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question about motor starting current
2

question about motor starting current

question about motor starting current

(OP)
i know i send two questions today, but i was curious about this and this is not my main area, we got a pump centrifugal pump, about 1,2 MW 4160V and it´s gonna move some water from a tank to a higher level, 50 m high, there´s no valves, this means that when you detect level enough pump would start and push the water 50 m high, the way it´s gonna be moved is with a Soft start, 3,5In, and with my calculations would take about 3,12 seconds for starting and reach 700 rpm nominal, this is induction motor.
  The question is: some guys says that the worse case is when you need to start without load, and on my opinion we always starts with load.... and about this, what do you think of the starting of the pump with a soft start, i see right,cause we are always given the same flow, and the soft start would be able to start the pump....do you think it is justify a VFD?
ok, thanks, i know it´s not a very clear post, this is cause i really don´t know exactly why they talk about that diferences and why they talk about a VFD, when with a soft start would be enough

RE: question about motor starting current

2

Quote:

i know i send two questions today...
Yes, sorry, there is a limit; 1 question per day. pipe

Not really, thumbsup2
You can ask a many as you want!

If your system cannot benefit from variable flow, and it appears on the surface that this application is one that cannot, then a VFD is a very expensive way to facilitate this. But please see my response in your other question about sizing the generator, sometimes it's the only way. As a general rule though, adding genset capacity at the design stage to allow the use of a soft starter tends to be less expensive than adding a VFD to allow for a smaller genset, especially in Medium Voltage where a VFD is extremely expensive.

Quote:

...some guys says that the worse case is when you need to start without load, and on my opinion we always starts with load...
I cannot imagine any situation where starting unloaded would present a "worse case" than starting with a load. It's exactly the opposite. but specifically to a centrifugal pump (most types), there is little load when you first start it, load increases with flow. No flow = no load, low flow = low load, max flow = max load. So when starting, loading doesn't occur until the motor gets to about 60-75% speed. Starting a centrifugal pump with a soft starter is very simple, probably 75% of all soft starter applications are for centrifugal pumps.
If you have already determined that your pump will start in 3.12sec. at 350% current limit, you are right in the breadbasket of a perfect soft starter application.

 

"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln  
For the best use of Eng-Tips, please click here -> FAQ731-376: Eng-Tips.com Forum Policies  

RE: question about motor starting current

Quote:

i don´t need to oversize the generator just in case i need to use VFD.

Maybe not, but you will have to oversize the generator to use a soft-starter.

How did you determine the current and accel time? Just curious because there are some widely published forumula that are basically useless to calculate such things.
 

RE: question about motor starting current

Jraef

In your response you allued to the fact that a larger capacity genset is required for a softsarter than would be for a VFD.  Or in other words when using a vfd as opposed to a softstarter why are you saying that you can get away with a smaller genset.

RE: question about motor starting current

If I am allowed to say a few words.

The fact that you can 'lead' the motor speed all the way from stillstand to desired speed and always stay on or close to rated torque and rated current is the reason why a motor that is ramped up by a VFD needs less current than a soft-starter. The latter works with mains frequency and it is inevitable that you have to pass the peak torque and the peak current points. They are reduced with the SS, but the VFD limits torque and current to rated values - if needed.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: question about motor starting current

Thanks Gunner

I understand what you are saying but can you please explain these two sentences?

"The latter works with mains frequency and it is inevitable that you have to pass the peak torque and the peak current points. They are reduced with the SS, but the VFD limits torque and current to rated values - if needed."

Are you saying that with a soft start the load torque will exceed the reduced peak breakdown toruqe at some point?  Can you explain it as if we were looking at a speed vs torque curve?  

RE: question about motor starting current

Quote (Skogsgurra ):


If I am allowed to say a few words.
As if we could stop you... wink

At great personal risk of stepping all over the venerable Skogsgurra;

A starting method does not change the torque requirement curve of the load. If the load torque requirement curve ever exceeds the motor torque output curve, you have defined a stall condition.

But what a VFD does is allow full torque at any speed as the motor is accelerating (or not for that matter) whereas a Soft Starter has significantly reduced torque in the beginning and slowly increases it. So by being able to apply full torque from the outset with a VFD, acceleration can occurr with less line current draw, usually less than or equal to the motor FLC.

A soft starter however will at some point require at least 200% FLC, usually a lot more, even if only for a short time as the load demands more torque when speed increases. So to be sure it is a very important application consideration in soft starter selection that the motor torque be allowed to always exceed the load torque curve requirements.

Bottom line, it takes more genset capacity to start with a soft starter compared to a VFD, but that is still less than if you tried to start it DOL.

"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln  
For the best use of Eng-Tips, please click here -> FAQ731-376: Eng-Tips.com Forum Policies  

RE: question about motor starting current

Yes. 'nuff said.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: question about motor starting current

Thanks Jraef

Let me see if I can put your explanation into my own words to see if I can understand it.

When starting a load with a VFD as you mentioned the VFD is able to supply full torque throughout the whole ramping process and therefore while supplying full torque the current stays right at full load as you mentioned.  The drive is able to supply the full torque by the fact that it varies the V/Hz ouput ratio of the drive throughout the whole ramping process in order to maintain the rated torque and current.

With a softstarter the SS does not vary the frequency but rather only reduces the voltage.  As a result of this the voltage is reduced and therefore the avaliable torque output from the softstarter is reduced by the square of the voltage reduction.  The initial starting current of the motor is less than a DOL start because the voltage to the motors equivelent circuit is reduced however as the load increases its speed it increases its torque requirement and the SS needs to be able to supply the required torque.  Because the SS is not able to vary the frequency to adjust the V/Hz ratio to apply rated torque, the SS must compensate by allowing the output to supply more current to the motor to provide the rated torque required by the load as it increases in speed.

Do I have the the basic idea correct?

RE: question about motor starting current

A slightly different twist on the same theme is that a motor operating at low slip has  tends to have much higher ratio of torque to current than a motor operating at low slip. (because at low slip we have very low power factor).

VFD accelerates at low slip and high ratio torque to current throughout.  Heating will be very small compared to the other methods.

For DOL or soft start, the majority of acceleration occurs at high slip. That means low torque per current... you've got to have higher currents or longer times or both... product of current times time will be longer and heating more.

There are differences in the torque profile vs time for each of these options and vfd may offer more flexibility through ramp rate. But again at a given speed early during acceleration the vfd can achieve a given torque with far less current than dol or soft start.

=====================================
(2B)+(2B)'  ?

RE: question about motor starting current

Correction in bold:

Quote (electricpete correction):

A slightly different twist on the same theme is that a motor operating at low slip has  tends to have much higher ratio of torque to current than a motor operating at high slip. (because at high slip we have very low power factor).

VFD accelerates at low slip and high ratio torque to current throughout.  Heating will be very small compared to the other methods.

For DOL or soft start, the majority of acceleration occurs at high slip. That means low torque per current... you've got to have higher currents or longer times or both... product of current times time will be longer and heating more.

There are differences in the torque profile vs time for each of these options and vfd may offer more flexibility through ramp rate. But again at a given speed early during acceleration the vfd can achieve a given torque with far less current than dol or soft start

=====================================
(2B)+(2B)'  ?

RE: question about motor starting current

(OP)
I understand your points, the way to calculate the time is to calculate the voltage on several points of the curves of the motor, with a 3,5 In limitation, and with this voltage calculate the torque and compare with the needed torque of the load, this must be higher.... and for the time, we divide the slip of the motor on the same pieces that we used for the voltage,and calculation of the torque, and with the calculated torque and the diference on speed i calculate the time, then i make sumation of all the times and this is the starting time + 0,8 seconds on safety,
  I said that the genset would be shorter on Soft start cause this guys tell me that i need to multiply by two the genset if i use a VFD, but now after reading you i understand that i don´t need and for sure, it would be a smaller gen set for VFD than for Soft start.
thanks everybody for your help

RE: question about motor starting current

Rockman7892,

Bingo thumbsup

"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln  
For the best use of Eng-Tips, please click here -> FAQ731-376: Eng-Tips.com Forum Policies  

RE: question about motor starting current

Thanks Jraef its crystal clear now!

RE: question about motor starting current

You can use the (supplied current / Full-voltage current) ^2 to calculate the torque reduction.

Did you use the motor torque minus the load torque as the accelerating torque when calculating the acceleration time?

Short cut formulas are often used to calculate the required motor current and accel time. In general, this works fine because no-one really cares that much if the calculated vs real world match. However, when you make assumptions for an application such as this it can get very expensive if the installed system fails to work.
 

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