Question 2.1: Consider a balanced three-phase delta load connected across ......
Consider a balanced three-phase delta load connected across an unbalanced three-phase supply system, as shown in Figure 2.7. The currents in lines a and b are given.
The currents in the delta-connected load and also the symmetrical components of line and delta currents are required to be calculated. From these calculations, the phase shifts of positive and negative sequence components in delta windings and line currents can be established. The line current in c is given by
I_{c} = -(I_{a} + I_{b})
= -30 + j6.0 A
The currents in delta windings are
I_{AB} =\frac{ 1}{3} (I_{a} – I_{b}) = -3.33 + j4.67 = 5.735 < 144.51° \ A
I_{BC} = \frac{ 1}{3}(I_{b} – I_{c}) = 16.67 – j5.33 – 17.50 < -17.7° \ A
I_{CA} = \frac{ 1}{3} (I_{c} – I_{a}) = -13.33 + j0.67 = 13.34 < 177.12° \ A
Calculate the sequence component of the currents I_{AB}. This calculation gives
I_{AB1} = 9.43 < 89.57° \ A
I_{AB2} = 7.181 < 241.76° \ A
I_{AB0} = 0 \ A
Calculate the sequence component of current I_{a}. This calculation gives
I_{a1} = 16.33 < 59.57° \ A
I_{a2} = 12.437 < 271.76° \ A
I_{a0} = 0 \ A
This shows that the positive sequence current in the delta winding is 1/\sqrt{3} times the line positive sequence current, and the phase displacement is +30°, that is,
I_{AB1} = 9.43 < 89.57° = \frac{I_{a1}}{\sqrt{3}} < 30° = \frac{16.33}{\sqrt{3}} < (59.57° + 30°) \ A
The negative sequence current in the delta winding is 1/\sqrt{3} times the line negative sequence current, and the phase displacement is -30°, that is,
I_{AB2} = 7.181 < 241.76° = \frac{I_{a2}}{\sqrt{3}} < -30° = \frac{12.437}{\sqrt{3}} < (271.76°- 30°) \ A
This example illustrates that the negative sequence currents and voltages undergo a phase shift that is the reverse of the positive sequence currents and voltages.