Variable Frequency Drive 1

[streeta]

Anyone
We are using a VFD on a 250KW motor and it is running at almost full load. motor rated at 418A. when running on the VFD motor will trip everytime when getting near 100%. however on fixed speed it will run just under the 418A rating. I belive this to be the voltage difference when using the VFD.
Does anybody know a formula for what the voltage drop may be when reducing the supply frequency thus increasing the amps to over the 418A. if we have a supply of 415V at 50Hz. what would the voltage be if we droped the suppply to the motor to 45Hz. Is there a way i can work this out.
And yes i know we will have to upgrade the motor at some stage. But the motor and pump have done this job before (pump has been overhauled as well)

Thanks

Greg

 

[rbulsara]

What is the O/L trip setting on the VFD? You can set it just above the normal load current say by 10%. Normal O/L setting is 115 to 125% of FLA.

Voltage has nothing to do with frequency. Reducing the frequecny will reduce speed, hence the power (HP or the pumping capacity) and the current. I don't udnerstand why you want to increase the amp over 418A, it will only ensure that the drive trips on overload.

Keeping the frequency (speed) and the load the same, you can increase the amps in direct proportion of the reduction in current. I=VA/(sqrt (3)*V) for 3 phase motors. No frequency is involved there.

 

[DougMSOE]

If I understand you correctly,
The motor must be kept at a constant Volts/Hz value such that magnetic saturation does not occur.

 

[dpc]

What protective device is tripping the motor when running on the VFD?

 

[engineerkribhco]

As mentioned by DougMose Voltage/frequency ration should always be maintained constant in variable frequency drive.Further,try to adjust the torque preset if it is adjustable in your variable frequency drive.(Assuming motor is tripping on overload and overload setting is correct.)

 

[gsimson]

Please check the sensitivity & accuracy of the OL protective relays, the old & the new.
Also U can check the actual frequency and voltage , both old & new.

 

[electricuwe]

streeta,

since your post doesn't state that clearly I'm going to assume the following:

You added a variable frequency drive to an existing motor branch consisting of a motor an an approbiate sized and set breaker and now this breaker trips on overcurrent.

Please consider the following:
Variable spped control doesn't come for free

-you also habe to supply the losses of the converter, efficiency eta may be in the range of 0.96 to 0.97
-the current drawn by the VFD is usually distorted and the total power factor lamda is about 0.75
-if you motor is an induction motor the power factor cosphi might be in the range of 0.8 at full load

Now make the following calculation:

Ivfd/Iind=cosphi/(lamda*eta) which is larger than 1 for the values given above

 

[jbartos]

Suggestion: The induced voltage is proportional to frequency.
Start with Faraday's Voltage Induction Law:
e=d(N x Phi)/dt
where
N are coil turns
Phi is a time varying flux linkage of the coil N x Phi
e is Faraday's Induced Voltage
More known relationship is:
E=(1/sqr2) x w x N x Phi = 4.44 x f x N x Phi,max
where
f is frequency
w is angular frequency
So that the voltage will be decreasing with decreasing frequency.
Reference: M.G. Say "Alternating Current Machines," John Wiley and Sons, 1978

 

[edison123]

Your VFD may have been programmed to trip the motor at high currents on lower speeds since the motor HP comes down with the speed (& frequency). So, if you are running at 418 Amps at 50 HZ, you cannot expect the same current capacity from the motor at 45 HZ. Both your voltage (to maintain V/F ratio) and current (to maintain ampacity of the windings) must come down at 45 HZ.

 

[gsimson]

Generally every pump develops more pressure than required to match the flow requirtements. One of the advantage of VFD for pump application is that we can run the motor at lower speed to develop the required pressure at the rated flow.
This will reduce the actual power requirement. As the voltage is reduced in proportion to the speed the current may remain marginally less only considering the losses in pump & motor in relation to speed.

 

[jbartos]

Suggestion: Reference:
1. Bimal K. Bose, "Modern Power Electronics and AC Drives", Prentice Hall PTR, 2002,

http://www.phptr.com,

page 44, Section 2.2.8 Constant Volt/Hz Operation
If the supply frequency is reduced at the rated supply voltage, the air gap flux PHIm will tend to saturate, causing excessive stator current and distortion of flux wave. Therefore, the region below the base or rated frequency (Wb) should be accompanied by the proportional reduction of stator voltage so as to maintain the air gap flux constant. The breakdown torque:
Tem=3(P/2)(Vs/We)^2 x Wslm Rr / [Rr^2 + (Wslm x Llr)^2] ....equation 2.42
remains approximately valid, except in low frequency region where the air gap flux is reduced by the stator impedance drop (Vm<Vs). Therefore, in this region, the stator voltage drop must be compensated by an additional boost voltage so as to restore the Tem value.