The armature resistance is 03 Ω and field resistance is 250 Ω It is desired to

# The armature resistance is 03 ω and field

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The armature resistance is 0.3 and field resistance is 250 . It is desired to reduce the speed of the motor to 800 rpm at 75% full load. Calculate (a) the resistance to be added in the armature circuit, (b) efficiency (c) power loss in the armature circuit and (d) speed at half full load with the additional resistance in the armature circuit. Assume that the flux remains constant. [(a) 2.69 (b) 73.67 % (c) 4.20 kW (d) 876.80 rpm] Solution Data: V L = 500 V; full load current I L = 52 A; N =1000 rpm; r a = 0.3 ; R s = 250 . V 485 3 . 0 50 500 and rpm; 1000 : load full At A 0 . 50 2 52 ; A 0 . 2 250 500 b sh l a sh E N I I I I (a) When the speed is 800 rpm the torque is 75% of the full load torque T FL . Now, T I a ; T FL 50; or T FL = k 50; and k = 50 FL T At reduced speed let the armature current be I a . Nagsarkar & Sukhija Basic Electrical Engineering,2/E Copyright © Oxford University Press , 2011 Then 0.75 T FL = k I a = 50 FL T I a ; I a = 50 0.75 = 37.5 A Since = constant, the back emf E b N ; 485 1000 Thus V 388 1000 800 485 ; 2 2 1 2 1 b b b E N N E E . If the resistance added to the armature circuit is r 500 – 37.5 ( r + 0.3) = 388; r = 2.69 . rpm 80 . 876 25 25 . 425 50 485 5 . 0 60 2 60 1000 2 V 25 . 425 99 . 2 25 500 A 25 2 50 I current armature load full half At (d) kW 2 . 4 69 . 2 3 . 0 37.5 circuit armature in loss Power (c) % 67 . 73 100 5 . 39 500 37.5 388 Efficiency (b) a 2 N T N T E FL FL b 9.32 A dc shunt motor has an armature resistance of 0.4 and is connected across a voltage supply of 300 V. An external resistance is connected in the armature circuit. Calculate the value of the resistance so that the armature current is limited to 60 A at starting. Calculate the magnitude of the generated e. m. f. when the armature takes a current of 30 A (with the additional resistance in the circuit) at constant speed. [4.6 , 150 V] Solution Data: r a = 0.3 ; V L = 300 V. Let the resistance to be added to the armature circuit be r . At starting E b = 0; then 300 – 60 (0.4 + r ); r = 4.6 . The back emf at constant speed E b = 300 – 30 (0.4 + 4.6) = 150 V. Nagsarkar & Sukhija Basic Electrical Engineering,2/E Copyright © Oxford University Press , 2011 9.33 A 300 V six poles dc shunt motor is wave connected and has 1332 conductors. The flux per pole is 3 mWb when the no load field current is 0.5 A. (a) What is the speed of the motor when the no load current is 2 A? (b) What is the current drawn from the supply when the motor develops a torque of 35 Nm? (c) Determine the speed of the motor at the torque developed in (b). (d) Calculate the percentage change in speed from no load to the speed in (c). Assume an armature resistance of 0.6 . [(a) 1497 rpm (b) 18.84 A (c) 1446 rpm (d) 3.4 %] Solution Data: V L = 300 V; a = 2; Z = 1332; Φ = 3 mWb; at no load I f0 = 0.5A; r a = 0.6 % 4 . 3 100 1497 1446 - 1497 speed in Change (d) rpm 09 . 1446 34 . 18 0 . 289 35 60 2 V 0 . 289 6 . 0 34 . 18 300 (c) A 84 . 18 5 . 0 34 . 18 current Line A 34 . 18 2 60 1332 6 10 3 35 60 2 Nm 35 of torque a develops it when rpm be motor the of speed Let the (b) rpm 1497 1332 6 10 3 2 60 299.1 speed load No (a) V 299 6 . 0 5 . 1 300 5 . 1 5 . 0 0 . 2 load, no At 1 1 0 1 3 1 1 3 - 0  #### You've reached the end of your free preview.

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• Summer '18
• MUKUL SHUKLA

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