Monday, April 18, 2011

Induction Machines-Performance

Induction Machines-Performance

  • In induction motor, the magnetizing current is much larger than the active current, so that the no-load power factor is low,0.1 or 0.25
  • If a load torque applied to the shaft, the active component of stator current increases and magnetizing current remains constant, the power factor is raised
  • The maximum effieciency ocurs roughly where the I2R losses in stator and rotor sum to ana amount equal to the no-load constant losses.
  • Usual reltio to maximum to full load torque is 2:1
  • Cage motors:
  • Motors with normal starting torque and current , having cages with low resistance and reactance, low full load slip, good efficiency and power factor
  • Motors with high starting torque and low starting current, using deep-bar rotor or double cage rotors.
  • Motors with high full load slip, using a comparatively high resistance cage , with large starting torque, low starting current and low efficiency. Used for fluctuating torque drives(punching machines)
  • Multispeed Motors:
  • variable-torque, power output proportional to square of speed-fans,centrifugal pumps
  • Constant torque, power output proportional to speed-conveyors,stokers, reciprocating compressors,printing presses
  • Inverse torque, power output rating constant-machine tools,laths,boring mills,drills,planners
  • Gear motors- high speed motors-conveyors,agitators,blowers and screens
  • A startor may either increase rotor resistance or decrease starting current in induction machine
  • The inclusion of rotor resistance will reduce the stator current at standstill and simultaneously raise its power factor.
  • The starting current of a cage motor can only be reduced by lowering the voltage applied to the stator
  • The output and torque for a given slip change as the square of the allied voltage and current reduces in proportionally
  • Direct online startor
  • For direct switching case, if TSC be the torque developed on short circuit at normal voltage with currents ISC then starting torque is TS = TSC and the starting current per phase is IS = ISC the ratio of starting to full load torque is TS/Tn=(ISC/In)2sn
  • Stator resistor startor:
  • The normal voltage per phase to the fraction x is applied to the termianls. The starting current is Is = xIsc and starting torque Ts = x2Tsc
  • The ratio of statig torque to full load torque is Ts/Tn = (Is/In)2sn = x2(Isc/In)2sn
  • Auto-trnasformer startor:
  • The auto-transformer starter is used to reduce the phase voltage to the fraction x of normal value. The starting current is Is = xIsc and starting torque Ts = x2Tsc
  • The ratio of statig torque to full load torque is Ts/Tn = (Is/In)2sn = x2(Isc/In)2sn
  • Star- Delta startor
  • A motor must be built to run normally with a mesh connected stator winding. At starting the winding is connected temporarily in star. The phase voltage is thus reduced to 1/√3 = 0.58 of normal.
  • The starting current per pahse is Is = 0.58Isc, the line current is (0.58)2Isc = 0.33Isc, the starting torque is one third of short circuit value.
  • the ratio to starting torque to full load torque is Ts/Tn = (1/3)(Isc/In)2sn
  • Cogging and Crawling:
  • If the number of stator slots S1 equal to the rotor slots S2, the machine may efuse to start at all, this is called cogging or teeth locking.
  • With other ratios S2/S1, the motor may exhibit a tendency to run stabnly at low speed (one-seventh of normal): this is called crawling.
  • The 5th harmonic r.m.f rotates agaonst the fundamental at one-fifth synchronous speed.
  • The 7th harmonic r.m.f rotates with the fundamental at one-seventh synchronous speed.
  • If the harmonic torques are sufficiently pronounced, the seventh harmonic may prevent the motor speed exceeding abiut one-seventh of normal, since the downward slope of the resultant torque at this speed is a stable running condition over the torque range between the maximum and minimum points. The motor therefore crawls.
  • If n1 is the synchronous speed of fundamental, and n the speed of the rotor, then the synchronous speed of the thirteenth harmonic is +n1/13, or relative to the rotor -(n1-n)/13. The rotor runs at a speed n, and so rotates its own thirteenth harmonic at a speed -(n1-n)/13+n relative to the stator. The sttor and rotor thirteenth harmonic fall into step when +n1/13 = -(n1-n)/13+n i.e. n= (1/7)n1. Thus the torque discontinuity at one seventh synchronous speed is here produced not by the seventh harmonic, but by the thirteen.

Induction Machines-Theory

Induction Machines-Theory

  • The induction motor consists of a fixed core, the stator, carrying a three phase winding in slots on its bore.
  • The stator winding is connected to the supply, and a uniform rotating magnetic filed is produced therein.
  • It works based on principle of Electro-magnetic interaction
  • Speed of the stator m.m.f is synchronous speed
  • The speed of rotation,n, of the rotor is less than the synchronous speed by a fraction,s known as the slip
  • A high ratio of resistance to reactance is desirable under starting conditions, but the resistance should be low for running near synchronous speed, the normal condition
  • The induction motor is a machine of substantially constant speed, and fills the same role as the plain shunt motor among the dc machines
  • The cage rotor is adaptable to any number of poles. whereas the slip-ring winding has to be made for one(or possibly two) definite values of p.
  • leakage flux is greater than transformer in induction machine due to its air gap
  • Power relations:
  • Power input to the stator per phase is P1 = V1I1cosØ1
  • Power delivered by transformer action to the rotor, P2
  • The mechanical power developed at the rotor, Pm = (1-s)P2
  • Rotor power loss is I22r2 = sP2 = s.rotor input
  • P2:Pm:I22r2 = 1:(1-s):s
  • If the rotor had no resistance it could develop no torque
  • condition for maximum torque is r2 = ±sx2
  • Slip values for different operations: If s is positive and between 0 and 1 : Motor, If s is negative and -1 to 0 : generator, If s greater than unity: braking
 




Solved Problems
1.A 220-V, three-phase, two-pole, 50-Hz induction motor is running at a slip of 5 percent. Find:
(a) The speed of the magnetic fields in revolutions per minute
(b) The speed of the rotor in revolutions per minute
(c) The slip speed of the rotor
(d) The rotor frequency in hertz

SOLUTION:

(a) The speed of the magnetic fields is

(b) The speed of the rotor is


(c) The slip speed of the rotor is


(d) The rotor frequency is



2.A three-phase, 60-Hz induction motor runs at 890 r/min at no load and at 840 r/min at full load.
(a) How many poles does this motor have?
(b) What is the slip at rated load?
(c) What is the speed at one-quarter of the rated load?
(d) What is the rotor’s electrical frequency at one-quarter of the rated load?
SOLUTION
(a) This machine has 8 poles, which produces a synchronous speed of

(b) The slip at rated load is


(c) The motor is operating in the linear region of its torque-speed curve, so the slip at ¼ load will be


The resulting speed is


(d) The electrical frequency at ¼ load is

Saturday, April 16, 2011

Transformers- Testing

Transformers- Testing

 
 
    No-load test:
  • As the name suggests, the secondary is kept open circuited and nominal value of the input voltage is applied to the primary winding and the input current and power are measured.
  • This test is performed to determine core or iron loss, Pi and no-load current I0.
  • This test is helpful in determination of magnetizing component Im’ energy component Ie and so no-load resistance R0 being given as V1/Ie and no-load reactance given as V1/Im.

  • In this test secondary (usually high voltage) winding is left open, all metering instruments (ammeter, voltmeter and wattmeter) are connected on primary side and normal rated voltage is applied to the primary (low voltage) winding, as illustrated below





  • Since no load current I0 is very small, therefore, pressure coils of watt meter and the volt meter should be connected such that the current taken by them should not flow through the current taken by them should not flow through the current coil of the watt meter




  • Short Circuit Test:
  • This test is performed to determine the full-load copper loss and equivalent resistance and reactance referred to secondary side.



  • In this test, the terminals of the secondary (usually the low voltage) winding are short – circuited, all meters (ammeter, voltmeter and wattmeter) are connected on primary side and a low voltage, usually 5 to 10 % of normal rated primary voltage at normal frequency is applied to the primary, as shown in fig below. 



  • The applied voltage to the primary, say Vs’ is gradually increased till the ammeter A indicates the full load current of the side in which it is connected. The reading Ws of the wattmeter gives total copper loss (iron losses being negligible due to very low applied voltage resulting in very small flux linking with the core) at full load. Le the ammeter reading be Is.





  • Sumpner’s Test or Back to Back Test
  • This test is mainly performed to determine temperature rise.



  • In this test two transformers are loaded fully, in similar way as two dc machines in a regenerative test, and the power required from the supply is just to meet the iron and full load copper losses of the two transformers.



  • It needs two identical transformers







  • Practice Objectives:

    1. In a single-phase transformer, the magnitude of leakage reactance is twice that of resistance of both primary and secondary. With secondary short-circuited, the input power factor is
    a. 1/2
    b. 1/5
    c. 2/5
    d. 1/3

    2. When short-circuit test on a transformer is performed at 25 V, 50 Hz, the drawn current ‘I1’ is at a lagging p.f. angle of Φ1. If the test is performed at 25 V, 25 Hz and the drawn current “I2’ is at a lagging p.f. angle of Φ2, then
    a. I2 > I1 and Φ2 < Φ1 b. I2 > I1 and Φ2< Φ1 c. I2 < I1 and Φ2 > Φ1
    d. I2 < I1 and Φ2 > Φ1

    3.A 20 kVA, 2000/200 V, 1-phase transformer has nameplate leakage impedance of 8% Voltage required to be applied on the high - voltage side to circulate full - load current with the low voltage winding short-circuited will be
    a. 16 V
    b. 56.56 V
    c. 160 V
    d. 568.68 V

    4. In a 100 kVA, 1100/220 V, 50 Hz single -phase trans former with 2000 turns on the high -voltage side, the open- circuit test result gives 220 V, 91 A, 5kW on low voltage side. The core-loss component of current is, approximately
    a. 9.1 A
    b. 22.7 A
    c. 45.0 A
    d. 91 A

    5. A 1 kVa, 200/100 V, 50 Hz, single-phase transformer gave the following test results on 50 Hz :
    OC (LV side) : 100 V, 20 watts
    SC (HV side) : 5 A, 25 watts
    It is assumed that no-load loss components are equally divided. The above tests were then conducted on the same transformer at 40 Hz.
    Tests results were:
    OC (HV):160 V, W1 watts.
    SC (LV): 10 A, W2 watts
    neglecting skin effect, W1 and W2 will be

    a. W1 = 16 watts, W2 = 25 watts
    b. W1 = 25 watts, W2 = 31.25 watts
    c. W1 = 20 watts, W2 = 20 watts
    d. W1 = 14.4 watts, W2 = 25 watts

    6. Match List I (Test) with List II (Quantities) and select the correct answer using the code given below the lists:
    List I
    A. O C Test
    B. S C Test
    C. Sumpner’s Test
    D. Load Test
    List II
    1. Copper loss and iron loss
    2. Total losses
    3. Iron loss
    4. Copper loss
    Codes;
    ---A B C D
    a. 3 4 1 2
    b. 2 1 4 3
    c. 3 1 4 2
    d. 2 4 1 3


    Solved Problems

    1. A single-phase 10-kVA 480/120-V transformer is to be used as an auto-transformer tying a 600-V distribution line to a 480-V load. When it is tested as a conventional transformer, the following values are measured on the primary (480-V) side of the transformer:
    Open-circuit test Short-circuit test
    VOC = 480 V VSC = 10.0 V
    IOC = 0.41 A ISC = 10.6 A
    VOC = 38 W PSC = 26 W
    Find the per-unit equivalent circuit of this transformer when it is connected in the conventional manner. What is the efficiency of the transformer at rated conditions and unity power factor? What is the voltage regulation at those conditions?

    Solution:
    The base impedance of this transformer referred to the primary side is
    The open circuit test yields the values for the excitation branch (referred to the primary side):

    The excitation branch elements can be expressed in per-unit as


    The short circuit test yields the values for the series impedances (referred to the primary side):

    The resulting per-unit impedances are

    The per-unit equivalent circuit is

    At rated conditions and unity power factor, the input power to this transformer would be PIN = 1.0 pu.
    The core losses (in resistor RC) would be


    The copper losses (in resistor REQ) would be

    The output power of the transformer would be

    and the transformer efficiency would be

    The output voltage of this transformer is


    The voltage regulation of the transformer is




Transformers-parallel operation

Transformers-parallel operation


  • Requirements for parallel operation of transformers:


  • a)the same voltage-ratio
    b)the same per-unit impedance
    c)the same polarity
    d)the same phase sequence and zero relative phase displacement.
    (c) and (d) are absolutely essential, (a) must satisfy to a close degree, in (b), more latitude. it decides load sharing

  • The currents carried by two transformers are proportional to their ratings if their ohmic impedance are inversely proportional to those ratings and their per-unit impedance are identical.

  • A difference in the per-unit impedance results in a divergence of phase angle of the two currents, i.e. two transformers will work at different power factors


  • Transformers can be operated in parallel:


  • Transformer 1: Yy Yd Yd
    Transformer 2: Dd Dy Yz

    Practice Objectives:
    1. If all other requirements for parallel operation of transformer are fulfilled, which one of following pairs of three phase transformers, with given VECTOR GROUPS, can be operated in parallel?
    a. Y d 1 and Yy 0
    b. Y d 1 and Dy 11
    c. Dd 6 and Dy 1
    d. Dd 0 and Dy 11

    2.Two transformers operating in parallel will share the load depending upon their
    a. Ratings
    b. Leakage reactance
    c. Efficiency
    d. Per unit impedance

    3.If per unit impedance of two transformers connected in parallel are not equal, then which one of the following statements is correct?
    a. The power factor of the two transformers will be different from that of the connected load
    b. Transformers will get overloaded
    c. Dead short circuit occurs
    d. The transformer with higher per unit impedance will share more load

    4.Two single phase transformers A and B are connected in parallel, observing all requirements of a parallel operation, except that the induced voltage Ea is slightly greater than Eb; Zea and Zeb being the equivalent impedances of A and B, both referred to the secondary side.



    Under this operating condition with the primary bus-bars being energized, a circulating current will flow:
    a. Only in the secondary windings of A and B
    b. In both the primary and the secondary windings of A and B
    c. In both the primary and the secondary windings of A and B, as well as in the primary side network
    d. In the primary and the secondary windings of A and B, and boost the voltages on the secondary side of both A and B

    Solved Problems
    1. A 500 kVA. transformer with 0.01pu resistance and 0.05 pu reactance is connected in parallel with a 250 kVA transformer with 0.015pu resistance and 0.04 pu reactance. Find how they share a load of 750 kVA, at a power factor 0.8 lagging.
    Solution:
    choose base kVA as 250 kVA.
    Z1 = 0.01+j0.05 = 0.051/78.70 Ω
    and Z2 = 2(0.015+j0.04 = 0.03+j0.08 = 0.085/69.40 Ω.
    Z1+Z2 = 0.04+j0.13 = 0.136/72.90 Ω.

    total load kVA is S = 750/-36.90 kVA;
    S1 = 750/-36.90 . 0.085/69.40 / 0.136/72.90 = 471/-40.40 kVA
    S1 = 471 kVA at power factor 0.762 lagging.

    S2 = 750/-36.90 . 0.051/78.70 / 0.136/72.90 = 281/-31.10 kVA
    S1 = 281 kVA at power factor 0.856 lagging.







    Basic Electrical Objectives

    1. In order for a 30 volt, 90 watt lamp to work properly in a 120 volt supply the required series resister in ohm is
    (A) 10 (B) 20
    (C) 30 (D) 40 ans 4
    2. According to Thévenin's theorem, any linear active network can be replaced by a single voltage source
    (A) in series with a single impedance
    (B) in parallel with a single impedance
    (C) in series with two impedances
    (D) in parallel with two impedances
    3. The internal resistance of ammeter is
    (A) very small (B) very high
    (C) infinite (D) zero
    4. Hay bridge is used mainly for the measurement of
    (A) resistance (B) inductance
    (C) conductance (D) capacitance
    5. Which of the following is true about series resonance
    (A) The reactance becomes zero and impedance becomes equal to resistance
    (B) The current in the circuit becomes maximum
    (C) The voltage drop across inductance and capacitance cancels each other
    (D) All of the above statements are correct
    6. A 3-Ф, 4 wire, 400/230 v feeder supplies 3-phase motor and an unbalanced lighting load. In this system
    (A) all four wires will carry equal current
    (B) neutral wire will carry no current
    (C) neutral wire will carry both motor current and lighting load current
    (D) neutral wire will carry current only when lighting load is switched on
    7. Equalizing connections are required when paralleling two
    (A) alternators (B) compound generators (C) series generators (D) both (B) and (C)
    8. An ideal transformer is one which
    (A) has a common core for its primary and secondary windings
    (B) has no losses and magnetic leakage
    (C) has core of stainless steel and windings of pure copper metal
    (D) has interleaved primary and secondary windings
    9. The principle of operation of a 3-phase induction motor is most similar to that of a
    (A) synchronous motor (B) repulsion-start induction motor (C) transformer with a shorted secondary
    (D) capacitor-start, induction-run motor
    10. In the forward region of its characteristic, a diode appears as
    (A) an OFF switch (B) a high resistance (C) a capacitor (D) an ON switch
    11. The common-emitter forward amplification factor βdc is given by
    (A) IC/IE (B) IC/Ib (C) IE/IC (D) IB/IF
    12. A common emitter amplifier is characterized by
    (A) low voltage gain
    (B) moderate power gain
    (C) signal phase reversal
    (D) very high output impedance
    13. After VDS reaches pinch-off value VP in a JFET, drain current IO becomes
    (A) zero (B) low (C) saturated (D) reversed
    14. An electronic oscillator
    (A) needs an external input
    (B) provides its own input
    (C) is nothing but an amplifier
    (D) is just a dc/ac converter
    15. In an SCR, the function of the gate is to
    (A) switch it off
    (B) control its firing
    (C) make it unidirectional
    (D) reduce forward breakdown voltage
    16. NAND and NOR gates are called ‘universal’ gates primarily because they
    (A) are available everywhere
    (B) are widely used in IC packages
    (C) can be combined to produce AND, OR and NOT gates
    (D) are the easiest to manufacture
    17. Registers and counters are similar in the sense that they both
    (A) count pulses
    (B) store binary information
    (C) are made from an array of flip-flops and gates integrated on a single chip
    (D) are in fact shift register
    18. A flip-flop
    (A) is a sequential logic device
    (B) is a combinational logic device
    (C) remembers what was previously stored in it
    (D) both (A) and (C)
    19. An operational amplifier
    (A) can be used to sum two or more signals
    (B) can be used to subtract two or more signals
    (C) uses to principle of feed back
    (D) all of the above
    20. TTL logic is preferred to DRL logic because
    (A) greater fan-out is possible
    (B) greater logic levels are possible
    (C) greater fan-in is possible
    (D) less power consumption is possible
    An electron rising through a potential of 250 V will acquire an energy of :
    (A) 250 eV
    (B) 800 eV
    (C) 250 J
    (D) 800 J
    If the amount of impurity, either donor type or acceptor type added to the intrinsic
    semiconductor is controlled to 1 part in one million, the conductivity of the sample :
    (A) increases by a factor 103
    (B) reduces by a factor 10−3
    (C) increases by a factor 106
    (D) reduces by a factor 10−6
    4. Laplace transform and Fourier integrals are related through :
    (A) frequency domain
    (B) time domain
    (C) both frequency and time domain
    (D) none
    A clamper circuit :
    (i) adds or subtracts a dc voltage to or from a waveform
    (ii) does not change the shape of the waveform
    (iii) amplifies the waveform
    (A) (i) and (ii) are correct
    (B) (i) and (iii) are correct
    (C) (ii) and (iii) are correct
    (D) (i), (ii) and (iii) are correct
    A ring counter consisting of five flip flop will have :
    (A) 5 states
    (B) 10 states
    (C) 32 states
    (D) infinite states
    Which one of the following can be used as parallel to series converter ?
    (A) Decoder
    (B) Encoder
    (C) Digital counter
    (D) Multiplexer
    An interrupt in which the external device supplies its address as well as the interrupt
    request, is known as :
    (A) vectored interrupt
    (B) maskable interrupt
    (C) polled interrupt
    (D) non-maskable interrupt
    An instruction that can be recognized and used without translation must be written
    in :
    (A) Source code
    (B) Machine code
    (C) Basic language
    (D) Assembly code
    The angle for which there is no reflection and the incident wave is vertically polarized
    is known as :
    (A) Steradian angle
    (B) Reflection angle
    (C) Brewster’s angle
    (D) Critical angle
    A PLL can be used to demodulate :
    (A) PAM signals(B) PCM signals(C) PM signals(D) DSB-SC signals
    The main function of balanced modulator is to :
    (A) produce balanced modulation of a carrier wave
    (B) produce 100 percent modulation
    (C) suppress carrier signal in order to create a single side band or double side band
    (D) limit noise picked up a receiver
    An SCR can be termed as :
    (A) DC switch(B) AC switch(C) Both DC and AC switch(D) Square wave switch
    Fiber optics communication offers the largest bandwidth in the range of :
    (A) 1010 Hz(B) 106 Hz(C) 1014 Hz(D) 1020 Hz
    Silicon photosensors have their maximum spectral response in the :
    (A) infrared region
    (B) ultraviolet region
    (C) visible region
    (D) X-ray region
    (1) Instrument is a device for determining
    (a) the magnitude of a quantity
    (b) the physics of a variable
    (c) either of the above
    (d) both (a) and (b)
    (2) Electronic instruments are preferred because they have
    (a) no indicating part
    (b) low resistance in parallel circuit
    (c) very fast response
    (d) high resistance in series circuit
    (e) no passive elements.
    (3) A DC wattmeter essentially consist of
    (a) two ammeters
    (b) two voltmeters
    (c) a voltmeter and an ammeter
    (d) a current and potential transformer
    (4) Decibel is a unit of
    (a) power
    (b) impedance
    (c) frequency
    (d) power ratio
    (5) A dc voltmeter may be used directly to measure
    (a) frequency
    (b) polarity
    (c) power factor
    (d)power
    (6) An accurate voltmeter must have an internal impedance of
    (a) very low value
    (b) low value
    (c) medium value
    (d) very high value
    (7) The insulation resistance of a transformer winding can be easily measured with
    (a) Wheatstone bridge
    (b) megger
    (c) Kelvin bridge
    (d) voltmeter
    (8) A 100 V voltmeter has full-scale accuracy of 5%. At its reading of 50 V it will give an error of
    (a) 10%
    (b) 5%
    (c) 2.5%
    (d) 1.25%
    (9) You are required to check the p. f. of an electric load. No p.f. meter is available. You would use:
    (a) a wattmeter
    (b) a ammeter, a voltmeter and a wattmeter
    (c) a voltmeter and a ammeter
    (d) a kWh meter
    (10) The resistance of a field coil may be correctly measured by using
    (a) a voltmeter and an ammeter
    (b) Schering bridge
    (c) a Kelvin double bridge
    (d) a Maxwell bridge
    (11) An analog instrument has output
    (a) Pulsating in nature
    (b) Sinusoidal in nature
    (c) Which is continuous function of time and bears a constant relation to its input
    (d) Independent of the input quantity
    (12) Basic charge measuring instrument is
    (a) Duddel's oscillograph
    (b) Cathode ray oscillograph
    (c) Vibration Galvanometer
    (d) Bailastic Galvanometer
    (e) Battery Charging equipment
    (13) A.C. voltage can be measured (using a d.c. instrument) as a value obtained
    (a) by subtracting the d.c. reading from it's a.c. reading.
    (b) Using the output function of the multimeter
    (c) By using a suitable inductor in series with it
    (d) By using a parallel capacitor with it
    (e) None of the above
    (14) A moving coil permanent magnet ammeter can be used to measure
    (a) D. C. current only
    (b) A. C. current only
    (c) A. C. and D. C. currents
    (d) voltage by incorporating a shunt resistance
    (e) none of these
    (15) Select the wrong statement
    (a) the internal resistance of the voltmeter must be high
    (b) the internal resistance of ammeter must be low
    (c) the poor overload capacity is the main disadvantage of hot wires instrument
    (d) the check continuity with multimeter, the highest range should be used.
    (e) In moving iron voltmeter, frequency compensation is achieved by connecting a capacitor across its fixed coil.
    (16) Which of the following instrument is suitable for measuring both a.c. and d.c.
    quantities.
    (a) permanent magnet moving coil ammeter.
    (b) Induction type ammeter.
    (c) Quadrant electrometer.
    (d) Moving iron repulsion type ammeter.
    (e) Moving iron attraction type voltmeter.
    (17) Swamping resistance is used in moving coil instruments to reduce error due to
    (a) thermal EMF
    (b) temperature
    (c) power taken by the instrument
    (d) galvanometer sensitivity.
    (18) A power factor meter is based on the principle of
    (a) electrostatic instrument
    (b) Electrodynamometer instrument
    (c) Electro thermo type instrument
    (d) Rectifier type instrument.
    (19) A potentiometer recorder is used for
    (a) AC singles
    (b) DC signals
    (c) both (a) and (b)
    (d) time varying signals
    (e) none of these.
    (20) Transformers used in conjunction with measuring instruments for measuring purposes are called
    (a) Measuring transformers
    (b) transformer meters
    (c) power transformers
    (d) instrument transformers
    (e) pulse transformers.
    (21) Leakage flux in an electrical machine is measure by
    (a) Ballistic galvanometer
    (b) Flux meter
    (c) Either (a) or (b)
    (d) Vibration galvanometer
    (e) CRO
    (22) A C.R.O. is used to indicate
    (a) supply waveform
    (b) magnitude of the applied voltage
    (c) B.H. loop
    (d) all of these
    (e) Magnitude of the current flowing in it.
    (23) An oscillator is a
    (a) an amplifier having feedback network
    (b) a high gain amplifier
    (c) a wide band amplifier
    (d) a untuned amplifier
    (e) None of these
    (24) Distortion can be measured by
    (a) Wave meter
    (b) Digital filters
    (c) Wein bridge circuit
    (d) Bridge T filter circuit
    (25) Series connected Q- meter is preferable for measurement of components having
    (a) high impedance
    (b) low impedance
    (c) both (a) and (b)
    (d) high frequency
    (e) low capacitance
    (26) A potentiometer is
    (a) an active transducer
    (b) a passive transducer
    (c) a secondary transducer
    (d) a digital transducer
    (e) a current sensing transducer
    (27) The basic components of a digital voltmeter are:
    (a) A/D converter and a counter
    (b) A/D converted and a rectifier
    (c) D/A converter and a counter
    (d) Ramp generator and counter
    (e) Comparator
    (28) Which of the following electrical equipment cannot convert ac into dc
    (a) diode
    (b) converter
    (c) transformer
    (d) mercury arc rectifier
    (29) Voltage measurement are often taken by using either a voltmeter or
    (a) an ammeter
    (b) an ohmmeter
    (c) an oscillator
    (d) a watt-meter
    (30)The electric device which blocks DC but allows AC is called:
    (a) capacitance
    (b) inductor
    (c) an oscilloscope
    (d) a watt-meter
    (31 ) The range of an ammeter can be extended by using a
    (a) shunt in series
    (b) shunt in parallel
    (c) multiplier in series
    (d) multiplier in parallel
    (32)A device that changes one form of energy to another is called
    (a) rheostat
    (b) oscillator
    (c) transducer
    (d) varicap
    (33) Aquadag is used in CRO to collect
    (a) primary electron
    (b) secondary electron
    (c) both primary and secondary
    (d) none of above
    (34) A vertical amplifier for CRO can be designed for
    (a) only a high gain
    (b) only a broad bandwidth
    (c) a constant gain time bandwidth product
    (d)all of the above
    (35) One of the following is active transducer
    (a) Strain gauge
    (b) Selsyn
    (c) Photovoltic cell
    (d) Photo emissive cell
    (36) The dynamic characteristics of capacitive transducer are similar to those of
    (a) low pass filter
    (b) high pass filter
    (c) band pass filter
    (d) band stop filter
    (37) Thermocouples are
    (a) passive transducers
    (b) active transducers
    (c) both active and passive transducers
    (d) output transducers
    (38) The size of air cored transducers as compare to iron core counter part are
    (a) bigger
    (b) smaller
    (c) same
    (39) From the point of view of safety, the resistance of earthing electrode should be:
    (a) low
    (b) high
    (c) medium
    (d) the value of resistance of electrode does not effect the safety
    (40) In CRT the focusing anode is located
    (a) between pre accelerating and accelerating anodes
    (b) after accelerating anodes
    (c) before pre accelerating anodes
    (d) none of above
    (41) Which transducer converts heat energy into electrical energy
    (a) I. V. D. T.
    (b) thermocouple
    (c) photoconductor
    (d) none of the above
    (42) Which of photoelectric transducer is used for production of electric energy by converting solar energy
    (a) photo emission cell
    (b) photo diode
    (c) photo transistor
    (d) both (b) and (c)
    (43) Which of the following instruments consumes maximum power during measurement?
    (a) induction instruments
    (b) hot wire instruments
    (c) thermocouple instruments
    (d) electrodynamometer instruments
    (44) Which of the following meters has the best accurancy
    (a) moving iron meter
    (b) moving coil meter
    (c) rectifier type meter
    (d) thermocouple meter
    (45) The function of the safety resistor in ohm meter is to
    (a) limit the current in the coil
    (b) increase the voltage drop across the coil
    (c) increase the current in the coil
    (d) protect the battery
    (46) Which of the following instruments is free from hysteresis and eddy current losses?
    (a) M.l. instrument
    (b) electrostatic instrument
    (c) electrodynamometer type instrument
    (d)all of these
    (47) The dielectric loss of a capacitance can be measured by
    (a) Wien bridge
    (b) Owen bridge
    (c) Schering bridge
    (d) Maxwell bridge
    (48) Reed frequency meter is essentially a
    (a) recording system
    (b) deflection measuring system
    (c) vibration measuring system
    (d) oscillatory measuring system
    (49) In measurements made using a Q meter, high impedance elements should preferably be connected in
    (a) star
    (b) delta
    (c) series
    (d) parallel
    (50) A digital voltmeter measures
    (a) peak value
    (b) peak-to-peak value
    (c) rms value
    (d) average value

    Friday, April 15, 2011

    Auto-transformers


    Auto-Transformers





  • An auto-transformer has windings common to primary and secondary. So that the input and output circuits are electrically connected as one continuous winding per phase.




  • In the common part of the winding, the input and output currents are superposed.




  • The principal application of the auto-transformer is in case where voltage ratio is not great.




  • Applications: Boosters,static balancers and induction motor startors.




  • Adavntages:saving in copper, cores and losses




  • As compared to an ordinary 2-winding transformer of same output, an auto-transformer has higher efficiency but smaller size. Moreover, its voltage regulation is also superior.




  • Weight of copper in auto-transformer=(1-K)×Weight of copper in ordinary 2-winding transformer. Here K=V2/V1




  • Saving is increase when K incrasing unity




  • Inductively transferred power=(1-K)×input




  • Conductively transferred power=K×input




  • Rating of auto transformer = K/(K∼1)× Rating of 2-winding transformer





  • Practice questions:
    1.A single-phase transformer has a rating of 15 k VA, 600/120V. It is reconnected as an auto- transformer to supply at 720 V from a 600 V primary source. The maximum load it can supply is
    a. 90 kVA
    b. 18 kVA
    c. 15 kVA
    d. 12 kVA

    2.An auto-transformer having a transformation ratio of 0.8 supplies a load of 10 kW. The power transferred inductively from the primary to the secondary is
    a. 10kW
    b. 8kW
    c. 2kW
    d. Zero

    3.In case of auto-transformers, which of the following statements are correct?
    1. An auto-transformer requires less copper as compared to a conventional, 2-winding transformer of the same capacity.
    2. An auto-transformer provides isolation between the primary and secondary windings.
    3. An auto-transformer has less leakage reactance as compared to the conventional, 2-winding transformer of the same capacity.
    Select the correct answer using .the codes given below:
    a. 1, 2 and 3
    b. 1 and 2
    c. 1 and 3
    d. 2 and 3

    4.In an auto-transformer of voltage ratio V1 / V2 and V1 > V2, the fraction of power transferred inductively is
    a.V1/(V1+V2 )

    b.V2/V1

    c.(V1-V2)/(V1+V2 )

    d.(V1-V2)/V1

    5.A single-phase, 10 kVA, 2000/200 V, 50 Hz transformer is connected to form an auto transformer as shown in the figure given above. What are the values of V1and I2 respectively?



    a. 2200 V, 55 A
    b. 2200 V, 45 A
    c. 2000 V, 45 A
    d. 1800 V, 45 A

    6. What is the efficiency of an autotransformer in comparison to that of a two winding transformer of the same rating?
    a. Slightly less than that of a two-winding transformer
    b. Same as that of a two-winding transformer
    c. More than that of a two-winding transformer
    d. As low as 1/5th of the efficiency of a two-winding transformer

    7. A two-winding transformer is converted into an auto-transformer. If we apply additive polarity and subtractive polarity for the connections, then the secondary voltage is 2640 V and 2160 V, respectively. What is the ratio of primary to secondary voltage on the original transformer?
    a. 66 : 54
    b. 54 : 66
    c. 10 : 1
    d. 1 : 10

    Solved problems
    1.A 5000-VA 480/120-V conventional transformer is to be used to supply power from a 600-V source to a 120-V load. Consider the transformer to be ideal, and assume that all insulation can handle 600 V.
    (a) Sketch the transformer connection that will do the required job.
    (b) Find the kilovoltampere(kVA) rating of the transformer in the configuration.
    (c) Find the maximum primary and secondary currents under these conditions.
    SOLUTION (a) For this configuration, the common winding must be the smaller of the two windings, and NSE=4NCThe transformer connection is shown below:
    (b) The kVA rating of the auto-transformer can be found from the equation
    (c) The maximum primary current for this configuration will be
    and the maximum secondary current is















    Saturday, April 9, 2011

    Semiconductor physics:

    • The charge, or quantity of negative electricity is 1.60*10-19 C.
    • The number of electrons per second represents current, I.
    • The charge of a positive ion is an integral multiple of the charge of the electron.
    • Hole is an effective charge carrier.
    • The force, f on a unit positive charge, q is an electric field is the electric field intensity, at that point (f = q ).
    • The potential V of point B at x with respect to point A at x1 is the work done against the field in taking a unit positive charge from A to B.

    • The electric field equals the negative gradient of the potential , ?= - dV/dx
    • Electron volt (eV) is the unit of work or energy(electric, mechanical, thermal, etc.,)
    • For germanium EG is 0.785 eV, for silicon 1.21 eV.
    • Semiconductors have property of negative temperature coefficient of resistance.
    • A semiconductor in which electrons and holes are solely created by thermal excitation is called a pure or intrinsic semiconductor.
    • In intrinsic semiconductor the number of holes is always equal to the number of electrons.
    • Hole may serve as a carrier of electricity comparable in effectiveness with free electron.
    • If impurities added to pure semiconductor then it is called extrinsic semiconductor. This process is called doping.
    • The current in a conductor is due to flow of electrons, whereas the current in a semiconductor results from the movement of both electrons and holes.
    • The transport of the charge in a crystal under influence of an electric filed result in drift current.
    • Diffusion current is the result of a non-uniform concentration gradient.
    • The drift velocity is proportional to applied electric filed intensity .
      = µ , where µ (square meters per volt second, m2/vs) is called mobility of the electrons.
    • The current desist J is the current per unit area of the conducting medium.
    • The above equations defines Ohm's law, the conduction current is proportional to the applied voltage.
    • Power density , power dissipated within the metal by electrons is J = ? 2
    • If the dopant has five valance electrons those will call donor or n-type impurities ex. Antimony, phosphorus and arsenic.
    • If intrinsic semiconductor doped with n type impurities, the electrons will increase, holes will decrease. This type material is called as n-type semiconductor.
    • If the dopant has three valance electrons those will call acceptor or p-type impurities ex. Boron, gallium and indium.
    • If intrinsic semiconductor doped with p type impurities, the holes will increase, electrons will decrease. This type material is called as p-type semiconductor.
    • The product of free negative and positive concentrations is a constant independent of the amount of donor and acceptor impurity doping. This is called mass-action law.
    • np = ni2
    • Metal is unipolar where as semiconductor is bipolar (two charge carrying particles).
    • If a specimen carrying a current I is placed in a transverse magnetic filed B, an electric filed is induced in the direction perpendicular to both I and B. This phenomenon, known as the Hall Effect.
    • Hall Effect is used to determine type of semiconductor (p or n) and carrier concentration.
    • By Hall Effect can measure mobility.
    • Hall Effect applications: magnetic field meter, Hall Effect multiplier.
    • Thermistor has a negative temperature coefficient of resistance.
    • Silicon and germanium not used as thermistors because their properties are too sensitive to impurities.
    • A heavily doped semiconductor can exhibit a positive temperature coefficient of resistance. Such device called sensistor.
    • If radiation falls upon a semiconductor, its conductivity increases.
    • In Hall Effect the output voltage produced across the crystal is due to movement of charge carriers toward one end.
    • The voltage measured between the two faces is called hall voltage.
    • Hall voltage is zero for intrinsic semiconductor.
    • Hall coefficient depends on the type of material.
    • Hall Effect probes are used for measurement of DC current in a wire.
    • Fermi level is a measure of probability of occupancy of electrons or holes in the allowed energy states.
    • In pure semiconductor the Fermi level is at the middle of the conduction and valence band.
    • In n type semiconductor the Fermi level is just below the conduction band.
    • In p type Fermi level is just above the valance band.
    • In a heavily doped n- type semiconductor the Fermi level is in the conduction band, similarly in a heavily doped p- type semiconductor the Fermi level is in the valance band.
    • Fermi Dirac function f(E) gives the probability that a quantum state with energy E is occupied by an electron.
    • Fermi level EF represents the energy state with 50% probability of being filled if no forbidden band exists.
    • If E >> EF then f(E) = 0, means there is zero probability of finding an occupied quantum state of energy greater than EF.
    • If E << EF then f(E) = 1, means all quantum levels with energy less than EF are occupied.
    • Silicon and germanium are called indirect band gap semiconductor where as gallium arsenide is called direct band gap semiconductor.

    Friday, April 8, 2011

    Diode fundamentals.

    • If a junction formed between a sample of p type and one of n type semiconductor, this combination possesses the properties of a rectifier (diode).
    • At this junction, one side has a high concentration of holes whereas the other side has high concentration of electronics due to this movement of charge carriers takes and this phenomenon is called diffusion.
    • The potential barrier (depletion region) is negative on p side and positive on n side which restricts the recombination of electrons with holes.
    • Within this narrow space charge layer there are no mobile carriers (no holes and no electrons).
    • If we want charge carriers to cross the junction, then an external voltage of appropriate polarity has to be applied in order to overcome the opposition of the barrier potential known as cut-in voltage.
    • In semiconductor both diffusion and conduction currents exist simultaneously.
    • The reverse saturation current depends on temperature and independent of magnitude of reverse bias.
    • Doping inversely proportional to the square root of the potential width.
    • Depletion region opposes flow of majority carrier and assist the flow of minority carriers across the junction (drift currents).
    • Charge density is depends on amount of doping.
    • Pn junction device s bipolar device.
    • Diode characteristics:
    • Volt equivalent of temperature is VT = T/11600 = 26 mV.
    • At high currents, the diode behaves more like a resistor than diode, and the current increases linearly rather than exponentially with applied voltage.
    • The electron in germanium 3 times faster than silicon, for high frequency applications germanium is suitable.
    • The value of reverse saturation current is about double in magnitude for every 10o C rise in temperature.
    • The generation and recombination process occur simultaneously. But generation rate is higher than the rate of recombination in extrinsic semiconductor.
    • In intrinsic semiconductor both rates same.
    • Static resistance R of a diode is defined as the ratio V/I of the voltage to the current. It is not useful parameter because of its wide variation due to V and I.
    • Dynamic resistance of a diode is defined as the slope of the V-I characteristics of the diode.
    • Dynamic resistance inversely proportional to the forward current.
    • Capacitive effect at junction due to uncovered charge with applied voltage is called transition or space charge capacitance, CT.
    • The CT is not constant, but depends upon the magnitude of reverse voltage.
    • The rate of change of injected charge with voltage called diffusion or storage capacitance CD in forward bias.
    • CD is much larger than CT in forward bias.
    • The CD is proportional to the current.

    Transformers_part2



    • Special alloy steel of high resistance and low hysteresis loss is used as transformer core

    • As the flux in the cores is a pulsating one, the magnetic circuit must be lamianted and the separate laminations insulated in order to retain the advantage
      of subdivision

    • Cut the transformer sheets as far as possible along the grain which is the direction in which the material has a higher permeability

    • For large three phase transformer, the core is five-limbed core than , which needs a cross section in the yokes less than that required in usual three limbed core but core losses increase



    • Cross over coils are suitable for currents not exceeding about 20A. They are used for HV windins in comparatively small transformers, and comprise wires of small circular section with double cotton covering

    • Disc coils are made up of a number of flat sections, generally with rectangular wire

    • In Disc coils every turn is contact with oil hence better cooling

    • Sandwich windings , commonly employed for shell-type transformers, allow of easy control over reactance.

    • Conservators are required to take up the expansion and contraction of the oil with changes of temperature in service without allowing the oil to come in contact with air , from which it is liable to take up moisture

    • The displacement of air due to change of oil volume takes place through a breather containing calcium chloride or silica gel, which extracts the moisture from the air.

    • Oil in transfomer serves the double purpose of cooling and insulating.

    • Sludging means slow formation of semi solid hydrocarbons, due to heat and oxidation. It will cause over heating of transformer by blocking the cooling ducts

    • Sludge formation is more in the presence of bright copper surfaces

    • Three phase transformer connections representations:

    • >Group 1: Zero phase displacement--Yy0,Dd0,Dz0
      >group 2: 1800 Phase displacement--Yy6,Dd6,Dz6
      >group 3: 300 lag Phase displacement--Dy1,Yd1,Yz1
      >group 4: 300 lead Phase displacement--Dy11,Yd11,Yz11
    • Star-Star connection is most economical connection for small, high voltages transformers. In this connection third harmonic voltages are absent from the line voltages; unless there is a fourth wire no third harmonic currents will flow. Best suitable for core type construction, for shell type need tertiary winding to stabilize the neutral condition.

    • Delta-Delta connection is an economical connection for large , low voltage transformers.

    • In Delta - Delta, even though one phase faulty the transformer can operate at 58 percent of normal. This connection called Vee(VV) connection.






    Transformers_part1


    • The physical basis of transformer (T/F) is mutual induction between two circuits linked by a common magnetic field. It is a static device.

    • Mutual inductance is the same irrespective of which circuit is primary and which circuit is secondary.

    • The voltage applied to the primary is almost completely concerned in opposing the induced e.m.f.

    • If the primary voltage is constant, the mutual flux remains approximately constant regardless of the load connected across the secondary coil.

    • Important tasks performed by T/F are

    • • Changing voltage and current levels in power system.
      • Matching source and load impedances for maximum power transfer
      • Electrical isolation
    • In core type, to avoid leakage flux, it is usual to have half the primary and half of the secondary winding side-by-side or concentrically on each limb; not primary on one limb and secondary on other limb.




    • The three phase core type, the principle that the sum of the fluxes in each phase in a given direction along the cores is zero.

    • In core type inspection of coils and core is easy.

    • The shell type is more robust mechanically.

    • In shell type core, the cooling is good.

    • The e.m.f induced in transformer given by: E = 4.44fΦmT volts

    • Volt/turn is same in primary winding and secondary winding provided only that it links the same flux in both windings.

    • Both primary and secondary e.m.f.’s are in phase.

    • The e.m.f’s lag by 900 in time on the flux.

    • The applied voltage V1 opposes E1, while E2 provides the secondary output voltage V2.

    • E1/E2 = I2/I1 = T1/T2



    • A small magnetizing current is needed to maintain the magnetic circuit or core in magnetized state, when the secondary is open



    • The m.m.f. of the primary on no load is of the order of 5 per cent of its m.m.f. on full load.

    • The no load current, Io has two components, magnetizing component, Im and loss component, Ir.



    • Leakage between primary and secondary could be eliminated if the winding could be made to occupy the same space.


    • Reductions in leakage flux can achieve by sectionalizing and interleaving the primary and secondary coils.


    • An equivalent circuit is useful for calculations of regulation, efficiency, parallel operation, etc.


    • I2R loss and per-unit reactance voltage in primary and secondary is same.


    • Core loss due to the pulsation of the magnetic flux in the iron producing eddy current and hysteresis loss.

    • I2R loss due to heating of the conductors by the passage of current.

    • Stray loss due to stay magnetic field causing eddy currents in the conductors or in surrounding metal (tank).

    • Dielectric loss in the insulating materials (oil and solid insulation of HT T/F).

    • Efficiency of transformer is given by:


    • Where S is full load kVA, x is per-unit load, Pi is iron loss, Pc is full load copper loss.
    • For maximum efficiency occurs when the variable loss is equal to the constant loss, i.e. x2 Pc= Pi.

    • Maximum efficiency occurs below full load.


    • Maximum efficiency point independent of power factor.

    • The regulation of a transformer refers to the change of secondary terminal voltage between no-load and load conditions: it is usually quoted as a percent value for full load at given power factor.

    • Per unit regulation:

    • Maximum regulation occurs when

    • Zero regulation occurs when
      i.e. leading p.f.

    • On account of the easier insulation facilities, the low-voltage winding is placed nearer to the core in the case of core type and on the outside positions in the case of shell type transformers.

    • Under no-load conditions, the “hum” developed by energized power transformer originates in the core, where the laminations tend to vibrate by magnetic forces.

    • The essential factors in noise production are magnetostriction, mechanical vibrations by the laminations.