armature when the load full At 14 77 250 resistance Armature b a E r

Armature when the load full at 14 77 250 resistance

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armature when the load full At 14 . 0 77 . 161 27 . 227 250 resistance Armature b a E r Nm. 351.18 unchanged, is current field the Since (e) rpm 3000 or 1000 22 . 77 18 . 351 60 2 (d) kW 77.22 161.77 477.35 output Power (c) T N N Nagsarkar & Sukhija Basic Electrical Engineering,2/E Copyright © Oxford University Press , 2011
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9.21 In problem .16, the field current is reduced by 25% while the armature voltage is held constant at 250 V. Calculate (a) power developed, (b) torque and (c) speed of the motor at full load. [(a) 36.78 kW (b) 263.53 Nm (c) 1333.5 rpm] Solution Data: Same as problem 9.16; field current reduced by 25% and armature voltage constant at 250 V. It is known that E b = 60 60 N K a PNZ ; and K as P , Z , and a are constant. Now, K = 13.64 (calculated in problem 9.16). When the flux is reduced by 25%, then K = 13.64 0.75 kW 36.78 161.77 227.35 developed Power (a) V 227.35 0.14 161.77 - 250 A; 77 . 161 : 9.16 problem In b a E I rpm 5 . 1333 263.52 2 60 1000 36.78 (c) Nm 52 . 263 77 . 161 2 10.23 (b) N T 9.22 Show that the speed of a dc motor is directly proportional to its back e. m. f. and inversely proportional to its flux per pole. There from derive an expression for the ratio of the speeds of a dc series motor (a) when the motor is operating on the linear portion of the magnetic characteristic and (b) the magnetic circuit is saturated. A dc series motor has a speed of 500 rpm when it draws a current of 50 A from a dc supply source of 300 V. Calculate the speed of the motor when it draws a current of 90 A. The series field and armature winding resistances are 0.2 and 0.15 respectively. Assume a voltage drop of 1 V per brush. ] 91 . 263 ; ) ( ) [( 2 1 2 1 1 2 2 1 2 1 rpm E E N N b I E I E N N a b b a b a b Solution Nagsarkar & Sukhija Basic Electrical Engineering,2/E Copyright © Oxford University Press , 2011
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Data: N = 500 rpm; supply voltage V = 300 V; I a = 50A; r se = 0.2 ; r a = 0.15 ; voltage drop/brush = 1.0 V 1 2 2 1 2 1 b 1 and , or Φ 60 60 Φ b b b b E E N N N E N PZ a E N a PNZ E (a) When the series motor is operating on linear portion of the magnetisation curve, field current I f = I a and I f 1 2 2 1 2 1 a a b b I I E E N N (b) When the magnetic circuit is saturated, then is independent of I f and I a . 1 2 2 1 a a I I N N rpm 91 . 263 90 5 . 280 50 5 . 266 500 or 50 5 . 266 90 5 . 280 V 5 . 266 0 . 1 2 15 . 0 2 . 0 90 300 V 5 . 280 0 . 1 2 15 . 0 2 . 0 50 300 2 1 2 2 1 2 1 2 1 N I E I E N N E E a b a b b b 9.23 Show that for a dc series motor the torque developed is proportional to the square of armature current. A dc series motor is run at 250 rpm with the series winding separately excited. The voltage across its terminals is 1500 V when the field current is 300 A. Calculate the speed and torque of the motor if it is run from a 600 V supply and it takes a current of 30 A.
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  • MUKUL SHUKLA

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