this lesson of this question design of power systems in the medium voltage systems :-

Material = Cu Cu = 56 [m/ohm.mm²] x = 0,3 (ohm/km) r=1000/q* = 1000/(q * 56) (ohm/km) U = 12 kV

Distances to L are in kilometers

the following system is a medium voltage distribution system ,the system is powered by A and B

Find the I _qa current and I_qb current and find the point where the Maximum voltage drop occurs.

the question will be solved like this and this is a example for the solve for the same puestion but with use another data he used aluminum element (al)

thanks alot

1 2 3 4 5 B A + 3 km 5 km 4 km 3 km 1.5 km 3 km 1000 S (kVA) Cosa 1600 0,95 0,86 1700 0,92 900 0,85 800 0,9 A 1 2 CA 3 4 5 B + + 3 cm 5 km 4 km 3 km 15 km 3 km S (KVA) Coso Sino P=S.Coso (kW Q=S.Sino (kVAr) S I= (A) 3. U 1.-I.Coso (A) 1.I.Sing (A) 1000 0,86 0,51 860 510,294 1600 0,95 0,31 1520 499,599 1700 0,92 0,39 1564 666,261 900 0,85 0,52 765 474,104 800 0,9 0,43 720 348,711 52,486 83,978 89,226 47,237 41,989 45,138 26,783 79,779 26,222 82,088 34,969 40,152 24,884 37,790 18,302 CURRENT CALCULATIONS 1., 211 +1 (L4 +L12) +1 (L41 +212 +223)+1,(LA1 + 112 + L23 +234) +1,(LA1 +212 +L23 +234 +245) Lu+12+ L23 +234 +245 +LSE 1. = 1000 45, 138.3 +79,779(3+5) + 82, 088(3+5+4) + 40, 152(3+5+4 + 3) + 37,790(3 + 5+ 4 +3+1,5) = 153,052 A 3 +5 +4+3 + 1,5 + 3 1 =1, +12 +1, +1 +1 -19 1 = 45, 138 + 79,779 +82,088 +40, 152 +37,790 - 153,052 = 131, 895 A 14. 19, L1 +14,(L41 +212) +14,(L11 +212 +223)+14. (441 +212 + L23 +132) + 14. (1+212 +223 +234 +245) 441 +12 +L23 + L34 +245 +LSE = 1000 26,783.3 +26, 222(3+5) + 34,969(3+5+4) +24,884(3+5 +4+3) + 18,302(3+5+4+3+1,5) 3 + 5 + 4 + 3 + 1,5 + 3 = 71,026 A 4 = 14, +1, +1 +14, +14.-14. U=11 kV (a.) U=11 kV 18 3 km 1 5 km 2 4 km 3 3 km 4 4 1,5 km 5 3 km @ 1801044 1168.730 4 A 1. = 144,957 1.-131895 I-601354 B 92.9474 1-86,7574 I. -33,3514 19,976 A -69774 - 7,129 A Ise 126749 115.2624 - 153052 75. 110A - 27,8394 -52.7234 -71,026 A I, = 52,486 A 12=83,978 A 1 79,7794 1 = 45,138 A 1 = 26,783 A 1 = 89,226 A 1 = 82,088 1 = 34,969 A 1. = 47,237 A 1. = 40,152 A 124,884 A Is = 41,989 A 1. = 37,790 A 1 = 18,302 A 1 = 26,222 A 1 pza + Ipaa = 123 6,977 + 75, 110 = 82,088 82,087 282,088 (proof of point a) 1922 +1943 = 192 7,129 + 27,839 = 34,969 34,968 34,969 (proof of point B) 1 2 3 4 5 B A + 3 km 5 km 4 km 3 km 1.5 km 3 km 1000 S (kVA) Cosa 1600 0,95 0,86 1700 0,92 900 0,85 800 0,9 A 1 2 CA 3 4 5 B + + 3 cm 5 km 4 km 3 km 15 km 3 km S (KVA) Coso Sino P=S.Coso (kW Q=S.Sino (kVAr) S I= (A) 3. U 1.-I.Coso (A) 1.I.Sing (A) 1000 0,86 0,51 860 510,294 1600 0,95 0,31 1520 499,599 1700 0,92 0,39 1564 666,261 900 0,85 0,52 765 474,104 800 0,9 0,43 720 348,711 52,486 83,978 89,226 47,237 41,989 45,138 26,783 79,779 26,222 82,088 34,969 40,152 24,884 37,790 18,302 CURRENT CALCULATIONS 1., 211 +1 (L4 +L12) +1 (L41 +212 +223)+1,(LA1 + 112 + L23 +234) +1,(LA1 +212 +L23 +234 +245) Lu+12+ L23 +234 +245 +LSE 1. = 1000 45, 138.3 +79,779(3+5) + 82, 088(3+5+4) + 40, 152(3+5+4 + 3) + 37,790(3 + 5+ 4 +3+1,5) = 153,052 A 3 +5 +4+3 + 1,5 + 3 1 =1, +12 +1, +1 +1 -19 1 = 45, 138 + 79,779 +82,088 +40, 152 +37,790 - 153,052 = 131, 895 A 14. 19, L1 +14,(L41 +212) +14,(L11 +212 +223)+14. (441 +212 + L23 +132) + 14. (1+212 +223 +234 +245) 441 +12 +L23 + L34 +245 +LSE = 1000 26,783.3 +26, 222(3+5) + 34,969(3+5+4) +24,884(3+5 +4+3) + 18,302(3+5+4+3+1,5) 3 + 5 + 4 + 3 + 1,5 + 3 = 71,026 A 4 = 14, +1, +1 +14, +14.-14. U=11 kV (a.) U=11 kV 18 3 km 1 5 km 2 4 km 3 3 km 4 4 1,5 km 5 3 km @ 1801044 1168.730 4 A 1. = 144,957 1.-131895 I-601354 B 92.9474 1-86,7574 I. -33,3514 19,976 A -69774 - 7,129 A Ise 126749 115.2624 - 153052 75. 110A - 27,8394 -52.7234 -71,026 A I, = 52,486 A 12=83,978 A 1 79,7794 1 = 45,138 A 1 = 26,783 A 1 = 89,226 A 1 = 82,088 1 = 34,969 A 1. = 47,237 A 1. = 40,152 A 124,884 A Is = 41,989 A 1. = 37,790 A 1 = 18,302 A 1 = 26,222 A 1 pza + Ipaa = 123 6,977 + 75, 110 = 82,088 82,087 282,088 (proof of point a) 1922 +1943 = 192 7,129 + 27,839 = 34,969 34,968 34,969 (proof of point B)