Thursday, 19 September 2013

EE-unit

EE-unit
PART-A

1.         What are the two types of transformer cores used?

Ø Core type    
Ø  Shell type.

2.         What is a transformer? Define an ideal transformer?

Transformer  is  a static electrical  device, which  transfers  electrical  energy one  circuit  to another circuit  without  any change in  frequency . It decreases or increases the input ac voltage.

An  Ideal   transformer  is  one which  does  noinvolve  any power  loss  and  does not   have  any  change in  frequency .  It decreases or increases the input ac voltage.

3.         What  are the  test  to  be connected  in  a transformer  to obtain  its   Equivalent  circuit?

Ø  Opecircuit  test         
Ø  Short  circuit  test

4.         Why transformer rating is  in KVA?

Copper  loss  of  a transformer  depends   on  current  and  iron  loss  depends  on  voltage. Hence total   transformer loss  depends   on  volt-ampere(VA)  and  not  on  phase angle  between  voltage and current   ie,  it  is  independent  of load  powefactor .  That is  why rating of  transformers  is  in  KVA  & not in  KW.

5.         Whais  the  necessity of  using steppecore ?
           
In  transformers  of medium  and  large  capacity, Only circular  coils  are used which  are mechanically  stronger.  Iis  more  economicato  place circular  shape  coils  around

Stepped   core: With  increase in  steps , the diameter  gets  reduced  & ultimately reduce the volume  of  the  copper  conductor   and  also  the  resistance of  the  coil.




6.         What  are  Instrument   transformers?

To  measure higcurrents  in  the    order  of  several  hundreamperes  and voltages  of several  kilo-volts,  instrument  transformers  are  used along with  ammeters  & voltmeters  of lowerange.

Type: Current  transformer - to measure high  current.
Potential  transformer - to  measure high  voltage.

7.         What  is  the  purpose of   constructing transformecore by silicon   content   steel   laminations?

Steel  and  steel  alloys  offers  less  reluctance to  the path  of  magnetic field.  Addition  of  3  to 5  percent  silicon with   steel  reduceHysteresis  loss  in  the core .  Laminating the  core reduces   eddy current   loss   occurring  in  the core.

8.         Name two important   electrical   performances  of  transformers .

The two  important  electrical   performances  of  transformers  are

Ø  Voltage regulation
Ø  Efficiency.

9.         Mention the difference between core and shell type transformers?

In core type, the windings surrounded the core considerably and in shell type the core surround the windings i.e winding is placed inside the core.


10.       What is the purpose of laminating  the core in a transformer?

To   reduc the   edd current   loss   in   the   cor of   the   transformer.

11.       Give the emf equation of a transformer and define each term?

E1 =4.44 fФmN1.
E2 =4.44 fФmN2.
f--->freq of AC input
Фm---->maximum value of flux in the core
N1,N2--->Number of primary & secondary turns.


12.       Does transformer draw any current when secondary is open? Why?

Yes,  it(primary) will draw the current from the main supply in order to magnetize the core and to supply for iron and copper losses on no load. There will not be any current in the secondary since secondary is open.

13.       Define voltage regulation of a transformer?

When a transformer is loaded with a constant primary voltage, the secondary voltage decreases for lagging PF load, and increases for leading PF load because of its internal resistance and leakage reactance. The change in secondary terminal voltage from no load to full load expressed as a percentage of no load or full load voltage is termed as regulation.

%regulation down=(V2noload-V2full load)*100/V2noload
%regulation up=(V2noload-V2 full load)*100/V2 full load

14.     Full load copper loss in a transformer is 1600W. what will be the loss at half load?

If x is the ratio of actual load to full load then copper loss = x2(F.L copper loss) here Wc = (0.5)2 1600=400W.

15.       Define all day efficiency of a transformer?

It  is  computed on the  basis of energy consumed during a certain period, usually a day of 24 hrs. all day efficiency=output in kWh/input in kWh tor 24 hrs.

16.       Why transformers are rated in kVA?

Copper loss of a transformer depends on current & iron loss on voltage. Hence total losses depends on Volt-Ampere and not on PF. That is why the rating of transformers are in kVA and not in kW.

17.       What are the typical uses of auto transformer?

Ø  To give small boost to a distribution cable to correct for the voltage drop.
Ø  As induction motor starter.

18.       What are the application of step-up & step-down transformer?

Step-up transformers are used in generating stations. Normally the generated voltage will be either 11kV. This voltage(11kV) is stepped up to 110kV or 220kV or 400kV snd transmitted through transmission lines(simply called as sending end voltage). Step-down transformers are used in receiving stations. The voltage are stepped down to 11kV or 22kV are stepped down to 3phase 400V by means of a distribution transformer and made available.

19.       Mention the difference between core and shell type transformers?

In core type the windings surround the core shell type the core surround the windings i.e winding is placed inside the core.

20.       What is the purpose of laminating the core in a transformer?

To reduce the eddy current loss in the core of the transformer.

21.       How transformers are classified according to their construction?

Ø  core type
Ø shell type.
In core type, the winding (primary and secondary) surround the core and in shell type, the core surround the winding.

22.       Explain on the material used for core construction?

The core is constructed  by sheet steel laminations assembled to provide a continuous magnetic path with minimum of air gap included. The steel used is of high silicon content sometimes heat treated to produce a high permeability and a low hysteresis loss at the usual operating fluxdensities. The eddy current loss is minimized by laminating the core, the laminations being used from each other by light coat of core-plate vanish or by oxide layer on the surface. The thickness of lamination varies from 0.35mm  for a frequency
of 50Hz and 0.5mm for a frequency of 25Hz. When will a Bucholtz relay operates and give an alarm.

23.       How does change in frequency affect the operation of a given transformer?

With  a  change  in  frequency,  iron  and  copper  loss,  regulation, efficiency & heating varies so the operation of transformer is highly affected.

24.       What is the angle by which no-load current will lag the ideal applied voltage?
In an ideal transformer, there are no copper & core loss  i.e loss free core. The no load current is only magnetizing current therefore the no load current lags behind by angle 90˚. However the winding possess resistance and leakage reactance and therefore the no load current lags the applied voltage slightly less than 90˚.

25. List the arrangement of stepped core arrangement in a transformer?

Ø  To reduce the space effectively
Ø  To obtain reduce length of mean turn of the winding
Ø  To  reduce  I2R  loss.  Why  are  breathers  used  in  transformers? Breathers are used to entrap the atmospheric moisture and thereby not allowing it to pass on to the transformer oil. Also to permit the oil inside the tank to expand and contract as its temperature increases and decreases.

26.       What is the function of transformer oil in a transformer?

Ø  It provides good insulation
Ø  Cooling.


27.       Can the voltage regulation go negative? If so under what condition?

Yes, if the load has leading PF.

28.       Distinguish power transformers & distribution transformers?

Power transformer are very high rating. They are used in generating and receiving stations. Sophisticated controls are required. Voltage ranges will be very high. Distribution transformers are used in receiving side. Voltage levels will be medium. Power ranging will be small in order of kVA. Complicated controls are not needed.

29.     What   is  the  purpose  of  providing  Taps  in  transformer  and   where  these  are provided?
In order to attain the required voltage, taps are provided, normally at high voltages side(low current).

30.         What is the function of a transformer?
            Transformer is a static device that
Ø  Transfers electric power from one circuit to another.
Ø  It does so without change of frequency.
Ø  It accomplishes this electro-magnetic induction.
31.       State the different types of single phase transformer based on construction?
Ø  Core type transformer
Ø  Shell type transformer.
32.       What is an ideal transformer?
            An ideal transformer is one that has
Ø  no winding resistance.
Ø  no leakage flux.
Ø  no iron losses in the core.
Ø  l2R Loss (copper loss)
33.       What are turns ratio and transformation ratio of transformer?
            The transformation ratio is defined as the ratio of the secondary voltage to primary voltage. It is denoted by K, K = E2/E1 = N2/N1 = I1/I2
34.       Write the e.m.f equation of two winding transformer.
E1=4.44fmN1 volts
E2= 4.44fmN2 volts
Where, m= Maximum flux in the core, in wb.
                          N1=no of turns in primary winding
                          N2=no of turns in secondary winding
                          F=supply frequency in Hz
35.       What is the purpose of laminating the core in a transformer?
            To reduce the eddy current loss, transformer core are laminated.
36.       Give the expression of secondary resistance as referred to primary.
           R’2=R2/K2

37.       How does change in frequency affect the operation of a given transformer?
Ø    Iron loss increases with decrease in frequency.
Ø    Since total loss is greater at lower frequency, the temperature is increased with decrease in frequency.
Ø    Reactive drop is affected, regulation at low power factors decrease with decrease in frequency.
38.       What load current of a transformer increases, how the input current increases to meet with the new condition does.
            Increase the load current secondary current to increase, thus reduce flux in the core, in order to maintain the flux the primary current increases.
39.       Write the name of material used for transformer core.
            Transformer core is made up of high grade of silicon steel.
40.       What are the losses in a transformer? How will you minimize them?
There are two losses occur transformer,
Ø  Iron Losses or core losses.
Ø  Copper losses.
Iron losses include Hysteresis loss and Eddy current loss. Hysteresis loss is reduced by making transformer core high grade of silicon steel. Eddy current loss is minimized by laminating the transformer core.
            Copper loss is minimized by reducing the leakage flux which is linked with both primary and secondary windings.
41.       What is the condition for maximum efficiency of a transformer?
            Copper loss = Iron Losses is the condition for maximum efficiency of transformer.
42.       Define the regulation of a transformer.
            The change in secondary terminal voltage from no load to full load is known regulation of a transformer
43.       Define all day efficiency?
All day efficiency = Output power in KWh / Input power in KWh  (for 24 hours)
44.       Mention three phase transformer connections.
Ø  star-star
Ø  star-Delta
Ø  Delta-Delta
Ø  Delta-star     
Ø  open-Delta   
Ø  scott connections.
45.       What is mean by leakage flux?
            Magnetic flux can be confined into a desired path. The greater portion of the flux (i.e., the mutual flux) remains confined to two core and links both the windings, but a small portion called the leakage flux completes its path through the air surrounding the coils.
46.       What is the condition for maximum regulation of transformer?
            Tan =X/R is the condition for maximum regulation transformer.
47.       What is step-up transformer?
            If the number of turns in secondary winding (N2) is greater than the number of turns in primary winding (N1). i.e., when the secondary voltage is greater than the primary voltage.
48.       Write short notes on impedance test.
            This is economical method for determine the following
Ø  Equivalent impedance (Z01 (or) Z02) Leakage reactance (L01 or L02) and total resistance (R01 or R02) of the transformer as referred to the winding.
Ø  Cu loss of full load.
Ø  Regulation of the transformer.

49.       Define a transformer.
            Transformer is a static electric device that transforms electrical power from one circuit to another circuit through magnetic induction and electrical isolation without change of frequency.
50.       When the load current of a transformer increases, how does the input current increases to meet with the new condition?
            Increase in load causes secondary current to increases thus reduces flux in the core, in order to maintain the flux, the primary current increases.
51.       What is the purpose of conducting open circuit and short circuit tests in transformer?
O.C. Test
            The purpose of condition open circuit test is to determine no load loss or core loss and no load current (I0) which is helpful in finding X0 and R0.
S.C. Test
            The short circuit test is conducted to determine the following
Ø  Equivalent impedance (Z01 or Z02), leakage reactance (X01 or X02 ) and total resistance (R01 or R02) of the transformer as referred to the winding.
Ø  Copper loss at full load.
Ø  Regulation of the transformer.
52.       What is the function of transformer oil in a transformer?
            Oil provides better insulation to the air and it is a better conductor of heat then air. It carries away the heat. Mineral oil is used for this purpose.
53.       Why are iron losses constant at all load in a transformer?
            Because Core flux remains constant for all loads, the core losses also remain constant.
54.       What are the applications of step up and step down transformer?
Step up transformer
Ø  Transmission system.
Ø  Current transformer.
Ø  Welding transformer.
Step down transformer
Ø  Distribution system
Ø  Potential transforms.
Ø  Electronic application (Television, Radio computer)

PART-B

1.            Explain in details the construction and operating principle of a transformer(DEC 2003, JUNE 2010)
Working Principle              =          2 Marks
Diagram                                =          4 Marks
Constructional Features   =          2 Marks
Theory                                   =          8 Marks
2.            Derive the EMF equation for a transformer (DEC 2003, JUNE 2010)( MAY 2011)
How EMF is developed      =          2 Marks
Derivation                            =          4 Marks
Final Eqn with units          =          2 Marks
3.            Derive the equivalent circuit of transformer (DEC 2007), (DEC 2010)
Diagram(4 Steps)                =          8 Marks
Equation                               =          4 Marks
Theory                                   =          4 Marks
4.            Explain with help of circuit diagrams how efficiency and regulation of single phase transformer are predetermined by conducting open and short circuit tests? (MAY 2011)
OC Test:
Diagram(2)                           =          4 Marks
Equation                               =          2 Marks
Theory                                   =          2 Marks
SC Test:
Diagram(1)                           =          4 Marks
Equation                               =          2 Marks
Theory                                   =          2 Marks
5.            Explain how the efficiency of a transformer may be found from the open circuit and short circuit tests(MAY 2008)
OC Test:
Diagram(2)                           =          4 Marks
Equation                               =          2 Marks
Theory                                   =          2 Marks
SC Test:
Diagram(1)                           =          4 Marks
Equation                               =          2 Marks
Theory                                   =          2 Marks

6.            Describe the tests to determine core loss and full load copper loss of a transformer (or) Explain the OC and SC tests performed on a transformer (Nov/Dec 2011)
OC Test:
Diagram(2)                           =          4 Marks
Equation                               =          2 Marks
Theory                                   =          2 Marks
SC Test:
Diagram(1)                           =          4 Marks
Equation                               =          2 Marks
Theory                                   =          2 Marks
7.            Explain the working of single phase transformer with neat sketch (Nov/Dec 2011)
Working Principle              =          2 Marks
Diagram                                =          4 Marks
Constructional Features   =          2 Marks
Theory                                   =          8 Marks
8.            Define voltage regulation. Draw the phasor diagram of lagging power factor and determine voltage regulation.
Voltage Regulation                        =          2 Marks
Phasor Diagram                  =          6 Marks
Theory                                   =          6 Marks
Equation                               =          2 Marks
9.            Write short notes on (DEC 2006)
a.            Transformer on no- load          b. Transformer on  load
Diagram                                =          4 Marks
Theory                                   =          8 Marks
Equation                               =          4 Marks

10.         Discuss about power losses in a transformer.(DEC 2009)
Theory                                   =          4 Marks
Equation                               =          4 Marks
11.         A 10KVA, 200/400V, 50Hz single phase transformer gave the following test results.
OC Test: 200V, 1.3A, 120W...............on L.V Side
SC Test: 22V, 30A, 200W..................on H.V side
Calculate (i). The magnetizing current and the component corresponding to core loss at normal frequency and voltage (ii)The magnetizing branch impedances and (iii)% voltage regulation when supplying full load at 0.8 pf leading.(DEC 2009).
Formula                                =          2 Marks
Steps                                      =          3 Marks
Final Answer                       =          2 Marks
Unit                                        =          1 Mark
12.         A single phase 20KVA, 2200/220V, 50Hz transformer. The O.C/S.C test results are as follows.
O.C test:  220V, 4.2A, 148W........ (L.V. Side)               S.C test:  86V,    10.5A, 360W...... (H.V.Side)
Determine (i).Regulation and efficiency at 0.8 p.f lagging at full load and (ii) Power factor on short circuit (iii) obtain the approximate equivalent circuit referred to H.V.side.
Formula                                =          2 Marks
Steps                                      =          3 Marks
Final Answer                       =          2 Marks
Unit                                        =          1 Mark
13.         A 600KVA single Ф transformer has an efficiency of 94% both at full load and half load at unity power factor. Determine the efficiency at 75% of full load at 0.9 power factor.
Formula                                =          2 Marks
Steps                                      =          3 Marks
Final Answer                       =          2 Marks
Unit                                        =          1 Mark
14.         The OC and SC tests on a 4KVA, 200/400V, and 50Hz single phase transformer gave the following results: OC test on LV side: 200V, 1A, 100W.SC test with LV side shorted: 15V, 10A, 85W.Determine the parameters of the equivalent circuit and draw the equivalent circuit referred to LV side.
Formula                                =          2 Marks
Steps                                      =          3 Marks
Final Answer                       =          2 Marks
Unit                                        =          1 Mark
15.         The parameters of a 2300/230V, 50-Hz transformer is given below:
R1=0.286Ω        R2’ =0.319Ω          Ro=250Ω
X1=0.73Ω           X2’=0.73Ω             X0=1250Ω
The secondary load impedance ZL= 0.387+j0.29. Solve the exact equivalent circuit with normal voltage across the primary.
Formula                                =          2 Marks
Steps                                      =          3 Marks
Final Answer                       =          2 Marks
Unit                                        =          1 Mark

                              

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