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Unformatted text preview: 34 (c) The magnetization current is a higher percentage of the fullload 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. 26. A 15kVA 8000/230V 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 79677610 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 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 = × =  27. A 5000kVA 230/13.8kV singlephase power transformer has a perunit resistance of 1 percent and a per unit reactance of 5 percent (data taken from the transformer’s nameplate). The opencircuit test performed on the lowvoltage 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 lowvoltage side of this transformer....
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This note was uploaded on 10/15/2009 for the course EE 255 taught by Professor Shaban during the Spring '05 term at Cal Poly.
 Spring '05
 Shaban
 Flux

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