Induction Motors and Undervoltage Conditions

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Induction motor 3ph-380V 50hz 15kW is running on its full-load, then suddenly voltage drop down to 3ph-300V balance. What will happen to motor? Ignore any protection devices.

Same case as above, but motor is solo-running without loads connected, what will happen to motors?

Hi,

In both cases above, Iwill answer "What should happen to the motor" instead of "What will hapen to the motor".

In the first case.......If the voltage drops to 300v , The motor speed should drop.If the motor speed does not drop and the voltage drops further, there comes a time when the speed WILL drop.For all practical purposes we can say that the motor speed will drop.Lets continue....When the motor speed drops, the line current will increase.Asthe line current increases, the temperature will increase. This current increase, if sufficently large enough the overload protection will trip and your motor is protected.If this current increase is NOT large enough to trip the overload protection, then the motor will slowly overheat and burn itself.

Should we ignore the protective device, then unless the maintenance guy turn the motor off, the motor temperature will increase and burn itself up.

In the 2nd case.........On no load the motor speed should not drop. For all practical purposes we can say The motor speed will not drop. Should the voltage drop low enough the speed will drop, the current will increase .......and the 1st case senario takes effect.

The motor will run normally. nothing will hapen to the motor

I have a question for you.

The speed of an induction motor is governed by the formula

n is equal to 120 f /p where

n is the speed of the motor in rpm

f is the line frequency

p is the number of poles.

The speed of an induction motor according to the formula is independant of the voltage. So why in case 1 above,the speed dropped when the voltage dropped.

Regards

The speed dropped because the rotor lag increased.In a normal motor , there is a certain amount of rotor "slip".If there is no slip, there is no current draw,and consequently, no work being performed.The more slip, the more the rotor current lags the field current, and the more current the motor draws, and under normal conditions, the more work is performed.However, when the voltage drops, rotor slip increases due to insufficient current, initially, and the motor perceives this as an increase in load.As the slip increases, the current increases, and the losses of the motor increase,along with the temperature of the windings.This damaging conditon will eventually lead to motor failure.Thermal damage is accumulative.30 minutes overtemp today plus 30 minutes overload tomorrow has the same effect as 1 hour of overtemp all at once.(given the same temp each time). It does not "heal" when the motor cools down.

If the electrical supply capacity is insufficient to supply the motor at full load, the voltage will drop.In the case of a bad bearing,combined with full load, this may occur if the system capacity is overloaded.

(Disregarding OL, short circuit, and Thermal protection)

Even though the motor is inherently designed for a certain RPM, described by the given formula, when the motor is overloaded by excessive load,or falsly overloaded by low voltage or improper frequency, the rotor slip exceeds a full revolution, and the RPM drops.

A motor actually is more efficient at 10% overvoltage than at 10% undervoltage.

Hope this helps

EZ Street

draws more amps--should maintain the speed because that's determined by phase, not voltage. Forcing more amps through wires designed for lower amperage will cause wires to heat up--not necessarily overheat, but they will get hot. Smell of burning rubber will enter the room as volatiles in insulation turn to gas. Insulation throughout the windings will become more brittle and less effective as complex molecules break down and more carbon is left in its place. Carbon is a conductor, so some power will begin leaking across the windings, causing the equipment to need even more amperage. At some point windings will begin to fuse, insulation will begin to smoke, and if you truly lack any safeguards like overloads or circuit breakers or fuses, you'll wind up with a fire on your hands, a shorted-out motor, and a big rewinding job. An 80 volt drop might not have such dire results, but it will weaken the motor and make it more vulnerable to failure. Engineers tend to design motors like this with only a 10% margin of error, which would have been exceeded after voltage dropped to 342v. Engineers, please make equipment 20-50% tolerant.

Good answer, and regarding your request; they did make such equipmet in the past decades, but it was oversized and non cost effective.

I can speak from experience, and theory. The speed will drop.It is the drop in speed of the rotor that causes the increase in amps.The induced field in the rotor is lagging the winding speed.This is called slip.

A 1200 rpm motor does not actually turn 1200 rpm, if it did, it would consume no power. All a/c motors are designed to have a certain amount of slip.

That is why the name plate will state:1175 rpm, 1725 rpm, etc.

The rotating magnetic field, however (the windings) are actually rotating at the ideal theoretical speed (1200 rpm, 1800 rpm, etc.).

A sychronous motor is an exception, but it must have external excitation.These motors are sometimes used for power factor correction.

need a bearing?

EZ STREET is correct.

Study A/C motor theory.

Learn.

Add a new wrinkle.

Live long and prosper.

Exactly right.

GA!

Im most cases the moter will over heat and burnup

True statement.

"It ain't what people don't know that makes them ignorant.It's what they know that just ain't so"

_ Abraham Lincoln

Dear Mr.AbuKhansa,

The Power developed will fall which can be calculated as (300/380)^2= 0.6232 OR 62.32% of the power at 380 Volts.

Speed will slightly fall, i.e. SLIP will INCREASE. Over Heating of Motor will take place. When Load is not on the Motor, NO SIGNIFICANT change will be noticed, except more warm of the motor since the Power Factor is LOW and Heating will be more.

DHAYANANDHAN.S

A 15 kW, 380 V, 50 Hz motor is running at its rated load/full load and drives a constant torque load. Suddenly system voltages drops from 380 V to 300 V. What would happen to the load?

Nameplate of the motor is: 15 kW, 380 V, 50 Hz, 31.5 A, P.F 0.85, 1465 RPM (Efficiency: 85 %)

When Voltage drops, following will change :-

· Motor Output power

· Motor Torque

· Motor speed

At 380 V and at Full Load:-

Mechanical Power delivered to the load: 15 kW

Electrical Power Consumed by motor: 17.64 kW

Load Torque: 97.82 N.m

Speed of motor at 380 V : 1465 RPM (Slip_380V = 1500 -1465/1500 = 2.33 %)

When Voltage Drops to 300 V:-

Torque of motor at 300 V:-

T_300V = T _380V (V_{300 V}/V_{380 V}) ²

T_300V = 97.82 (300/380) ²

T_300V = 60.966 N.m

Motor is able to deliver only 60.966 N.m. It means Torque delivering capability of motor is reduced by 37.68 %.

Speed of Motor at 300 V:

Increase in Slip is

Slip _300V = 1/(V_300V/V_{380 V}) ²

Slip _300V = 1.60 or 160 % = 1.6 x 2.33 = 6.06 %

Speed at 300 V = 1409 RPM

Due to increase in Slip, RPM of motor reduces from 1465 RPM to 1410 RPM.

Mechanical Power Delivered to Load at 300 V:-

P_{M at 300 V} = T_{300 V} x Speed _300V

P_{M at 300 V} = 8.997 kW

Mechanical Power delivering capability of motor reduces (40 %) from 15 KW to 9 KW.

Summary:-

Motor could not drive the load and trips on overload.

"abi bhi jis ko sumhaj na aye ooo khutyy da putur hae"

A1) One cannot <...ignore...protection devices....>; there is every chance that the <...motor...> will stall and a <...protection device...> will disconnect it from the supply.

A2) Probably nothing.