# Who’s Right?

Q. At my facility, we have several 4160V asynchronous motor applications. A situation arose where a grinding line would blow fuses when anyone tried to start an autogenous mill driven by dual 3000-hp motors. The motors are started across-the-line asynchronously, and the field contactor picks up within 96% of synchronous speed. Because we had some voltage dips, someone said the motors would draw more

Q. At my facility, we have several 4160V asynchronous motor applications. A situation arose where a grinding line would blow fuses when anyone tried to start an autogenous mill driven by dual 3000-hp motors. The motors are started across-the-line asynchronously, and the field contactor picks up within 96% of synchronous speed. Because we had some voltage dips, someone said the motors would draw more current with a lower line voltage because “they are going to produce the full 3000 hp no matter what the line voltage is.” I beg to differ. The voltage dips may cause fuses to blow; but it would be because the motors took longer to accelerate because they could not create enough starting torque. Who’s right? —J.T.

A. J. T. and his associate are partly right, both having part of the truth as to why the 3000-hp motors are blowing fuses upon starting. If the motors are loaded, running current will increase if the voltage dips. The power input to the motor is given by the equation: Pin4 =32I2power factor (PF) . Assuming Pin remains constant (It may increase due to decreased efficiency of the motor), the product of I2PF must increase for a decrease in supply voltage. Typically, PF will increase slightly, and current will increase significantly under this condition.

On the other hand, the starting torque will be proportional to the square of the applied voltage. So, a small reduction in voltage will have significant impact on the starting torque. (A 10% reduction in voltage results in about 20% reduction in starting torque.) This increases the acceleration time required to reach synchronous speed, exposing the fuses to starting current longer. Protective devices respond to I2t, so you may expect protective device’s activation under this condition.

J.T. should consider that higher running current and longer acceleration times are overload conditions and should affect the overload device—not the short-circuit protective device, i.e. the fuse. Take a careful look at the selection of overload and short-circuit protective devices utilized in this application. When properly selected, these devices should protect the motor against overload and short-circuit conditions without encroaching on the damage curve for the motor. —C.D.H.

A. Both views have an element of truth. An AC motor rotates at a designed speed relative to a presumed frequency (usually 60 Hz) and draws the nameplate current at its designed load and voltage. However, accomplishing this demands a certain amount of power—the product of the voltage times the amps. With the designed frequency, but a lower-than-designed voltage, the motor will try to turn at its designed speed. This compensates for the lower voltage with an increase in the current portion of the power equation to produce enough torque to meet the demand of the load.

This is true whether the motor is already at full speed or in a starting situation. However, if it is starting up, the problem will be exacerbated by the several-times-normal starting current. In either case, I am not surprised the fuse blew, which should signal a problem much more costly in terms of dollars and downtime than fuse replacement, such as shortened motor winding life caused by the greater heat resulting from the greater-than-designed current. —R.L.

TAGS: Design