can u show work plz The following details are known about a 200 hp, 500 V, 435 r

Question

can u show work plz

The following details are known about a 200 hp, 500 V, 435 rpm dc shunt motor: full-load current is 330 A, insulation class is H, weight is 3400 kg. external diameter of the frame is 915 mm and length of the frame is 1260 mm. If the motor is running at full-load, calculate: the total losses and efficiency. the approximate shunt field exciting current if the shunt field causes 15% of the total losses. the value of the armature resistance as well as the counter-emf. knowing that 45% of the total losses at full-load are due to armature resistance.

Explanation / Answer

output powe=200*746=149200W

input power=VI=500*330=165000W

total losses=165000-149200=15.8KW

effeciency=(Pout/Pin)*100=90.4 %

b)

shunt current=330*.15*.15=7.425A

c)

45% loss of total losses=7110W

armature resistance=500/330=1.5151ohm

counter E.MF=(500-1.515*330)=128.78

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