Service current, drive, lower Hz

[W@ttson]

Suppose I have a motor that when ran at 60HZ its full load current is 100A. The motor is fed by a flux vector drive. The load is a constant torque load.

The input to the drive would be something on the order of Iout x Drive input PF. Say the input to the drive had a pf of 0.95, then the current input to the drive is 95A.

The drive makes up 98% of the load in the distribution system. So I assume that the current at the service will be more or less 100A, factoring in some current for some other misc loads.

Now suppose I run that motor at a maximum Hz of 30hz. This being a constant torque load my torque will be the same so my torque producing current will be the same. I assume my magnetizing current will change a bit due to the change in impedances at the lower frequency. My overall current should be more or less 100A at the output. Perhaps a little higher since my inductance will be slightly higher.

Again I assume that my input to my drive would be around 0.95 of the output current and I should assume that my service current largely stays the same.

Am I getting all of the above correct?

 

[W@ttson]

Perhaps a little higher since my inductance will be slightly higher.

Inductance would be lower, current will be higher due to the lower impedance.

 

[petersonra]

The input current to the drive will reflect the actual power used by the motor and will be close to a unity PF.

It is somewhat counterintuitive but generally the input current to a drive is less than the output current to the motor becasue the output current has a PF shift off of 1, while the input PF is close to 1.

 

[W@ttson]

The input to the drive would be something on the order of Iout x Drive input PF. Say the input to the drive had a pf of 0.95, then the current input to the drive is 95A

It is somewhat counterintuitive but generally the input current to a drive is less than the output current to the motor becasue the output current has a PF shift off of 1, while the input PF is close to 1.

Yes, I was trying to account for that in the above statement. When at 60Hz the output of the drive is 100A, the input is at ~95A.

Given the load is constant torque, what will it be if the motor is run at 30hz?
The output of the drive = ?
The input of the drive = ?

 

[petersonra]

Yes, I was trying to account for that in the above statement. When at 60Hz the output of the drive is 100A, the input is at ~95A.

Given the load is constant torque, what will it be if the motor is run at 30hz?
The output of the drive = ?
The input of the drive = ?

the input current is a reflection of the output power being used.

HP = Torque X RPM/5252

If the RPM is half and the torque stays the same the HP will be half. So roughly the input current will be half what it was at 60 Hz.

 

[W@ttson]

the input current is a reflection of the output power being used.

HP = Torque X RPM/5252

If the RPM is half and the torque stays the same the HP will be half. So roughly the input current will be half what it was at 60 Hz.

That was my first thought!

However, how do the two concepts reconcile (first concept of HP being less so less current and second concept of producing the same torque requires the same current)?

If torque is the same and torque is directly proportional to current, wouldn't the output current of the drive be the same?

If the output current of the drive is the same, wouldn't the input current of the drive be the same say 0.95*output current or so?

I tried referencing this post to get the concept of what happens to current at the output of the drive at lower frequencies: post: https://forums.mikeholt.com/threads/current-output-on-a-vfd.43357/ and then tried to combine that with this post that explores what the input of the drive is as compared to the output of the drive: https://forums.mikeholt.com/threads/input-amps-vfd-overload-condition.140350/

 

[W@ttson]

I guess what it comes down to is that you end up being in the Constant Torque region so the HP is changing but the torque is remaining the same. You have the same V/Hz ratio so your magnetic flux is constant. Since the magnetic flux is constant you maintain the torque and therefore current? Speed is directly proportional so since the voltage is half, the the speed will be half. Since the speed is half, the HP will be half. All this with the same magnetic flux.

 

[winnie]

The drive can be considered a fancy 'buck converter'. The output current is actually circulating through IGBTs (or diodes) on the same bus rail for a chunk of time, and not drawing current from the DC link capacitors. The output current can easily exceed the supply current to the drive.

So at 30 Hz the motor is running at full torque but half voltage. The output current is the same. The IGBTs are running at a lower duty cycle to supply the lower voltage and between supply pulses the current just circulates on the output. Since the output current spends more time circulating, the input current drops.

Look up a buck converter circuit. This is a DC in, DC out circuit with higher output current then input current. Much easier to understand when you don't have to deal with 3 phase AC output.

Jonathan

 

[W@ttson]

The drive can be considered a fancy 'buck converter'. The output current is actually circulating through IGBTs (or diodes) on the same bus rail for a chunk of time, and not drawing current from the DC link capacitors. The output current can easily exceed the supply current to the drive.

So at 30 Hz the motor is running at full torque but half voltage. The output current is the same. The IGBTs are running at a lower duty cycle to supply the lower voltage and between supply pulses the current just circulates on the output. Since the output current spends more time circulating, the input current drops.

Look up a buck converter circuit. This is a DC in, DC out circuit with higher output current then input current. Much easier to understand when you don't have to deal with 3 phase AC output.

Jonathan

Well lucky for me, I recently found my Power Electronics book.

 

[W@ttson]

The drive can be considered a fancy 'buck converter'. The output current is actually circulating through IGBTs (or diodes) on the same bus rail for a chunk of time, and not drawing current from the DC link capacitors. The output current can easily exceed the supply current to the drive.

So at 30 Hz the motor is running at full torque but half voltage. The output current is the same. The IGBTs are running at a lower duty cycle to supply the lower voltage and between supply pulses the current just circulates on the output. Since the output current spends more time circulating, the input current drops.

Look up a buck converter circuit. This is a DC in, DC out circuit with higher output current then input current. Much easier to understand when you don't have to deal with 3 phase AC output.

Jonathan

When you say input current drops, what would the factor be? Would be by a factor of ~2?

 

[winnie]

When you say input current drops, what would the factor be? Would be by a factor of ~2?

Roughly, yes.

Power is being delivered to the DC bus via the input rectifier. Power is being supplied to the motor. When power to the motor drops, power drawn from the supply to the DC bus must also drop.

Jonathan

 

[W@ttson]

Roughly, yes.

Power is being delivered to the DC bus via the input rectifier. Power is being supplied to the motor. When power to the motor drops, power drawn from the supply to the DC bus must also drop.

Jonathan

That is extremely interesting.

If ran at 60Hz, the input current is roughly the output current (difference by input pf). When the motor is ran slower, and assuming the same load, the input power no longer equals roughly the output power. The output power stays the same but the input power reduces.

This is really interesting. So I could be looking at the following:
60Hz
The output of the drive = 100A
The input of the drive = 95A

30Hz
The output of the drive = 100A
The input of the drive = ~55A

And I think you are absolutely right on this but just could not justify it.

 

[wwhitney]

If ran at 60Hz, the input current is roughly the output current (difference by input pf). When the motor is ran slower, and assuming the same load, the input power no longer equals roughly the output power. The output power stays the same but the input power reduces.

No, that's not correct. Energy is conserved, and so if energy is not being stored somewhere, power is conserved.

When you run the motor at 30 Hz instead of 60 Hz, isn't the output voltage also halved? Constant V/Hz. In which case if the output current is the same, the output power is halved. Whereas the input power is halved by halving the current.

Cheers, Wayne

 

[winnie]

When the motor is ran slower, and assuming the same load, the input power no longer equals roughly the output power. The output power stays the same but the input power reduces.

Not quite.

When you slow the motor down, the output power drops.

Mechanical power is torque times speed. So your constant torque load draws less power as it slows down.

A rough approximation is that motor voltage is proportional to speed and motor current is proportional to torque.

The only reason input current and output current are roughly equal at 60Hz is because that is the frequency that the motor is designed to operate at 460V. Input voltage =output voltage so input current = output current.

If you had a 480V inverter running a 230V motor, then at 60Hz the output current would be about 2x the input current.

 

[W@ttson]

If ran at 60Hz, the input current is roughly the output current (difference by input pf). When the motor is ran slower, and assuming the same load, the input power no longer equals roughly the output power. The output power stays the same but the input power reduces.

Sorry I was so excited I miss typed. In the above where I say "power" I meant to say current.

Power in should = Power out leaving losses out.