Induction Motor Starting

Why is 738A not a "safe value" in this scenario? Is the supply cable too small?

yes first of all the cable is of 1.5m length and the motor specification is 3*27V, 5kw, 1500rpm .

If the supply cable is undersized, a VFD won't make any difference. Concentrate on selecting the correct size of supply cable and fuses. British Standard 7671 gives a methodology for correct selection.

So if it is the cable that is limited, what is its size/material/current rating/rated temperature??

What is the start time?

How often does it start?

In the usual way of motors, the cable does not have to be rated to more than the rated motor full load current, increased to the fuse rating - if fuse rating has to be higher to survive the starting current and time.

If the start is quick, the cable temperature rise during start is often negligible, despite the increased current. Start time has a considerable effect on the fuse required, a 125 amp B.S. 88 fuse in BS7671 is indicated with 1 second operate time at 1000 amps, but 20 seconds at 500 amps.

One question , i know this is a little childish question, but want to know what if a voltage of 33V is applied on a 24V motor? please don't underestimate this question....

Well, it will accelerate quicker than it would if operating on 24V - provided it doesn't burn out in so doing and provided the mechanical load will let it. The overload current trip setting must now be set for the correct full load current at this higher voltage otherwise its demotion to a convector heater can be assured [non-underestimated answer].

Now a childish question from me - 27V for 5KW motor looks to be a bit abnormal are you sure?

Such a small motor carrying FLA of 125A ? How one is going to accommodate the thick conductors? while managing the iron area?

Now as far as your question is concerned. the VFD s would automatically regulate the current so that usually the FLA is not exceeded and that is done by the varying the voltage but not by increasing it.

Since to keep the constant flux V'f is to be maintained constant hence the starting (lower frequencies) Voltage is automatically lower.

The control is achieved by the control on slip which had lesser effect on lower supply frequencies.

Increasing the voltage to 33V - why do you want to do it? The motors are rated to only 10% over/under voltage. If the name plate shows 27V, then may be you might go upto (just below) 30V.

Over voltage would have effect on

a) Insulation

b) magenetic losses since V/f and hence the flux density would increase, the iron losses too would and worse a portion or the whole magnetic circuit may reach saturation.

Anonymous Poster #1

Actually the motor is fed by inverter ( having control through CAN bus) . The inverter is fed from battery of 48V dc. According to a previous discussion about controllers /VFD its output voltage must be 70% of input voltage ie. 3* 33V.

This 33V is fed to the motor whose name plate is given of 3* 27V.

So, if i provide this supply voltage then

1) what according to you will be the effect?

2)And does it automatically regulate the inrush current?

If i m overthinking for a simple solution pardon me , but please correct me......

There are still too many unanswered questions about this installation to provide sensible guidance.

Higher up, the cable was stated as being too small. The job of a VFD is not to make up for deficiencies in the cable sizing. It is there to vary the speed of the motor.

So why is the cable too small? What is being done to correct it?

Tackle that first, then add the VFD later.

Please state the maker and part number of the VFD.

Still the question is unanswered- Are you sure that the 3 Phase 27V motor is 5KW rated?

The inverter is fed by a 48V DC but the output of it could always be stepped up (being AC) through a simple transformer even if it isn't done already electronically. And that could be done to any voltage.

Are you feeding with an inverter (simple one) or through a VFD which is an inverter but not a simple one.

Is the motor designed for a VFD? Most of the motors aren't.

As far as your questions are there

1) Already answered. Not advisable.

2) if it is VFD, the inrush current is controlled. (Already answered) If it is a simple inverter then it isn't and you are going to land up in a burnt- no not cable - but inverter (Its semiconductors would fail sooner than the cable)

You dont need to reduce the voltage in order to reduce the current, in fact it creates the opposite effect if you drop the voltage too much, and will burn your motor. To reduce the high starting current, use an acceleration ramp, which will start at a low frequency and then go up to the required value for the application ( I believe you are working with an electric vehicle).

Anonimous poster #1 and PWSlack are right, the first thing you should focus on and correct is your wiring caliber to the drive and motor.

As per the reduction on voltage, the drive takes care of that on its own if you enter the motor ratings and application duty when progarmming it.

Regards

Mr Yahlasit,

you got me , you r right about the application.

I want to know the control of starting current by varying acceleration ramp since i require it.

I need to know everyones suggestion.

HAS THE CABLE SIZING BEEN ADDRESSED YET? <sigh>

What do you know about the load being driven by the motor?

Have you got the moment of inertia of motor and, especially, its load device?

Is there any large load torque to be driven, additional to accelerating torque?

Do you have a speed versus torque curve for the load?

I was thinking of using 70mm square cable for a current consumption of 123A rms FLA.

Why don't you step up to a proper voltage and then use a lower ampere motor? I checked up a website and that uses a 240V motor (but the voltage is increased at a post control stage so that the driver is not shocked .

Does that thought come via the methodology contained in British Standard 7671 - yes or no, please?

Yes it does

Then the VSD can now be added to the system without fear of the inrush current.

If an inverter/vfd is supplied by a battery of 48V, and is its output is given to 3 phase traction motor drawing 123A FLA. Does the inverter output current and input current remain the same, if so how should be the cable sizing?

The inverter overload is set to the full load current of the motor.

The cables are sized according to the local codes. In the UK, it would be British Standard 7671.

Is the message getting through yet?

Does this system really exist? - one minute it is supplied by 27V, then 33V, now it's 48V. Please eliminate the confusion.

According to formula given BS7671, clause 434 (which considers only heat capacity of copper and leaves out heat loss from cable), it takes more than 30 minutes for a 70 sq.mm copper cable [ hot, with 70 Celsius max normal temperature - PVC insulation] to reach its limit temperature of 160 Celsius [90 degrees rise] for fuse protection operation when working at 180 amps. In any case, the rating of such a cable at 30 Celsius ambient is about 200 amps continuous.

So use 70 sq.mm for inverter DC input and measure the r.m.s. current in the 30 minutes or more before cable gets too hot - then re-assess cable rating.

Since we know nothing about your inverter and its efficiency (and the effect of its waveform on motor losses and input) I do not think anyone can suggest better.

But this is a traction application, so it is very unlikely you would run motor at full power for more than a few minutes in service - it is quite possible the battery will be discharged in 30 minutes at full power.