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9. Braking of dc Motors. Regenerative. F. m. F. m. Speed. w. w. 3. F. 2. 1. F. l. F. F. l. mg. mg. T(. T(downhill). Torque. Speed. 1. T(uphill). Torque. Review of Basic Equations. Example.
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F m F m Speed w w 3 F 2 1 F l F F l mg mg T( T(downhill) Torque Speed 1 T(uphill) Torque
Example • A 440 V dc motor has a rated armature current of 76 A at a speed of 1000 rev/min. The armature resistance of the motor is 0.377 W, and the rotational losses is 1 kW. The load of the motor is bi-directional.
a. Calculate the no load speed of the motor. b. If the armature current is 60 A during a regenerative braking, calculate the motor speed c. Calculate the load torque during regenerative braking d. Calculate the speed during regenerative braking e. Calculate the power delivered by the source under normal motor operation. f. Calculate the generated power during regenerative braking.
Speed V 1 w 2 o1 3 1 V 2 w o3 4 Torque T 1 Regenerative Braking during Speed Reduction
Ra Ib Rb Vf If Ea w
R b2 R R b3 b1 B I I I b1 b2 b3 Speed A I a Current
Constant Torque
Example • A 440 V dc Separately Excited motor has a rated armature current of 76 A at a speed of 1000 rev/min. The armature resistance of the motor is 0.377. Assume that the load torque is gravitational. A dynamic braking technique employing a braking resistance of 2 is used. Calculate the new steady state speed.
Solution Compute
Counter Current Braking • Plugging • Terminal voltage reversl
Speed V 1 w 2 o1 3 1 V 4 2 Torque w T o3 sc 5 T 1
Speed V 1 w 2 o1 3 1 V 4 2 Torque w T o3 sc 5 T 1
Speed 2 1 Torque 3 Current (I) 4 T l T l
Speed 2 1 Torque 3 Current (I) 4 T l T l
Example 9.5 • A dc motor has an armature resistance of 1 , and K = 3 Vsec. When the motor’s terminal voltage is adjusted to 320 V, the motor speed is 1000 r/min. A TVR braking is applied, calculate the value of the braking resistance that would reduce the maximum braking current to twice the rated current.