ETAP Motor Acceleration Study

Are you sure it's a motor and not a generator, Mildred?

Yes, I do. It is a inducción motor.

Does this seem correct to you?

Please explain your answer.

No, I dont know if It is correct

What do you mean by "at the report"?

Have you measured the voltage on the bus, and, at the terminals of the motor?

When you finish the motor acceleration study you can print a report where apperas the %V at the terminal an at the bus.

At the bus there is 0.4 kV and at the motor terminal 0.399kV

From you original post: "%Voltage at the terminal motor is higher than the %Voltage in the bus" or %Vm > %Vb Where Vm is Terminal motor and Vb is Terminal bus.

From your last post: "At the bus there is 0.4 kV and at the motor terminal 0.399kV" or %Vm (0.399kV) < %Vb (0.4kV).

Please decide which of these statements is true, and then inform us please.

Both. If you put a voltimeter there is 0.399kV at the motor terminal. But in the report of the motor acceleration when the motor is running the %Voltage at the motor is 98% and at the bus 94%

You write "motor acceleration when the motor is running".

If you want to accelerate a motor with VFD, the frequency setting must be ramped-up. So VFD frequency is greater than motor shaft "speed" in Hz plus steady state slip. V/Hz control ⇒volts will be greater than 50 Hz volts. You have not mentioned how much inertia you are accelerating ⇒more torque =more slip = higher frequency = higher volts.

You have not clarified this 50Hz out is with motor steady state or accelerating.

It may seem an elementary point, but is the shaft torque greater than rated for which a VFD might increase voltage above nominal. Or have you set output speed for e.g. 1500 rev/min instead of 1492?

And are you comparing identical waveforms at mains and motor? Measuring rms in both cases?

Is the motor current the same as when it is connected direct to mains with same shaft speed/torque? Same in all phases? VFD Harmonics may cause negative torque needing more slip to compensate.

PWM

http://www.vfds.com/blog/what-is-a-vfd

Can you explain?

"AC motors are designed for a magnetic field (flux) of constant strength. The magnetic field strength is proportional to the ratio of voltage (V) to frequency (Hz), or V/Hz. But a VFD controls the motor speed by varying the frequency of the applied voltage, according to the synchronous speed equation:

N = 120*f / P

Where:

N = motor speed (RPM)

f = input voltage frequency

P = number of motor poles

V/Hz control maintains a constant ratio between voltage (V) and frequency (Hz).
Image credit: Square D

Varying the voltage frequency affects both the motor speed and the strength of the magnetic field. When the frequency is lowered (for slower motor speed), the magnetic field increases, and excessive heat is generated. When the frequency is increased (for higher motor speed), the magnetic field decreases, and lower torque is produced. In order to keep the magnetic flux constant, the V/Hz ratio must remain constant. This keeps torque production stable, regardless of frequency.

V/Hz control of a VFD drive avoids this variation in the magnetic field strength by varying the voltage along with the frequency, in order to maintain a constant V/Hz ratio. The appropriate V/Hz ratio is given by the motor’s rated voltage and frequency. For example, a motor rated for 230 V and 60 Hz will operate best at a V/Hz ratio of 3.83 at all times (230/60 = 3.83)."

http://www.designworldonline.com/faq-vhz-control-mode-ac-drives/

Thank you. But the VFD is working at 50Hz as the bus. The bus is working at 50Hz.

Yes but what is the frequency at the motor terminals? - loss of efficiency of the VFD?

WTF?

Now you explain how the "bus" is at 50Hz!!

At the bus there is 0.4 kV and at the motor terminal 0.399kV

That's pretty close. Let's see:

1. Were these measured with separate meters? Could the meters not be in exact calibration?
2. If measured by the same meter, how much time between measurements? Could the voltage have changed in that time?

they are separate 100m. The measurements are at the starting point of the motor. 98% voltage at motor terminal and 94%voltage at bus.

Maybe the %Voltage at terminal motor indicarles another thing? Not exactly the voltage?

Sorry, I guess you'll need to just ask someone there who can look at what you are seeing.

Nobody know the exactly meaning at the report of the motor acceleration what is the meaning of %voltage kVb terminal y %bus Voltage?

Following other posts, specially #13, I'd say the difference between 0.4 and 0.399kV is too small to worry about.

Why do you say %Voltage? What is the significance of %?

You say it's a motor acceleration study. Where does acceleration come into this?

Why do you want to "study" acceleration anyway?

Presumably you measured acceleration. Was the motor bare shaft or driving a load?

How did the result compare with the figure calculated from torque and moment of inertia?

I repeat the study, this are the result. You can see that the %Terminal voltage at the motor is higuer than the %Voltage at the bus

I can't read that well enough to comment, but I'm still curious about what the study is for. Can you answer my questions?

Cut table in half along vertical mid-line; make black & white images to cut file sizes.

Insert it as two pictures.

It will probably be readable.

When you write "study", do you mean "test on real motor" or a computer simulation?

A 1V drop across 100m seems very low to me, but, it is a drop not a gain as you earlier implied.

VFD stands for "Variable Frequency Drive", so to say it's working at 50Hz doesn't seem very likely.

https://etap.com/motor-starting/motor-starters

click on the demo button.

You state....

"At the bus there is 0.4 kV and at the motor terminal 0.399kV"

First.... what do you understand as "the bus" and in the diagram below show your test point(s) or reply with our own diagram with test points clearly marked..

Second.... why are you referring to voltage in Kv?

In my opinion, it depends where is located the motor side measuring point=before the VFD[point1]or after[point2]. If it was in point 2 no connection will be between the two considered voltages.

The measurement is at the point 2.

http://ecmweb.com/ops-amp-maintenance/troubleshooting-your-vfds

It is likely due to the convergence tolerance of the multiple simultaneous equations, (real & reactive power). The solutions stopped when the convergence reached the tolerance. You probably have the option to force the calculations to a tighter tolerance, or past the point where they stopped.

The steady state motor voltage is the same or lower, if the lead resistance was modeled.

The other unlikely explanation is that at the point of reducing the accelerating torque, the motor momentarily acts like a generator, which will create a higher voltage at the motor terminals. Seems unlikely this imperfect operation of a VFD would be modeled.

Why you found it necessary to model a motor start with a VFD is curious.

Agree with your last line. If the AP told us what the point of the test is it might help us to help him. I've asked more than once.

Is it possible that we are looking at a "non-problem"? I'm not familiar with ETAP, because I use a different software package. However, a VFD can output a voltage marginally above the incoming bus voltage, depending on the actual waveform. Make sure that you are using a "True RMS" reading voltmeter because of the very large amount of high-frequency noise that usually is present on the output of a VFD.

--JMM

In my opinion, even the supply voltage will be only 94% of rated[400*0.94=376 V] the VFD could

achieve 0.98*400-and even more. The d.c. voltage will be between 0.94*400*sqrt(2)=531.7 and

0.866*0.94*400*sqrt(2)=460.5 V D.C.

That means the inverter could still achieve 400 V A.C.- at least.