Lecture 2 DC MACHINE CONSTRUCTION ARMATURE WINDING o

Lecture # 2

DC MACHINE CONSTRUCTION - ARMATURE WINDING o The windings of armatures are connected in different ways (Lap winding and wave winding ) depending on the requirements of the machine. o The difference in how the coils are interconnected at the commutator segments and the number of segments used. 19

DC MACHINE CONSTRUCTION - ARMATURE WINDING Lap Wound o Lap Wound Armatures: are used in machines designed for low voltage and high current armatures are constructed with large wire because of high current. o Example: It is used in the starter motor of almost all automobiles. o The windings of a lap wound armature are connected in parallel. o This permits the current capacity of each winding to be added and provides a higher operating current. o No. of parallel current paths, a = P ; P = No. of poles (simplex lap winding ) o No. of parallel current paths, a = 2 P ; P = No. of poles (duplex lap winding ) o In general, a = P plex Lap wound armatures 20

DC MACHINE CONSTRUCTION - ARMATURE WINDING Wave Wound o Wave Wound Armatures are used in machines designed for high voltage and low current. o Their windings connected in series. o When the windings are connected in series, the voltage of each winding adds, but the current capacity remains the same. o Example: small generator in hand-cranked mega-ohmmeters. Wave wound armatures

DC MACHINE CONSTRUCTION - ARMATURE WINDING Wave Wound o For a simplex wave winding there will always be two parallel paths regardless of the number of poles (a=2). o A duplex wave winding will double the number of parallel paths present in the simplex (a = 2 2). o In general a = 2 plex o The number of coils on an armature is not important. o What is important is the total number of conductors (Z) on the armature and the number of parallel paths. o The number of conductors per path is the total number of armature conductors divided by 2.

PRINCIPLES OF VOLTAGE AND TORQUE GENERATION Voltage induced in a coil: Recall Eg = ZPϕω 2π = NPϕω π In order to include the effect of the number of parallel paths (a), the above equation is modified as follows ZPϕω NPϕω Eg = = 2πa πa Or Eg = ka ϕω ; where ZP NP ka = = 2πa πa where ka is the armature constant, N is the number of coil turns, and Z is the total number of conductors. (Z = 2 N)

MACHINE MAIN WINDINGS OVERVIEW Winding Armature Wave a = 2 plex Lap a = P plex Field Permanent Magnet Self excited Compound Differential Compound Shunt Separately Excited Series Commutative Compound

IDENTIFYING WINDINGS o The windings of a dc motor can be identified with an ohmmeter. o The shunt field winding can be identified by the fact that it has a high resistance as compared to the other two windings. o The series field and armature windings have a very low resistance. They can be identified, however, by turning the motor shaft. o When the ohmmeter is connected to the series field and the motor shaft is turned, the ohmmeter reading will not be affected. When the ohmmeter is connected to the armature winding and the motor shaft is turned, the reading will become changeable as the brushes make and break contact with different commutator segments.

PRINCIPLES OF VOLTAGE AND TORQUE GENERATION Example 1: An armature is wound simplex lap in a 6 -pole generator. The total number of conductors is 600. The resistance per conductor is 0. 04 and the induce voltage per conductor is 0. 48 V. If the wire is rated 15 A, find: Sol. Given: Plex=1 , P=6 , ZT=600 cond , Rz=0. 04 , Vz=0. 48 V , Iz=15 A (a) The armature resistance Ra For simplex lap, a = P x Plex = 6 paths Each path must have Zpath =ZT/a = 600/6 =100 cond. Rpath = Zpath x Rz = 100 0. 04 = 4 Ra = Rpath / a = 4/6 = 0. 667 26

PRINCIPLES OF VOLTAGE AND TORQUE GENERATION Example 1: An armature is wound simplex lap in a 6 -pole generator. The total number of conductors is 600. The resistance per conductor is 0. 04 and the induce voltage per conductor is 0. 48 V. If the wire is rated 15 A, find: Sol. (b) The armature voltage Eg Eg = Epath = Zpath x Vz = 100 0. 48 = 48 V (c) The rated armature current Ia Ia = a x Iz = 6 15 = 90 A (d) The rated power of the armature Po = Eg x Ia = 48 90 = 4320 W 27

• Example 2: • An armature is wound simplex wave in a 6 -pole generator. The total number of conductors is 600. The resistance per conductor is 0. 04 and the induce voltage conductor is 0. 48 If Vthe wire Given: Plex=1 , P=6 per , ZT=600 cond. , Rz=0. 04 , Vz. V. =0. 48 , Iz=15 A is rated 15 A, find: (a) The armature resistance Ra • Sol. For simplex wave, a = 2 x Plex = 2 paths Each path must have Zpath =ZT/a = 600/2 =300 cond. Rpath = Zpath x Rz = 300 0. 04 = 12 Ra = Rpath / a = 12/2 = 6 28

PRINCIPLES OF VOLTAGE AND TORQUE GENERATION Example 2: An armature is wound simplex wave in a 6 -pole generator. The total number of conductors is 600. The resistance per conductor is 0. 04 and the induce voltage per conductor is 0. 48 V. If the wire is rated 15 A, find: Sol. (b) The armature voltage Eg Eg = Epath = Zpath x Vz = 300 0. 48 = 144 V (c) The rated armature current Ia Ia = a x Iz = 2 15 = 30 A (d) The rated power of the armature Po = Eg x Ia = 144 30 = 4320 W 29

PRINCIPLES OF VOLTAGE AND TORQUE GENERATION Example 3: An 8 -pole generator has a total of 300 conductors. It is driven at 180 rad/s and the flux/pole is 2 m. Wb. Find the average voltage generated by the armature if it is wound: (a) simplex lap, (b) simplex wave. Sol. (a) For simplex lap; a = P = 8 ZP 300 × 8 150 ka = = = 2πa 2π × 8 π Eg = k a ϕ ω Eg = 150 × 2 × 10− 3 × 180 π Eg = 17. 2 V (b) For simplex wave; a = 2 Eg = ZPϕ ω 2πa 300 8 2 10 3 180 Eg 2 2 Eg 68. 75 V 30

PRINCIPLES OF VOLTAGE AND TORQUE GENERATION Example 4: A 32 -pole generator has 800 conductors. It is driven at 55 rpm. Find the flux per pole necessary to generate an average voltage of 120 V, if the armature is wound: (a) simplex lap, (b) simplex wave. Sol. (a) For simplex lap; a = P = 32 2π ω = 55 × = 5. 76 rad/s 60 2πa. Eg ZPϕω Eg = ⇒ϕ = 2πa ZPω 2π × 32 × 120 ϕ= = 164 m. Wb 800 × 32 × 5. 76 For simplex wave; a = 2 Eg = k a ϕω ZP 800 × 32 6400 = ka = = 2πa 2π × 2 π Eg 120 × π k a ω 6400 × 5. 76 ϕ = 10. 24 m. Wb ϕ= =

PRINCIPLES OF VOLTAGE AND TORQUE GENERATION Example 5: An eight -pole generator has a total of 640 conductors and a flux per pole of 5. 5 m. Wb. Find the speed it must be driven at to generate 460 V if the armature is wound: (a) simplex lap, (b) simplex wave. Sol. (a) For simplex lap; a = P = 8 Eg k a Eg = Z P 640 8 320 ka 2 a 2 8 E � g ka 460 320 5. 5 10 3 (b) For simplex wave; a = 2 ZPϕω 2πa ⇒ω = 2πa. Eg ZPϕ 2πa. Eg 2π × 2 × 460 = ω= ZPϕ 640 × 8 × 5. 5 × 10− 3 821 rad / s ω = 205. 3 rad/s

PRINCIPLES OF VOLTAGE AND TORQUE GENERATION Example 6: A 16 -pole generator has a total of 2240 armature conductors, a flux per pole of 5 m. Wb and is driven at 955 rpm. Find the generated voltage if the armature is wound: (a) simplex lap, (b) simplex wave. Sol. (a) For simplex lap; a = P = 16 a ZP k = 2πa 955 2240 × 16 = 2π × 16 2 100 60 1120 = π rad / s Eg = k a ϕ ω E g 1120 5 10 3 100 178. 3 V (b) For simplex wave; a = 2 ZP 2240 × 16 8960 = ka = = π 2πa 2π × 2 Eg = k a ϕ ω E g 8960 5 10 3 100 1426 V

PRINCIPLES OF VOLTAGE AND TORQUE GENERATION Example 7: A certain simplex wave-wound dc generator, running at a speed of 300 rpm is to generate an induced emf of about 535 V. the flux/pole being 55. 5 m. Wb. Determine the number of poles if the number of conductors is 648. Sol. For simplex wave; a = 2 Eg = ZPϕω 300 2 2πa 31. 42 rad / s 60 2πa. Eg 2π × 2 × 535 P= = = 5. 95 ⇒ P = 6 Zϕ ω 648 × 0. 0555 × 31. 42 The exact induced emf for 6 -pole generator is ZPϕω 648 × 6 × 55. 5 × 10− 3 × 31. 42 Eg = = = 539. 2 V 2πa 2π × 2

Extra Problems 1. An 8 -pole armature is wound simplex wave. It has a total of 1200 conductors, each having a resistance of 35 m. The average generated voltage per conductor is 0. 56 V. If the conductors are rated 15 A, find: (a) the armature resistance, (b) the armature voltage, (c) the rated armature current, and (d) the rated power of the armature. 2. The armature in an eight-pole generator is wound simplex lap. It is rated 230 V, 5 k. W. If it has a total of 1200 conductors and the total armature resistance is 0. 8 , find: (a) the rated current per conductor, (b) the generated voltage per conductor, and (c) the resistance per conductor. 3. Repeat problem 2 if the armature is simplex wave wound. 4. A six-pole generator has a total of 900 conductors. When it is driven at 60 rad/s, its armature generates 230 V. Find the required flux per pole if the armature is wound (a) Simplex lap, (b) Simplex wave. 5. A four-pole generator has a total of 1440 conductors and a flux per pole of 4. 6 m. Wb. When it is driven at 64 rad/s the generated voltage is 265 V. Is the armature wound simplex lap, simplex wave, or neither?