60 where Now 45 flux V 255 1 45 300 1 1 a T T k a PZ k k

# 60 where now 45 flux v 255 1 45 300 1 1 a t t k a pz

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60 where 1000 45 60 1000 45 255 Now 45 flux ; V 255 0 . 1 45 300 1 1 a T T k a PZ k k a Z P I E b 2.45 1.0 - 3.45 added be to Resistance 3.45 circuit. armature motor the of resistance total the is where 38.86 275 . 4 4.275 38.86 - 300 Hence, A 86 . 38 275 . 4 750 0057 . 0 750 18 . 82 60 750 2 0 5 7 then , is current motor armature the If Nm 18 . 82 1000 750 109.58 (I), g Usin rpm, 750 ue When torq (a) 2 2 2 2 r r I r E I I I I I k I E I k E I T T a b a a a a a a b a b a 3.64 1.0 - 4.64 added be to Resistance 64 . 4 65 . 33 275 . 4 65 . 33 300 Then, A 645 . 33 or, 64 . 61 60 750 2 275 . 4 Nm 64 . 61 75 . 0 58 . 109 750 and 1000 58 . 109 speed When (b) 2 2 2 2 2 2 2 r r I I T T T a a Nagsarkar & Sukhija Basic Electrical Engineering,2/E Copyright © Oxford University Press , 2011
1.41 1.0 - 2.41 added be to Resistance 2.41 44.87 4.275 44.87 - 300 Then, A 87 . 44 58 . 109 60 750 2 4.275 Nm 109.58 to equal is and constant is (c) 2 a r r I I T a 9.27 A 300 V dc series motor has 6 poles and a wave wound armature. The number of conductors per slot is 4 and the total number of slots is 300. If the armature and field winding resistance is 0.5 and 0.25 respectively and the flux per pole is 0.025 Wb, calculate (a) speed of the motor, (b) BHP output, (c) efficiency and (d) force on the pulley if the diameter is 0.5m. The line current is 50 A. Assume that 1kW is lost in windage and friction. [(a) 175 rpm (b) 16.485 HP (c) 80.83 % (d) 2646.52 N] Solution Data: V L = 300 V; P = 6; a = 2; conductors/slot = 4; number of slots = 300; r a = 0.5 , R s = 0.25; = 0.025 Wb; pulley diameter = 0.05 m; windage and friction loss = 1.0 kW. N 52 . 2646 2 0.5 661.63 pulley on the Force Nm 63 . 661 1000 125 . 12 60 175 2 (d) % 83 . 80 100 15.0 12.125 Efficiency kW 15 50 300 power Input (c) hp metric 485 . 16 735.5 1000 12.125 pulley at the Output kW 12.125 1.0 - 13.125 losses - power Armature Output kW 125 . 13 50 5 . 262 power Armature (b) rpm 175 1200 6 0.025 2 60 262.5 (a) 1200 4 300 Z conductors of Number V 5 . 262 25 . 0 5 . 0 50 300 T T I E N r r I V E a b se a a L b Nagsarkar & Sukhija Basic Electrical Engineering,2/E Copyright © Oxford University Press , 2011
9.28 The armature winding of a 6 a pole dc shunt motor is wave wound and has 666 conductors. It takes a full load current of 60 A at 600 V. If the armature resistance is 0.3 , flux per pole is 0.025 Wb and the voltage drop in each brush is 1.5 V, determine (a) speed at full load, (b) armature torque, (c) output in kW and (d) efficiency. Assume windage and friction loss equal to 1 kW. Calculate the above quantities for a motor whose armature is lap wound.

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• Summer '18
• MUKUL SHUKLA

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