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Today I saw a setup in a factory in which a 2.2 KW 3 phase induction motor is connected to a 1.5 KW VFD.

The motor plate: enter image description here

the VFD plate: enter image description here

The motor is connected to a gearbox (the transmission ratio is 1/20 I guess) to control a roller in a fabric rolling machine.

My question is how is it possible to connect a 1.5 KW VFD to a 2.2 KW motor?

Additional Information

I found out that the star connection is used for the motor which, accordong to the motor's plate, demands 5.73A which is lower than the vfd's maximum current (7.5A) does that make sense?

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It is possible that the driven machine does not require 2.2 kW. Oversized motors are sometimes selected for use with VFDs to prevent the motor from overheating during sustained operation at low speed. It was once very common to use an oversized motor. Now motors that are rated for VFD use have better thermal capability and it is preferable to select a motor that is rated for the output power that is is expected to be required.

It is probably not necessary to set the current limit to 7.5 amps. The VFD probably has the ability to provide as much as 150% of rated current for a short time, typically 60 seconds. Check the detailed specifications for the VFD. The VFD should protect itself from excessive duty time at current exceeding the rated current.

Added Information re Comments

The motor is designed to operate at 400 volts and 50 Hz in the star connection. The VFD rating plate indicates that it can produce no more than 240 volts output. At 240 volts the motor should operate 30 Hz (8 volts per hertz). The frequency at which the maximum output voltage occurs can be configured over a wide range in most VFDs. If rated V/Hz can be provided over some range of frequency, rated torque can be provided over that range. VFDs are usually configured for the nameplate V/Hz of the motor for as much of the speed range as possible. The VFD could be configured for rated V/Hz at 30 Hz and below. That would provide the capability to deliver rated torque at about 0.6 of rated speed or 60% of rated power, 1.32 kW.

If the V/Hz ratio is less than 8 V/Hz, the motor slip will be higher at any given torque and the maximum torque that the motor can produce will be reduced. That will mean that the motor is less efficient. If the losses in the motor are within the normal range, it is ok to operate at higher than normal slip. The VFD must supply the power required to turn the load plus the motor losses. The continuous operating capacity would be exceeded if either the power or the current supplied exceeds the VFD ratings: 1.5 kW or 7.5 amps.

Above 30 Hz, the motor can operate on a constant power, declining torque curve up to perhaps 50 Hz. If the motor is actually driving the center of a winder that winds up fabric at constant fabric tension and surface speed, the required torque would follow a constant power, declining torque curve.

Depending on the load characteristics, it may be possible that the overload protection, current limit, etc. can be set so that this system is safe and reliable. Even with an optimum selection of equipment, the VFD settings would need to be set properly for the system to be safe and reliable. Having matching motor and VFD ratings will not guarantee that the system is safe and reliable. The settings need to be correct. However the default settings would likely be adequate if the ratings match.

The motor will consume the power required to turn the load at the operating speed. The torque capability can be 100% of rated torque at 30 Hz if the VFD is adjusted for rated motor V/Hz. There are no guarantees. The speed vs torque curve under which the motor can reasonably be expected to operate can be estimated based on VFD configuration. The actual achievable operating speeds depend on the torque vs. speed demand curve of the load.

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  • \$\begingroup\$ I found out that the star connection is used for the motor which, accordong to the motor's plate, demands 5.73A which is lower than the vfd's maximum current (7.5A) does that make sense? \$\endgroup\$ – Macit May 2 '17 at 0:52
  • \$\begingroup\$ To recap what I understood, 1. The motor is going to relatively consume 1/3 of its nominal power (as in Marko Buršič's answer) 2. The maximum guaranteed speed is about 830 RPM (1382*30/50) which can be achieved when the frequency is set at 30 Hz. Am I right? \$\endgroup\$ – Macit May 2 '17 at 11:18
  • \$\begingroup\$ and 3. The motor wouldn't work at maximum speed if it was wired in delta connection because the VFD wouldn't deliver the nominal power that the motor needs. \$\endgroup\$ – Macit May 2 '17 at 11:27
  • \$\begingroup\$ I edited the question to include the above information re star connection. I deleted my comments and revised my answer to include my responses to comments. \$\endgroup\$ – Charles Cowie May 2 '17 at 19:37
  • \$\begingroup\$ Please as a side question in this case should the motor voltage parameter of the VFD be set at 230V or 400V? \$\endgroup\$ – Macit May 2 '17 at 20:08
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It would not be legal here in the US, but I don't know about wherever you are. Here you are required by code to use equipment that is "suitable for the intended use".

But from a strictly functional standpoint, the VFD can be programmed for "Current Limit" and in this case, you could set that limit at 7.5A, the rating of the VFD. You would not get full power out of the motor this way; the FLC of that motor is 9.9A, so in theory you can get about 75% power from it. Depending on the load on that motor, you may never know, but for sure it will not allow full speed operation at full load.

What a VFD does to limit the current is that it will override your commanded speed once the current hits the limit and stop any further acceleration. So if you are telling that motor to run at 75% speed or less, the VFD is likely fine with that. If you tell it to go 100% speed, it may ignore you in deference to the Current Limit setting.

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The motor is oversized, but there are some tricks you should know:

The VFD has the input of 230VAC, meanwhile the motor can be wired in delta (230VAC @ 50HZ) or star (400VAC @ 50Hz). From your description, the motor is wired in star, thus the output power is 1/3 of nominal. So, 2.2/3 = 0.73 kW. it is not known why one would put such large motor and then set it up to work like a small motor. Probably the output shaft dimensions, mechanical stress, or whatever determined the output shaft dimension, that matched the specific reducer size and with it also the motor size.

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  • \$\begingroup\$ @CharlesCowie VFD has 200-240VAC input, the motor is star wired, the maximum performance of such combination as well the power is all known, no need for additional data. \$\endgroup\$ – Marko Buršič May 2 '17 at 15:08
  • \$\begingroup\$ @Charles Cowie, Marko Buršič's answer seems logical to me. I mean in general if you take the VFD out of the equation we all know that for the same power supply (230 V in this case) and the same motor if you replace a delta connected motor with a star connected one the consumed power is going to drop by 1/3. The question is "Does this stay valid with VFDs?" I think it should stay valid. \$\endgroup\$ – Macit May 2 '17 at 15:51
  • \$\begingroup\$ @Charles Cowie Actually I know the concept behind the V/Hz ratio and now I do understand that the motor can rotate with 100% torque at 30 Hz but here we are comparing the power consumption between the two connection types. How is the V/Hz ratio is related to this? \$\endgroup\$ – Macit May 2 '17 at 17:42
  • \$\begingroup\$ @CharlesCowie You are partially correct, except that OP said that motor is star connected (4.6V/Hz). Partially, because the VFD would be overloaded outputting 230V/30Hz @ delta connected, it can feed not even 230/50Hz, because it is rated only 1.5kW while the motor is 2.2kW. \$\endgroup\$ – Marko Buršič May 2 '17 at 18:47
  • \$\begingroup\$ @CharlesCowie Excuse me, you are correct saying: motor is star connected and V/f characteristics can be adapted, so that VFD at 30Hz outputs full voltage 230V. The nominal current should be 5.7A. And power is 2.2kW(30/50)= 1.32kW \$\endgroup\$ – Marko Buršič May 2 '17 at 18:57

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