Dula \u0394 Y Connection Current ratio of each phase of the transformer a I \u00d8S I\u00d8P

Dula δ y connection current ratio of each phase of

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Flaviano D. Dula Δ -Y Connection Current ratio of each phase of the transformer a = I ØS /I ØP Line - to – current ratio of the 3Ø transformer a T = I S /I P = I ØS /(√3I ØP ) = I S /(I P /√3 ) = (√3I S )/I P
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Three-Phase Transformer
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Flaviano D. Dula Problem 70 Three 10:1 transformers are connected Δ-Y for stepping up the 2300- volts 3-Ø source. Calculate the secondary line voltage. Ans. V S = 39.84kV
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Flaviano D. Dula Solution V P = 2300V V ØP V P /a √3V P a Primaries Secondaries ∆ - Y step-up transformer = V S
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Flaviano D. Dula Solution V S √3 V P a √3(2300) 1/10 = = 39.84 kV Answer Or a V ØP V ØS = V P V S / √3 Ratio of transformation of each transformer: a V S √3 V P a then = = =
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Flaviano D. Dula Problem 71 A 2000-kW, 2400-V, 75% power factor load is to be supplied from a 34,500 -volt, 3Ø line through a single bank transformers. Give the secondary/primary line currents in amperes for the Y-Y connections. REE April 2006 A.50/200 B. 48/650 C. 60/800 D. 642/45
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Flaviano D. Dula Solution I P P load √3 V P cosθ P = 2000kW √3(34.5kV)(0.75) = 44.626A Answer Primary line current: I P =
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Flaviano D. Dula Solution I S P load √3 V S cosθ = 2000kW √3(2.4kV)(0.75) = 641.5A Answer Secondary line current: I S I S = aI P = (34500/2400) x 44.626A = 641.5A or =
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Flaviano D. Dula Problem 72 A 13.8-kV/480-V, 10-MVA 3-Ø transformers has 5% impedance. What is the impedance in ohms referred to primary? REE March 1998 A. 0.952Ω B. 0.03Ω C. 5.125Ω D. 9.01Ω
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Flaviano D. Dula Solution Per Unit Quantity = Actual Quantity Base Quantity Z p.u. = Z actual Z base Z e (V base ) 2 /S base = = Z e S base (V base ) 2 or Per unit impedance: Z p.u.
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Flaviano D. Dula Solution Z p.u. Z e I rated V rated = Z e S rated (V rated ) 2 %Z Z(MVA) (kV) 2 Percent impedance: %Z Per unit impedance: Z p.u = Multiply each by V rated = X 100%
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Flaviano D. Dula Solution herefore, the impedance in ohms referred to primar Z P %Z(kV L ) 2 (MVA ) x 100% (0.05) ( 13.8 ) 2 10 = 0.952 Ω Note: Z actual = Z p.u . Z base , Z base = (kV) 2 /MVA Actual impedance = =
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Flaviano D. Dula Problem 73 A transformer rated 2000-kVA, 34500/240- volts has 5.75% impedance. What is the per- unit impedance? REE April 2004 A. 0.0635 B. 0.0656 C. 0.0575 D. 34.2
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Flaviano D. Dula Solution Z p.u. = %Z/100 = 5.75% / 100 = 0.0575 Answer Per unit impedance: Z p.u
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Flaviano D. Dula Problem 74 A Y- Δ transformation is used to take power from a 3-phase supply at a line voltage of 13,200-volts and deliver a 3-Ø load at a line voltage of 480-volts. The total load is 2500-kVA. At rated load: determine: (i) ratio of transformation a , (ii) high and low sides line and phase currents, (iii) the transformer coil (or winding) voltages, (iv) kVA rating of each transformer, and (v) bank or overall ratio.
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Flaviano D. Dula Solution V P = 13200V V P 3 I P I ØS = a I ØP a ( √3I P ) = I S V P a√3 Primaries Secondaries I ØP = I P 2500 kVA Load, 480V (3Ø) = V S V ØP =
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Flaviano D. Dula Solution (i) a V ØP V ØS = V P /√3 V S = 13200/√ 3 480 = 15.877 Answer = Ratio of transformation of each transformer or phase transformation ratio: a Y P S
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Flaviano D. Dula Solution (ii) High or primary rated line current: I P I P S Load √3V P 2500 x 1000VA √3 x 13200V = = 109.35A Answer = Y P
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Flaviano D. Dula Solution Low or secondary rated line current: I S I S S Load √3V S 2,500 , 000VA √3 x 480V = = 3007.03A Answer a(√ 3 I P ) = 13200/√3 480 (√3 x 2500 x 1000 √3 x 13200 ) = 3007.03A or = = Y-Δ s
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Flaviano D. Dula Solution = I ØS a I ØP = I P = 109.35A and and high or primary rated phase current: I ØP I ØP S Load 3V ØP 2,500 , 000 3(13200/√3) = = 109.35A Answer Note: Y-connection: I L = I Ø or = Y P
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