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### Transcript of Soln Assignment 6 - U of S Solutions for EE 343.3 Assignment 6 Instructor: N. Chowdhury Marker:...

• Solutions for EE 343.3 Assignment 6

Instructor: N. Chowdhury Marker: Kiran Raj Timalsena

For the 3-phase full-wave controlled rectifier circuit (Figure 1), ��� = 208√2 sin(120��) � Vc = 100 V, L = 9 mH, R = 71. Ω and delay angle, � = 30 degrees. Assume that the load current is continuous. Sketch the three line currents for one cycle and calculate the following:

a) Average output current. b) RMS output current. c) Average and RMS thyristor currents. d) RMS input line current. e) 5th harmonic RMS input line current. f) Power factor at the ac source. g) The kVA size of the transformer.

��� = 208 � �� = 100 � � = 1.7 Ω � = 9 mH � = 30

57.3 = 0.524 ���

��� = 208√2 sin(120��) � � = 377 ���

�� �� ��

�� �� ��

• (a) Calculation of average output current

�� = �

3 ��� �� =

2�

3 ���

������ = 6

2� � ���(�)�

����

����

� = 243.265 �

������ = ������ − ��

� = 84.274 �

• (b) RMS output current

�(�) = 6

� � ���(�)sin (��)

����

����

��

�(�) = 6

� � ���(�)cos (��)

����

����

��

�(6) = −48.154 � �(6) = 13.901 � �(12) = 23.572 � �(12) = −3.402 � �(18) = −15.654 � �(18) = 1.506 �

�(�) = ��(�)� + �(�)� �(�) = ��� + (���)�

�(�) = �(�)

�(�) ��(�) =

�(�)

√2

�(6) = 50.12 � �(6) = 20.429 Ω �(6) = 2.453 A ��(6) = 1.735 A �(12) = 23.816 � �(12) = 40.751 Ω �(12) = 0.584 A ��(12) = 0.413 � �(18) = 15.726 � �(18) = 61.098 Ω �(18) = 0.257 A ��(18) = 0.182 �

������ = ������� � + ��(6)

� + ��(12) � + ��(18)

� = 84.293 �

(c) Average and RMS thyristor current

������ = ������

3 = 28.091 �

������ = ������

√3 = 48.666 �

(d) RMS line current

��������� = √2 ������ = 68.825 �

• (e) 5th harmonic input RMS line current

Each thyristor conducts for 120 degrees and a line current is made up of two thyristor currents that are 180 degrees apart from each other.

�����(�) = 1

� �� ������ sin (��)

�� �

� �

�� + � −������ sin (��)

��� �

�� �

���

�����(1) = 92.925 �

�����(3) = 0 �

�����(5) = −18.585 �

�����(7) = −13.275 �

�����(�) = 1

� �� ������ cos (��)

�� �

� �

�� + � −������ cos(��)

��� �

�� �

���

�����(1) = 0 �

�����(3) = 0 �

�����(5) = 0 � �����(7) = 0 �

�����(�) = ������(�) � + �����(�)

� �����(5) = 18.585 �

�����(5) = �����(�)

�(�) = 1.09 A

Rms value of 5th harmonics= �.��

√� = 0.771 A

(f) Power factor at the ac source

�� = � ������ � + �� ������ = 20506.43 �

�� = √3 ��� ��������� = 24795.35 ��

�� = �� ��

= 0.827

(g) The KVA size of the transformer

�� = 24.79 ���