II/IV B.Tech(Regular) DEGREE EXAMINATION. Electronics & Instrumentation Engineering

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1 SCHME OF EVALUTION II/IV B.Tech(Regular) DEGREE EXAMINATION JUNE,2016 EI ET 403 Electrical Technology Electronics & Instrumentation Engineering Max.Marks :60 marks Answer all questions. Answer one question from each unit. (1 12=12Marks) (4 12=48Marks) (a) Laminated the armature core to reduces the eddy current losses in DC machines. (b) convert alternating emf into unidirectional emf and vice varsa (c) Variable losses = Constant losses (d) Uniform Magnetic flux pass through primary winding to secondary. (e) Capacitive load leads to negative regulation in transformer. (f) Wattmeter reads only iron losses in open circuit test variable losses are negelected.i.e why wattmeter reds in oc test only core losses. (g) With increase in rotor resistance. Hence rotor flux will reduce. (h) 3-Phase supply phase sequence is changed for example RBY. (i) Tape recorders hysteresis motors used,tele printers stepper motors (j) Two types 1.sailent pole rotor 2. Non-sailent pole rotor (k) applications of stepper motors convers,tape readers,floppys,disk drives,cnc lathes,paper feeder on printers (l) p.f correction, higher effiency compare to 3-phase induction motors,compressors,paper mills centrifigul pumps,blowers line shaft etc UNIT-I 2 (a) EMF EQUATION OF DC GENERATOR [6M]

EMF Equation formula=1m Derivation =5M 2(b) [6M] Calculation of Emf=3M Calculation of Flux=3M OR 3 (a) When a motor is connected to the supply,heavy current flows througharmature.

2 EMF Equation formula=1m Derivation =5M 2(b) [6M] Calculation of Emf=3M Calculation of Flux=3M OR 3 (a) When a motor is connected to the supply,heavy current flows througharmature.since there is no back emf at the time of starting,this may damage thearmature. Hence,a starter is necessity which allows less armature current at the time of starting and gradually allows the rated current to the armature. Three Point Starter: Construction:

3 Working Of Three Point Starter: 1) Initially when a DC supply is switched ON with handle in the OFF position. 2) The handle is now moved clockwise to the first stud, the shunt field winding is directly connected across the supply while the whole starting resistance is inserted in series with the armature circuit. 3) As the handle is gradually moved over to the final stud, the starting resistance is cut out of the armature circuit in steps. The handle is now held magnetically by the no volt release coil which is energized by shunt field current. 4) If the supply voltage is suddenly interrupted or if the field excitation is accidently cut, the no volt release coil is demagnetized and the handle goes back to the OFF position under the pull of the spring. If no volt coil were not used, then in case of failure of supply. The handle would remain on the final stud. If then supply is restored, the motor will be directly connected across the supply, resulting in an excessive armature current. Limitations Of The Three Point Starter: Unnecessary tripping of starter if a field regulator is connected in series with the field windings for speed control operation. [7M] 3-Point starter diagram =4M Explanation =3M 3 (b) Torque Equation of a D.C. Motor [5M] The turning or twisting force about an axis is called torque. Consider a wheel of radius R meters acted upon by a circumferential force F newtons as shown in the Fig.

4 The wheel is rotating at a speed of N r.p.m. Then angular speed of the wheel is, ω = (2πN)/60 rad/sec So workdone in one revolution is, W = F x distance travelled in one revolution = F x 2 R joules And P = Power developed = Workdone/Time = (F x 2πR) / (Time for 1 rev) = (F x 2πR) / (60/N) = (F x R) x (2πN/60)... P = T x ω watts Where T = Torque in N - m ω = Angular speed in rad/sec. Let T a be the gross torque developed by the armature of the motor. It is also called armature torque. The gross mechanical power developed in the armature is E b I a, as seen from the power equation. So if speed of the motor is N r.p.m. then, Power in armature = Armature torque x ω... E b I a = x (2N/60) but E b in a motor is given by, E b = (ΦPNZ) / (60A)... (ΦPNZ / 60A) x Ia = T a x (2πN/60) Torque equation formula=1m Derivation =4M UNIT-II 4 (a) Losses In Transformer [4M] (I) Core Losses Or Iron Losses Eddy current loss and hysteresis loss depend upon the magnetic properties of the material used for the construction of core. Hence these losses are also known as core losses or iron losses. Hysteresis loss in transformer: Hysteresis loss is due to reversal of magnetization in the transformer core. This loss depends upon the volume and grade of the iron, frequency of magnetic reversals and value of flux density. It can be given by, Steinmetz formula: W h = ηb 1.6 max fv (watts)

where, η = Steinmetz hysteresis constant V = volume of the core in m 3 Eddy current loss in transformer: In transformer, AC current is supplied to the primary winding which sets up alternating

5 where, η = Steinmetz hysteresis constant V = volume of the core in m 3 Eddy current loss in transformer: In transformer, AC current is supplied to the primary winding which sets up alternating magnetizing flux. When this flux links with secondary winding, it produces induced emf in it. But some part of this flux also gets linked with other conducting parts like steel core or iron body or the transformer, which will result in induced emf in those parts, causing small circulating current in them. This current is called as eddy current. Due to these eddy currents, some energy will be dissipated in the form of heat. (Ii) Copper Loss In Transformer Copper loss is due to ohmic resistance of the transformer windings. Copper loss for the primary winding is I 2 1 R 1 and for secondary winding is I 2 2 R 2. Where, I 1 and I 2 are current in primary and secondary winding respectively, R 1 and R 2 are the resistances of primary and secondary winding respectively. It is clear that Cu loss is proportional to square of the current, and current depends on the load. Hence copper loss in transformer varies with the load. Classification=2M Explanation=2 M 4 (b) Given data:- [8M]

So only changes in drawing the phasor diagram is to draw I 2 lagging V 2 by Φ 2 in step 5 discussed earlier.

6 OR 5 (a) Lagging Power Factor Load, cos Φ 2 [6M] As load power factor is lagging cosφ 2, the current I 2 lags V 2 by angle Φ 2. So only changes in drawing the phasor diagram is to draw I 2 lagging V 2 by Φ 2 in step 5 discussed earlier. Accordingly direction of I 2 R 2, I 2 X 2, I 2 ', I 1, I 1 R 1 and I 1 X 1 will change. Remember that whatever may be the power factor of load, I 2 X 2 leads I 2 by 90 o and I 1 X 1 leads I 1 by 90 o.

7 Explanation phasor =3M Phasor diagram =3M 5(b) [6M] UNIT-III 6 (a) Basic Working Principle Of an Induction Motor:- [5M] In a DC motor, supply is needed to be given for the stator winding as well as the rotor winding. But in an induction motor only the stator winding is fed with an AC supply.

Alternating flux is produced around the stator winding due to AC supply. This alternating flux revolves with synchronous speed. The revolving flux is called as "Rotating Magnetic Field" (RMF).

8 Alternating flux is produced around the stator winding due to AC supply. This alternating flux revolves with synchronous speed. The revolving flux is called as "Rotating Magnetic Field" (RMF). The relative speed between stator RMF and rotor conductors causes an induced emf in the rotor conductors, according to the Faraday's law of electromagnetic induction. The rotor conductors are short circuited, and hence rotor current is produced due to induced emf. That is why such motors are called as induction motors. Now, induced current in rotor will also produce alternating flux around it. This rotor flux lags behind the stator flux. The direction of induced rotor current, according to Lenz's law, is such that it will tend to oppose the cause of its production. As the cause of production of rotor current is the relative velocity between rotating stator flux and the rotor, the rotor will try to catch up with the stator RMF. Thus the rotor rotates in the same direction as that of stator flux to minimize the relative velocity. However, the rotor never succeeds in catching up the synchronous speed. Explanation of 3-phase induction motor = 3M Constructons,operating principle =3M 6 (b) [7M]

OR 7(a) Torque Equation of Three Phase Induction Motor [5M] The torque produced by three phase induction motor depends upon the following three factors: Firstly the magnitude of rotor current,

9 OR 7(a) Torque Equation of Three Phase Induction Motor [5M] The torque produced by three phase induction motor depends upon the following three factors: Firstly the magnitude of rotor current, secondly the flux which interact with the rotor of three phase induction motor and is responsible for producing emf in the rotor part of induction motor, lastly the power factor of rotor of the three phase induction motor. Combining all these factors together we get the equation of torque as- Where, T is the torque produced by induction motor, φ is flux responsible for producing induced emf, I 2 is rotor current, cosθ 2 is the power factor of rotor circuit.the flux φ produced by the stator is proportional to stator emf E 1. i.e φ E 1 We know that transformation ratio K is defined as the ratio of secondary voltage (rotor voltage) to that of primary voltage (stator voltage). Rotor current I 2 is defined as the ratio of rotor induced emf under running condition, se 2 to total impedance, Z 2 of rotor side, and total impedance Z 2 on rotor side is given by, Putting this value in above equation we get, s= slip of Induction motor We know that power factor is defined as ratio of resistance to that of impedance. The power factor of the rotor circuit is Putting the value of flux φ, rotor current I 2, power factor cosθ 2 in the equation of torque we get, Combining similar term we get, Removing proportionality constant we get,

10 Where n s is synchronous speed in r. p. s, n s = N s / 60. So, finally the equation of torque becomes, Derivation of K in torque equation. In case of three phase induction motor, there occur copper losses in rotor. These rotor copper losses are expressed as P c = 3I 2 2 R 2 We know that rotor current, Substitute this value of I 2 in the equation of rotor copper losses, P c. So, we get The ratio of P 2 : P c : P m = 1 : s : (1 - s) Where, P 2 is the rotor input, P c is the rotor copper losses, P m is the mechanical power developed. Substitute the value of Pc in above equation we get, On simplifying we get, The mechanical power developed P m = Tω, Substituting the value of P m We know that the rotor speed N = N s (1 - s) Substituting this value of rotor speed in above equation we get, N s is speed in revolution per minute (rpm) and n s is speed in revolution per sec (rps) and the relation between the two is Substitute this value of N s in above equation and simplifying it we get

Comparing both the equations, we get, constant K = 3 / 2πn s Equation of Starting Torque of Three Phase Induction Motor Starting torque is the torque produced by induction motor when it is started.

11 Comparing both the equations, we get, constant K = 3 / 2πn s Equation of Starting Torque of Three Phase Induction Motor Starting torque is the torque produced by induction motor when it is started. We know that at start the rotor speed, N is zero. So, the equation of starting torque is easily obtained by simply putting the value of s = 1 in the equation of torque of the three phase induction motor, torque is also known as standstill torque. The starting Derivation of torque equation=4m Explain torque characteristics=3m 7 (b) The various Starting methods of a Single Phase Induction motor are as follows:- Split Phase Motor Capacitor Start Motor Capacitor Start Capacitor Run Motor or Two value capacitor motor Permanent Split Capacitor (PSC) or Single value capacitor motor Shaded Pole Motor. Split Phase or Resistance Start [8M] This method is majorly employed in simple industrial duty motors. These motors consist of two sets of windings, namely, start winding and main or run winding. The start winding is made from smaller wire with which it offers high resistance to electrical flow compared to run winding. Due to this high resistance, magnetic field is developed in start winding by the current earlier than run winding magnetic field development. Thus, two fields are 30 degrees apart, but this small angle itself is enough to start the motor.

12 UNIT-IV 8 (a) E.M.F. EQUATION OF AN ALTERNATOR OR AC GENERATOR Let, P= No. of poles Z= No. of Conductors or Coil sides in series/phase i.e. Z= 2T Where T is the number of coils or turns per phase (Note that one turn or coil has two ends or sides) f = frequency of induced e.m.f in Hz ф = Flux per pole (Weber) N = rotor speed (RPM) K d = Distribution factor K c or K P = Cos α/2 If induced e.m.f is assumed sinusoidal then, K f = Form factor = 1.11 In one revolution of the rotor i.e. in 60/N seconds, each conductor is cut by a flux of Pф Webers. dф= фp and also dt= secionds60/n then induced e.m.f per conductor ( average) = dф/ dt = Pф/(60/N) =P N ф/60..(a) But We know that f = PN/120 or N= 120f/P Putting the value of N in Equation (a) We get the average value of e.m.f per conductor = Pф/60 x 120 f/p = 2f ф Volts. à {N= 120f/P} If there are Z conductors in series per phase, then average e.m.f per phase = 2fфZ Volts= 4fфT Volts.{Z=2T} Also we know that Form factor= RMS Value/Average Value = RMS value= Form factor x Average Value, = 1.11 x 4fфT = 4.44fфT Volts. ( Note that is exactly the same equation as the e.m.f equation of the transformer) And the actually available voltage per phase = 4 K c K d fфt =4 K f K c K d fфt Volts EMF Equation of Alternator=1M Derivation of Alternator =4M 8 (b) [7M]

13 OR 9 (a) Consider the rotating magnetic field as equivalent to physical rotation of two stator poles N 1 and S 1. [6M] Consider an instant when two poles are at such a position where stator magnetic axis is vertical, along A-B as shown in the Fig. 1(a). At this instant, rotor poles are arbitrarily positioned as shown in the Fig. 1. At this instant, rotor is stationary and unlike poles will try to attract each other. Due to this rotor will be subjected to an instantaneous torque in anticlockwise direction as shown in the Fig. 1(a). Now stator poles are rotating very fast i.e. at a speed N s r.p.m. Due to inertia, before rotor hardly rotates in the direction of anticlockwise torque, to which it is subjected, the stator poles change their positions. Consider an instant half a period latter where stator poles are exactly reversed but due to inertia rotor is unable to rotate from its initial position.this is shown in the Fig.1(b). At this instant, due to the unlike poles trying to attract each other, the rotor will be subjected to a torque in clockwise direction. This will tend to rotate rotor in the direction of rotating magnetic field. But before this happen, stator poles again change their position reversing the direction of the torque exerted on the rotor. Note : As a result, the average torque exerted on the rotor is zero. And hence the synchronous motor is not self starting. Explanation of synchronous motor = 3M Why synchronous motor not self starting =3M 9 (b) Synchronous Impedance Method or E.M.F. Method [6M] The method is also called E.M.F. method of determining the regulation. The method requires following data to calculate the regulation. 1. The armature resistance per phase (R a ).

14 2. Open circuit characteristics which is the graph of open circuit voltage against the field current. This is possible by conducting open circuit test on the alternator. 3. Short circuit characteristics which is the graph of short circuit current against field current. This is possible by conducting short circuit test on the alternator. Let us see, the circuit diagram to perform open circuit as well as short circuit test on the alternator. The alternator is coupled to a prime mover capable of driving the alternator at its synchronous speed. The armature is connected to the terminals of a switch. The other terminals of the switch are short circuited through an ammeter. The voltmeter is connected across the lines to measure the open circuit voltage of the alternator. The field winding is connected to a suitable d.c. supply with rheostat connected in series. The field excitation i.e. field current can be varied with the help of this rheostat. The circuit diagram is shown in the Fig. 1. Fig. 1 Circuit diagram for open circuit and short circuit test on alternator 1.1 Open Circuit Test Procedure to conduct this test is as follows : i) Start the prime mover and adjust the speed to the synchronous speed of the alternator. ii) Keeping rheostat in the field circuit maximum, switch on the d.c. supply. iii) The T.P.S.T switch in the armature circuit is kept open. iv) With the help of rheostat, field current is varied from its minimum value to the rated value. Due to this, flux increasing the induced e.m.f. Hence voltmeter reading, which is measuring line value of open circuit voltage increases. For various values of field current, voltmeter readings are observed. The observation for open circuit test are tabulated as below : Observation table for open circuit test : From the above table, graph of (V oc ) ph against I f is plotted. 1.2 Short Circuit Test After completing the open circuit test observation, the field rheostat is brought to maximum position, reducing field current to a minimum value. The T.P.S.T switch is closed. As ammeter has negligible resistance, the armature gets short circuited. Then the field excitation is gradually increased till full load current is obtained through armature winding. This can be observed on the ammeter connected in the armature circuit. The

graph of short circuit armature current against field current is plotted from the observation table of short circuit test. This graph is called short circuit characteristics, S.C.

15 graph of short circuit armature current against field current is plotted from the observation table of short circuit test. This graph is called short circuit characteristics, S.C.C. This is also shown in the Fig. 2.Observation table for short circuit test : The S.C.C. is a straight line graph passing through the origin while O.C.C. resembles B-H curve of a magnetic material. Note : As S.C.C. is straight line graph, only one reading corresponding to full load armature current along with the origin is sufficient to draw the straight line. 1.3 Determination of From O.C.C. and S.C.C. The synchronous impedance of the alternator changes as load condition changes. O.C.C. and S.C.C. can be used to determine Z s for any load and load p.f. conditions. In short circuit test, external load impedance is zero. The short circuit armature current is circulated against the impedance of the armature winding which is Z s. The voltage responsible for driving this short circuit current is internally induced e.m.f. This can be shown in the equivalent circuit drawn in the Fig. 3. From the equivalent circuit we can write, Z s = E ph / I asc Now value of I asc is known, which can observed on the alternator. But internally induced e.m.f. can not be observed under short circuit condition. The voltmeter connected will read zero which is voltage across short circuit. To determine Z s it is necessary to determine value of E which is driving I asc against Z s. Now internally induced e.m.f. is proportional to the flux i.e. field current I f. E ph α Φ α I f... from e.m.f. equation So if the terminal of the alternator are opened without disturbing I f which was present at the time of short circuited condition, internally induced e.m.f. will remain same as E ph. But now current will be zero. Under this condition equivalent circuit will become as shown in the Fig. 4. It is clear now from the equivalent circuit that as I a = 0 the voltmeter reading (V oc ) ph will be equal to internally induced e.m.f. (E ph ).

16 This is what we are interested in obtaining to calculate value of Z s. So expression for Z s can be modified as, So O.C.C. and S.C.C. can be effectively to calculate Z s. The value of Z s is different for different values of I f as the graph of O.C.C. is non linear in nature. So suppose Z s at full load is required then, I asc = full load current. at full load General steps to determine Z s at any load condition are : i) Determine the value of (I asc ) ph for corresponding load condition. This can be determined from known full load current of the alternator. For half load, it is half of the full load value and so on. ii) S.C.C. gives relation between (I asc ) ph and I f. So for (I asc ) ph required, determine the corresponding value of I f from S.C.C. iii) Now for this same value of I f, extend the line on O.C.C. to get the value of (V oc ) ph. This is (V oc ) ph for same I f, required to drive the selected (I asc ) ph. iv) The ratio of (V oc ) ph and (I asc ) ph, for the same excitation gives the value of Z s at any load conditions. The graph of synchronous impedance against excitation current is also shown in the Fig Regulation Calculations From O.C.C. and S.C.C., Z s can be determined for any load condition. The armature resistance per phase (R a ) can be measured by different methods. One of the method is applying d.c. known voltage across the two terminals and measuring current. So value of R a per phase is known. So synchronous reactance per phase can be determined. No load induced e.m.f. per phase, E ph can be determined by the mathematical expression derived earlier.

17 where V ph = Phase value of rated voltage I a = Phase value of current depending on the load condition cosφ = p.f. of load Positive sign for lagging power factor while negative sign for leading power factor, R a and X s values are known from the various tests performed. The regulation then can be determined by using formula, Explanation and methods of synchronous motor of EMF Method=3M Diagrams synchronous motor of EMF Method =3M

2014 ELECTRICAL TECHNOLOGY

SET - 1 II B. Tech I Semester Regular Examinations, March 2014 ELECTRICAL TECHNOLOGY (Com. to ECE, EIE, BME) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~~~~~