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264

ELECTRICAL ENGINEERING

Therefore

N = A1 - B1T

The machinewillbehave as a shunt machine. The seriesmachinecharacteristicsfor speed vs la and torque vs la are shown in Fig. 6.19. Initially, under unsaturated condition when la is small, Tala2 and after large value ofla when the machine gets saturated T a la

Speed

i

Torque

 

torque

 

Fig. 6.19. Speed V518 and torque V/a characteristics for series motor.

The speedtorque curve ofa de series motoris very suitable for electric traction and cranes. The starting torque is very high (TaF). Therefore, heavy masses can be accelerated very fast. The decrease in speed withincrease intorque is a stable operatingconditionforthis motor. When the motor has to go up hill, the torquerequired increases, hence the speed automatically reduces.

6.7.3DC Motor Starter

We know that the current drawn by a de motor is limited by its back emfas this emf opposes the supply voltage and hence the resultant voltage is responsible for the current. This back e.m.f. is proportional to the speed. Hence when the motor is to be started, initially the sped is zero and hence the back emfis zero and hence at start thecurrentis limited onlyby the armature resist-

ance which usually is in terms of friction of an ohm and the current I8T = RV where V is the

supply voltage andthe current becomes in terms ofhundreds ofamperes whichamay damage the armature winding. Hence a device known as a de motor starter is invariably used for starting these motors. The starting current can be reduced by inserting a resistance in series with the armature. As the motor is switched on to supply, the current is limited to suitable value by the external resistance andthemotorstartsrunning. Asthe motorpicksup speed, back emfis induced whichopposes the supplyvoltage andtheexternalresistanceis reducedgraduallyorin steps such that whenthe motorhas reached its normalrated speed the external resistance is completelycut off. The starterhas two additionalfeatures, the no-voltcoil and over load release coilto protectthe motorfrom damage when operatingunder abnormal conditions. Out ofthevarious starters avail­ able, three-point starter is more commonly used and is discussed next.

Three-PointStarter. The starterconsists ofa numberofresistances connected between studs. The arm ofthe starter makes contact with these studs and also to two metal arcs. These arcs are connected to the no-volt coil andtheoverloadreleaseto the starter arm. Initiallythe arm is in the extreme left position and all the resistance elements are in the circuit i.e., in series with the armature circuit. As the motor picks up speed, the arm is gradually moved towards right, cutting out the resistances in step. In the final position the arm is in the extreme right position whenthe whole ofexternal resistance has been cut out and the motor is directly connected on to the supply. The connection ofthe starter to the motor is shown in Fig. 6.20. The internal details ofthe starter are also shown.

DC MACHINES

265

Overload release

Fig. 6.20. Three point starter for d.c. shunt motor.

The starterhas two important featuresthe no-volt release andover load release. Suppose the motor is running under normal condition and suddenly the supply fails. If the starter arm remains in the extreme right condition when no external resistance is inserted in the armature circuit and inthe mean time the supply is restored, the armature ofthe motor will get damaged. However,hereno-voltrelease comes to therescue ofthe motor. As soon as thesupplyfails no volt coil gets de-energised and the starter arm goes back to extreme left position due to spring action.

Similarly whenunder operatingconditionifthe current drawn by the motor is more than 130% ofthe full load current the overload release attracts an armature which short circuits the no volt coil and the starter arm is brought back to its extreme leftposition (offposition) and the supply to the motor is disconnected. Thus the motor is saved from overheating.

6.8LOSSES IN DC MACHINES

Mainly following losses take place in a de machine :

1. Copper loss. These are the l2R loss and take place in armature conductors (la2Ra),

series fieldwinding (FRs) and shunt

fieldwinding

V2

Rsh

or

Ish

2Rsh

.

2.Rotational losses. In a de machine, the core loss is a true rotational loss. Except in the pole faces, the fluxinthe statorremains constant in steadystateoperation. The flux in the armature core is constantly alternating as the machine rotates and at zero speed the core loss is zero. The rotational loss then consist ofthe sum ofthe friction windage and core losses. In a generator orconstant speed motor these are considered part ofthe constant losses.

3.Brush drop loss. Brushcontactloss occurs due to voltagedropacross thebrush andis equal to 2Vdlawhere Vdis thevoltage drop across one brush.

4.Strayload loss. We calculate armaturecopperloss onthebasis ofderesistance whereas the current in the armature winding is alternating. An incremental loss should be included to account for skin effect, eddy currents in the conductors andcore loss due to

DC MACHINES

269

Armature copper losszat any load current Ia2 Ra, v

Field copper loss = -

Rr,

Total loss at load = Mechanical andironlosses+Armaturecopperloss+ FieldCopperloss + Stray Loss

Swinburnetestcanthusbeutilisedtoobtainefficiencyofthe machine at differentloadings. However, the test does notprovide anyinformationregardingthe temperaturerise ofthe windings ofthe machine.

Example 6.11. Swinburne test gave the following results on a de shunt motor:

Supply voltage 500 V, no load current 5A, Armature resistance 0.5.Qand Field resistance 250 .D. Determine the efficiency ofthe machine (i) as a generator delivering 100A at 500 V(ii) as a motor having a line current of 100 A at 500 V Neglect temperature rise during operation. Assume stray losses at 1 % of output.

Solution. Input at no load

500 x 5 = 2500 watts

Field current

 

I =

500

= 2 A

 

 

 

 

 

250

 

 

 

 

Hence armature current

5

- 2I= 3 A= 22 x 250 = 1000 watts

Field copper loss

 

Armaturecopper loss

 

 

= 32 x 0.5 = 4.5 watts

Mechanical and iron loss

 

 

 

2500 - 1000 - 4.5

 

 

 

= 1495.5 watts

(z) Working as a generator

 

=

500 x 100

 

 

Output power

 

 

 

 

50,000 watts

Armature current

 

100 + 2 = 102 A

 

 

Armaturecopperloss

 

 

= 1022 x 0.5 = 5202 watt

 

 

=

1

 

=

 

Stray loss =

x 50,000 = 500 watts

 

500 x 100 =

--

 

 

 

 

 

 

 

100

 

 

 

Total losses = 5202 + 1000 + 1495.5 + 500

 

 

 

= 8197.5 watts

 

 

. .

 

50000

 

 

Hence efficiency = 58197.5 = 0

.86 or 860Yo Ans.

(iz) Working as a motor

 

 

 

 

 

 

 

Input to the motor

 

 

 

50,000 watts

Armature current

 

100 - 2 = 98 A

 

 

 

Armaturecopper loss

 

 

= 982 x 0.5 = 4802 watts

Output = 50,000 - 4802 - 1000 - 1495.5 = 42702.5 watts

Stray loss = 427 watts

x 50 = 12.5 volts
= 45.5 A

270

 

ELECTRICAL ENGINEERING

Hence net output = 42702.5

- 427 = 42275.5 watts

. .

42275.5

 

Hence efficiency =

50,000

x 100 = 84.55% Ans.

Example 6.12. A 100 KW belt driven shunt generator running at 300 rpm on 220 V bus bars continues to run as a motor when the belt breaks then taking 10 KW. Determine its speed. Assume armature resistance 0.025 D, field resistance 60 Q contact drop under each brush 1 V. Ignore armature reaction.

Solution. Current taken bv the motor lO,OOO220

Initially current delivered bythe generator 455 A Hence the induced emf 220 + 455 x 0.025 = 231.375

The back emfwhile motoring 220 - 45.5 x 0.025 = 218.87 V

Add 2 volts contact drop in case ofgenerator and subtract 2 volts in case ofmotor, we have

Total induced emffor generator 233.375 and for the motor back emf216.87

Hence the speed under motoring

condition

 

 

216·87

x 300

=

279 ·p

m

Ans.

233.375

 

 

r

 

Example 6.13. A series generator, having an external characteristic which is a straight line through zero to 50 V at 200 A, is connected as a booster between a station bus bar and a feeder of 0.3 Q resistance. Determine the voltage between the far end of the feeder and the bus bar at a current of (a) 1 60 A (b) 50 A.

Solution. (a) Since the generator characteristic is a straight line passing through the origin, at 160 A the bus bar voltage willbe

20050 x 160 = 40 volts

and the drop in the feeder due to the flow of 160 Ais 160 x 0.3 = 48 volts

and hence the difference ofvolts between the bus bar and the far end ofthe feeder is 48 - 40 = 8 volts Ans.

(b) Similarly for 50 A, the bus bar voltage is

50

200

andthe drop inthe feeder due to 50A current flowis 50 x 0.3 = 15 volt

and hence the difference in voltage is

15 - 12.5 = 2.5 V Ans.

Example 6.14. A long shunt compound generator delivers a load current of 50 A at 500 V and has armature, series field and shunt field resistances of 0.05 !1, 0.03 Q and 250 Q respec­ tively. Determine the induced emf and the armature current. Allow 1 volt per brush for contact drop.

DC MACHINES

 

 

 

 

 

 

271

Solution. The long shunt figure is given here

 

 

The shunt field current

5oo

= 2 A

 

 

 

 

 

Hence armature current

250

= I +

]sh

= 50

+ 2 = 52 A

 

 

Ia

Ans

.

L

 

 

 

 

 

 

 

 

50 A

 

 

500 v

zz

 

 

 

 

 

 

Fig. E6.14

 

 

 

 

 

The drop in the armature and the series field resistance

 

 

 

 

 

52(0.05

 

 

0.03) = 4.16

 

 

 

 

 

The brush drop 2 volts. Hence the e.m.f. induced is 500 + 2 + 4.16 =

 

 

volts. Ans.

 

 

 

+

 

 

 

 

 

506.16 '.:::'506..2

Example 6.15. A 250 Vshunt motor on no load runs at 1000 rpm and takes 5A. The total

armature and shunt field resistances are respectively 0.2 .Q and 250

 

Determine the speed

when loaded taking a current of 50A, if armature reaction weahens the field by

 

Solution. The field current is

 

250 = 1 A

 

 

.Q.

 

 

3%.

 

 

250

 

 

 

 

 

Under no load condition the armature current is 5 - 1 = 4 A. Hence back emfis

 

250 - 4 x 0.2

 

 

 

249.2 volts

 

 

 

 

 

Since the armature current is 4 A which is very small as compared to 50 A as far as

armature reaction effect inconcerned, hence we neglect the effect.

 

 

 

 

 

 

 

=

 

 

 

 

 

 

 

 

The back emfwhen 50 A is drawn by the motor

 

 

 

 

 

Now

250 - (50 - 1) x 0.2 = 240.2 volts

 

 

 

 

 

249.2 =Kqi x 1000

 

 

 

 

 

 

240.2

 

=

 

0.97 KN

 

 

 

 

 

 

Hence

0.97 KNcp

 

 

240.2

 

 

 

 

 

 

Kcp x 1000

 

249.2

 

 

 

 

 

 

or

N=

 

240.2 x

1000

- 993_7

 

 

 

 

 

 

 

 

 

 

249.2

0.97

 

 

 

 

 

'.:::'994rpm. .

Example 6.16. A shunt generator delivers 50 KWat 250 Vand 400 rpm. The armature and field resistances are 0.02 .Q and 50 .Q respectively. Determine the speed of the machine running as a shunt motor and taking 50KWinput at 250 V Allow 1 Vper brush for contact drop.