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EE 255 - Chapter 2b

EE 255 - Chapter 2b - (c T he magnetization current is a...

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34 (c) The magnetization current is a higher percentage of the full-load current for the 50 Hz case than for the 60 Hz case. This is true because the peak flux is higher for the 50 Hz waveform, driving the core further into saturation. 2-6. A 15-kVA 8000/230-V distribution transformer has an impedance referred to the primary of 80 + j 300 Ω . The components of the excitation branch referred to the primary side are Ω = k 350 C R and Ω = k 70 M X . (a) If the primary voltage is 7967 V and the load impedance is Z L = 3.2 + j 1.5 Ω , what is the secondary voltage of the transformer? What is the voltage regulation of the transformer? (b) If the load is disconnected and a capacitor of – j 3.5 Ω is connected in its place, what is the secondary voltage of the transformer? What is its voltage regulation under these conditions? S OLUTION (a) The easiest way to solve this problem is to refer all components to the primary side of the transformer. The turns ratio is a = 8000/230 = 34.78. Thus the load impedance referred to the primary side is ( ) ( ) 2 34.78 3.2 1.5 3871 1815 L Z j j = + Ω = + Ω The referred secondary current is ( ) ( ) 7967 0 V 7967 0 V 1.78 28.2 A 80 300 3871 1815 4481 28.2 S j j ∠ ° ∠ ° = = = ∠ - ° + Ω + + Ω ° Ω I and the referred secondary voltage is ( )( ) 1.78 28.2 A 3871 1815 7610 3.1 V S S L Z j = = ∠ - ° + Ω = ∠ - ° V I The actual secondary voltage is thus 7610 3.1 V 218.8 3.1 V 34.78 S S a ∠ - ° = = = ∠ - ° V V The voltage regulation is 7967-7610 VR 100% 4.7% 7610 = × = (b) The easiest way to solve this problem is to refer all components to the primary side of the transformer. The turns ratio is again a = 34.78. Thus the load impedance referred to the primary side is ( ) ( ) 2 34.78 3.5 4234 L Z j j = - Ω = - Ω The referred secondary current is ( ) ( ) 7967 0 V 7967 0 V 2.025 88.8 A 80 300 4234 3935 88.8 S j j ∠ ° ∠ ° = = = ° + Ω + - Ω ∠ - ° Ω I and the referred secondary voltage is ( )( ) 2.25 88.8 A 4234 8573 1.2 V S S L Z j = = ° - Ω = ∠ - ° V I The actual secondary voltage is thus
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35 8573 1.2 V 246.5 1.2 V 34.78 S S a ∠ - ° = = = ∠ - ° V V The voltage regulation is 7967 8573 VR 100% 7.07% 8573 - = × = - 2-7. A 5000-kVA 230/13.8-kV single-phase power transformer has a per-unit resistance of 1 percent and a per- unit reactance of 5 percent (data taken from the transformer’s nameplate). The open-circuit test performed on the low-voltage side of the transformer yielded the following data: V OC kV = 138 . A 1 . 15 OC = I kW 9 . 44 OC = P (a) Find the equivalent circuit referred to the low-voltage side of this transformer. (b) If the voltage on the secondary side is 13.8 kV and the power supplied is 4000 kW at 0.8 PF lagging, find the voltage regulation of the transformer. Find its efficiency. S OLUTION (a) The open-circuit test was performed on the low-voltage side of the transformer, so it can be used to directly find the components of the excitation branch relative to the low-voltage side. EX 15.1 A 0.0010942 13.8 kV C M Y G jB = - = = ( )( ) 1 1 OC OC OC 44.9 kW cos cos 77.56 13.8 kV 15.1 A P V I θ - - = = = ° EX 0.0010942 77.56 S 0.0002358 0.0010685 S C M Y G jB j = - = ∠ - ° = - 1 4240 C C R G = = Ω 1 936 M M X B = = Ω The base impedance of this transformer referred to the secondary side is ( ) 2 2 base base base 13.8 kV 38.09 5000 kVA V Z S = = = Ω so ( )( ) EQ 0.01 38.09 0.38 R = Ω = Ω and ( )( ) EQ 0.05 38.09 1.9 X = Ω = Ω . The resulting equivalent circuit is shown below:
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36 Ω = 38 . 0
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