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With any variable frequency drive application, the critical parameters are TORQUE - what's the worst-case torque? - and SPEED.Your application is 0.55kW over a span of 50-300RPM. Assuming your mechanical variable speed drive is either a variable belt drive or a disc drive, it is a torque-multiplying drive where the available output power is constant as the speed changes. 0.55kW at 300RPM is 17.5 N-m, while the same 0.55kW at 50RPM is 105N-m. Then it gets worse. A three-phase AC motor is capable of producing much more torque at start than it can output continuously.

Exactly how much more is dependent on the design of the motor, but generally 150-250% of full-load torque is available at start. Not taking this into account results in a lot of misapplied inverter/motor systems because that torque is not available from an inverter unless the inverter is sized to deliver a lot more current.The question then becomes "what is the worst-case condition?" That is, where do you need the most torque when operating your machine? Let's say your equipment is an industrial mixer, and stipulate that it MUST be started with the mechanical variable speed drive at its slowest setting when loaded with materials or it will fail to start.

The worst-case torque condition then would be (assuming 200% starting torque available) 2 x 105 N-m, or 210 N-m. To properly size a VFD/motor combination, your system MUST be able to produce 210 N-m. Typically, an industrial-duty VFD unit is rated to put out a certain amount of current continuously, and 150% of that rating for 60 seconds.

A drive designed for pump and fan applications may only be rated for 110% overload for 60 seconds. When operating a motor on a VFD, output current is (for all intents and purposes) equivalent to torque.So, what size motor do we need to produce 210 N-m at start? If 210 N-m is 150% of the motor's full-load rating, the motor full-load torque would be 140 N-m.

Assume we're going to use a four-pole 50Hz rated motor with a nameplate rated speed of 1475RPM. One hundred forty N-m at 1475RPM means 21.6kW. The closest standard rating is 22kW.Wait, what? The old system uses a 0.55kW motor, and the new one requires a 22kW motor? Welcome to the wonderful world of mechanical torque multiplication.

Remember, your maximum speed is 300 RPM. You don't want to operate an induction motor at full-load current over a maximum range of 50-300RPM if you can possibly avoid it. You don't want just a motor, you want a motor with a gearbox.

Such as a motor/gearbox combination with a rated output speed of 300 RPM and a rated continuous output torque of 140 N-m. Plug those numbers into the equations and the motor rating drops to 4.4kW. The closest standard rating is 5.0kW. Yes, the motor is almost 10x the size of the original, but that is what is required to ensure that the system will start under the worst possible condition.

But wait! There's more! Once the system is started, the torque required to run it falls off rather dramatically, since the current system with the 0.55kW motor can spin the output shaft at 300 RPM while producing no more than 17.5 N-m of torque. One advantage of variable-frequency drives is that you can operate the motor at OVER the nameplate frequency, so long as the motor is mechanically capable of handling the overspeed condition, and the motor does not draw more than the nameplate full-load current.

How about a motor with a gearbox rated 140 N-m at 200 RPM? Now the motor size drops to 2.9kW, and the closest standard motor is 3.0kW. Were it me, that's the size I'd select. To get 50-300 RPM out of the system, the VFD would operate between 12.5Hz and 75Hz. However, at 75Hz the motor/gearbox combination would be capable of providing 95 N-m.

In other words, the motor would be very lightly loaded. Still not what I'd consider ideal, but getting closer. Play with the gear ratios, possibly the motor full-load speed, and an optimal solution can be found.

Now, what if the application is a conveyor belt? A conveyor is a constant-torque application. That is, it requires the same amount of torque regardless of the speed, with the singular exception of - once again - starting. However, let's say in this application the conveyor can be started with the mechanical variable speed drive in its highest speed setting, so now the worst-case torque is 2 x 17.

5 N-m, or 35 N-m.

Again, your maximum shaft speed is 300 RPM and minimum shaft speed is 50RPM, so some gear reduction is needed to keep the new motor operating somewhere reasonably near nameplate speed. Again, let's assume a four-pole motor with a 1450RPM nameplate rating at 50Hz. 1450/300 yeilds 4.83:1 reduction, so either a gearbox, V-belts and sheaves or a chain drive can be configured to get us 5:1 reduction (and 5x torque multiplication).

With a starting torque requirement of 35 N-m at the conveyor, we divide by 5 to get the starting torque requirement at the motor shaft - 7 N-m. Seven Newton-meters represents 150% of the motor full-load rating, so 7/1.5 = 4.67 N-m rated at 1450RPM, or 0.71kW. The closest standard rating is 0.75kW. So, a 0.75kW motor, a 5:1 gear reduction and a 0.75kW inverter are required. So we have two applications for your 0.55kW mechanical variable speed drive, and the inverter/motor replacements are VASTLY different.

Application is everything, because application defines the worst-case torque requirements. I've been doing this for a living for more than 25 years now, and I've replaced a LOT of misapplied drive systems where worst-case torque was not properly considered. Hope this helps.What is the best practice or strategy to replace a mechanical variable speed drive asynchronous motor with a regular asynchronous motor and a frequency inverter?.

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Inertia. Those kinds of doors are typically very heavy, thick tempered glass so they can't be easily broken. It takes a lot of energy to start one moving. The sensor can't sense you when you're a long way from the door and start opening early, because it would detect random passers-by and open unnecessarily.

That would waste electricity and let the heat out. It could be done with more powerful motors, but it's probably not worth the expense.Why don't automatic doors ever open quickly enough so that I can walk through without breaking my stride?.

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The principle of this device is that, when you make the call to the DTMF device, it automatically receives the call.Then the DTMF tones generated from that device are received by the microcontroller. When the call is connected, you can type the door opening as well as the locking code from the keypad of your phone.

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