 2020 D.Rajesh Asst.Prof. EEE Department 55
  • 56.  This is on account of energy lost mainly in the conductor due to 𝐼2R losses.  Assuming a constant current in the conductor throughout the year,  The energy lost in the conductor is proportional to resistance. where R = ρ l/ π‘Ž  The energy lost in the conductor is inversely proportional to area of X-section. energy lost Ξ± 1 π‘Ž  Thus, the annual cost of energy wasted in an overhead transmission line can be expressed as : Annual cost of energy wasted = 𝑃3 π‘Ž where P3 is a constant. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 56
  • 57.  Total annual cost, C=𝑃1 + 𝑃2π‘Ž+ 𝑃3 π‘Ž only area of X-section a is variable. Therefore, the total annual cost of transmission line will be minimum if differentiation of C w.r.t. a is zero i.e. ⅆ𝐢 β…†π‘Ž =0 β…† β…†π‘Ž (𝑃1 + 𝑃2π‘Ž+ 𝑃3 π‘Ž )=0 𝑃2π‘Ž βˆ’ 𝑃3 π‘Ž2=0 𝑃2π‘Ž = 𝑃3 π‘Ž2 i.e. Variable part of annual charge = Annual cost of energy wasted Therefore Kelvin’s Law can also be stated The most economical area of conductor is that for which the variable part of annual charge is equal to the cost of energy losses per year. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 57
  • 58.  The straight line shows the relation between the annual charge (i.e., 𝑃1 + 𝑃2π‘Ž) and the area of X- section a of the conductor  The rectangular hyperbola gives the relation between annual cost of energy wasted ( 𝑃3 π‘Ž ) and X-sectional area a.  Curve 3 shows the relation between total annual cost of transmission line and area of X-section a. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 58
  • 59.  (i) It is not easy to estimate the energy loss in the line without actual load curves, which are not available at the time of estimation.  (ii) The assumption that annual cost on account of interest and depreciation on the capital outlay is in the form 𝑃1 + 𝑃2π‘Ž is strictly speaking not true. For instance, in cables neither the cost of cable dielectric and sheath nor the cost of laying vary in this manner.  (iii) This law does not take into account several physical factors like safe current density, mechanical strength, corona loss etc.  (iv) The conductor size determined by this law may not always be practicable one because it may be too small for the safe carrying of necessary current.  (v) Interest and depreciation on the capital outlay cannot be determined accurately. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 59
  • 60. A 2-conductor cable 1 km long is required to supply a constant current of 200 A throughout the year. The cost of cable including installation is Rs. (20 a + 20) per meter where β€˜a’ is the area of X-section of the conductor in π‘π‘š2 . The cost of energy is 5P per kWh and interest and depreciation charges amount to 10%. Calculate the most economical conductor size. Assume resistivity of conductor material to be 1Β·73 ΞΌ Ξ© cm. economical conductor size a =? Energy lost per annum=2𝐼2 𝑅𝑑 Resistance of one conductor=ρ l/ π‘Ž = 1.73βˆ—10βˆ’6βˆ—105 π‘Ž = 0.173 π‘Ž Ξ© Energy lost per annum= 2βˆ— 2002 βˆ— 0.173 π‘Ž βˆ— 8760 = 121238.4 π‘Ž kWh 6 October 2020 D.Rajesh Asst.Prof. EEE Department 60
  • 61.  Annual cost of energy lost = Cost per kWh Γ— Annual energy loss =Rs 0.05* 121238.4 π‘Ž kWh=Rs 6062 π‘Ž The capital cost (variable) of the cable is given to be Rs 20 a per metre. Therefore, for 1 km length of the cable, the capital cost (variable) is Rs. 20 a Γ— 1000 = Rs. 20,000 a. Variable annual charge = Annual interest and depreciation on capital cost (variable) of cable = Rs 0Β·1 Γ— 20,000 a = Rs 2000 a According to Kelvin’s law, for most economical X-section of the conductor, Variable annual charge = Annual cost of energy lost 2000 a = 6062 π‘Ž a= 6062 2000 =1.74 π‘π‘š2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 61
  • 62. A 2-wire feeder carries a constant current of 250 A throughout the year. The portion of capital cost which is proportional to area of X-section is Rs 5 per kg of copper conductor. The interest and depreciation total 10% per annum and the cost of energy is 5P per kWh. Find the most economical area of X-section of the conductor. Given that the density of copper is 8Β·93gm/ π‘π‘š3 and its specific resistance is 1Β·73 Γ— 10βˆ’8 Ξ© m. Consider 1 metre length of the feeder. Let a be the most economical area of X-section of each conductor in π‘š2 Energy lost per annum=2𝐼2 𝑅𝑑 Resistance of one conductor=ρ l/ π‘Ž = 1.73βˆ—10βˆ’8βˆ—1 π‘Ž = 1.73βˆ—10βˆ’8 π‘Ž Ξ© Energy lost per annum= 2βˆ— 2502 βˆ— 1.73βˆ—10βˆ’8 π‘Ž βˆ— 8760 = 18,94,350 π‘Ž βˆ— 10βˆ’8 kWh 6 October 2020 D.Rajesh Asst.Prof. EEE Department 62
  • 63.  Annual cost of energy lost = Cost per kWh Γ— Annual energy loss =Rs 0.05* 18,94,350 π‘Ž βˆ— 10βˆ’8 kWh=Rs 94,717.5 π‘Ž βˆ— 10βˆ’8 Mass of 1 metre feeder = 2 (Volume Γ— density) = 2 Γ— a Γ— 1 Γ— 8Β·93 Γ— 103kg = 17Β·86 Γ— 103 a kg Capital cost (variable) = Rs 5 Γ— 17Β·86 Γ— 103 a = Rs 89Β·3 Γ— 103 a Variable annual charge = 10% of capital cost (variable) = 0Β·1 Γ— 89Β·3 Γ— 103 a = Rs 8930 a According to Kelvin’s law, for most economical X-section of the conductor, Variable annual charge = Annual cost of energy lost 8930 a = 94,717.5 π‘Ž βˆ— 10βˆ’8 a= 94,717.5 8930 βˆ— 10βˆ’8 =3.25βˆ— 10βˆ’4 π‘š2 = 3.25 π‘π‘š2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 63
  • 64. Determine the most economical cross-section for a 3-phase transmission line, 1 km long to supply at a constant voltage of 110 kV for the following daily load cycle : 6 hours 20 MW at p.f. 0Β·8 lagging 12 hours 5 MW at p.f. 0Β·8 lagging 6 hours 6 MW at p.f. 0Β·8 lagging The line is used for 365 days yearly. The cost per km of line including erection is Rs (9000 +6000 a) where β€˜a’ is the area of X- section of conductor in cm2. The annual rate of interest and depreciation is 10% and the energy costs 6P per kWh. The resistance per km of each conductor is0Β·176/a. Power transmitted P= 3𝑉𝐼cos Ο† Energy lost per annum=3R(𝐼2 𝑑) 6 October 2020 D.Rajesh Asst.Prof. EEE Department 64
  • 65.  The load currents at various loads are :  At 20 MW, 𝐼1= 𝑃 3𝑉cos Ο† = 20βˆ—106 3βˆ—110βˆ—103βˆ—0.8 = 131.2𝐴  At 5 MW, 𝐼1= 𝑃 3𝑉cos Ο† = 5βˆ—106 3βˆ—110βˆ—103βˆ—0.8 = 32.8𝐴  At 6 MW, 𝐼1= 𝑃 3𝑉cos Ο† = 6βˆ—106 3βˆ—110βˆ—103βˆ—0.8 = 39.36𝐴  Energy loss per day in 3-phase line =3* 0.176 π‘Ž (131.22 βˆ— 6+ 32.82 βˆ— 12+ 36.362 βˆ— 6) = 66.26 π‘Ž kWh Energy lost per annum = 66.26 π‘Ž *365= = 24,184.9 π‘Ž kWh Annual cost of energy=Rs 6 100 βˆ— 24,184.9 π‘Ž = = 1451.09 π‘Ž Variable annual charge = 10% of capital cost (variable) of line = Rs 0Β·1 Γ— 6000 a = Rs 600 a According to Kelvin’s law, for most economical X-section of the conductor, Variable annual charge = Annual cost of energy 600 a= 1451.09 π‘Ž a= 1451.09 600 =1.56 π‘π‘š2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 65
  • 66. The conveyance of electric power from a power station to consumers’ premises is known as electric supply system.  Electric supply system consist of three principle components: 1) Power station 2) Transmission lines 3) Distribution system  The electric supply system can be broadly classified into: 1) DC or AC system 2)Overhead or underground system 6 October 2020 D.Rajesh Asst.Prof. EEE Department 66
  • 67. Different blocks of typical acpowersupply scheme are asper following:: 1)Generating Station: Ingenerating station power is producedby three phase alternators operating inparallel. The usual generation voltage is 11kV. Foreconomy in the transmission of electric power, the generation voltage is stepped upto 132 kVor more at generating station with the help of three phase transformer. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 67
  • 68. 2)Primarytransmission: Theelectric power at 132 kVis transmitted by 3-phase,3 wire overhead system to the out 3)Secondary transmission: Atthe receiving station the voltage is reduced to33kV by step down transformer. 4)Primary distribution: Atthe substation voltage is reduced 33kV to 11kV.The11kV line run along important roadsides to city. This forms the primary distribution.
  • 69. 5)Secondarydistribution: Theelectric power form primary distribution line is delivered to distribution substation. The substation is located nearthe consumers location and step down the voltage to 400V, 3- phase ,4-wire for secondary distribution. The voltage between any two phases is 400V and between any phase and neutral is 230V.
  • 70. Basis for Comparison AC Transmission Line DC Transmission Line Definition The ac transmission line transmit the alternating current. The dc transmission line is used for transmitting the direct current. Number of Conductors Three Two Inductance & surges Have Don’t Have Voltage drop High Low Skin Effect Occurs Absent Need of Insulation More Less Interference Have Don’t Have Corona Loss Occur Don’t occur Dielectric Loss Have Don’t have Synchronizing and Stability Problem No difficulties Difficulties Cost Expensive Cheap Length of conductors Small More Repairing and Maintenance Easy and Inexpensive Difficult and Expensive Transformer Requires Not Requires 6 October 2020 D.Rajesh Asst.Prof. EEE Department 70
  • 71. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 71
  • 72. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 72
  • 73. Which distributes electric power for local use is known as distribution system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 73
  • 74.  (i) Feeders. A feeder is a conductor which connects the sub-station to the area where power is to be distributed.  (ii) Distributor. A distributor is a conductor from which tappings are taken for supply to the consumers.  (iii) Service mains. A service mains is generally a small cable which connects the distributor to the consumers’ terminals. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 74
  • 75. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 75
  • 76. Primary distribution system 6 October 2020 D.Rajesh Asst.Prof. EEE Department 76 Secondary distribution system
  • 77. 2-wire d.c. system 6 October 2020 D.Rajesh Asst.Prof. EEE Department 77 3 -wire d.c. system
  • 78. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 78
  • 79. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 79
  • 80.  (i) Radial System. In this system, separate feeders radiate from a single substation and feed the distributors at one end only. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 80
  • 81.  Limitations  (a) The end of the distributor nearest to the feeding point will be heavily loaded.  (b) The consumers are dependent on a single feeder and single distributor. Therefore, any fault on the feeder or distributor cuts off supply to the consumers who are on the side of the fault away from the substation.  (c) The consumers at the distant end of the distributor would be subjected to serious voltage fluctuations when the load on the distributor changes. Due to these limitations, this system is used for short distances only. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 81
  • 82.  (ii) Ring main system. In this system, the primaries of distribution transformers form a loop. The loop circuit starts from the substation bus-bars, makes a loop through the area to be served, and returns to the substation. βœ“ substation supplies to the closed feeder LMNOPQRS. βœ“ The distributors are tapped from different points M, O and Q of the feeder through distribution transformers. ➒ There are less voltage fluctuations at consumer’s terminals. ➒ The system is very reliable as each distributor is fed via two feeders. ➒ In the event of fault on any section of the feeder, the continuity of supply is maintained. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 82
  • 83.  (iii) Interconnected system. When the feeder ring is energized by two or more than two generating stations or substations, it is called inter-connected system. (a) It increases the service reliability. (b) Any area fed from one generating station during peak load hours can be fed from the other generating station. This reduces reserve power capacity and increase efficiency of the system. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 83
  • 84.  (i) Proper voltage.  One important requirement of a distribution system is that voltage variations at consumer’s terminals should be as low as possible.  The changes in voltage are generally caused due to the variation of load on the system.  Low voltage causes loss of revenue, inefficient lighting and possible burning out of motors.  High voltage causes lamps to burn out permanently and may cause failure of other appliances.  Thus, if the declared voltage is 230 V, then the highest voltage of the consumer should not exceed 244 V while the lowest voltage of the consumer should not be less than 216 V. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 84
  • 85.  (ii) Availability of power on demand. Power must be available to the consumers in any amount that they may require from time to time.  (iii) Reliability. Modern industry is almost dependent on electric power for its operation.  The reliability can be improved to a considerable extent by (a) interconnected system (b) reliable automatic control system (c) providing additional reserve facilities. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 85
  • 86. (i) Distributor fed at one end (ii) Distributor fed at both ends (iii) Distributor fed at the centre (iv) Ring distributor. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 86
  • 87.  (i) Distributor fed at one end 6 October 2020 D.Rajesh Asst.Prof. EEE Department 87  (ii) Distributor fed at both ends
  • 88.  (iii) Distributor fed at the centre 6 October 2020 D.Rajesh Asst.Prof. EEE Department 88  (iv) Ring distributor
  • 89. (i) Concentrated loading (ii) Uniform loading (iii) Both concentrated and uniform loading. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 89
  • 90.  2-wire d.c. distributor AB fed at one end A and having concentrated loads 𝐼1, 𝐼2, 𝐼3 and 𝐼4 tapped off at points C, D, E and F respectively.  Let π‘Ÿ1 , π‘Ÿ2, π‘Ÿ3 π‘Žπ‘›π‘‘ π‘Ÿ4 be the resistances of both wires (go and return) of the sections AC, CD, DE and EF of the distributor respectively.  Current fed from point A = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4 Current in section AC = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4  Current in section CD = 𝐼2 + 𝐼3 + 𝐼4 Current in section DE = 𝐼3 + 𝐼4  Current in section EF = 𝐼4  Voltage drop in section AC = π‘Ÿ1 (𝐼1 + 𝐼2 + 𝐼3 + 𝐼4)  Voltage drop in section CD = π‘Ÿ2 (𝐼2 + 𝐼3 + 𝐼4) Voltage drop in section DE = π‘Ÿ3 (𝐼3 + 𝐼4)  Voltage drop in section EF = π‘Ÿ4 𝐼4  ∴ Total voltage drop in the distributor =π‘Ÿ1 (𝐼1 + 𝐼2 + 𝐼3 + 𝐼4)+π‘Ÿ2 (𝐼2 + 𝐼3 + 𝐼4)+π‘Ÿ3 (𝐼3 + 𝐼4)+ π‘Ÿ4 𝐼4 6 October 2020 D.Rajesh Asst.Prof. EEE Department 90
  • 91.  2-wire d.c. distributor AB fed at one end A and loaded uniformly with i amperes per metre length.  The load tapped is i amperes. Let 𝑙 metres be the length of the distributor and r ohm be the resistance per metre run. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 91
  • 92.  Consider a point C on the distributor at a distance x metres from the feeding point A  Then current at point C is = i π‘™βˆ’ i x amperes = i (π‘™βˆ’ x) amperes Now, consider a small length dx near point C. Its resistance is r dx and the voltage drop over length dx is dv = i (𝑙 βˆ’ x) r dx = i r (𝑙 βˆ’ x) dx Total voltage drop in the distributor upto point C is v =‫׬‬0 π‘₯ ir (l βˆ’ x)dx =ir 𝑙π‘₯ βˆ’ π‘₯2 2 The voltage drop upto point B (i.e. over the whole distributor) can be obtained by putting x = 𝑙 in the above expression. ∴ Voltage drop over the distributor AB = ir 𝑙 βˆ— 𝑙 βˆ’ 𝑙2 2 = 1 2 π‘–π‘Ÿπ‘™2 = 1 2 𝑖𝑙 π‘Ÿπ‘™ = 1 2 𝐼𝑅 Thus, in a uniformly loaded distributor fed at one end, the total voltage drop is equal to that produced by the whole of the load assumed to be concentrated at the middle point. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 92
  • 93.  The two ends of the distributor may be supplied with (i) equal voltages (ii) unequal voltages. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 93
  • 94.  Consider a distributor AB fed at both ends with equal voltages V volts and having concentrated loads 𝐼1, 𝐼2, 𝐼3, 𝐼4 and 𝐼5 at points C, D, E, F and G respectively  As we move away from one of the feeding points, say A, p.d. goes on decreasing till it reaches the minimum value at some load point, say E, and then again starts rising and becomes V volts as we reach the other feeding point B.  All the currents tapped off between points A and E (minimum p.d. point) will be supplied from the feeding point A while those tapped off between B and E will be supplied from the feeding point B.  The current tapped off at point E itself will be partly supplied from A and partly from B. If these currents are x and y respectively, then, x+y= 𝐼3  Therefore, we arrive at a very important conclusion that at the point of minimum potential, current comes from both ends of the distributor.  Voltage drop from feeding point A to the point of minimum potential difference is equal to the voltage drop from feeding point B to the point of minimum potential difference 6 October 2020 D.Rajesh Asst.Prof. EEE Department 94
  • 95.  Distributor AB fed with unequal voltages ; end A being fed at 𝑉1volts and end B at 𝑉2 volts.  In order to determine the point of minimum potential, assume of 𝐼𝐴 amperes is supplied from feeding point A  Let r be the resistance of each conductor per unit length(l=1) for both go & return  Resistance of portions AC,CD, DE,EF,FB are π‘Ÿ1 , π‘Ÿ2, π‘Ÿ3 , π‘Ÿ4, π‘Žπ‘›π‘‘ π‘Ÿ5 respectively  Voltage drop between A and B = Voltage drop over AB 𝑉1 βˆ’ 𝑉2 = πΌπ΄π‘Ÿ1+(𝐼𝐴- 𝐼1) π‘Ÿ2 + (𝐼𝐴- 𝐼1- 𝐼2) π‘Ÿ3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) π‘Ÿ4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 βˆ’ 𝐼4) π‘Ÿ5 The load point at which currents are coming from both sides of the distributor will be the point of minimum potential 6 October 2020 D.Rajesh Asst.Prof. EEE Department 95
  • 96.  A 2-wire d.c. street mains AB, 600 m long is fed from both ends at 220 V. Loads of 20 A, 40 A, 50 A and 30 A are tapped at distances of 100m, 250m, 400m and 500 m from the end A respectively. If the area of X-section of distributor conductor is 1π‘π‘š2 , find the minimum consumer voltage. Take ρ = 1Β·7 Γ— 10βˆ’6 Ξ© cm.  Voltage drop between A and B = Voltage drop over AB R= ρ l/A  Resistance of section AC of distributor=2Γ— 1Β·7 Γ— 10βˆ’6βˆ—100βˆ—102 1 = 3Β·4 Γ— 10βˆ’2 Ξ©  Resistance of section CD of distributor =0.051 Ξ©  Resistance of section DE of distributor =0.051 Ξ©  Resistance of section EF of distributor =0.034 Ξ©  Resistance of section FB of distributor =0.034 Ξ©  Voltage drop between A and B = Voltage drop over AB  𝑉𝐴 βˆ’ 𝑉𝐡 = πΌπ΄π‘Ÿ1+(𝐼𝐴- 𝐼1) π‘Ÿ2 + (𝐼𝐴- 𝐼1- 𝐼2) π‘Ÿ3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) π‘Ÿ4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 βˆ’ 𝐼4) π‘Ÿ5  220-200=0.034 𝐼𝐴+(𝐼𝐴-20)0.051+ (𝐼𝐴-60)0.051+(𝐼𝐴-110)0.034+(𝐼𝐴-140)0.034 𝐼𝐴=61.7A 6 October 2020 D.Rajesh Asst.Prof. EEE Department 96
  • 97.  Actual distribution of currents in the various sections of the distributor is  Current in section EF, 𝐼𝐸𝐹 = πΌπ΄βˆ’110 =61Β·7βˆ’110 =βˆ’48Β·3A from E to F = 48Β·3 A from F to E  Current in section DE, 𝐼𝐷𝐸 = 𝐼𝐴 βˆ’60=61.7-60=1.7A from D to E 6 October 2020 D.Rajesh Asst.Prof. EEE Department 97 It is clear that currents are coming to load point E from both sides i.e. from point D and point F. Hence, E is the point of minimum potential. ∴ Minimum consumer voltage, 𝑉𝐸 = 𝑉𝐴-(𝐼𝐴𝐢 π‘Ÿ1+ 𝐼𝐢𝐷 π‘Ÿ2+ 𝐼𝐷𝐸 π‘Ÿ3) =220-(61Β·7 Γ— 0Β·034 + 41Β·7 Γ— 0Β·051 + 1Β·7 Γ— 0Β·051] =215.69V
  • 98. A two-wire d.c. distributor AB, 600 metres long is loaded as under : Distance from A (metres) : 150 300 350 450 Loads in Amperes : 100 200 250 300 The feeding point A is maintained at 440 V and that of B at 430 V. If each conductor has a resistance of 0Β·01 Ξ© per 100 metres, calculate : (i) the currents supplied from A to B, (ii) the power dissipated in the distributor. Let 𝐼𝐴amperes be the current supplied from the feeding point A. Then currents in the various sections of the distributor are Resistance of 100 m length of distributor (both wires) = 2 Γ— 0Β·01 = 0Β·02 Ξ© Resistance of section AC, π‘Ÿ1 = 0Β·02 Γ— 150/100 = 0Β·03 Ξ© Resistance of section CD, π‘Ÿ2 = 0Β·02 Γ— 150/100 = 0Β·03 Ξ© Resistance of section DE, π‘Ÿ3= 0Β·02 Γ— 50/100 = 0Β·01 Ξ© Resistance of section EF, π‘Ÿ4= 0Β·02 Γ— 100/100 = 0Β·02 Ξ© Resistance of section FB, π‘Ÿ5= 0Β·02 Γ— 150/100 = 0Β·03 Ξ© 6 October 2020 D.Rajesh Asst.Prof. EEE Department 98
  • 99.  Voltage drop between A and B = Voltage drop over AB  𝑉𝐴 βˆ’ 𝑉𝐡 = πΌπ΄π‘Ÿ1+(𝐼𝐴- 𝐼1) π‘Ÿ2 + (𝐼𝐴- 𝐼1- 𝐼2) π‘Ÿ3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) π‘Ÿ4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 βˆ’ 𝐼4) π‘Ÿ5  440-430=0.03 𝐼𝐴+(𝐼𝐴-100)0.03+ (𝐼𝐴-300)0.01+(𝐼𝐴-550)0.02+(𝐼𝐴-850)0.03 𝐼𝐴=437.5A  The actual distribution of currents in the various sections of the distributor are  Current supplied from end A, 𝐼𝐴 = 𝐼𝐴𝐢 = 437Β·5 A  Current supplied from end B, 𝐼𝐡 = 𝐼𝐹𝐡 = 𝐼𝐴 -850= - 412Β·5 A  𝐼𝐢𝐷 = 337.5𝐴 𝐼𝐷𝐸 = 137.5𝐴 𝐼𝐸𝐹 = βˆ’112.5𝐴  Therefore E is the point of minimum potential.  Power loss in the distributor =𝐼𝐴𝐢 2 π‘Ÿ1+ 𝐼𝐢𝐷 2 π‘Ÿ2 + 𝐼𝐷𝐸 2 π‘Ÿ3+ 𝐼𝐸𝐹 2 π‘Ÿ4+ 𝐼𝐹𝐡 2 π‘Ÿ5 =(437.5)2 βˆ— 0.03 + 337.5 2 βˆ— 0.03 + 137.5 2 βˆ— 0.01 + 112.5 2 βˆ— 0.02 + (412.5)2 βˆ— 0.03 =14.705kW 6 October 2020 D.Rajesh Asst.Prof. EEE Department 99
  • 100.  (i) Distributor fed at both ends with equal voltages.  Consider a distributor AB of length 𝑙 metres, having resistance r ohms per metre run and with uniform loading of 𝑖 amperes per metre run.  Let the distributor be fed at the feeding points A and B at equal voltages, say V volts.  The total current supplied to the distributor is 𝑖𝑙.  As the two end voltages are equal, therefore, current supplied from each feeding point is 𝑖𝑙 2 i.e. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 100
  • 101.  Consider a point C at a distance x metres from the feeding point A. Then current at point C is = 𝑖𝑙 2 - 𝑖π‘₯= 𝑖( 𝑙 2 -π‘₯) Now, consider a small length dx near point C. Its resistance is r dx and the voltage drop over length dx is dv = i ( 𝑙 2 βˆ’ x) r dx = i r ( 𝑙 2 βˆ’ x) dx Total voltage drop in the distributor upto point C is v =‫׬‬ 0 π‘₯ ir ( 𝑙 2 βˆ’ x)dx =ir 𝑙π‘₯ 2 βˆ’ π‘₯2 2 = π‘–π‘Ÿ 2 (𝑙π‘₯- π‘₯2) Obviously, the point of minimum potential will be the mid-point. Therefore, maximum voltage drop will occur at mid-point i.e. where x = 𝑙 2 Max. voltage drop= π‘–π‘Ÿ 2 (𝑙π‘₯- π‘₯2)= π‘–π‘Ÿ 2 (𝑙 𝑙 2 - ( 𝑙 2 )2)= 1 8 π‘–π‘Ÿπ‘™2= 1 8 𝑖𝑙 π‘Ÿπ‘™ = 1 8 IR Minimum voltage=V- 1 8 IR volts 6 October 2020 D.Rajesh Asst.Prof. EEE Department 101
  • 102.  (ii) Distributor fed at both ends with unequal voltages.  Consider a distributor AB of length l metres having resistance r ohms per metre run and with a uniform loading of i amperes permetre run. Let the distributor be fed from feeding points A and B at voltages 𝑉𝐴 and 𝑉𝐡 respectively.  Suppose that the point of minimum potential C is situated at a distance x metres from the feeding point A. Then current supplied by the feeding point A will be i x.  ∴ Voltage drop in section AC= 1 2 π‘–π‘Ÿπ‘₯2 volts  As the distance of C from feeding point B is (l βˆ’ x), therefore, current fed from B is i (l βˆ’ x).  ∴ Voltage drop in section BC = 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 volts 6 October 2020 D.Rajesh Asst.Prof. EEE Department 102
  • 103.  Voltage at point C, 𝑉𝐢 = 𝑉𝐴 βˆ’ Drop over AC = 𝑉𝐴 - 1 2 π‘–π‘Ÿπ‘₯2 Voltage at point C, 𝑉𝐢 = 𝑉𝐡 βˆ’ Drop over BC = 𝑉𝐡 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 From above equations we get, 𝑉𝐴 - 1 2 π‘–π‘Ÿπ‘₯2 = 𝑉𝐡 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 𝑉𝐴- 𝑉𝐡 = 1 2 π‘–π‘Ÿπ‘₯2 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 = 1 2 π‘–π‘Ÿπ‘₯2 - 1 2 π‘–π‘Ÿπ‘™2 - 1 2 π‘–π‘Ÿπ‘₯2 + 1 2 π‘–π‘Ÿ2𝑙π‘₯ = π‘–π‘Ÿπ‘™ {π‘₯- 1 2 𝑙 } π‘₯= π‘‰π΄βˆ’ 𝑉𝐡 π‘–π‘Ÿπ‘™ + 𝑙 2 The point on the distributor where minimum potential occurs can be calculated 6 October 2020 D.Rajesh Asst.Prof. EEE Department 103
  • 104.  Expression for the power loss in a uniformly loaded distributor fed at both ends  current supplied from each feeding point is 𝑖𝑙 2  Consider a small length dx of the distributor at point P which is at a distance x from the feeding end A.  Resistance of length dx = r dx  Current in length dx = 𝑖𝑙 2 - 𝑖π‘₯= 𝑖( 𝑙 2 -π‘₯)  Power loss in length dx = (current in dx)2 Γ— Resistance of dx = 𝑖( 𝑙 2 βˆ’π‘₯) 2 Γ—rdx  Total power loss in the distributor is=‫׬‬ 0 𝑙 𝑖( 𝑙 2 βˆ’π‘₯) 2 Γ—rdx =𝑖2 π‘Ÿ ‫׬‬ 0 𝑙 𝑙2 4 βˆ’ 𝑙π‘₯ + π‘₯2 𝑑π‘₯ = 𝑖2π‘Ÿ 𝑙2π‘₯ 4 βˆ’ 𝑙π‘₯2 2 + π‘₯3 3 𝑙 0 P= 𝑖2π‘Ÿπ‘™3 12 6 October 2020 D.Rajesh Asst.Prof. EEE Department 104
  • 105. A 2-wire d.c. distributor AB 500 metres long is fed from both ends and is loaded uniformly at the rate of 1Β·0 A/metre. At feeding point A, the voltage is maintained at 255 V and at B at 250 V. If the resistance of each conductor is 0Β·1 Ξ© per kilometre, determine : (i) the minimum voltage and the point where it occurs (ii) the currents supplied from feeding points A and B Let the minimum potential occur at a point C distant x metres from the feeding point A. π‘₯= π‘‰π΄βˆ’ 𝑉𝐡 π‘–π‘Ÿπ‘™ + 𝑙 2 Resistance of distributor/m, r = 2 Γ— 0Β·1/1000 = 0Β·0002 Ξ© π‘₯= 255βˆ’ 250 1βˆ—0.0002βˆ—500 + 500 2 =300m i.e. minimum potential occurs at 300 m from point A. Minimum voltage, 𝑉𝐢 = 𝑉𝐴 - 1 2 π‘–π‘Ÿπ‘₯2 = 𝑉𝐡 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 255- 1βˆ—0.0002βˆ— (300)2 2 =246V Current supplied from A = i x = 1 Γ— 300 = 300 A Current supplied from B = i (𝑙 βˆ’ x) = 1 (500 βˆ’ 300) = 200 A 6 October 2020 D.Rajesh Asst.Prof. EEE Department 105
  • 106. A 800 metres 2-wire d.c. distributor AB fed from both ends is uniformly loaded at the rate of 1Β·25 A/metre run. Calculate the voltage at the feeding points A and B if the minimum potential of 220 V occurs at point C at a distance of 450 metres from the end A. Resistance of each conductor is 0Β·05 Ξ©/km.  Minimum voltage, 𝑉𝐢 = 𝑉𝐴 - 1 2 π‘–π‘Ÿπ‘₯2 = 𝑉𝐡 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2  For 𝑉𝐴 𝑉𝐢 = 𝑉𝐴 - 1 2 π‘–π‘Ÿπ‘₯2 𝑉𝐴=220+ 1.25βˆ—0.0001βˆ— (450)2 2 =232.65V  For 𝑉𝐡 𝑉𝐢 = 𝑉𝐡 - 1 2 π‘–π‘Ÿ(𝑙 βˆ’ π‘₯)2 𝑉𝐡 =220+ 1.25βˆ—0.0001βˆ— (800βˆ’450)2 2 =227.65V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 106
  • 107.  (i) A uniformly loaded distributor is fed at the centre. Show that maximum voltage drop = I R/8 where I is the total current fed to the distributor and R is the total resistance of the distributor.  (ii) A 2-wire d.c. distributor 1000 metres long is fed at the centre and is loaded uniformly at the rate of 1Β·25 A/metre. If the resistance of each conductor is 0Β·05 Ξ©/km, find the maximum voltage drop in the distributor.  ∴ Max. voltage drop = Voltage drop in half distributor = 1 2 ( 𝑖𝑙 2 )( π‘Ÿπ‘™ 2 ) = 1 8 𝑖𝑙 π‘Ÿπ‘™ = 1 8 IR where i l = I, the total current fed to the distributor r l = R, the total resistance of the distributor  (ii) Total current fed to the distributor is  I = i l = 1Β·25 Γ— 1000 = 1250 A  Total resistance of the distributor is  R = r l = 2 Γ— 0Β·05 Γ— 1 = 0Β·1 Ξ©  Max. voltage drop = 1 8 IR = 1 8 *1250*0.1 =15.62V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 107
  • 108.  A two-wire d.c. distributor cable 1000 metres long is loaded with 0Β·5 A/metre. Resistance of each conductor is 0Β·05 Ξ©/km. Calculate the maximum voltage drop if the distributor is fed from both ends with equal voltages of 220 V. What is the minimum voltage and where it occurs ?  Current loading, i = 0Β·5 A/m  Resistance of distributor/m, r = 2 Γ— 0Β·05/1000 = 0Β·1 Γ— 10βˆ’3 Ξ©  Length of distributor, l = 1000 m  Total current supplied by distributor, I = i l = 0Β·5 Γ— 1000 = 500 A  Total resistance of the distributor, R = r l = 0Β·1 Γ— 10βˆ’3 Γ— 1000 = 0Β·1 Ξ©  ∴ Max. voltage drop = 1 8 IR = 6Β·25 V  Minimum voltage will occur at the mid-point of the distributor and its value is  = 220 βˆ’ 6Β·25 = 213Β·75 V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 108
  • 109. The total drop over any section of the distributor is equal to the sum of drops due to concentrated and uniform loading in that section. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 109
  • 110. Two conductors of a d.c. distributor cable AB 1000 m long have a total resistance of 0Β·1 Ξ©. The ends A and B are fed at 240 V. The cable is uniformly loaded at 0Β·5 A per metre length and has concentrated loads of 120 A, 60 A, 100 A and 40 A at points distant 200 m, 400 m, 700 m and 900 m respectively from the end A. Calculate (i) the point of minimum potential (ii) currents supplied from ends A and B (iii) the value of minimum potential. Distributor resistance per metre length, r = 0Β·1/1000 = 10βˆ’4 Ξ© (i) Point of minimum potential. The point of minimum potential is not affected by the uniform loading of the distributor. Therefore, let us consider the concentrated loads. Suppose the current supplied by end A is 𝐼𝐴. Then currents in the various sections will be 6 October 2020 D.Rajesh Asst.Prof. EEE Department 110
  • 111.  Voltage drop between A and B = Voltage drop over AB  𝑉𝐴 βˆ’ 𝑉𝐡 = πΌπ΄π‘Ÿ1+(𝐼𝐴- 𝐼1) π‘Ÿ2 + (𝐼𝐴- 𝐼1- 𝐼2) π‘Ÿ3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) π‘Ÿ4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 βˆ’ 𝐼4) π‘Ÿ5  0= 10βˆ’4( ) 200 𝐼𝐴+(πΌπ΄βˆ’120)200+ (πΌπ΄βˆ’180)300+(πΌπ΄βˆ’280)200+(πΌπ΄βˆ’320)100 𝐼𝐴=166A The actual distribution of currents in the various sections of the distributor due to concentrated loading. It is clear from this figure that D is the point of minimum potential. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 111
  • 112. (ii) The feeding point A will supply 166 A due to concentrated loading plus 0Β·5 Γ— 400 = 200 A due to uniform loading. ∴ Current supplied byA, 𝐼𝐴 = 166 + 200 = 366 A The feeding point B will supply a current of 154 A due to concentrated loading plus 0Β·5 Γ— 600 = 300 A due to uniform loading. ∴ Current supplied by B, 𝐼𝐡 = 154 + 300 = 454 A (iii) ∴ Minimum potential, 𝑉𝐷 = 𝑉𝐴 βˆ’ Drop in AD due to conc. loading – Drop in AD due to uniform loading 𝑉𝐷 Now, Drop in AD due to conc. loading = 𝐼𝐴𝐢 𝑅𝐴𝐢 + 𝐼𝐢𝐷 𝑅𝐢𝐷 = 166 Γ— 10βˆ’4 Γ— 200 + 46 Γ— 10βˆ’4 Γ— 200 = 3Β·32 + 0Β·92 = 4Β·24 V Drop in AD due to uniform loading = 1 2 π‘–π‘Ÿπ‘™2 = 1 2 0.5 βˆ— 10βˆ’4 βˆ— (400)2 =4V ∴ 𝑉𝐷 = 240 βˆ’ 4Β·24 βˆ’ 4 = 231Β·76 V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 112
  • 113. A d.c. 2-wire distributor AB is 500m long and is fed at both ends at 240 V. The distributor is loaded as shown in Fig. The resistance of the distributor (go and return) is 0Β·001Ξ© per metre. Calculate (i) the point of minimum voltage and (ii) the value of this voltage. Let D be the point of minimum potential and let x be the current flowing in section CD. Then current supplied by end B will be (60 βˆ’ x). 6 October 2020 D.Rajesh Asst.Prof. EEE Department 113
  • 114.  (i) If r is the resistance of the distributor (go and return) per metre length, then, Voltage drop in length AD = 𝐼𝐴𝐢 𝑅𝐴𝐢 + 𝐼𝐢𝐷 𝑅𝐢𝐷 =(100+x)*100r+x*150r Voltage drop in length BD= 1 2 π‘–π‘Ÿπ‘™2 +(60-x)*250r = 1 2 1 βˆ— π‘Ÿ βˆ— (200)2 +(60-x)*250r Since the feeding points A and B are at the same potential (100+x)*100r+x*150r= 1 2 1 βˆ— π‘Ÿ βˆ— (200)2 +(60-x)*250r X=50A The actual directions of currents in the various sections of the distributor are currents supplied by A and B meet at D. Hence point D is the point of minimum 6 October 2020 D.Rajesh Asst.Prof. EEE Department 114
  • 115.  (ii) Total current = 160 + 1 Γ— 200 = 360 A  Current supplied by A, 𝐼𝐴 = 100 + x = 100 + 50 = 150 A  Current supplied by B, 𝐼𝐡 = 360 βˆ’ 150 = 210 A  Minimum potential, 𝑉𝐷 = 𝑉𝐴 βˆ’ (𝐼𝐴𝐢 𝑅𝐴𝐢 + 𝐼𝐢𝐷 𝑅𝐢𝐷 ) = 240 βˆ’ 150 Γ— (100 Γ— 0Β·001) βˆ’ 50 Γ— (150 Γ— 0Β·001) = 240 βˆ’ 15 βˆ’ 7Β·5 = 217Β·5 V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 115
  • 116.  A distributor arranged to form a closed loop and fed at one or more points is called a ring distributor.  For the purpose of calculating voltage distribution, the distributor can be considered as consisting of a series of open distributors fed at both ends.  The principal advantage of ring distributor is that by proper choice in the number of feeding points, great economy in copper can be affected. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 116
  • 117.  Sometimes a ring distributor has to serve a large area. In such a case, voltage drops in the various sections of the distributor may become excessive.  In order to reduce voltage drops in various sections, distant points of the distributor are joined through a conductor called interconnector. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 117
  • 118.  The ring distributor ABCDEA. The points B and D of the ring distributor are joined through an interconnector BD.  Network can be obtained by applying Thevenin’s theorem.  (i) Consider the interconnector BD to be disconnected and find the potential difference between B and D. This gives Thevenin’s equivalent circuit voltage 𝐸0.  (ii) Next, calculate the resistance viewed from points B and D of the network composed of distribution lines only. This gives Thevenin’s equivalent circuit series resistance 𝑅0.  (iii) If 𝑅𝐡𝐷 is the resistance of the interconnector BD, then Thevenin’s equivalent circuit will be ∴ Current in interconnector BD = 𝐸0 𝑅0+𝑅𝐡𝐷  Therefore, current distribution in each section and the voltage of load points can be calculated. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 118
  • 119. A 2-wire d.c. distributor ABCDEA in the form of a ring main is fed at point A at 220 V and is loaded as under : 10A at B ; 20A at C ; 30A at D and 10 A at E. The resistances of various sections (go and return) are : AB = 0Β·1 Ξ© ; BC = 0Β·05 Ξ© ; CD = 0Β·01 Ξ© ; DE = 0Β·025 Ξ© and EA = 0Β·075 Ξ©. Determine :  (i) the point of minimum potential  (ii) current in each section of distributor Let us suppose that current I flows in section AB of the distributor. Then currents in the various sections of the distributor are According to Kirchhoff’s voltage law, the voltage drop in the closed loop ABCDEA is zero i.e. 𝐼𝐴𝐡𝑅𝐴𝐡+𝐼𝐡𝐢𝑅𝐡𝐢+ 𝐼𝐢𝐷𝑅𝐢𝐷+ 𝐼𝐷𝐸𝑅𝐷𝐸 + 𝐼𝐸𝐴𝑅𝐸𝐴 =0 0.1𝐼+(𝐼-10)0.05+ (𝐼-30)0.01+(𝐼-60)0.025+(𝐼-70)0.075=0 0.26 𝐼 = 7.55 𝐼 = 29.04A 6 October 2020 D.Rajesh Asst.Prof. EEE Department 119
  • 120.  (ii) Current in section AB = I = 29Β·04 A from A to B  Current in section BC = I βˆ’ 10 = 29Β·04 βˆ’ 10 = 19Β·04 A from B to C  Current in section CD = I βˆ’ 30 = 29Β·04 βˆ’ 30 = βˆ’ 0Β·96 A = 0Β·96 A from D to C  Current in section DE = I βˆ’ 60 = 29Β·04 βˆ’ 60 = βˆ’ 30Β·96 A = 30Β·96 A from E to D  Current in section EA = I βˆ’ 70 = 29Β·04 βˆ’ 70 = βˆ’ 40Β·96 A = 40Β·96 A from A to E 6 October 2020 D.Rajesh Asst.Prof. EEE Department 120 ∴ C is the point of minimum potential.
  • 121.  A d.c. ring main ABCDA is fed from point A from a 250 V supply and the resistances (including both lead and return) of various sections are as follows : AB = 0Β·02 Ξ© ; BC = 0Β·018 Ξ© ; CD = 0Β·025 Ξ© and DA = 0Β·02 Ξ©. The main supplies loads of 150 A at B ; 300 A at C and 250 A at D. Determine the voltage at each load point. If the points A and C are linked through an interconnector of resistance 0Β·02 Ξ©, determine the new voltage at each load point.  Without Interconnector.  Let us suppose that a current I flows in section AB of the distributor. Then currents in various sections of the distributor will be  According to Kirchhoff’s voltage law, the voltage drop in the closed loop ABCDA is zero i.e.  𝐼𝐴𝐡𝑅𝐴𝐡+𝐼𝐡𝐢𝑅𝐡𝐢+ 𝐼𝐢𝐷𝑅𝐢𝐷+ 𝐼𝐷𝐴𝑅𝐷𝐴 =0  0.02𝐼+(𝐼-150)0.018+ (𝐼-450)0.025+(𝐼-700)0.02=0 0Β·083 I = 27Β·95 ∴ I = 27Β·95/0Β·083 = 336Β·75 A 6 October 2020 D.Rajesh Asst.Prof. EEE Department 121
  • 122. The actual distribution of currents are ..  Current in section AB = I = 336.75 A from A to B  Current in section BC = I βˆ’ 150 = 336.75 βˆ’ 150 = 186.75 A from B to C  Current in section CD = I βˆ’ 450 = 336.75 βˆ’ 450 = βˆ’ 113.25 A = 113.25 A from D to C  Current in section DA = I βˆ’ 700 = 336.75 βˆ’ 700 = βˆ’ 363.25A = 363.25 A from A to D Voltage drop in AB = 336Β·75 Γ— 0Β·02 = 6Β·735 V Voltage drop in BC = 186Β·75 Γ— 0Β·018 = 3Β·361 V Voltage drop in CD = 113Β·25 Γ— 0Β·025 = 2Β·831 V Voltage drop in DA = 363Β·25 Γ— 0Β·02 = 7Β·265 V ∴ Voltage at point B = 250 βˆ’ 6Β·735 = 243Β·265 V Voltage at point C = 243Β·265 βˆ’ 3Β·361 = 239Β·904 V Voltage at point D = 239Β·904 + 2Β·831 = 242Β·735 V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 122
  • 123.  With Interconnector.  The ring distributor with interconnector AC. The current in the interconnector can be found by applying Thevenin’s theorem. 𝐸0 = Voltage between points A and C = 250 βˆ’ 239Β·904 = 10Β·096 V 𝑅0 = Resistance viewed from points A and C = 0.02+0.018 0.02+0.025 0.02+0.018 + 0.02+0.025 = 0.02Ξ© 𝑅𝐴𝐢 = Resistance of interconnector = 0Β·02 Ξ©  From Thevenin’s equivalent circuit Current in interconnector AC is 𝐸0 𝑅0 + 𝑅𝐴𝐢 = 10.096 0.02 + 0.02 = 252.4𝐴 π‘“π‘Ÿπ‘œπ‘š 𝐴 π‘‘π‘œ 𝐢 Let us suppose that current in section AB is 𝐼1 . Then current in section BC will be 𝐼1 βˆ’ 150. As the voltage drop round the closed mesh ABCA is zero, 0.02𝐼1+(𝐼1-150)0.018-0.02*252.4=0 0.038 𝐼1=7.748 𝐼1=203.15A 6 October 2020 D.Rajesh Asst.Prof. EEE Department 123
  • 124. The actual distribution of currents are ..  Current in section AB = 𝐼1 = 203.15A from A to B  Current in section BC = 𝐼1 βˆ’ 150 = 203.15 βˆ’ 150 = 53.15 A from B to C  Current in section CD = 𝐼1 βˆ’ 450+252.4 = 203.15βˆ’ 450 +252.4= 5.55A C to D  Current in section DA = 𝐼1 βˆ’ 700 = 203.15+252.5 βˆ’ 700 = βˆ’ 244.45A = 244.45A from A to D  Drop in AB = 203Β·15 Γ— 0Β·02 = 4Β·063 V  Drop in BC = 53Β·15 Γ— 0Β·018= 0Β·960 V  Drop in AD = 244Β·45 Γ— 0Β·02 = 4Β·9 V  ∴ Potential of B = 250 βˆ’ 4Β·063= 245Β·93 V  Potential of C = 245Β·93 βˆ’ 0Β·96 = 244Β·97 V  Potential of D = 250 βˆ’ 4Β·9 = 245Β·1 V It may be seen that with the use of interconnector, the voltage drops in the various sections of the distributor are reduced. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 124
  • 125.  A.C. distribution calculations differ from those of d.c. distribution in the following respects :  (i) In case of d.c. system, the voltage drop is due to resistance alone. However, in a.c. system, the voltage drops are due to the combined effects of resistance, inductance and capacitance.  (ii) In a d.c. system, additions and subtractions of currents or voltages are done arithmetically but in case of a.c. system, these operations are done vectorially.  (iii) In an a.c. system, power factor (p.f.) has to be taken into account. Loads tapped off form the distributor are generally at different power factors. There are two ways of referring power factor viz (a) It may be referred to supply or receiving end voltage which is regarded as the reference vector. (b) It may be referred to the voltage at the load point itself. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 125
  • 126.  The power factors of load currents may be given (i) w.r.t. receiving or sending end voltage or (ii) w.r.t. to load voltage itself. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 126
  • 127.  (i) Power factors referred to receiving end voltage. Consider an a.c. distributor AB with concentrated loads of 𝐼1and 𝐼2 tapped off at points C and B. Taking the receiving end voltage 𝑉𝐡 as the reference vector, let lagging power factors at C and B be cos Ο†1 and cos Ο†2 w.r.t. 𝑉𝐡 . Let 𝑅1, 𝑋1 and 𝑅2, 𝑋2 be the resistance and reactance of sections AC and CB of the distributor. Impedance of section AC, 𝑍𝐴𝐢= 𝑅1 + j 𝑋1 Impedance of section CB,𝑍𝐢𝐡 = 𝑅2 + j 𝑋2 Load current at point C, 𝐼1 = 𝐼1 (cos Ο†1 βˆ’ j sin Ο†1) Load current at point B, 𝐼2 = 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) Current in section CB, 𝐼𝐢𝐡= 𝐼2 = 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) Current in section AC, 𝐼𝐴𝐢 = 𝐼1+ 𝐼2 =𝐼1 (cos Ο†1 βˆ’ j sin Ο†1)+𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) 6 October 2020 D.Rajesh Asst.Prof. EEE Department 127
  • 128. Ο†1  Voltage drop in section CB,  𝑉𝐢𝐡= 𝐼𝐢𝐡 𝑍𝐢𝐡= 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) (𝑅2 + j 𝑋2 )  Voltage drop in section AC,  𝑉𝐴𝐢= 𝐼𝐴𝐢 𝑍𝐴𝐢=(𝐼1+ 𝐼2) 𝑍𝐴𝐢  =𝐼1 (cos Ο†1 βˆ’ j sin Ο†1)+𝐼2 (cos Ο†2 βˆ’ j sin Ο†2)(𝑅1 + j 𝑋1 )  Sending end voltage, 𝑉𝐴= 𝑉𝐡 + 𝑉𝐢𝐡 + 𝑉𝐴𝐢  Sending end current, 𝐼𝐴= 𝐼1+ 𝐼2 6 October 2020 D.Rajesh Asst.Prof. EEE Department 128 𝑉𝑩 𝐼𝟐 𝐼𝟏 𝐼𝑨π‘ͺ Ο†2 ΞΈ
  • 129. Ο†1  (ii) Power factors referred to respective load voltages.  Suppose the power factors of loads are referred to their respective load voltages. Then Ο†1 is the phase angle between V𝐢 and I1 and Ο†2 is the phase angle between V𝐡 and I2. The vector diagram under these conditions is  Voltage drop in section CB,  𝑉𝐢𝐡= 𝐼2 𝑍𝐢𝐡= 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) (𝑅2 + j 𝑋2 )  Voltage at point C = 𝑉𝐡 + Drop in section CB = V𝐢 ∠ Ξ± (say)  Now 𝐼1 = 𝐼1 ∠ βˆ’ Ο†1 w.r.t. voltage V𝐢  ∴ . 𝐼1 = 𝐼1 ∠ βˆ’ (Ο†1 βˆ’ Ξ±) w.r.t. voltage V𝐡  i.e. 𝐼1 = 𝐼1 [cos (Ο†1 βˆ’ Ξ±) βˆ’ j sin (Ο†1 βˆ’ Ξ±)]  Now 𝐼𝐴𝐢 = 𝐼1+ 𝐼2 =𝐼1 [cos (Ο†1 βˆ’ Ξ±) βˆ’ j sin (Ο†1 βˆ’ Ξ±)]+ 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2)  Voltage drop in section AC, 𝑉𝐴𝐢= 𝐼𝐴𝐢 𝑍𝐴𝐢  ∴ Voltage at point A = V𝐡 + Drop in CB + Drop in AC 6 October 2020 D.Rajesh Asst.Prof. EEE Department 129 𝑉𝑩 𝐼𝟐 𝐼𝟏 𝐼𝑨π‘ͺ Ο†2 Ξ±
  • 130. A single phase a.c. distributor AB 300 metres long is fed from end A and is loaded as under : (i) 100 A at 0Β·707 p.f. lagging 200 m from point A (ii) 200 A at 0Β·8 p.f. lagging 300 m from point A The load resistance and reactance of the distributor is 0Β·2 Ξ© and 0Β·1 Ξ© per kilometre. Calculate the total voltage drop in the distributor. The load power factors refer to the voltage at the far end. Impedance of distributor/km = (0Β·2 + j 0Β·1) Ξ© Impedance of section AC, 𝑍𝐴𝐢 = (0Β·2 + j 0Β·1) Γ— 200/1000 = (0Β·04 + j 0Β·02) Ξ© Impedance of section CB, 𝑍𝐢𝐡 = (0Β·2 + j 0Β·1) Γ— 100/1000 = (0Β·02 + j 0Β·01) Ξ© Taking voltage at the far end B as the reference vector, we have, Load current at point B, 𝐼2 = 𝐼2 (cos Ο†2 βˆ’ j sin Ο†2) = 200 (0Β·8 βˆ’ j 0Β·6) = (160 βˆ’ j 120) A Load current at point C, 𝐼1 = 𝐼1 (cos Ο†1 βˆ’ j sin Ο†1) = 100 (0Β·707 βˆ’ j 0Β·707) = (70Β·7βˆ’ j 70Β·7) A Current in section CB,𝐼𝐢𝐡= 𝐼2 = (160 βˆ’ j 120) A Current in section AC,𝐼𝐴𝐢= 𝐼1+ 𝐼2 = (70Β·7 βˆ’ j 70Β·7) + (160 βˆ’ j 120)= (230Β·7 βˆ’ j 190Β·7) A Voltage drop in section CB, 𝑉𝐢𝐡= 𝐼𝐢𝐡 𝑍𝐢𝐡 = (160 βˆ’ j 120) (0Β·02 + j 0Β·01) = (4Β·4 βˆ’ j 0Β·8) volts Voltage drop in section AC, 𝑉𝐴𝐢= 𝐼𝐴𝐢 𝑍𝐴𝐢 = (230Β·7 βˆ’ j 190Β·7) (0Β·04 + j 0Β·02) = (13Β·04 βˆ’ j 3Β·01) volts Voltage drop in the distributor = 𝑉𝐴𝐢 + 𝑉𝐢𝐡 = (13Β·04 βˆ’ j 3Β·01) + (4Β·4 βˆ’ j 0Β·8) = (17Β·44 βˆ’ j 3Β·81) volts Magnitude of drop = 17.44 2 + 3.81 2 = 17Β·85 V 6 October 2020 D.Rajesh Asst.Prof. EEE Department 130
  • 131.  A single phase ring distributor ABC is fed at A. The loads at B and C are 20 A at 0.8 p.f. lagging and 15 A at 0.6 p.f. lagging respectively ; both expressed with reference to the voltage at A. The total impedance of the three sections AB, BC and CA are (1 + j 1), (1+ j2) and (1 + j3) ohms respectively. Find the total current fed at A and the current in each section. Use Thevenin’s theorem to obtain the results. 6 October 2020 D.Rajesh Asst.Prof. EEE Department 131