Motor Rating

I have a curly little prob here. I have a 300kw motor that I have been asked to work out its total torque on start up because it is oprating on a conveyor that empties a chute.

Now I calculate the motor being 300kw, 4pole, 3.3kv has a FLC of 91A running unloaded. Now FLC on startup should be roughly five times that so 455A.

Motor torque should be T=HP * 5250/rpm so roughly torque = 1400 Ft lbs. But this is unloaded torque.

In theory, does the motor have the ability to produce five times this torque value on start up for a few seconds as it does with starting FLC?

TFC

Hi down there opposite side of globe.
Unfortunately Cosine(some Greek letter as Fii) is small, maybe 0.3 to
0.4 (my quess) and that for reactive part of current don't do torque efect. Maybe it is not so simple, but explanations anyhow. For normally motors it is 1.5-2.5 times nominal torque and vith VFD we can get maximum (kip ? I dont know English word) torque from motor at start, even more.

I wouldn't advise using current for calculating torque when you know the KW (HP) and speed.

Given 300KW, Hp would be 300/.746 or 402 hp. Assuming a nameplate speed of 1770 rpm, torque in ft-lbs would be

torque = hp x 5252/rpm = 402 x 5252/1770 = 1193 ft-lbs

Now its important to understand that nameplate data on a motor is FULL LOAD DATA. Therefore, when this motor is loaded to 1193 ft-lbs torque, the speed will sag from 1800 to 1770rpm and the calculated hp will be 300kw or 402 hp and the amps should be whatever the nameplate FLA or FLC says it is.

Don't know what you might have meant by no load torque---no load means zero torque.

Now, to get starting torque, you must consult the speed/torque curve for the motor. In the United States, most motors are designed to NEMA Design B specs which pretty well defines the torque/speed characteristic curve from no-load down to stall. What you might have in Australia is unknown to me but very important for you to understand.

If this motor is on an inverter, then you must use the same speed/torque curve but you must identify precisely where on the curve you are operating under starting conditions. Normally, you would not stray very far from the full load operating point on the curve due to the limited ampacity of the inverter and the ability (hopefully) of the inverter to manage output voltage and frequency properly to keep the motor operating in near-synchronous mode.

Going back to your original question, if you are starting across-the-line, you can expect to see the 5x nameplate current you mentioned but, unfortunately, the torque is far less than that. The problem is that, on start, an induction motor is standing still (equal to 0Hz) and is hit immediately with 50 or 60Hz. This places the motor far outside its synchronous operating range and its ability to convert amps into shaft torque is severely impaired. This confused state inside the motor exists until the rotor struggles up to typically 90% or so of its sync speed where it transitions to near-synchronous operation. When that happens, the amps fall dramatically, efficiency improves dramatically as well, and the expected correlation between motor load and amps occurs.

Exactly how much torque the motor can develop when operating dis-synchronously is entirely a matter of design compromises built into the motor. That's why a speed/torque curve is essential to answer your question.

As Dick points out, the starting or stall torque is different from full load torque, and the ratio varies with the design of the motor. For a typical motor, NEMA Design B in the US, starting torque is around 150% of full load or nominal torque. The maximum torque actually occurs at a higher speed and this "breakdown torque" is about twice full load torque. You can get special high starting torque motors, like Design D, that have starting torque of 250% full load or more.

Until you get at running speed, and above 50% load, amperage is a lousy way to determine motor load. No load current may be higher than 50% load current.

I use 166% of full load torque as the average torque during accelleration for a standard NEMA Design B motor. If you are worried about breaking things, get a motor curve. If you are calculating acceleration time, I use the following:

Time = (WKsquared x Delta_rpm)/(308 x T_ave)

Time = seconds to accelarate
WKsquared = rotational inertia lb-ftsquared
T_ave = average torque, lb-ft, which is 1.66 x full load torque for Des. B

We have a lot of large motor where I work large mixers for 20 tonne machines, but 300kW for a conveyor is huge! What is going along this conveyor? maybe a lauch pad for a space shuttle?.......

Anyway I would say for the unloaded conveyor the start torque for an average motor would be about 40%. This would equate to a starting current of 335% of normal Full load current.

Maybe I am way off the mark here, but it does vary vastly between motors and what you are using the motor for!

Of is it a standard induction motor? Do you have a soft starter or a clutch (Yuck) attached to the motor?

Regards

Barry

When considering a soft starter, remember, starting torque is defined by the load, not by the motor or starting technique. If all available motor torque is required to break away the load, you will gain nothing by using a soft starter.

On the other hand, if starting (and accelerating torque) is dictated by rotating mass only, then a soft starter will be useful as long as you can take longer to accelerate up to full speed.

Since a soft starter is a 50 or 60hz device only, you still suffer from the inefficiencies etc. I mentioned earlier while the motor is struggling up to near sync speed. Under ideal conditions of very low breakaway torque and long accel time, you will still need about 3x FLA to get the job done.

It is used for a 1km long conveyor carrying ore. Starting is by secondary reistance starter.

Another conveyor feeds this one with ore into a chute. When that conveyor faults it still travels for around 5 seconds because of its huge length. Ops want to know if the motor is capable of starting the conveyor that carries ore out of the chute. The mechanical guys have done the belt calculation for tension of a full chute. I have been asked to find out full torque. I would tell them name plate torque, but I am new to calculating starting torque so I might tell them 150%.

For me this is another interesting topic.
Thanks for the help so far.

That is a big enough motor and a critical enough question that I would make a call to the motor manufacturer and find out stall torque and break down torque. If you have the nameplate data they should be able to give you the answer easily.

Tom Jenkins has it exactly right. Do what he says.

I suspect your comment "starting by secondary resistance starter" means that this is a wound rotor motor. All the more reason to follow Tom's advice.

Yes Dick you are correct, it is a wound rotor motor.
Tbanks for the help guys, the manufacturer will be called. I wanted to have a go myself to try and learn a little.

Thanks again!