Current Reduction if the VFD is Connected to Motor

Is VFD a varible frequency Drive ?

http://www.google.co.uk/search?q=VFD&sourceid=ie7&rls=com.microsoft:en-gb:IE-SearchBox&ie=&oe=&rlz=1I7DKUK_en-GB&redir_esc=&ei=kP0DTMS9DIz-0gTykZ33Ag

that depends on the installation height difference of motor and VFD. 1meter = 10%age.

I assume the load and the RPM remain same. In this case, the short answer is 'no reduction of current'. The motor will draw the same current with or without the VFD.

I think your question or concern is about the saving energy with using VFD. In one of the earlier post (http://cr4.globalspec.com/thread/52637), I explained it which is copied here again:

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The intention of using VFD is to decrease starting current, control speed and save energy. However, saving energy by VFD is sometimes misused and misunderstood, VFD can save energy only when it is used for controlling the flow with varying the speed of centrifugal pumps (for liquid) or fans (for air) and the flow varies in wide range and remains at lower end most of the time.

How VFD saves energy?

VFD saves energy based on the affinity law which indicates

Flow ∞ RPM

Head ∞ RPM²

Power ∞ RPM³

Let me explain with this example: In a system, a centrifugal pump delivers 1000 CFM at 1800 RPM and 500 CFM at 900 RMP. The flow of the system varies from 1000 CFM to 500 CFM and most of times the flow remains at 600 CFM.

In this case, if you use a DOL starter, the motor runs at 1800 RPM all the times and the theoretical power consumption of the motor is proportional to the cube of 1800 (as per the affinity law). If a VFD is used here, the theoretical power consumption is mostly proportional to the cube of 600. So, VFD saves power in this case.

Let's consider the same example but most of times the flow remains at 900 CFM (instead of 600 CFM).

In this case, the theoretical power consumption by the VFD is mostly proportional to the cube of 900 and the power saving is not significant as compared to the previous case.

(Note: I mentioned 'theoretical power consumption' because the calculation is not exactly this in practical – there are other things to consider as well)

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See also http://ww1.microchip.com/downloads/en/AppNotes/00887a.pdf (page 17)

- MS

Hi,

Thanks for detailed explanation. I have a different question related to VFD. We have many conveyers which are connected with 1400 RPM motors, but since the speed required for the conveyer is very little, we are reducing the speed with mechanical gear system. The desired speed is ~50 rpm.

If we use vfd to reduce this speed, can we expect energy saving.

Regards,

M

No, but you may burn up the motors.

When you reduce speed with a gear / belt reduction system, you are INCREASING torque at the final shaft in inverse proportion to the speed reduction. When you reduce speed with a VFD, you are maintaining the SAME torque as the speed reduces.

If your motors and gear reducers were selected based upon the final output shaft torque at the drive coupling, removing the gears will leave you with only the original motor torque available and it will most likely be inadequate to overcome friction and keep the load moving. You would need to calculate the final shaft torque after the gear box, then work that backwards into selecting a larger motor.

For example, lets say you have a 1kW motor right now. At 1400RPM, the gear reduction must be 28:1, so to remove the gear box and have the same torque at the drive coupling, you will need a 28kW motor. A larger motor means more fixed losses in the motor; i.e. more energy wasted.

Also when running at a 28:1 speed turn-down, the motor cooling will be non-existent. You will not be able to use standard motors, you must replace them with Inverter Duty motors that are rated for at least that 28:1 speed ratio. Usually they jump from 10:1 to 100:1, so you would need to buy the 100:1 rated units, which also means they will likely have separately powered constant speed cooling blowers.

Any energy losses you might save in eliminating the gear box will be more than made up for by the other new losses from using a VFD.

If you run your motor at 50Hz, and it is a 50Hz motor, you will INCREASE the current consumption bt the losses in the VFD, sometimes as much as 5%.

The VFD can reduce the starting current of the motor by "ramping" it up to speed gradually. If the operating requirements allow for less than full speed, the VFD can reduce the motor current. If the motor must be run at full nominal speed, the VFD will slightly increase the current drawn, due to the losses mentioned already.

I think it save to assume that the motor in question is connected to 460 Vac system @ 60 Cycles guaging by the full load current of 45 Amps. Power factor on a 30 kW induction motor should run in the region of 0.89 lagging. The connecting of a VSD should offer local power factor correction at the motor terminals by virtue of the DC Bus capacitors resulting in a figure closer to unity of 0.96 lagging.

If we consider the Power, voltage, current and power factor relationship calculation in a 3 Phase system, it is easy to determine that there will be an improvement in the load current.

Power (Watts) = 1.72 x Volts x Current x Power Factor

Note: When applying this calculation it is important to take the motor efficiency into consideration as name plate data represents the mechanical kW at the motor shaft. An efficiency 2 motor should be in the region of 92 % efficient which means that taking all losses into consideration, the motor will consume 32.6 kW of electrical power to produce 30 kW mechanical.

If I am correct in assuming that you are using a 460 Vac system at 60 Hz, you have indicated a reduction in motor speed by reducing the Frequency to 50 Hz.

Shaft speed can be determined by the following calc: Speed in Rpm = Hz x 60 Sec / Pole pairs - Slip. Although you haven't mentioned the speed of the motor lets assume a 4 Pole motor.

The speed @ 50 Hz in Rpm = 50 x 60 / 2 - Slip of 3 % (varies according to design and load conditions.)

Rpm = 3000 / 2 - slip

Rpm = 1500 - slip @ 3%

Rpm = 1500 x 0.97

Rpm = 1455 Rpm

This reduction in speed can definitely have a positive effect on the load current depending on the application. If the application follows square or cube laws for power loading, such as certain pump and fan applications you will notice a distinct reduction in load amperage.

In selecting VSD's it is important to know the load characteristics of the application. As drives can be split into two catagories, constant torque and variable torque. Vector type control been ideal for Constant torque applications and V/F control for variable torque applications such as pumps and fans.

Which current for VFD do you mean? Motor or supply current? If motor current, VFD and mains will be similar if your supply is similar to motor rating and tolerance.

If supply current we need more information to estimate -

Does VFD have any AC or DC reactor (Choke)?

What is the supply condition and level? What is the motor nameplate data re voltage and FLC and also cos phi?

This will help in trying to make an estimate.