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Transformer tests • The performance of a transformer can be calculated on the basis of its equivalent circuit which contains the 4 main parameters: 1. Equivalent resistance R01(or R02) 2. Equivalent leakage reactance X01 (or X02) 3. Core loss resistance R0 4. Magnetizing reactance X0. • These parameters are determined from the following tests: a) Open circuit test b) Short circuit test Open circuit test I0 W A V1 V V2=E2 Low Voltage winding High Voltage winding Open Circuit Test • As the primary no-load current I0 is small, Copper loss is negligibly small in primary & nil in secondary. Hence, wattmeter reading represents the core loss under no load condition. • Core loss = W = input power on no-load = V1I0cosΦ0 => cosΦ0 = W/(V1I0) Hence, Iw = I0 cos Φ0 & Iμ=I0sin Φ0 Also, X0 = V1/ Iμ & R0 = V1/ Iw Short Circuit test • This test is conducted to determine: 1. Full-load copper loss 2. Equivalent resistance & reactance referred to metering side. W A LV supply V High Voltage winding Low Voltage winding • A low voltage (5 -10% of normal primary voltage) at correct frequency is applied to the primary winding & is continuously increased till full load currents flow both in primary & secondary. Since applied voltage is small, flux linking with core is very small & hence, iron loss can be neglected & reading of wattmeter gives total copper losses at full load. • If Vsc is the voltage required to circulate rated load currents, then Vsc Z 01 ;W I sc R01 X 01 Z 01 R01 2 2 2 I sc Regulation of a transformer • Voltage regulation is defined as: “the change in secondary voltage when rated load at a specified power is removed”. • Let, E2 = sec. terminal voltage at no load V2 = sec. terminal voltage at full load E 2 V2 Then, % regulation = 100 V2 Approximate Equivalent circuit with primary impedances transferred to secondary I1/k R02 = R2 +k2R1 X02= X2+ k2X1 I2 I0/k kV1=E2 k2R0 k2X0 ZL V2 I2 Regulation expressed in terms of secondary values E2=kV1 I2 Z02 θ2 I2 X02 cosθ2 δ I2 R02 cosθ2 0 I2 R02 sinθ2 θ2 V2 θ2 I2 X02 I2 I2 R02 I2 X02 sinθ2 E 2 V2 I 2 R 02 cosθ 2 I 2 X 02 sinθ 2 I 2 X 02 cosθ 2 I 2 R 02 sinθ 2 2 2 2 V2 I 2 R 02 cos2 θ 2 I 2 X 02 sin 2 θ 2 2V2 I 2 R 02 cosθ 2 2 2 2 2 2 2I 2 R 02 X 02 sinθ 2 cosθ 2 2V2 I 2 X 02 sinθ 2 I 2 X 02 cos2 θ 2 I 2 R 02 sin 2 θ 2 2 2 2 2 2 2I 2 R 02 X 02 sinθ 2 cosθ 2 2 V2 I 2 R 02 I 2 X 02 2V2 I 2 R 02 cosθ 2 2V2 I 2 X 02 sinθ 2 2 2 2 2 2 E 2 V2 I 2 R 02 I 2 X 02 2V2 I 2 R 02 cosθ 2 2V2 I 2 X 02 sinθ 2 2 2 2 2 2 Efficiency of a transformer • Losses in a transformer – 1. Core or Iron loss = Hystersis loss + Eddy Current loss ( from OC test) 2. Copper loss = I 2 R1 I 2 R2 I 2 R 01 I 2 R 02 1 2 1 2 (from SC test) Output Output Efficiency Input Output Losses Output Output Iron loss Cu loss Condition for maximum efficiency • Cu loss = I12 R01 Iron loss = Wh We Wi considering primary side Primary input = V1 I1 cos 1 V1 I1 cos1 losses V1 I1 cos1 V1 I1 cos1 I1 R01 Wi 2 V1 I1 cos1 I1 R01 Wi 1 V1 cos 1 V1 I1 cos 1 Differentiating both sides w.r.t I1, we get dη R 01 Wi 0 dI1 V1cosφ1 V1 I 21cosφ1 dη For η to be maximum, dI1 0 R 01 Wi 0 0 V1cosφ1 V1 I 2 1 cosφ1 R 01 Wi V1cosφ1 V1 I 2 1 cosφ1 I 2 1 R 01 Wi Cu loss Iron loss The current corresponding to maximum efficiency is: I 1 W i R01 I1 Wi I2 k R02 Efficiency of Transformer at any load x Full Load kVA p.f η 100% x Full Load kVA p.f Iron Loss x 2 Full Load Copper Loss x Ratio of Actual to Full Load kVA Ex: A 2300/208 V, 500kVA, 60Hz transformer has the following data: Test V A W Side used O.C 208 85 1800 L.V S.C 95 217.4 8200 H.V Calculate efficiency of the transformer at 0.8 pf lagging & 1/4th rated load. Solution x Full Load kVA p.f η 100% x Full Load kVA p.f Iron Loss x Full Load Copper Loss 2 1 500 0.8 4 1 1 ( 500 0.8) ( ) 2 8.2 1.8 4 4 97.74%

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posted: | 4/15/2010 |

language: | English |

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