Basic_Electrical_Engineering_4th_edition
.pdfDC 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
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.