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Bangladesh University of Engineering & Technology Assignment on DC beat back Course none : EEE271 Group no.: 01 Roll: 0608001 to 0608015 train: 2, Term: 1 Dept.: I.P.E. entry Date: 26-10-2008 Submitted To: Tanzina Khalaque Dept. of E.E.E., BUET. Question 29.13(d). A five hundredV D.C. shunt motor draws a zephyr veritable of 5A on leisurely- shoot. If armature apology is 0.15ohm and electron orbit resistance is 200ohm, determine the faculty of the machine running as a generator delivering a load up-to-the-minute of 40A. Solution: It is presumption that, The supply potential drop of the shunt motor, V = 500V Armature resistance, Ra = 0.15ohm bowl resistance, Rf = 200ohm On wanton load the motor draws line current, Io = 5A When the configuration whole kit and caboodle as a generator, efficiency = ? The field current, If = = = = 2.5A As the light loaded line current Io consists of armature current Iao and field current If Io = Iao + If Iao = Io - If = (5 - 2.5)A = 2.5A At light load, Input power, Po = V Io = 500 * 5 = 2500watt As in the situation of light load or no load the motor rotates freely and at the time it represents no output power. So solely comment power ultimately turn into injuryes. Variable deprivation, armature Cu dismission (It changes with change in Iao) = Iao2 * Ra = 2.52 * 0.15 = 1 watt. Constant loss (including field Cu loss, no load mechanical loss, crystalline lens nucleus loss) = 2500 1 = 2499 watt When the machine runs as a generator, the load current is at present 40A. I = 40A As it is constant as supply voltage and Rf is constant, so at a time the armature current, Ia1 = I + If = 40 + 2.5 = 42.5 A Because straightway load current consis! ts of armature current and field current, I = Ia - If So, at once new armature Cu loss = Ia2 * Ra...If you want to get a full essay, order it on our website: OrderEssay.net

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