Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN: 9781305632134
Author: J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher: Cengage Learning
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- A single-phase step-down transformer is rated 13MVA,66kV/11.5kV. With the 11.5 kV winding short-circuited, rated current flows when the voltage applied to the primary is 5.5 kV. The power input is read as 100 kW. Determine Req1andXeq1 in ohms referred to the high-voltage winding.arrow_forwardConsider the three single-phase two-winding transformers shown in Figure 3.37. The high-voltage windings are connected in Y. (a) For the low-voltage side, connect the windings in , place the polarity marks, and label the terminals a, b, and c in accordance with the American standard. (b) Relabel the terminals a, b, and c such that VAN is 90 out of phase with Va for positive sequence.arrow_forwardThe ratings of a three-phase three-winding transformer are Primary(1): Y connected 66kV,15MVA Secondary (2): Y connected, 13.2kV,10MVA Tertiary (3): A connected, 2.3kV,5MVA Neglecting winding resistances and exciting current, the per-unit leakage reactances are X12=0.08 on a 15-MVA,66-kV base X13=0.10 on a 15-MVA,66-kV base X23=0.09 on a 10-MVA,13.2-kV base (a) Determine the per-unit reactances X1,X2,X3 of the equivalent circuit on a 15-MVA,66-kV base at the primary terminals. (b) Purely resistive loads of 7.5 MW at 13.2 kV and 5 MW at 2.3kV are connected to the secondary and tertiary sides of the transformer, respectively. Draw the per- unit impedance diagram, showing the per-unit impedances on a 15-MVA,66-kV base at the primary terminals.arrow_forward
- An ideal transformer has no real or reactive power loss. (a) True (b) Falsearrow_forward(a) An ideal single-phase two-winding transformer with turns ratio at=N1/N2 is connected with a series impedance Z2 across winding 2. If one wants to replace Z2, with a series impedance Z1 across winding 1 and keep the terminal behavior of the two circuits to be identical, find Z1 in terms of Z2. (b) Would the above result be true if instead of a series impedance there is a shunt impedance? (c) Can one refer a ladder network on the secondary (2) side to the primary (1) side simply by multiplying every impedance byat2 ?arrow_forwardFor an ideal 2-winding transformer, an impedance Z2 connected across winding 2 (secondary) is referred to winding 1 (primary) by multiplying Z2 by (a) The turns ratio (N1/N2) (b) The square of the turns ratio (N1/N2)2 (c) The cubed turns ratio (N1/N2)3arrow_forward
- The rated voltages of a single-phase transformer are 2000 V and 200 V in its primary and secondary respectively. The shunt impedance can be modelled with a resistance of 14000 ohms in parallel with a reactance of 1100 ohm, both refer to primary side. The primary and secondary series impedances are (0.12+j 0.35) ohms and (0.0012+j 0.0035) ohms respectively. A load is connected to the secondary side of transformer, draws 810 KVA with a lagging power factor of 0.85. The load voltage is same as rated voltage, that is 200 V. ▪ Use per unit method for analysis and find the voltage on primary side in pu.arrow_forwardSingle phase non-ideal transformer. Turn Ratio= 3000/600. R₁ = 0.9Q R₂= 36 mQ Ro= 3000 Q X₁₁ = 0.6Q X12= 0.0024 Q X0= 6000 Q Input voltage-1440 V (also known as rms voltage) Load impedance of secondary winding is z₂= 0.12 + j0.3 Q a) Secondary winding current phasor 12 (NOT to be confused for I'₂) b) Load voltage phasor V2 (NOT to be confused for V'2) c) Primary winding current phasor ī₁ → X11 X12 R₁ R'2 www V₂ V₁ ī₁ R m ×°arrow_forwardA three-phase transformer bank is connected wye–delta. The primary voltage is 12,470 V, and the secondary voltage is 480 V. The total capacity of the transformer bank is 450 kVA. One of the three transformers that form the three-phase bank develops a shorted primary winding and becomes unusable. A suggestion is made to reconnect the bank for operation as an open-delta. Can the two remaining transformers be connected open-delta? Explain your answer as to why they can or why they cannot be connected as an open-delta. If they can be reconnected open-delta, what would be the output capacity of the two remaining transformers?arrow_forward
- Single phase non-ideal transformer. Turn Ratio= 3000/600. R₁ = 0.9Q R2= 36 mQ Ro= 3000 Q X₁₁ = 0.60 X12= 0.0024 02 X= 6000 Input voltage= 1440 V (also known as rms voltage) Load impedance of secondary winding is Z₂= 0.12 + j0.3 a) Copper losses on primary and secondary windings b) Core loss c) Transformer efficiency V₁ Ī₁ Ro ww m Xu R₁ R'2 ww m ww I₂ X12 marrow_forward(e) MaximumfAux- 13. A 1000 VA transformer has core loss of 15W and copper loss at half load of 5W. Calculate the full load efficiency at 0.9 pf lagging.arrow_forwardThe full-load efficiency of transformer at 0.85 lagging p.f. is 97%. Its efficiency at full load 0.85 leading power factor will be *arrow_forward
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