AC Distribution

AC Distribution 

The AC distribution system is the electrical system between the step-down substation fed by the transmission system and the consumers' meters. AC Distribution is two types , Here discuses  two type of AC distribution system.

• Methods of Solving AC Distribution Problems

A single phase ac distributor AB is 300 m long and is fed as 100 A at 0.707 pf lag 200 m from A, 200 A at 0.8 pf lag 300 m from A The resistance and reactance of the distributor is 0.2 Ω and 0.1 Ω per km. Calculate the total voltage drop in the distributor.

Zline = 0.2 + j0.1 W/ km
ZAC = 0.2(0.2 + j0.1) = 0.04 + j0.02 = W 0.0447 Ð27.560W
ZCB = 0.1(0.2 + j0.1) = 0.02 + j0.01 = W 0.0224Ð27.560W
I1 =100- Ð 45o A = 70.7 - j70.7 A
I2 = 200- Ð36.87o A =160 - j120 A
I = I1 + I2 = 230.7 - j190.7 A = 299.3- Ð39.6o A
VCB = I2ZCB = 4.48- Ð 9.31= 4.42 - j0.73V
VAC = IZ AC =13.38- Ð12.04 =13.09 - j2.79V
VAB = VCB +VAC =17.51- j3.52 =17.86- Ð13.36o V

Example 14.6: A 3-phase ring main ABCD fed at A at 11 kV supplies balanced load of 50 A
at 0.8 pf lag at B, 120 A at unity pf at C and 70 A at 0.866 pf lag at D. The
impedances of the various sections are

AB:1+ j0.6W BC:1.2+ j0.9W CD: 0.8+ j0.5W DA:3+ j2W
Calculate the currents in various sections and system bus bar voltages at B, C and D

Let the current in section AB is x + jy
So,             IBC = x + jy - 50(0.8 - j0.6) = (x - 40) + j(y + 30)
                    ICD = (x - 40) + j(y + 30) - (120 + j0) = (x -160) + j(y + 30)
                    IDA = (x -160) + j(y + 30) -70(0.866 - j0.5) = (x - 220.6) + j(y + 65)

VAB = IABZAB = [x + jy][1+ j0.6] = (x - 0.6y) + j(0.6 + jy)
VBC = IBCZBC = [(x - 40) + j(y + 30][1.2 + j0.9] = (1,2x - 0.9y - 75) + j(0.9x +1.2y)
VCD = ICDZCD = [(x -160) + j(y + 30][0.8 + j0.5] = (0.8x - 0.5y -143) + j(0.5x + 0.8y -56)
VDA = IDAZDA =[(x - 220.6) + j(y + 65][3+ j2] = (2x - 2y -791.8) + j(2x +3y - 246.2)

Applying KVL around ABCDA,
                        (x -0.6y) + j(0.6+ jy) + (1,2x -0.9y -75) + j(0.9x +1.2y) +
                        (0.8x -0.5y -143) + j(0.5x + 0.8y -56) +(2x - 2y -791.8) + j(2x +3y - 246.2) = 0

 

 

à	We get , 6x - 4y -1009.8 = 0	and   4x + 6y -302.2 = 0
Solving,	x =139.7	y - =42.8
So,	IAB =139.7 - j42.8 ICD = (x -160) + j(y + 30) - =20.3- j12.8 A IDA = (x - 220.6) + j(y + 65) - =80.6 + j22.2 A	A IBC = (x - 40) + j(y + 30) = 99.7 - j12.8 A

 

Per phase voltages are

A =11000 / 3 = 6351
VB =VA - IABZAB = (6351 + j0) - (139.7 - j42.8)(1+ j0.6) = 6185.6 - j41.0 V
VC =VB - IBCZBC = 6185.6 - j41.0- (99.7 - j12.8)(1.2 + j0.9) = 6054.5 - j115.4 V
VD =VC - ICDZCD = (6054.5 - j115.4) - (-20.3- j12.8)(0.8 + j0.5) = 6064.3- j95 V

Phase Unbalanced Loads

Wire Star-Connected Unbalanced Loads

Example: Non-reactive loads of 10 kW, 8 kW and 5 kW are connected between the
neutral and the red, yellow and blue phases of a 3-phase 4-wire system. If the line
voltage is 400 V, calculate the current in each line and in neutral wire.

Phase voltage = 400 / 3 = 231V
I A I A I A
R =10 ´103 / 231 = 43.3 Y = 8´103 / 231 = 34.6 B = 5´103 / 231 = 21.65
I I I I A
N R Y B
Ð =0 - Ð +120 Ð +120 = 43.3Ð0 +34.6- Ð120 + 21.65Ð120 =18.87 - Ð 36.60

14.5 Ground Detectors
Lamp circuit is used to detect ground fault in underground ac systems. Lamps of same
capacity are connected as shown in the figure. If ground fault occurs at any line, the lamp
connected to that line will be dim while other lamps will be brighter.

Fig:- Ground detector circuit