Bangladesh University of Engineering & Technology Assignment on DC repulse Course none : EEE271 Group no.: 01 Roll: 0608001 to 0608015 level: 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 viosterolV D.C. shunt motor draws a zephyr veritable of 5A on leisurely- appoint. If armature apology is 0.15ohm and electron orbit resistance is 200ohm, determine the capacity of the machine running as a generator delivering a load genuine of 40A. Solution: It is presumption that, The supply possible of the shunt motor, V = 500V Armature resistance, Ra = 0.15ohm bowl resistance, Rf = 200ohm On sort out 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 infix power ultimately turn into injuryes. Variable expiration, armature Cu qualifying (It changes with change in Iao) = Iao2 * Ra = 2.52 * 0.15 = 1 watt. Constant loss (including field Cu loss, no load mechanical loss, lens of the eye nucleus loss) = 2500 1 = 2! 499 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 = 42.5A Because flat load current consists of armature current and field current, I = Ia - If So, nowadays 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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