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Chapter 6: DC & AC Machine

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Presentation on theme: "Chapter 6: DC & AC Machine"— Presentation transcript:

1 Chapter 6: DC & AC Machine
MOHD ASRI BIN JUSOH

2 Chapter 6.1: DC Machine, DC Motor, DC Generator

3 Introduction An electrical machine is link between an electrical system and a mechanical system. Conversion of energy: - mechanical to electrical : generator - electrical to mechanical : motor

4 Introduction (Cont…) Types of electrical machine:
- AC machines (generator or motor in AC electrical system) - DC machines (generator or motor in DC electrical system)

5 DC Machine DC machine can be divided with 2 types: - DC motor
- DC generator

6 DC Machine (Cont…) Construction: Cutaway view

7 DC Machine (Cont…) Construction: (cont…) Cutaway view

8 DC Machine (Cont…) Construction: (cont…) Rotor

9 DC Machine (Cont…) Construction: (cont…) Stator

10 DC Machine (Cont…) Fundamentals:
- Stator : stationary part. Carrie field winding that is used to produce required magnetic field by DC excitation. - Rotor : rotating part. Carries a distributed winding where e.m.f is induced. - Field winding : wound on stator poles to produce magnetic field in the air gap. - Armature winding : coil placed in armature slot.

11 DC Machine (Cont…) Fundamentals: (cont…)
- Commutator : copper bars connected to the armature winding and insulated from each other. - Brush : place against commutator surface. Used to connect the armature winding to external circuit.

12 DC Machine (Cont…) Fundamentals: (cont…)
Conversion energy from electrical to mechanical for or vice versa follow concept: - Generator : when conductor moved in a magnetic field, voltage is induced in the conductor. - Motor: when a current carrying conductor is placed in magnetic field, the conductor experience a mechanical forces.

13 DC Motor Basic operation:
- stator poles are supply by dc current which produce dc magnetic field. - rotor supply dc current through the brushes, commutator and coils. - interaction of magnetic field and rotor current generate a force that drive the motor.

14 DC Motor (Cont…) Basic operation: (cont…)
- magnetic field enter the rotor from north to south pole. - interaction between field and current produce force. - the generated force turn the rotor until coil reaches neutral point. - however, inertia drive motor beyond neutral zone where direction of magnetic field is reverse. - commutator change current direction to avoid reversal force.

15 DC Motor (Cont…)

16 DC Motor (Cont…) DC equivalent circuit: - Armature circuit :
present by voltage and resistance connected series (armature resistance, RA). - Field circuit : present by winding that generate magnetic field and a winding resistance (RF)in series.

17 DC Motor (Cont…) Φ = flux/pole (Weber)
DC equivalent circuit: (cont…) - EA = back e.m.f. Φ = flux/pole (Weber) Z = total number of armature conductors = number of slots x number of conductor/slot P = number of poles A = number of parallel paths in armature [A = 2 (for wave winding), A = P (for lap winding)] N = armature rotation (rpm)

18 DC Motor (Cont…) DC equivalent circuit: (cont…) Term:
- VT = supply voltage - EA = internal generated voltage / back e.m.f - RA = armature resistance - RF = field / shunt resistance - RS = series resistance - IL = load current - n = speed - IF – field current - IA = armature current

19 DC Motor (Cont…) DC equivalent circuit: (cont…)
1. Separately excited DC motor

20 DC Motor (Cont…) DC equivalent circuit: (cont…) 2. Shunt DC motor

21 DC Motor (Cont…) DC equivalent circuit: (cont…) 3. Series DC motor

22 DC Motor (Cont…) DC equivalent circuit: (cont…) 4. Compound DC motor

23 DC Motor (Cont…) Example 1:
A 250 V, DC shunt motor takes a line current of 20 A. resistance of shunt field winding is 200 Ω and resistance of armature is 0.3 Ω. Find the armature current and the back e.m.f.

24 DC Motor (Cont…) Example 2:
A 50 hp, 250 V, 1200 r/min DC shunt motor with compensating winding has an armature resistance (including brushes, compensating windings and inter poles) of 0.06 Ω. Its field circuit has a total resistance Radj + RF of 50 Ω which produces a no load speed of r/min. there are 1200 turns per pole on the shunt field winding. Find speed of this motor when its input current is: (a) 100 A (b) 200 A

25 DC Motor (Cont…) Example 3:
The motor in example 2 is now connected in separately excited circuit as shown in figure. The motor initially running at speed, n = 1103 r/min with VA = 250 and IA = 120 A, while supplying a constant torque load. If VA is reduced to 200 V, determine: (a) internal generated voltage, EA (b) the final speed of this motor, n2

26 DC Motor (Cont…) Example 3: (cont…)

27 DC Motor (Cont…) Example 4:
A DC series motor is running with a speed of 800 r/min while taking a current of 20 A from the supply. If the load is changed such that the current drawn by the motor is increased to 50 A, calculate the speed of the motor on new load. The armature and series filed winding resistances are 0.2 ohm and 0.3 ohm respectively. Assume the flux produced is proportional to the current and voltage supply is 250 V.

28 DC Generator Generating of AC voltage:
- The voltage generated in any DC generator inherently alternating and only becomes DC after it has been rectified by the commutator.

29 DC Generator (Cont…) Generating of DC voltage:

30 DC Generator (Cont…) Φ = flux/pole (Weber)
DC equivalent circuit: (cont…) - EA = back e.m.f. Φ = flux/pole (Weber) Z = total number of armature conductors = number of slots x number of conductor/slot P = number of poles A = number of parallel paths in armature [A = 2 (for wave winding), A = P (for lap winding)] N = armature rotation (rpm)

31 DC Generator (Cont…) DC equivalent circuit: (cont…)
1. Separately excited DC generator

32 DC Generator (Cont…) DC equivalent circuit: (cont…)
2. Shunt DC generator

33 DC Generator (Cont…) DC equivalent circuit: (cont…)
3. Series DC generator

34 DC Generator (Cont…) DC equivalent circuit: (cont…)
4. Compound DC generator

35 DC Generator (Cont…) Example:
A DC shunt generator has shunt field winding resistance of 100 ohm. It is supplying a load of 5 kW at a voltage of 250 V. if its armature resistance is 0.02 ohm, calculate the induced e.m.f. of the generator.


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