VFD open loop speed control

Hello Everyone

I have a problem with determining motor speed driven open loop from a VFD.

Motor Data: 100 hp 480V 3 phase, 109A FLA 1780RPM

This motor is running a fan that can not exceed 2000RPM. It is currently pulling around 80amps so I know we are not exceeding the Hp of the motor.

The VFD is set at 66Hz

So I know assuming the 1780 is a good number that the motor is running at 1958RPM. My question is two fold
1. Is using full rated load RPM to calculate speed a good practice at loads well below rated load?

2. I followed the threads explaining that using motor current draw to calculate Hp is not a good idea. My AB Power Flex drive displays Flux amps and another amp variable. I am assuming the variable who’s name escapes me right now is the current in phase and therefore a good current number to calculate real work being done by the motor.

Flux amps were 38amps
the other amp variable was 64amps

Does this mean I can ignore motor eff and power factor to get real Hp ratings using in phase current draw?

( E * I * 1.73 )/ 746

I have to get this right. If that fan explodes…they will use my body as replacement fan blades....I run with a rough crowd


Steve D

Its not HP that changes so much as it is Torque but there is a correlation
Torque=(HPx5250)/rpm
RPM=(120xfrequency)/number of poles.

In this case 120x66/4=1980
Torque= (100x5250)/1980 = 265.15

At 60HZ:
120x60/4=1800
100x5250/1800=291.67

Maybe these formulas will help:
http://www.patchn.com/motorformula.htm

Maybe DickDV or someone can explain it better but its my understanding that below 90HZ the motor is still close enough to its HP and torque output to not need derating, assuming this is a inverter duty motor.

First, RS, your formula gives the nominal motor speed. Assuming a normal squirrel cage inductive motor, for actual speed the formula should read:

RPM = [(120xfrequency)/number of poles] - SLIP

The slip varies as a function of load, with slip nearly zero at low load and maximum at full rated load. Typical slip values are 2% to 4% of nominal rpm.

Steve, you say the fan can't exceed 2,000 rpm, and you are measuring the motor rpm, so I assume this is either a direct coupled fan or a 1:1 belt drive. If that isn't the case, you have to include the belt ratio in your max motor speed.

Take heart - the 2,000 rpm max speed almost surely isn't an exact limit or absolute number like the speed of light or pi. In other words, it isn't going to be a case of 2,000 rpm OK and 2,001 rpm the fan explodes! Limiting the maximum frequency of the drive, which can be set from the keypad or using the A-B software, is going to give an accurate enough speed limit for this purpose. If you want, your drive can be set up with the motor nameplate full load speed, and it will internally calculate and display the motor speed taking into account slip, percent of rated load, motor current draw, and so on. This answer will be quite accurate - certainly close enough for your purpose. (But remember to inlcude the ratio between the motor speed and fan speed if necessary.)

"Does this mean I can ignore motor eff and power factor to get real Hp ratings using in phase current draw?"

No - but the power in kW shown on the drive includes correction for the drive efficiency, and depending on the configuration may also include nominal motor efficiency and so on. You should look at the VFD manual to determine what parameters are able to be displayed and how they are calculated.

Is your fan direct connected or a gear ratio?

I have done controls in FPM (you need RPM), then scale the output (Hz) to match. We would verify with a hand held tach. if you need better feed back mount a tach on the fan shaft.

Brian

OK, Steve. Here's how you figure this!

From the motor nameplate data, you can see that this motor is wound with four poles. So, at 60Hz, the speed would be 1800rpm. (The 1780rpm is full-load speed)

Since you want to limit the fan to 2000rpm and it cannot be turning at that speed without being loaded, we need to set the max Hz so, at the fan load, the speed is 2000hz.

Assuming that the fan torque is equal to the available torque from the motor, the amount of speed slip is 20rpm (1800-1780=20). So, we want the Hz that will give 2020rpm at no load.

Using simple proportions, 60x2020/1800=67.3333Hz. I would round this down to 67Hz on the longshot chance that fan torque is less than full rated motor torque.

Clearly, your method which produces 66Hz is not much different, but, if you want to do this with some precision, the above is the way to do it. Just for illustration, 66Hz will give 1980rpm no load and 1960rpm at full load.

Further and an apology

Sorry guys, for trampling on your posts. I didn't see them as they came in while I was typing my response. There was a lot of good info there that I wouldn't have had to repeat.

To explain about motor current, the current in the motor leads is the vector sum of magnetizing amps and torque-producing amps. They are 90 degrees apart from each other.

This 100hp motor nameplate shows 109amps at full load. This is a nice tight efficient motor (only 20rpm slip) so I would expect no-load amps to be around 22% of full-load amps. That would be 24 amps.

To solve for the vector components, picture a right triangle with one leg being the magnetizing amps, the other leg being the torque amps, and the hypotenuse(spelling?)being total amps in the leads. We know that total amps at full load is 109 and magnetizing amps is 24amps from my estimate above.

Solving for the missing side (torque amps)

total amps squared = mag. amps squared + torque amps squared

That would be 109x109 = 24x24 + TxT

Solving for TxT gives 11881 - 576 = 11305

T = square root of 11305 which is 106.3

So 106.3 torque amps represents full load.

Now, Steve tells us that at 66Hz the total amps is 80amps.

Solving the vector equation again using 80 amps for total amps and the same 24amps for magnetizing amps, we get 76.3 torque amps.

Dividing 76.3 by 106.3 we get .718 which means that the load is 71.8% of full rated torque.

See how tough this is using motor lead amps! The easy way is to measure slip speed directly with a handheld tach or a strobe light. The amount of slip is directly proportional to torque output. At the above 71.8% load you should be able to measure 20rpm x .718 =14.4rpm slip with a strobe and get the same answer in a hurry.

Incidently, there is a small error in what I've done above. Since 66Hz is above the base speed of the motor, the magnetizing amps start to change some. The error, in this case, is small enough to make the illustration valid anyway.

I also said earlier that a fan cannot unload so we can assume it is always loaded. This might be a bit reckless if intake vent vanes are still in place. Since we are using variable speed fan control to adjust flow, I would not expect suction-side flow-control dampers to be operable. If they are, the fan could operate in a partial vacuum with resultant unload and slight overspeed. Better to be sure the vanes are fixed open.