(a) The efficiency of a 320 KVA, single phase transformer is 98.5 % when delivering full load at 0.75 power factor and 99.46 % at 35 % of load at unity power factor. Calculate (i) the iron loss (ii) the full load copper loss. (b) With neat diagram explain current transformer (any 4 points) and also write an application of this transformer.. PC = Pi=

(a) The efficiency of a 320 KVA, single phase transformer is 98.5 % when delivering full load at 0.75 power factor and 99.46 % at 35 % of load at unity power factor. Calculate (i) the iron loss (ii) the full load copper loss. (b) With neat diagram explain current transformer (any 4 points) and also write an application of this transformer.. PC = Pi=

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.15P: Rework Problem 3.14 if the transformer is delivering rated load at rated secondary voltage and at...

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Similar questions

An ideal transformer has no real or reactive power loss. (a) True (b) False
In developing per-unit circuits of systems such as the one shown in Figure 3.10. when moving across a transformer, the voltage base is changed in proportion to the transformer voltage ratings. (a) True (b) False
Rework Problem 3.14 if the transformer is delivering rated load at rated secondary voltage and at (a) unity power factor, (b) 0.8 power factor leading. Compare the results with those of Problem 3.14. -
The symbol shown is a(n) a. iron core transformer. b. auto transformer. c. current transformer. d. air core transformer.
For an ideal 2-winding transformer, the ampere-turns of the primary winding, N1I1 is equal to the ampere-turns of the secondary winding, N2I2 (a) True (b) False
A 23/230-kV step-up transformer feeds a three-phase transmission line, which in turn supplies a 150-MVA,0.8 lagging power factor load through a step-down 230/23-kV transformer. The impedance of the line and transformers at 230kVis18+j60. Determine the tap setting for each transformer to maintain the voltage at the load at 23 kV.
The ideal transformer windings are eliminated from the per-unit equivalent circuit of a transformer. (a) True (b) False
Consider 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.
A single-phase, 50-kVA,2400/240-V,60-Hz distribution transformer has the following parameters: Resistance of the 2400-V winding: R1=0.75 Resistance of the 240-V winding: R2=0.0075 Leakage reactance of the 2400-V winding: X1=1.0 Leakage reactance of the 240-V winding: X2=0.01 Exciting admittance on the 240-V side =0.003j0.02S (a) Draw the equivalent circuit referred to the high-voltage side of the transformer. (b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.
Three single-phase, two-winding transformers, each rated 450MVA,20kV/288.7kV, with leakage reactance Xeq=0.10perunit, are connected to form a three-phase bank. The high-voltage windings are connected in Y with a solidly grounded neutral. Draw the per-unit equivalent circuit if the low-voltage windings are connected (a) in with American standard phase shift or (b) in Y with an open neutral. Use the transformer ratings as base quantities. Winding resistances and exciting current are neglected.
For an ideal transformer, the efficiency is (a) 0 (b) 100 (c) 50
A single-phase l0-kVA,2300/230-volt,60-Hz two-winding distribution transformer is connected as an autotransformer to step up the voltage from 2300 to 2530 volts (a) Draw a schematic diagram of this arrangement, showing all voltages and currents when delivering full load at rated voltage. (b) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the rated values as a two-winding transformer. How much of this kVA rating is transformed by magnetic induction? (C) The following data are obtained from tests carried out on the transformer when it is connected as a two-winding transformer: Open-circuit test with the low-voltage terminals excited: Applied voltage =230V, input current =0.45A, input power =70W. Short-circuit test with the high-voltage terminals excited: Applied voltage =120, input current =4.5A, input power =240W. Based on the data, compute the efficiency of the autotransformer corresponding to full load, rated voltage, and 0.8 power factor lagging. Comment on why the efficiency is higher as an autotransformer than as a two-winding transformer.