calculate amps on each phase for a motor

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larsahl

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Hi,

I need some help with calculating what amps it is on each phase to dimension the wire area.
It is a frequency converter with following specs on marking plate max input of 20 amps, 3-phase 400V or can be configured 480Y/277 V. The load is a motor of 5.5kW with the following specification as seen in the picture below.

Capture.JPG

Its a three wire system without neutral wire. Three wire input to frequency converter and three wire out to motor.
What will the ampere be on the input wires to the frequency converter on each wire and total?
What will the ampere be on the output wires from frequency converter on each wire and total?

I'm thinking about the I = Iphase * sqr(3) formula used in school for many years ago but havent all in my memory to figure it out if i'm thinking right.
 
larsahl said:
Hi,

I need some help with calculating what amps it is on each phase to dimension the wire area.
It is a frequency converter with following specs on marking plate max input of 20 amps, 3-phase 400V or can be configured 480Y/277 V. The load is a motor of 5.5kW with the following specification as seen in the picture below.

View attachment 9593

Its a three wire system without neutral wire. Three wire input to frequency converter and three wire out to motor.
What will the ampere be on the input wires to the frequency converter on each wire and total?
What will the ampere be on the output wires from frequency converter on each wire and total?

I'm thinking about the I = Iphase * sqr(3) formula used in school for many years ago but havent all in my memory to figure it out if i'm thinking right.
Click to expand...
Looks like a motor nameplate so you are given the motor current.
VSD input current would normally be lower than that owing to power factor differences but, to be code compliant, input conductors may have to be rated in accordance with VSD rated input current plus any mandatory margin if applicable.

The "in total" and sqrt(3) don't apply. The three phase currents are what they are and one would expect them to be balanced.
 
larsahl said:
I'm thinking about the I = Iphase * sqr(3) formula used in school for many years ago . . . .
Click to expand...
That formula will tell you the line current, given the phase current in a delta-connected generator or load. It is not relevant to what you are asking here, because the parameters you are given are all external to the motor windings. We don?t know, or care, what currents are flowing in the motor windings.
larsahl said:
What will the ampere be on the input wires to the frequency converter on each wire and total?
Click to expand...
For starters, I look forward to the day when people cease forever all questions that speak of total currents in a three phase system. The notion is meaningless, but I doubt that this practice will cease during my active career. :happysad: Sigh.

What the name plate tells me is that you have a 5.5 kW load. I will assume a 80% power factor. 5500 watts divided by 0.8 gives you 6875 VA. Divide that by a 480 volt supply, and divide again by the square root of three, and you get a line current of 8.3 amps. As a cross check, table 430.250 tells me that for a 460 volt motor rated at 7.5 hp (slightly higher than that shown on the nameplate), the current will be 11 amps.

For planning purposes, I would use the 11 amp value.

 
charlie b said:
That formula will tell you the line current, given the phase current in a delta-connected generator or load. It is not relevant to what you are asking here, because the parameters you are given are all external to the motor windings. We don?t know, or care, what currents are flowing in the motor windings.
For starters, I look forward to the day when people cease forever all questions that speak of total currents in a three phase system. The notion is meaningless, but I doubt that this practice will cease during my active career. :happysad: Sigh.

What the name plate tells me is that you have a 5.5 kW load. I will assume a 80% power factor. 5500 watts divided by 0.8 gives you 6875 VA. Divide that by a 480 volt supply, and divide again by the square root of three, and you get a line current of 8.3 amps. As a cross check, table 430.250 tells me that for a 460 volt motor rated at 7.5 hp (slightly higher than that shown on the nameplate), the current will be 11 amps.

For planning purposes, I would use the 11 amp value.
Click to expand...

What do you make of the 20.6A on the nameplate?
 
larsahl said:
Hi,

I need some help with calculating what amps it is on each phase to dimension the wire area.
It is a frequency converter with following specs on marking plate max input of 20 amps, 3-phase 400V or can be configured 480Y/277 V. The load is a motor of 5.5kW with the following specification as seen in the picture below.

View attachment 9593
Click to expand...
Looking closely at the nameplate, I am not sure that this is in fact a normal AC motor.
The allowed frequency is 0-200Hz (includes DC!) and it refers to a PWM SOURCE VOLTAGE....(rest is out of the photo.)

If I took the rating information literally, I would say that this is a three phase DC traction motor which is designed to be connected only to a pulsed controller to provide the three phases.
It is probably not a universal motor (AC/DC) since I do not know how you would make a three phase one of those.

If my guess is wrong, then the nameplate is flawed and I would have a problem determining anything from it without more information from the manufacturer.

Bottom line: Things are not always what they seem.
 
charlie b said:
Starting current? Is the nearby word "peak" associated with the 20.6A value? That is my question.
Click to expand...
The nameplate rather suggests a variable frequency inverter as the source in which case starting current wouldn't be applicable. And for DOL starting it would likely be rather more that 20.6A.
The OP needs to come back with more information.
 
I am guessing that this is a three phase brushless DC motor. This guess is primarily based on the fact that there is no mention of slip, and that the Ke factor and stator resistance are given, but no mention is made of rotor resistance or inductance. Since I don't to BLDC machines, the below is just an extension of my guessing ;)

The motor is rated for a maximum of 3000 RPM, and a maximum continuous stall torque of 237 lb in. The product of this torque and speed exceeds the 5.5 KW rating of the system, implying that maximum torque is only available to a fraction of max speed.

The motor Ke implies that the output voltage of the VFD is lower than the input, and the output current is higher. At 3000 RPM the back-EMF of the motor is 540V peak to peak, or 311V RMS. (Note: the back EMF might not be sinusoidal!) Since full torque cannot be achieved at full speed, the highest current will not be seen at full speed; so you can't even size this as if it were a 311V 5.5KW motor.

At stall, depending on rotor angle, you could have DC running in only part of the stator. I am guessing that the 20.6A is the rated peak current (rather than RMS) that the stator can handle on a continuous basis, to give the rated peak continuous torque.

The input to the VFD would be sized for its input ratings. The conductors between VFD and motor would be sized to continuously carry 20.6A.

-Jon
 
winnie said:
I am guessing that this is a three phase brushless DC motor. This guess is primarily based on the fact that there is no mention of slip, and that the Ke factor and stator resistance are given, but no mention is made of rotor resistance or inductance. Since I don't to BLDC machines, the below is just an extension of my guessing ;)
Click to expand...

I don't think a DC motor would list an RMS voltage.:?
 
Oh, and as a BLDC machine, an inverter would be required for operation and matched to the machine. No such thing as DOL starting here.

A BLDC machine uses the same sort of switching transistors as are seen in VFD systems for other AC motors, but the transistor commutation or PWM pattern is controlled by rotor position sensors. In a sense a BLDC machine is an ordinary brush commutated DC motor where the brushes and commutation have been replaced by rotor position sensors and transistors.

Some BLDC machines are controlled by adjusting the DC voltage going in to the transistors. The motor here is fancier, with a voltage source inverter and something else going on to match the commutation to rotor position.

-Jon
 
david luchini said:
I don't think a DC motor would list an RMS voltage.:?
Click to expand...

The term 'brushless DC motor' is something of a misnomer.

Keep in mind that there are very few true DC motors in use. Most of the things called DC motors are machines which are DC powered, but which use an internal device to convert DC to AC: the mechanical commutator. The current flowing in the rotor bars is clearly _AC_.

A 'brushless DC motor' is really a permanent magnet AC synchronous motor. What makes people call it brushless DC is the control paradigm: you run the machine with switching transistors, and you control the gates using rotor position sensors. This makes the machine act like a common brush commutated DC motor, in the sense that the stator rotating field is always held in proper synchronism with the rotor magnetic field.

I should note that in answering David's comment I will change my guess from BLDC motor to PM synchronous motor :)

-Jon
 
winnie said:
The motor is rated for a maximum of 3000 RPM, and a maximum continuous stall torque of 237 lb in. The product of this torque and speed exceeds the 5.5 KW rating of the system, implying that maximum torque is only available to a fraction of max speed.
Click to expand...

Good spot.
Yes. Speed and torque maximums can't be simultaneous values.
 
It is a frequency driver connected to a Low Inertia Servo Motor.
It's a three phase AC motor.
And the manuals says for the motor, premium permanent magnet rotary servo motor.

The motor manual says also the following:
Continuous Stall Current Amperes (0-peak): 20,6 A
Peak Stall Current Amperes (0-peak): 68,0 A
Rated Output: 5.5kW

I = ampere
P = watts

I = P / (srq(3)*400*cos fi)
I do not find any information about cos fi.

But if we say I = 5500 / (1,732*400*1) = 7,94 A
 
larsahl said:
It is a frequency driver connected to a Low Inertia Servo Motor.
It's a three phase AC motor.
And the manuals says for the motor, premium permanent magnet rotary servo motor.
Click to expand...

That clears things up.

larsahl said:
The motor manual says also the following:
Continuous Stall Current Amperes (0-peak): 20,6 A
Peak Stall Current Amperes (0-peak): 68,0 A
Rated Output: 5.5kW
Click to expand...

If the motor manual says this next to the motor specifications, then I would take it to mean that the motor _mechanical_ output is 5.5kW

larsahl said:
I = ampere
P = watts

I = P / (srq(3)*400*cos fi)
I do not find any information about cos fi.

But if we say I = 5500 / (1,732*400*1) = 7,94 A
Click to expand...
While you may not know the power factor (cos fi) for your current equation to work, a much bigger unknown is the voltage, which you have assumed to be 400.

Even if the supply to the system is 400V, you can be quite certain that the voltage being supplied to the motor is much less than 400V.

The control electronics is reducing the voltage supplied to the motor to that required for operation at any given speed. At the same time, current can circulate between the control electronics and the motor, so that the current to the motor will be greater than the supply current.

In order to calculate the current to the motor at any given time, you will need to know the voltage being used at that time, as well as the actual power being delivered.

-Jon
 
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