Motor control via VFD using V/F ratio in Squared mode 4

[bamia]

Hello everyone,

This is from ABB ACH580 firmware manual:

In squared mode (default), the ratio of the voltage to frequency increases as the square of the frequency below the field weakening point. This is typically used in centrifugal pump or fan applications. For these applications, the torque required follows the square relationship with frequency. Therefore, if the voltage is varied using the square relationship, the motor operates at improved efficiency and lower noise levels in these applications. Thus using squared mode saves energy.

Can you please break it down for me in the simplest way possible.
Say for the following motor: 600 VAC, 20HP, 19.2FLA, 1770 RPM., 60Hz

In linear mode: 600/60 = 10v/Hz. So, at 10 Hz the drive will supply the motor with 100 volts and at 20 Hz, 200 volts, etc.
In Squared mode it is difficult to determine where the model starts, but let's say at 10 Hz the drive supplied 100 volts then at 20 Hz the drive will supply 20-10=2 squared = 4 x 100 volts = 400 volts.
Is that how it's suppose to work? Does it work that way from 1 volt and keeps getting squared every increment? We're going to run out of voltage pretty quick.
Clearly I'm missing the point here?

Thanks

 

[LittleInch]

The key word is below the field weakening point.

At synchronous speed you haven't reached that point.

So I read it as that this speed , 1770, is the end point, I.e.10V/ Htz.

Below that the squared rule operates.

 

[waross]

Clearly I'm missing the point here?

Think of it as voltage root mode.
Start with rated voltage at rated frequency and work backwards down from there.
Eg: 50% Hertz = 25% Volts.

 

[edison123]

The key word is below the field weakening point.

At synchronous speed you haven't reached that point.

So I read it as that this speed , 1770, is the end point, I.e.10V/ Htz.

Below that the squared rule operates.

At around 25 Hz, you already hit the field weakening point but achieved only at 42% synch speed.

Hope jeff weighs in. Never heard of this squared mode.

lps for OP for starting this thread.

 

[LionelHutz]

Your math isn't correct.

Make 0-60Hz = 0-100% frequency or 0-1 PU frequency. The voltage is the square of this frequency range.

The field weakening point happens when the VFD runs out of voltage. This typically occurs around 100% speed.

 

[Gr8blu]

Almost all motor-driven process applications fall into one of three possible torque profiles.
1) CP (constant power). The motor produces the SAME power at all speeds.
2) CT (constant torque). The motor produces power that is LINEARLY proportional to the speed.
3) VT (variable torque). The motor produces power that in NONLINEAR with respect to speed (e.g. affinity laws: k = HP/RPM^2 or k = HP^3/RPM^2)

A motor may have multiple profiles, depending on the operating speed.
One of the most common is to have a constant (or variable) torque range followed by a constant power range.
Typically, the constant (or variable) torque range is from zero to some frequency; most cases limit the upper end to the prevailing AC line frequency (50 or 60 Hz).
The constant power range then extends into the "field weakened" range, which corresponds to frequencies above the nominal line (> 50 or 60 Hz).

Most drives are essentially current limiting devices. This means that the voltage will be proportional to the power output required. For a "square law" relationship such as a pump, the PER UNIT voltage will be proportional to the PER UNIT power and the SQUARE of the PER UNIT speed.

All ABB is telling you is that they pre-programmed a curve in their drive (e.g. "squared mode") to mimic the specific affinity law where:
k = per unit HP/ per unit RPM^2 = per unit V/ per unit HZ^2

 

[jraef]

Think of it as voltage root mode.
Start with rated voltage at rated frequency and work backwards down from there.
Eg: 50% Hertz = 25% Volts.

This. You were starting at a low speed and trying to go up. You need to start at the full speed / full voltage point and reduce the voltage at the square root of the speed change.

The REASON for this is because on CENTRIFUGAL (quadratic) loads like some pumps and fans, as the speed reduces, the actual LOAD required from the motor reduces at the CUBE of the speed change and with less load, the motor requires less torque. Since the motor is not required to deliver full torque, it does not need to be fully fluxed, which would waste energy as heat. So by to reducing the V/Hz ratio as you lower the speed, you reduce the losses in the motor itself, saving additional energy. You also reduce the torque capability at that speed, but again, you don’t need it for THAT kind of load.

Note: Anything other than a centrifugal machine and the motor will likely stall, so don’t use that. The ACH is sold as an “HVAC DRIVE”, for pumps and fans, presuming centrifugal ones, hence this being the default setting.

 

[bamia]

This. You were starting at a low speed and trying to go up. You need to start at the full speed / full voltage point and reduce the voltage at the square root of the speed change.

The REASON for this is because on CENTRIFUGAL (quadratic) loads like some pumps and fans, as the speed reduces, the actual LOAD required from the motor reduces at the CUBE of the speed change and with less load, the motor requires less torque. Since the motor is not required to deliver full torque, it does not need to be fully fluxed, which would waste energy as heat. So by to reducing the V/Hz ratio as you lower the speed, you reduce the losses in the motor itself, saving additional energy. You also reduce the torque capability at that speed, but again, you don’t need it for THAT kind of load.

Note: Anything other than a centrifugal machine and the motor will likely stall, so don’t use that. The ACH is sold as an “HVAC DRIVE”, for pumps and fans, presuming centrifugal ones, hence this being the default setting.

Thank you jraef.

A couple of days after I posted this thread I was working on another one of those drives on an exhaust fan application., small motor: 1.5HP 575V 2A.
It ran well with the default control setting for V/H "Squared". Then as I was testing it in Hand mode, I ran it to 50Hz (drew about 1.97 amps) and let run for a minute or so, then stopped the motor and as the motor was slowing down, maybe about 25Hz (did not actually record the speed it was at, but roughly that), I started the motor again, still set to go to 50Hz. The motor pretty much stalled, the current went up to above 3.25 Amps and it got quite loud.
I stopped it and changed the drive setting to V/Hz in "Linear" mode and retested it the same way and it worked well.
So, to apply what I understood from your post, as the speed dropped the drive lowered the torques (voltage) by the square of the speed and it probably got to a point where the voltage was too low to produce enough torque to restart the drive. Am I on the right track?

 

[dvd]

Just tagging along because I am interested in VFDs, but strictly a mechanical engineer.

Is there some way to correlate the mechanical (HP = T x RPM/63025) with electrical (HP = 1.732 x amps x volts x power factor x etc /746 watts/HP)? Is the torque/amperage constant in this discussion? Power has to equal power.

 

[edison123]

jeff

"So by to reducing the V/Hz ratio as you lower the speed, you reduce the losses in the motor itself, saving additional energy."

Say at 50% speed, the power is X watts

With V/Hz, voltage at 50% speed is 0.5 V => Current is X/0.5 V = 2X => Copper loss 4X^2

With V^2/Hz, voltage at 50% speed is 0.25 V => Current is X/0.25 V = 4X => Copper loss 16X^2

While iron loss (which is originally lower than copper loss) is reduced by squared voltage, the copper loss is now 4 times along with the risk of torque loss. How is this power saving?

 

[Gr8blu]

Muthu
The drive acts to limit current - which means it is typically set to maintain a specific value.
The copper loss does not change (much) when operating at reduced V/Hz because the CURRENT DOES NOT CHANGE.

The POWER at a given speed point changes but that is because the VOLTAGE is changing. (P = V * I)

At full load, 1 pu power = 1 pu volt * 1 pu amp
At half speed on constant v/hz: 0.5 pu volt * 1 pu amp = 0.5 pu power. Copper loss is 1 pu amp * 1 pu amp * R = I*I*R
At half speed on v2/hz: 0.5 * 0.5 pu volt * 1 pu amp = 0.25 pu power. Copper loss is 1 pu amp * 1 pu amp * R = I*I*R

 

[edison123]

Gr8blu

"At half speed on constant v/hz: 0.5 pu volt * 1 pu amp = 0.5 pu power"

"At half speed on v2/hz: 0.5 * 0.5 pu volt * 1 pu amp = 0.25 pu power"

At half speed, the pump will demand the same power, isn't it? So the current has to double at v2/Hz and hence copper loss is 4 times.

I am assuming here the current limit does not act, which would make it worse.

 

[Gr8blu]

Muthu - Affinity laws.
For a given pressure or velocity, we either have k = (power squared) / (speed cubed) OR k = (power) / (speed squared).

At half speed (= 0.5 pu), the required power for a typical centrifugal fan is (0.5 * 0.5 = 0.25 pu).

 

[edison123]

ok, let's go with your 0.25 pu power at half speed.

what are the currents at V and V^2?

 

[waross]

what are the currents at V and V^2?

you must express V as a PU value of rated V.
Generally less than one.
The actual current and the actual losses depend on the load.
All the calculations in the world will not change the mechanical load.

 

[jraef]

With V^2/Hz, voltage at 50% speed is 0.25 V => Current is X/0.25 V = 4X => Copper loss 16X^2

Nope. You are assuming the LOAD is the same at 50% speed, that’s what is different about a CENTRIFUGAL machine. The actual LOAD on the motor at 50% speed is reduced by the CUBE root (Affinity speed law), so it is 1/8th (12.5%) what it was at full speed. Current follows load. So the idea of current becoming 4X is incorrect, ergo copper losses are not increased, they are DECREASED as well.

The energy savings are really coming from the speed reduction of that centrifugal load (compared to running at full speed and throttling the output mechanically). The V/Hz^2 issue only saves a LITTLE bit more in the reduced iron losses, but for some, every bit counts.

Yes, torque is reduced, but again, BECAUSE it is a centrifugal machine, the torque REQUIRED at the lower speed is reduced as well. So having “constant torque”, as you would want on another type of load like a conveyor, machine tool spindle, even a positive displacement pump, is inherently unnecessary. So the loss of torque at the V/Hz^2 mode is irrelevant.