Rony Parvej s EEE. Lecture 3 & 4: Transformer. Update: 30 April, fecabook.com/ronyiut

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1 Rony Parvej s EEE Lecture 3 & 4: Transformer Update: 30 April, 2015 fecabook.com/ronyiut

2 1 2 TRANSFORMER What is the voltage at secondary side of a transformer having a turn ratio of 1:10 if 440V dc is applied at primary? (a) 0 V (b) 4400 V (c) 44 V (d) 4.4 V What would be your comment regarding the size, of a 1 kva. 400 Hz transformer and a 1 kva, 50 Hz transformer? (a) 400 Hz transformer is smaller in size (b) 50 Hz transformer is smaller in size (c) both the transformer are of equal size (d) no comment can be made without seeing the transformer (e) none of the above DPDC-14 MCQ BPDB Draw an equivalent circuit of transformer referred to secondary. DWASA-14 4 Draw the vector diagram of a single phase step down transformer. BPDB-11 A 400 V/100 V, 500 VA single phase transformer has an equivalent impedance of 0.5 NWPGCL-14 5 p.u. If the primary winding impedance is 0.1 ohm,find the secondary winding impedance in ohm both referred to secondary. 6 What are the no load and full load losses of transformer? PGCB-14 7 What are the load and no load losses of a transformer? What is all day efficiency of a transformer? BUET M.Sc Maximum efficiency of a transformer is 98%. It runs at 50% efficiency for 14 hours PGCB-14 and at no load for remaining 10 hours. Find it s all day efficiency. 9 A 30 kva, 2400/240 V, 50 Hz single phase transformer, operating at no load, draws BPDB W at a p.f. of 0.21 lagging. Determine i) the exciting current of the transformer ii) the core loss component of the exciting current and iii) the magnetizing component of the exciting current. 10 A single phase 1000/200 V takes a no load current of 3A at 0.2 power factor lagging. EGCB-2014 Calculate the primary current and power factor when secondary current is 280 A at power factor 0.8 lagging. 11 A single phase transformer with a ratio of 100/200 V takes a no-load current of 3A at 0.2 power factor lagging. If the secondary supplies a current of 280 A at a p.f. of 0.8 lagging, estimate the current taken by the primary. BPDB-2015, EGCB The equivalent circuit impedances of a 20kVA, 8000/240V, 60Hz transformer are to be determined. The open circuit test and the short circuit test were performed on the primary side of the transformer, and the following data were taken: Open circuit test (primary) Short circuit test V OC = 8000 V V SC = 489 V I OC = A I SC = 2.5 A P OC = 400 W P SC = 240 W BPDB-14 (FF) 13 Find the impedance of the approximate equivalent circuit referred to the primary side, and sketch the circuit. [Example 2.2, Chapman 4 th edition] In short circuit test of a transformer in which side supply voltage is applied? (a) Low voltage side (b) High voltage side In short circuit test of single phase transformer, the wattmeter reading roughly gives DPDC MCQ- 14 MCQ

3 14 (a) Copper loss (b)core loss (c) Both (d)none PGCB-14 In open circuit test of transformer power supply is given in MCQ 15 (a) low voltage side (b) high voltage side (c) (d) PGCB-14 Practically in open circuit test of a transformer, the wattmeter measures DPDC MCQ- 16 (a) Copper loss (b) Core loss 14 (c) A slight copper loss (d) both (b) & (c) 17 Why transformer is rated in KVA? What are the instrument transformers? PGCL Write down the conditions of parallel operation of transformers. BUET M.Sc. Unknown Show the direction of the current in primary and polarity of the load initially: PGCB Which of the following is not an example of transformer connection? (a) Open delta - Y (b) (c) (d) MCQ- DWASA Dy1 means- (a) Primary is delta and secondary is star connected and secondary leads primary by 30º (b) Primary is delta and secondary is star connected and primary leads secondary by 30º What do you understand by DY1 (a) High voltage side is leading low voltage side by 30º (b) Low voltage side is leading high voltage side by 30º MCQ EGCB-14 MCQ EGCB In a 3-phase transformer DY1 means- MCQ EGCB-12 UNIT transformer at power generation is used as an- MCQ 24 (a) energy meter (b)step up transformer (c) auxiliary transformer PGCB-14 Why do we ground Y-neutral in D-Y transformer? (i) for flowing zero sequence current MCQ BPDB- 25 (ii) 14 (FF) 26 What are the disadvantages of 3-phase Y-Y connected transformer? How this problem can be solved? 27 A 500KVA, 13200/480 V distribution transformer has four taps 2.5% on its primary winding. What are the voltage ratios of this transformer at each tapings? BUET M.Sc.12 DWASA-14

What is the voltage at secondary side of a transformer having a turn ratio of 1:10 if DPDC-14 1 440V dc is applied at primary? (a) 0 V (b) 4400 V (c) 44 V (d) 4.

4 What is the voltage at secondary side of a transformer having a turn ratio of 1:10 if DPDC V dc is applied at primary? (a) 0 V (b) 4400 V (c) 44 V (d) 4.4 V A transformer cannot raise or lower the voltage of a D.C. supply because Faraday's laws of electromagnetic induction are not valid since the rate of change of flux is zero. As transformers require an alternating magnetic flux to operate correctly, transformers cannot therefore be used to transform or supply DC voltages or currents, since the magnetic field must be changing to induce a voltage in the secondary winding. In other words, Transformers DO NOT Operate on DC Voltages. So, the output is zero. Reference: Basic Electrical Engineering U.A. Bakshi What would be your comment regarding the size, of a 1 kva. 400 Hz transformer and a 1 kva, 50 Hz transformer? (a) 400 Hz transformer is smaller in size 2 (b) 50 Hz transformer is smaller in size (c) both the transformer are of equal size (d) no comment can be made without seeing the transformer (e) none of the above MCQ BPDB- 13 Due to some mechanical constraints, generally transformer size is inversely proportional to the operating frequency. Thus 50 hz, transformer will have max. size. The impedance of the transformer is greater for higher frequencies (as X=wL). So for constant voltage to keep same current rating (because for constant kva) we have to reduce the inductance of the winding. And to reduce the inductance we have to reduce the no. of turns in the windings of the transformer. By reducing the no. of turns the size of the transformer will be reduced for higher frequencies. This is a short cut method based on analysis. generally power transformers use lower frequency (50hz) and we know in electronic ckts like oscillators producing very high frqs in range of KHz we use only SMALL trfrs than pwr trfrs... so on analysing we can conclude tat higher frquency transformers have less volume... NOTE: this is only shortcut method, actual reason depends on the FORMULA only for determining the size

5 3 Draw an equivalent circuit of transformer referred to secondary. DWASA-14 Figure: Equivalent transformer circuit referring to the primary Figure: Equivalent transformer circuit referring to the secondary Reference: Electric Machinary Fundamentals - Chapman 4 Draw the vector diagram of a single phase step down transformer. BPDB-11 Figure: vector diagram of a single phase step down (V P >V S ) transformer at lagging power factor.

6 Figure: vector diagram of a single phase step down (V P >V S ) transformer at leading power factor. Figure: vector diagram of a single phase step down (V P >V S ) transformer at unity power factor. : Figure: vector diagram of a single phase step down (V P >V S ) transformer at no load.

7 A 400 V/100 V, 500 VA single phase transformer has an equivalent impedance of p.u. If the primary winding impedance is 0.1 ohm, find the secondary winding impedance in ohm both referred to secondary. p.u এর ম য থ কর র সম য় এট করব NWPGCL-14 6 What are the no load and full load losses of transformer? PGCB-14 No Load Losses: Iron loss: Hysteresis and eddy current loss Very small amount of copper loss also occurs in primary Full Load losses: At full load, Copper loss is largest and a very small amount of eddy current loss is there. 7 What are the load and no load losses of a transformer? What is all day efficiency of a transformer? BUET M.Sc. 13 A transformer has mainly two types of losses, these are, iron losses and copper losses. Iron loss, which is also referred as core loss, consists of hysteresis loss and eddy current loss. These two losses are constant when the transformer is charged. That means the amount of these losses does not depend upon the condition of secondary load of the transformer. In all loading condition, these are fixed. But the copper loss which is also referred as I 2 R loss entirely depends upon load I. A distribution transformer cannot be run with constant load throughout 24 hours. At day peak time it s loading is high, whereas in night lean time its loading may be negligible. So selecting a transformer depending upon its conventional efficiency is not practical and economical, too. As a solution of these problems, the concept of all day efficiency of distribution transformer came into the picture. In this concept, we use the ratio of total energy delivered by the transformer to the total energy fed to the transformer, during a 24 hrs span of time instead of ratio of power output and input of the transformer. Hence, all day efficiency is determined as, total KWh at the secondary of the total KWh at the primary of the transformer for a long specific period preferably 24 hrs. i.e, This is very much useful to judge the performance of a distribution transformer, whose primary is connected to the system forever, but secondary load varies tremendously throughout the day. 8 PGCB-14

9 9 A 30 kva, 2400/240 V, 50 Hz single phase transformer, operating at no load, draws 138 W at a p.f. of 0.21 lagging. Determine i) the exciting current of the transformer ii) the core loss component of the exciting current and iii) the magnetizing component of the exciting current. BPDB-13

10 10 A single phase 1000/200 V takes a no load current of 3A at 0.2 power factor lagging. Calculate the primary current and power factor when secondary current is 280 A at power factor 0.8 lagging. EGCB-2014 ১১ ন এর অন র প 11 A single phase transformer with a ratio of 100/200 V takes a no-load current of 3A at 0.2 power factor lagging. If the secondary supplies a current of 280 A at a p.f. of 0.8 lagging, estimate the current taken by the primary. BPDB-2015, EGCB The equivalent circuit impedances of a 20kVA, 8000/240V, 60Hz transformer are to be determined. The open circuit test and the short circuit test were performed on the BPDB-14

11 primary side of the transformer, and the following data were taken: (FF) Open circuit test (primary) Short circuit test V OC = 8000 V V SC = 489 V I OC = A I SC = 2.5 A P OC = 400 W P SC = 240 W Find the impedance of the approximate equivalent circuit referred to the primary side, and sketch the circuit. [Example 2.2, Chapman 4 th edition] In short circuit test of a transformer in which side supply voltage is applied? DPDC MCQ-

12 13 (a) Low voltage side (b) High voltage side 14 Open Circuit Test: The main purpose of this test is to find the iron loss and no load current which are useful in calculating core loss resistance and magnetizing reactance of the transformer. In O.C. test primary winding is connected to a.c. supply, keeping secondary open. Sometimes a voltmeter may be connected across secondary as voltmeter resistance is very high & voltmeter current is negligibly small so that secondary is treated as open circuit. Usually low voltage side is used as primary and high voltage side as secondary to conduct O.C. test. Short-Circuit Test: The main purpose of this test is to find full load copper loss and winding parameters (R01 &X01 or R02 & X02) which are helpful for finding regulation of transformer. (Reference: Chapman + Lab Manual ofshri Ramdeobaba College of Engineering and Management, Nagpur) ট পস মন র খ - She is Low Short Circuit Test Supply is at High voltage side -- Open Circuit Test - Supply is at Low voltage side In short circuit test of single phase transformer, the wattmeter reading roughly gives (b) Copper loss (b)core loss (c) Both (d)none In open circuit test of transformer power supply is given in (a) low voltage side (b) high voltage side (c) (d) Practically in open circuit test of a transformer, the wattmeter measures (a) Copper loss (b) Core loss (c) A slight copper loss (d) both (b) & (c) MCQ PGCB-14 MCQ PGCB-14 DPDC MCQ Why transformer is rated in KVA? What are the instrument transformers? PGCL-11

Cu loss of a transformer depends on current and iron loss on voltage. Hence, total transformer loss depends on voltampere (VA) and not on phase angle between voltage and current i.e. it is independent of load power factor.

13 Cu loss of a transformer depends on current and iron loss on voltage. Hence, total transformer loss depends on voltampere (VA) and not on phase angle between voltage and current i.e. it is independent of load power factor. That is why rating of transformers is in kva and not in kw. (Reference: Theraja topic 32.23) 18 Write down the conditions of parallel operation of transformers. BUET M.Sc. Unknown 1. Same voltage ratio of transformer. 2. Same percentage impedance. 3. Same polarity. 4. Same phase sequence. (Reference: electrical4u.com) 19 Show the direction of the current in primary and polarity of the load initially: PGCB-14 Answer: Tips: Based on the dot convention, a current entering the dotted terminal of one coil, will induce positive voltage at the dotted terminal of the second coil. Alternatively, if the current leaves the dotted terminal of the coil, it will induce negative voltage at the dotted terminal of the second coil. Referring to the circuit diagrams below: The circuit polarity signs '+' and '-' indicate example applied and resultant, relative voltage polarities. The instantaneous polarities of the voltages across each inductor with respect to the dotted terminals are the same. The circuit arrows indicate example applied and resultant relative current directions. The instantaneous directions of the current entering the primary inductor at its dotted end and the current leaving of the secondary inductor at its dotted end are the same.

14 Subtractive polarity transformer designs are shown in the upper circuit diagrams. Additive polarity transformer designs are shown in the lower circuit diagrams. 20 Which of the following is not an example of transformer connection? (a) Open delta - Y (b) (c) (d) MCQ- DWASA-14 (Reference: Chapman s Page-118) 21 Dy1 means- (a) Primary is delta and secondary is star connected and secondary leads primary by 30º (b) Primary is delta and secondary is star connected and primary leads secondary by 30º MCQ EGCB-14 Winding connection designations: First Symbol: for High Voltage: Always capital letters. D=Delta, Y=Star, Z=Interconnected star, N=Neutral Second Symbol: for Low voltage: Always Small letters. d=delta, y=star, z=interconnected star, n=neutral. Third Symbol: Phase displacement expressed as the clock hour number (1,6,11) Example Dyn11 Transformer has a delta connected primary winding (D) a star connected secondary (y) with the star point brought out (n) and a phase shift of 30 deg leading (11).

15 The digits (0, 1, 11 etc) relate to the phase displacement between the HV and LV windings using a clock face notation. The phasor representing the HV winding is taken as reference and set at 12 o clock. Phase rotation is always anti-clockwise. (International adopted). Use the hour indicator as the indicating phase displacement angle. Because there are 12 hours on a clock, and a circle consists out of 360, each hour represents 30.Thus 1 = 30, 2 = 60, 3 = 90, 6 = 180 and 12 = 0 or 360. The minute hand is set on 12 o clock and replaces the line to neutral voltage (sometimes imaginary) of the HV winding. This position is always the reference point. Example: Digit 0 =0 that the LV phasor is in phase with the HV phasor * Digit 1 =30 lagging (LV lags HV with 30 ) because rotation is anti-clockwise. Digit 11 = 330 lagging or 30 leading (LV leads HV with 30 ) Digit 5 = 150 lagging (LV lags HV with 150 ) Digit 6 = 180 lagging (LV lags HV with 180 ) 22 What do you understand by DY1 (a) High voltage side is leading low voltage side by 30º (b) Low voltage side is leading high voltage side by 30º MCQ EGCB In a 3-phase transformer DY1 means- MCQ EGCB UNIT transformer at power generation is used as an- (a) energy meter (b)step up transformer (c) auxiliary transformer Why do we ground Y-neutral in D-Y transformer? (i) for flowing zero sequence current (ii) MCQ PGCB-14 MCQ BPDB- 14 (FF) 26 What are the disadvantages of 3-phase Y-Y connected transformer? How this problem can be solved? BUET M.Sc.12

What are the voltage ratios of this transformer at each tapings?

16 (Reference: Chapman s page 119) 27 A 500KVA, 13200/480 V distribution transformer has four taps 2.5% on its primary winding. What are the voltage ratios of this transformer at each tapings? DWASA-14 (Reference: Chapman s example 2.6) Examples from Electric Machinery Fundamentals by Stephen J. Chapman At a Glance

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Module 7. Transformer. Version 2 EE IIT, Kharagpur

Module 7. Transformer. Version 2 EE IIT, Kharagpur Module 7 Transformer Lesson 28 Problem solving on Transformers Contents 28 Problem solving on Transformer (Lesson-28) 4 28.1 Introduction. 4 28.2 Problems on 2 winding single phase transformers. 4 28.3