CHAPTER 1DC MACHINES

1.1 INTRODUCTION

Energy is needed in different forms:

Light bulbs and heaters - electrical energy Fans and rolling miles - mechanical energy Need for energy converters

Figure 1.1

DC generator – supply current to the load DC motor – need current from the supply AC electric supply - AC machines (synchronous and asynchronous) DC electric supply - DC machines

Advantages of DC Machine (i) Adjustable motor speed over wide ranges(ii) Constant mechanical output (torque)(iii)Rapid acceleration or deceleration(iv) Responsive to feedback signals

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Figure 1.2 : DC machine Construction

1.2 GENERATION OF AC SIGNAL

Generator needs something to rotate or move itself. It is called the prime mover. The shaft of the prime mover will be coupled to the shaft of the generator so that the

generator can rotate once the prime mover rotates. In a big system, they normally used 3-phase induction motor. But, in small scale machine, they can use a mechanism to rotate the generator which

then generated electricity.

carbon brushFigure 1.3: Physical Arrangement of AC Generating Method

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coil

Slip ring:used to generate ac signalCarbon brush: used to collect output voltage (ac signal)

Figure 1.4 : Emf generated at various instant (angle position)

1.3 GENERATION OF DC SIGNAL

The generating of DC signal can be done by replacing the slip ring with commutator so that rectification process can happen.

Commutator can generate dc signal. This is done through rectification or commutation process; which converts ac signal

into dc mechanically. Therefore a commutator is called a mechanical rectifier.

Figure 1.5 : Physical Arrangement of DC Generating Method

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commutator

Carbon brush

commutator:used to generate dc signalCarbon brush: used to collect output voltage (dc signal)

Figure 1.6: AC is converted into DC Signal through Commutation Process

1.4 EMF GENERATED ON DC MACHINE

The EMF generated:-

where z = no of conductors in the armature circuitc = no. of parallel pathN = speed in rpmΦ = flux/pole (Wb)P = no. of pole pair

Figure 1.7 : Permanent Magnet Dc Machine

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Example 1

If the no-load voltage of a separately excited generator is 135 V at 850 rpm, what will be the voltage if the speed is increased to 1000 rpm? Assume constant field excitation.

Solution

Constant field excitation:-

Example 2

A separately excited generator has no-load voltage of 140 V when the field current is adjusted to 2 A. The speed is 900 rpm. Assume a linear relationship between the field flux and current. Calculate:(i) the generated voltage when the field current is increased to 2.5 A. Assume N1=N2.

(ii) the terminal voltage when the speed is increased to 1000 rpm with the field current set at 2.2 A.

Solution

(i)

(ii)

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1.5 DC MACHINE CONSTRUCTION

rotor/armature pole

Figure 1.8 : DC Machine Construction

DC motor principles DC motors consist of rotor-mounted windings (armature) at the rotor side and

stationary windings (field poles) at the stator side.

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In all DC motors, except permanent magnet motors, current must be conducted to the armature windings by passing current through carbon brushes that slide over a set of copper surfaces called a commutator, which is mounted on the rotor.

The commutator bars are soldered to armature coils. The brush/commutator combination makes a sliding switch that energizes particular

portions of the armature, based on the position of the rotor. This process creates north and south magnetic poles on the rotor that are attracted to

or repelled by north and south poles on the stator, which are formed by passing direct current through the field windings.

It's this magnetic attraction and repulsion that causes the rotor to rotate.

Figure 1.9 : DC Machine Construction

1.6 MACHINE WINDINGS

Machine winding can be divide into 2:-(i) armature winding (rotor side)(ii) field winding (stator side)

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Figure 1.10: Winding Connection in DC Machine

SELF-EXCITED FIELD WINDING

In self-excited dc machine, there are three types of excitation method namely:(i) Series Excitation: the field winding is connected in series with the armature

circuit.(ii) Shunt Excitation: the field winding is connected in parallel with the

armature circuit.(iii) Compound Excitation: the field winding are connected in series and parallel

with the armature circuit.

The schematic diagrams for the three types of these machines are illustrated in Figure 1.11.

The difference between dc motor and dc generator is in terms of the current direction.

(a) DC Series machine

Machine Winding

Field winding Armature Winding

**Self excited-direct connection between armature circuit and the field circuit

Separately excited-no direct connection between armature circuit and the field circuit

Series excitation

Shunt excitation Compound excitation

In dc generator: armature current, Ia is supplied by the armature.In dc motor: armature current, Ia is received by the armature.

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1.0

k

1.0m +-

DC

loa

d

Ra

Eg

Ia

RfIL

VT

1.0

k

1.0m +-

5.0

Ra

EC

Ia

RfIL

dc supplyVT

Generator: Eg = VT + Ia(Ra + Rf) Motor: Ec = VT - Ia(Ra + Rf)

(b)DC Shunt machine

1.0

k 1

.0m

+-

DC

lo

ad

Ra

Eg

Ia Rf

IfIL

VT

1.0

k 1

.0m

+-

5.0

Ra

Ec

Ia Rf

IfIL

dc supplyVT

Generator: Eg = VT + IaRa Motor: Ec = VT - IaRa

(c) DC Compound machine

1,0m

1,0

m 1

,0m

+-

DC

M1

DC

load

RaILIF

Rf2

Rf1VT

Eg

Ia 1,0m

1,0

m 1

,0m

+-

DC

M1 1,0

RaILIF

Rf2

Rf1dc supply VTEc

Ia

Generator: Eg = VT + Ia(Ra + Rf 2) Motor: Ec = VT - Ia(Ra + Rf 2)

Figure 1.11

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** Eg = generated emf Ec = counter emf

1.7 POWER FLOW DIAGRAM

Power flow diagram is normally represented as a fish bone.

POWER FLOW DIAGRAM FOR DC GENERATOR

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Input Power = Output + LossesLosses can be divided into 2:-

(i) Copper losses (Armature copper loss , Pca) and (Field copper loss,Pcf)(ii) Iron losses, Pµ (friction, stray, windage,mechanical losses)

For DC Generator: Pin = Pout + Total losseswhere Pout = VTIL

For DC Motor: Pin = Pout + Total losseswhere Pin = VTIL

Total losses = Pca + Pcf + Pµ

POWER FLOW DIAGRAM FOR DC MOTOR

DC SERIES Pµ Pcf

Pin Pout= VTIL

Pca

DC SHUNT Pµ Pcf

Pin Pout= VTIL

Pca

DC COMPOUND Pµ Pcf1

Pin Pout= VTIL

Pca Pcf2

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1.8 MOTOR TORQUE

For load torque@shaft torque@net torque@output torque:

For mechanical torque:

For loss torque:

DC SERIES Pca Pµ

Pm Pout Pin=VTIL

Pcf

DC SHUNT Pca Pµ PmPin=VTIL Pout

Pcf

DC COMPOUND Pca Pcf2

Pin = VTIL Pm Pout Pµ Pcf1

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1.9 EFFICIENCY

Generally efficiency is:

For dc generator:

For dc motor:

Example 3

A short-shunt compound generator delivers 50 A at 500 V to a resistive load. The armature, series field and shunt field resistances are 0.16, 0.08 and 200 Ω, respectively. Calculate the generated EMF and armature current, if the rotational losses are 520 W, determine the efficiency of the generator.

Solution

1,0m

1,0

m 1

,0m

+- DC

M1

load

0.16 ohm

0.08 ohm

200 ohm 500V

Eg

50 AILIF

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Example 4

A 150 V shunt motor has the following parameters:Ra = 0.5 Ω, Rf = 150 Ω and rotational loss 250 W. On full load the line current is 19.5 A and the motor runs at 1400 rpm. Determine:(i) the developed power/developed mechanical power(ii) the output power(iii) the output torque(iv) the efficiency at full load

Solution

IL = 19.5AN = 1400rpm

(i)

OR :-

(ii)

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(iii)

(iv)

Tutorial 1

1. A 300 V compound motor has armature resistance 0.18 Ω, series field resistance 0.3 Ω and shunt field resistance 100 Ω. The rotational losses are 200 W. On full load the line current is 25 A and the motor runs at 1800 rpm. Determine:

(i) the developed mechanical power(ii) the output power

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(iii) the output torque(iv) the efficiency at full load

2. A 120 V series motor has 0.2 Ω field resistance. On full load, the line current is 16.5 A. The output power is 1500 W and rotational loss is 150 W. Find the value of armature resistance.

3. Briefly explain the difference between motor and generator of DC machine.

4. A compound DC motor rated at 415 V, 6 HP, 2000 rpm has armature resistance 0.18 Ω, series field resistance 0.3 Ω and shunt field resistance 100 Ω. The rotational losses are 200 W. The full load line current is 40 A.

(i) Find the developed mechanical power.(ii) Find the output power.(iii) Find the load torque.(iv) Find the efficiency of the motor.(v) Draw the power flow diagram for this type of motor.

5. A DC series generator delivers 100 kW at 10 kV to a load. The armature resistance is 20 Ω and the field resistance is 50 Ω. Calculate:

(i) the generated emf, Eg

(ii) the input power if the stray and friction losses are 400 W.

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