Full load amps is found on the motor's nameplate, and is unique to that particular brand and model number. A Magnetec 2hp motor will draw different amps than a Baldor 2hp, etc. There are high efficiency motors, and not so high as well. Consult the nameplate. If you want a "rule of thumb" to estimate the RLA (Running Load Amps - this is the term most often found on the nameplate - same as full load amps), then use this: Figure about 1400 watts per horsepower. Divide Watts by volts to get Amps, for example: * 3hp motor, 120V * 3hp X 1400 W = 4200 W * 4200 W / 120 V = 35 A This formula is only for single-phase motors! I looked up a typical 3hp 120 V Baldor motor in the Grainger's catalog, and the RLA for it was 32.0, so this will get you in the ballpark. I have seen 3hp 120 V motors as low as 25 A and as high as 40, so DO NOT use the rule of thumb to size branch circuit conductors or overcurrent protection! For 3-phase motors: Figure about 1100 watts per horsepower (3-phase is a little more efficient). Divide Watts by Volts, then divide the answer by 1.73 to get Amps, for example: * 3hp motor, 208V 3Ph * 3hp X 1100 W = 3300 W * 3300 W / 208 V = 15.86 * 15.86 / 1.73 = 9.16 A This formula is only for 3-phase motors! My Grainger's catalog motor with the same Voltage and HP had a nameplate rating of 8.34 A.
First, calculate the load on the motor in watts, kW, or MW, depending on the size of the motor. Then compare to the motor curves provided by the manufacturer.
If you really want to calculate this, you'll need significant information from the manufacturer about the motor - characteristic impedances, time constants, etc. It's not a simple plug and chug operation. That's why they provide the curves.
Answer :
Power = 1.73 x Voltage x Current x Power factor.
Current = Power / (1.73 x Voltage x Power factor)
If you want to add efficiency factor you may do so, say 90% efficiency for example.
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Motors are rated in horse power in North America. Different size HP motors have different voltages and full load amperages.
If you know the HP of the motor then use the following formula to find the motors full load amperage; Amps = HP x 746/1.73 x Volts x % eff x pf.
A standard motor's % eff increases as the HP increases. From 10 to 100 HP the scale increases from .86 to .92.
A standard motor's power factor increases as the HP increases. From 1 to 100 HP the scale increases from .84 to .91
It should be in the manufacturers' data sheet, but it is not a very important parameter for a motor that is usually run with a mechanical load.
dnt suck..have a beer
hi.. this is pankaj working as power engineer. we have a 11 kw induction motor which is used in fluidising blower. its no load current i sarround 9.5 to 10.6 amp. and under load its ampere does nt exceed 15 amp. its winding resistance is 1.4 ohm.
As the load increases on a motor the amperage rises. Increasing the load amps over the motors full load amp rating can shorten the motor's life because of excessive heat. Motors should have overload protection in their circuits to prevent this condition from happening by tripping the motor contactor and opening the motor's voltage supply. <<>> Gets very warm .... vibrations., possible melt, seize, much energy expended.
A motor that is rated at 250 full load amps will need 400 amp time delay fuses or a 500 amp frame breaker with adjustable amperage trips. The wire size for this motor has to be 125% of the motors FLA. 250 x 125% = 312 amps. A 350 MCM copper conductor with an insulation factor of 90 degrees C is rated at 325 amps.
There are many types of overload protection these days and all are based on the FLA (full load amps) of the motor.
Motor: 12v AMP usage: 1.2amp Watt: 12x1.2=14.4W
Sizing circuit breakers for motors falls into a different category. Because of the inrush of current when the motor is in the stopped position, the breaker has to be sized to allow for this. Breakers usually are 250% of the nameplate full load amp rating. The wires that feed the motor are sized to 125% of the full load amps of the motor.
hi.. this is pankaj working as power engineer. we have a 11 kw induction motor which is used in fluidising blower. its no load current i sarround 9.5 to 10.6 amp. and under load its ampere does nt exceed 15 amp. its winding resistance is 1.4 ohm.
You should not exceed 80% of the breaker rating for a continuous load. Therefore, the continuous load should not exceed 16 amps.
As the load increases on a motor the amperage rises. Increasing the load amps over the motors full load amp rating can shorten the motor's life because of excessive heat. Motors should have overload protection in their circuits to prevent this condition from happening by tripping the motor contactor and opening the motor's voltage supply. <<>> Gets very warm .... vibrations., possible melt, seize, much energy expended.
A #10 copper conductor with an insulation rating of 90 degrees C is rated at 30 amps. The only time that a breaker larger that the load maximum capacity can used is in motor connections. A breaker can be used sized at 250% of the motor's full load amperage. In this case a 50 amp breaker can be used on a #10 conductor if the motor's full load amps falls within the amperage's of 20 - 22 amps.
Start by checking the motor current when the motor is running under load with a clamp on amp meter. If it is over nameplate rating, check and find out if the load the motor is driving is somehow dragging and creating more load that normal. 98% of the time it is the load seizing up that causes motors to trip. Drop the load off the motor and see if the motor runs under nameplate amperage. If it does that is good. Check for loose connection in the motor contactor and around the overload heater block. Check and make sure the overload heaters are set to the FLA of the motor. As a last resort megger the motor to make sure one of the phases has not grounded out.
A fuse rating is normally the 120% of the rated full load current. So, 4 amps times 120% is (4x120%) = 4.8amps
It depends on how powerful the motor is, and what type of load it is driving.
A motor that is rated at 250 full load amps will need 400 amp time delay fuses or a 500 amp frame breaker with adjustable amperage trips. The wire size for this motor has to be 125% of the motors FLA. 250 x 125% = 312 amps. A 350 MCM copper conductor with an insulation factor of 90 degrees C is rated at 325 amps.
the three phase dc motor never seen yet motor eaither 3phase m/r or dc motor.
If we consider the rpm to be 1450, then the full load current of 3 phase, 415V motor is approx 190 amps. Now, the starting current of a induction motor is 5-7 times the full load current. Hence the starting current will be approx 950 - 1330 amps. Similar calculations can be made for below mentioned ratings also: 110 KW, 220 V ------F.L. current: 350 amps 110KW, 240 V-------F.L. current: 325 amps
not smart to do it cant carry the load of the circuit it will blow. in emergency maybe but never go with a higher rating