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An induction motor draws very high starting current during its acceleration to full rated speed, typically 6 to 10 times the full load current. Reduced starting current is desirable where the electrical grid is not of sufficient capacity, or where the driven load cannot withstand high starting torque.
From this test, short-circuit current at normal voltage, power factor on short circuit, total leakage reactance, and starting torque of the motor can be found. It is very important to know a motor's starting torque since if it is not enough to overcome the initial friction of its intended load then it will remain stationary while drawing an ...
Now, if this motor is fed with current of 2 A and assuming that back-EMF is exactly 2 V, it is rotating at 7200 rpm and the mechanical power is 4 W, and the force on rotor is = N or 0.0053 N. The torque on shaft is 0.0053 N⋅m at 2 A because of the assumed radius of the rotor (exactly 1 m).
Inrush current, input surge current, or switch-on surge is the maximal instantaneous input current drawn by an electrical device when first turned on. Alternating-current electric motors and transformers may draw several times their normal full-load current when first energized, for a few cycles of the input waveform.
As an example, a 250 kVA motor–generator operating at 300 ampere of full load current will require 1550 ampere of in-rush current during a re-closure after 5 seconds. This example used a fixed mounted flywheel sized to result in a 1 ⁄ 2 Hz per second slew rate. The motor–generator was a vertical type two-bearing machine with oil-bath ...
As an example, consider the use of a 10 hp, 1760 r/min, 440 V, three-phase induction motor (a.k.a. induction electrical machine in an asynchronous generator regime) as asynchronous generator. The full-load current of the motor is 10 A and the full-load power factor is 0.8. Required capacitance per phase if capacitors are connected in delta:
Fixed-speed loads subject the motor to a high starting torque and to current surges that are up to eight times the full-load current. AC drives instead gradually ramp the motor up to operating speed to lessen mechanical and electrical stress, reducing maintenance and repair costs, and extending the life of the motor and the driven equipment.
The Load factor is the ratio of the load that a piece of equipment actually draws (time averaged) when it is in operation to the load it could draw (which we call full load). For example, an oversized motor - 15 kW - drives a constant 12 kW load whenever it is on. The motor load factor is then 12/15 = 80%.