Question

a two - conductor street main AB , 500 meters in length is fd from both ends at 250 volts . Load of 50A , 60A , 40A and 30A are tapped at distances of 100m , 250m , 350m and 400m from A , respectively.If the x - section of the conductors be 1 cm2 and specific resistance of the material of the conductors is 1.7x10-8 - m .determine the minimum consumer voltage

Answer 1

Given:- Loads : 50A GOA чоА 30 A distances : 100m, 250m, 350 4oom Length of the Line = 500 mo X-section of conductor I cm² Specific resistance s' = 1-7 Xioam solution: Below Figure shows the distributor with it tapped currents. Let 'IÁ amperes be the current supplied from the feeding end Å. Then currents to the various sections of the distributor are shown on Figure IA c (IA-50) D (FA-110) E (IA-150) F (IA-180)8 loom 150 100 100 250V 250 V 50A GOA HOA 30 A Resistance of 1m length of distributor 2 x 1.7 X10-8 2*0 s R - = 0.034 241.7 Xionem x 100 = 3.48104 Resistance of section AC, RAC = (3.4x104)x100 Resistance of section CD, Red (3.4x104) *150 = 0.051 v Resistance of section DG, RDG= 3,4% 1014 x 100 20.0344 Resistance of section EF REF = 3.4x164 x 50 = 0,017.1

Resistance of section FB, RFB = 3.48104 x100= 0.0342 Voltage at B = Voltage at A - Doop over length AB ie VB = VA- [IA. RAC +CIA-50) RCD + (FA-110) RDE +(TA-150 reft (JA-180) RFB] >> 250 - 250 [0.034x IA+ 0.051 (IA-50) +0.034 CIA-110) + 0.017 [IA-150] +0.034[TA-180]] 250-[ 0.17 TA -[2.55 +3.74 +2.55 +6.12 = 250-[0.17 IA 14.96) 35 O.ITIA = 14.96 IA 14.96 = 88 A 0.17 IA E 88 A

current In Section CD, ICD 2 IA - 50 = 88.50 = 38 A from C to D. Pleast Carrent 4 section DE, IDEE IA-110 = 88-llo = - 22 A from D to E. } It is clear that currents Ć from both sides. are coming to load point . Minimum Consumer voltage, VA-[FACE RAC + Fed. RCD] VE С D II E 5 22A F. < 92 A 88A 38A 62A 250V 250V 1 50A GOA чо А 30 A

VES 250 - [887 0.0345+ 38 x 0.051 51 스 250 C 2.992 +1.938 A = 250 4.93 VE = 245.07 Volts