PESIT Bangalore South Campus Hosur road, 1km before Electronic City, Bengaluru -100 Department of Electronics & Communication Engineering

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1 INTERNAL ASSESSMENT TEST 3 Date : 15/11/16 Marks: 0 Subject & Code: BASIC ELECTRICAL ENGINEERING -15ELE15 Sec : F,G,H,I,J,K Name of faculty : Mrs.Hema, Mrs.Dhanashree, Mr Nagendra, Mr.Prashanth Time : AM Note: Answer FIVE full questions, selecting any ONE full question from each part. PART 1 1 Derive with the help of vector diagram that two watt meters are sufficient to measure power in 3phase AC. Also derive the equation to find power factor. Draw nece ssary vector diagram Derive with the help of vector diagram the relationship between line and phase values of voltages in 3 phase system when the load is STAR connected. PART 3 A 3 phase delta connected load consumes a power of 100KW taking a lagging current of A at a line voltage of 00 V, 50 Hz. Find the parameters of each phase. What would be the power consumed if the load were connected in star. Each of two wattmeters connected to measure the input to a 3 phase circuit reads KW on a balanced load when the power factor is unity. What does each instrument read when the power factor falls to 0.866lag,the total 3 phase power remain unchanged. PART 3 5 A 6 pole, 3 phase alternator has 1 slots /pole and conductors /slot. The winding is 08 5/6 th of full pitch. A flux of 5 mwb is sinusoidally distributed along the air gap. Determine the line emf if alt is star connected 6 With neat diagrams explain the construction of a Salient pole Synchronous Generator. 08 PART 7 What is rotating magnetic field in 3 phase induction motor also explain the working 08 principle of it. 8 Derive EMF equation of Synchronous Generator. Write the assumptions made while 08 arriving at final emf equation. PART 5 9 a. Differentiate between shell type and core type transformer. 0 b. A 1 pole 3 phase alternator is coupled to an engine running at 500 RPM. It supplies an 0 induction motor which has full load speed of 10 RPM. Find % slip and number of poles of the motor. 10 a. Derive EMF equation of transformer. 0 Marks b A 6 pole, 3 phase, 50 Hz alternator has 1 slots/pole & conductors/slot. A flux of 5mw is sinusoidally distributed in air gap. Determine the line emf if the alternator is star connected. Assume full pitch winding & distribution factor of

2 SCHEME AND SOLUTION INTERNAL TEST-III Subject & Code : Basic Electrical Engineering, 15ELE15 Semester: I Section: F,G,H,I,J,K Name of faculty: Dhanashree Bhate Q. NO Marks Part Derive with the help of vector diagram that two watt meters are sufficient to measure power in 3 phase AC. Also derive the equation to find power factor. Draw necessary vector diagram. A cc1 W1 Ia LOAD pc1 Z B Ib N Z C pc Z cc Ic W Consider the circuit above where the load is star connected. The wattmeter W1 & W are connected to measure the power. Assume that Vab = VL 0 0, Vbc = VL -10 0, Vca = VL -0 0, The phase voltages Van =Vph -30 0, Vbn =Vph , Vcn =Vph We assume Z 0, we assume inductive load. Current lags by The reading of 1 st wattmeter W1 = current through cc1 X voltage across pc 1 X cosine of angle between current & voltage W1 = Vab X Ia cos (30+ )----- from vector diagram W1 = VL I L cos (30+ )-----eqn 1

3 The reading of nd wattmeter W = current through cc X voltage across pc X cosine of angle between current & voltage W = Vca X I c cos (30- )----- from vector diagram W = VL I L cos (30+ )-----eqn Ref vector diagram W1 + W = VL I L cos (30+ ) + VL I L cos (30- ) After solving above equation W1+ W = 3 VL I IL cos ( ) = power consumed in 3 phase AC Vca Van Ia Ic Vab Vcn Vbc Ib Vbn

4 . With the help of vector diagram, derive the relationship between line and phase values of voltages in 3 phase system when the load is STAR connected. A Van Vcn C B Vbn Vab Vca Vbc. Let Van = Vph 90 0 Vbn = Vph Vcn = Vph These are called phase voltage. Vab,Vbc,Vca are called line voltages Applying KVL to the loop ABNA we get Vab Van -Vbn ref Vcn Vbn From parallelogram law VL = Vph + Vph + Vph Vph cos (60 0 ) VL = 3 Vph We observe that the Vab is at 10 0 Vab = VL 10 0 Vbc = Vph 0 0 Vca = Vph 0 0 From circuit diagram it is clear that the line current is equal to phase current. IL = Iph

5 3. Part - A 3 phase delta connected load consumes a power of 100KW taking a lagging current of 00 A at a line voltage of 00 V, 50 Hz. Find the parameters of each phase. What would be the power consumed if the load were connected in star. Delta connected load Since the current is lagging each phase consist of R & L P = 3 VL I L cos 100 (1000) = 3 00( 00 ) cos cos = 0.71 = 3.8 In delta connection Iph = IL / 3 = 00/ 3 = A Z= 00/ = R=.5 XL =.397 Therefore L = 7.63 mh Iph = Vph / Z Iph = 00 3 / 3.6 = A P = 3 VL I L cos P= 3 (00) (66.67) cos( 3.8) P= 33.33KW. Each of two wattmeters connected to measure the input to a 3 phase circuit reads 10 KW on a balanced load when the power factor is unity. What does each instrument read when the power factor falls to 0.866lag,the total 3 phase power remain unchanged. W1 = W =10 KW Total power = 0 KW When power falls to 0.866, it is given that the total power is same. W1 + W = 0 KW Cos = = 30 0 tan = 3 ( W1-W ) / (W1+W) = 3 ( W1-W ) / 0 W1 W = 6.66KW W1+W = 0 KW W1 = 6.66KW W1 = KW W = = 6.66 KW

6 5. Part -3 A 6 pole, 3 phase alternator has 1 slots /pole and conductors /slot. The winding is 5/6 th of full pitch. A flux of 5 mwb is sinusoidally distributed along the air gap. Determine the line emf if alt is star connected 1 slots/pole & conductors /slot Therefore total conductors = 1( 6) () T= 1(6)()/6 T= 8 8 Winding = 180 (5)/6 = = = 30 0 = / no.of slots /pole = / 1 = 15 0 Kp = cos ( /) = cos( 30/) = cos (15) = Kd = / Kd = sin ( (15)/) / sin( 15/) = Erms /phase =. f T kp kd =. ( 5 X10-3 )( 50) (8)( )( ) = 6.5 V Eline = 3 ( 6.5) = 6.87 V

7 6. With neat diagrams explain the construction of salient pole Synchronous Generator In D.C. machines the field system (the pole) are stationary (stator) and the armature is rotating (rotor). In large commercial alternators the armature is designed to be stationary and the field system to be rotating for the following reasons. Large commercial alternators have rated voltage of 11kV or 16 kv. At this high voltages, if the armature is made the rotor The insulation will be subjected to high mechanical stresses in addition to normal electric stress. Also at these high voltages, it will be difficult to take out power from a rotating armature through slip ring and brushes. Stator: Stator is an assembly of silicon steel laminations attached to the yoke. In the slots of laminations the conductors are placed with proper insulation. Theses conductors are properly interconnected to form a balanced 3 phase star connected winding. Figure: Rotor: There are two types of rotors (i) Salient Pole Rotor or Projecting pole type of rotor (ii) Non salient type of Rotor Salient Pole Rotor or Projecting pole type of rotor In this type of rotor the poles project out of the rotor body. The poles are made up of silicon steel laminations and are fixed to the rotor body by dovetail joints. The field coils are wound over the poles. These coils are interconnected to form the field winding. Salient pole rotors are characterised by large diameters and short axial length. These kind of rotors are used in low and medium speed alternators. Diagram

8 7. Part - What is rotating magnetic field in three phase induction motor also explain the working of it. Three phase induction motor are the most widely used ac motors. Theirs advantages are simple and rugged construction, easy maintenance, good efficiency and low cost. Production of rotating magnetic field When a 3 phase balanced voltage is applied to a 3 phase balanced winding ( 3 three identical windings whose axes are physically 10 electrical degrees apart), three currents flow in the three windings. Hence three magnetic fields are produced. The resultant of three magnetic fields is a single magnetic field that s magnitude is constant but the field changes its direction. The resultant field is called Rotating Magnetic Field. The speed of rotation is called Synchronous speed Ns and is given by Ns = 10 f/p Where f is the frequency of applied voltage in Hz and P is number of poles for which the winding is wound. Principle of operation of 3 phase induction motor The 3 phase induction motor has a stator and a rotor. The stator contains a balanced 3 phase winding. To this winding a 3 phase balanced voltage is applied. This produces a rotating magnetic field, rotating at a synchronous speed Ns. This is called stator field. The rotor has short circuited copper bars or windings. Initially the rotor is at rest, hence there is a speed difference or relative speed of Ns-0 between the stator field and rotor. This induces an emf called rotor induced emf in the rotor winding. Since the rotor windings are shorted a current called rotor current flows in the rotor conductors. This means that current carrying rotor conductors are kept in the stator field. Therefore, the rotor conductors experience a force and torque is produced. The rotor starts rotating in the same direction as that of the field. (The direction of rotation is given by Lenz's law which states that the result opposes the cause. The cause is the relative speed and the result is rotation. To oppose the cause, the rotor should rotate in the same direction as that of the stator field, thus reducing the relative speed). However the rotor cannot catch up with the speed of stator field because in that case relative speed between the stator field and rotor is zero. Therefore, no rotor induced emf, no rotor current and no torque.

9 8. Derive EMF equation of Synchronous Generator. Write the assumptions made while arriving at final emf equation. Let N be the number of poles. Let be the flux per pole. The rotor rotates with the speed of N rpm. Z be the total number of conductors in the stator per phase. T is turns per phase Z= T/ When the rotor completes one revolution each conductor cuts P amount of flux. Average value of emf induced = flux cut / time taken Eavg / conductor = Eavg /phase = Z Eavg/phase = P Z N / 60 But f = PN /10 Eavg/ phase = f Z Eavg /phase = f T But Z = T Erms /phase =. f T Assumptions 1. We assume that the flux distributed in the air gap is sinusoidal in nature. So we multiply with 1.11 which is form factor for the sine wave.. Pole pitch is the distance between two similar points on the adjacent poles and it is defined to be electrical. Coil pitch or coil span is the distance between the two adjacent sides of a coil. If the armature winding is so wound that the coil pitch equal to the pole pitch then it is called as full pitched winding( ref diagram in class notes). But for practical reasons we make the coil span less than the pole pitch by an angle where this angle is called chording angle. Then the winding is called short pitch winding. Due to this the emf induced is reduced by a factor Kp the pitch factor. Kp = 3. Under the influence of each pole Z/P conductors belong to one phase. All these Z/P conductors cannot accommodated in one armature slot and we have to distribute them over two or more slots. This again reduces the induced emf by a factor Kd. Kd = / Where m = no of slots per pole per phase = 180/ umber of slots /pole With the following assumptions the emf equation we can arrive at Erms /phase =. f T kp kd

9.a Part -5 Differentiate between the core type and shell type transformer Core type of transformer: In core type a large part of core is surrounded by the windings.

10 9.a Part -5 Differentiate between the core type and shell type transformer Core type of transformer: In core type a large part of core is surrounded by the windings. Primary and secondary windings are wound on opposite limbs The coils used are of cylindrical type and are form wound in Helical layers. In general coils may be circular or oval or rectangular. Usually half of the primary and half the secondary windings are placed side by side on each limb so as to reduce the leakage flux. The layers are insulated from each other Shell type of transformer In shell type both the primary and secondary winding are placed on the central limb. The coils are wound to form layer.

11 9.b A 1 pole 3 phase alternator is coupled to an engine running at 500 RPM. It supplies an induction motor which has full load speed of 10 RPM. Find %slip and number of poles of the motor. P=1 N=500 RPM F= PN /10 = 1 X 500 / 10 = 50 Hz Nm= 10 RPM P = is the only possibility to have better slip in IM Therefore synchronous speed = 10 f/p = 10X 50 / = 1500 RPM 10 a Derive emf equation of a transformer. N 1 = No. of turns in primary N = No. of turns in secondary m = Maximum flux in core in webers = B m x A f = Frequency of a.c. input in Hz. Since is due to AC supply = m Sin wt The flux increases from it's zero value to maximum value m in one quarter of the cycle i.e. in 1/ f second. Therefore, r.m.s value of e.m.f./turn =. m volts Now, r.m.s value of induced e.m.f in the whole primary winding = ( induced e.m.f. / turn ) x No. of primary winding E 1 =. f N 1 m ( i ) Similarly, r.m.s. value of e.m.f. induced in secondary is, E =. f N m ( ii )

12 10 b. A 6 pole, 3 phase, 50 Hz alternator has 1 slots/pole & conductors/slot. A flux of 5mwb is sinusoidally distributed in air gap. Determine the line emf if the alternator is star connected. Assume full pitch winding & distribution factor of P =6, f = 50 Hz Z = 1 slots X 6 X = 88 = 5 mwb T= 88/6 = 8 E rms /phase =. f T kp Kd =.( 5 /1000) 50 ( 8) 1(0.957) = 5.95 V El = 1.57 V.

3. What is hysteresis loss? Also mention a method to minimize the loss. (N-11, N-12)

3. What is hysteresis loss? Also mention a method to minimize the loss. (N-11, N-12) DHANALAKSHMI COLLEGE OF ENGINEERING, CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING EE 6401 ELECTRICAL MACHINES I UNIT I : MAGNETIC CIRCUITS AND MAGNETIC MATERIALS Part A (2 Marks) 1. List