zxing barcode reader java download The motor starting current is directly proportional to the starting voltage, so in Software

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The motor starting current is directly proportional to the starting voltage, so
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348 V I L2 = I L1 = (0756)I L1 = (0756)(7 I rated ) = 5296 I rated 460 V
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The input power to this motor is
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PIN =
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(50 hp )(746 W/hp) = 4099 kW
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The rated current is equal to
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I rated =
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PIN ( 4099 kW ) = 591 A = 3 VT PF 3 ( 460 V )( 087 )
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Therefore, the motor starting current is
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I L 2 = 5843 I rated = (5296)(591 A ) = 313 A
The turns ratio of the autotransformer that produces this starting voltage is
N SE + N C 460 V = = 132 348 V NC
so the current drawn from the supply will be
I line =
7-23
I start 313 A = = 237 A 132 132
A wound-rotor induction motor is operating at rated voltage and frequency with its slip rings shorted and with a load of about 25 percent of the rated value for the machine If the rotor resistance of this machine is doubled by inserting external resistors into the rotor circuit, explain what happens to the following: (a) Slip s (b) Motor speed nm (c) The induced voltage in the rotor (d) The rotor current (e) ind (f) Pout (g) PRCL (h) Overall efficiency SOLUTION (a) (b) (c) (d) The slip s will increase The motor speed nm will decrease The induced voltage in the rotor will increase The rotor current will increase
(e) The induced torque will adjust to supply the load s torque requirements at the new speed This will depend on the shape of the load s torque-speed characteristic For most loads, the induced torque will decrease
(f) (g) (h) 7-24
The output power will generally decrease: POUT = ind m The rotor copper losses (including the external resistor) will increase The overall efficiency will decrease
Answer the following questions about a 460-V -connected two-pole 100-hp 60-Hz starting code letter F induction motor: (a) What is the maximum current starting current that this machine s controller must be designed to handle (b) If the controller is designed to switch the stator windings from a connection to a Y connection during starting, what is the maximum starting current that the controller must be designed to handle (c) If a 125:1 step-down autotransformer starter is used during starting, what is the maximum starting current that will be drawn from the line SOLUTION (a) The maximum starting kVA of this motor is
Sstart = (100 hp )( 560) = 560 kVA
Therefore,
I start =
S 560 kVA = = 703 A 3 VT 3 (460 V )
(b) The line voltage will still be 460 V when the motor is switched to the Y-connection, but now the phase voltage will be 460 / 3 = 2656 V Before (in ):
I , =
(RTH
460 V + R2 ) + j ( X TH + X 2 )
I L , = 3 I , =
After (in Y):
(RTH
797 V + R2 ) + j ( X TH + X 2 )
I L ,Y = I ,Y =
(RTH
2656 V + R2 ) + j ( X TH + X 2 )
Therefore the line current will decrease by a factor of 3! The starting current with a -Y starter is
I start =
703 A = 234 A 3
(c) A 125:1 step-down autotransformer reduces the phase voltage on the motor by a factor 08 This reduces the phase current and line current in the motor (and on the secondary side of the transformer) by a factor of 08 However, the current on the primary of the autotransformer will be reduced by another factor of 08, so the total starting current drawn from the line will be 64% of its original value Therefore, the maximum starting current drawn from the line will be
I start = (064 )(703 A ) = 450 A
7-25 When it is necessary to stop an induction motor very rapidly, many induction motor controllers reverse the direction of rotation of the magnetic fields by switching any two stator leads When the direction of rotation of the magnetic fields is reversed, the motor develops an induced torque opposite to the current direction of rotation, so it quickly stops and tries to start turning in the opposite direction If power is removed from the stator circuit at the moment when the rotor speed goes through zero, then the motor has been stopped very rapidly This technique for rapidly stopping an induction motor is called plugging The motor of Problem 7-19 is running at rated conditions and is to be stopped by plugging (a) What is the slip s before plugging (b) What is the frequency of the rotor before plugging (c) What is the induced torque ind before plugging (d) What is the slip s immediately after switching the stator leads (e) What is the frequency of the rotor immediately after switching the stator leads (f) What is the induced torque ind immediately after switching the stator leads SOLUTION (a) (b) (c) The slip before plugging is 0038 (see Problem 7-19) The frequency of the rotor before plugging is f r = sf e = ( 0038)( 60 Hz ) = 228 Hz The induced torque before plugging is 411 N m in the direction of motion (see Problem 7-19)
(d) After switching stator leads, the synchronous speed becomes 1800 r/min, while the mechanical speed initially remains 1732 r/min Therefore, the slip becomes
(e) (f)
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