Type of Starter Needed.

Worst case: 18 x 17.5 = 315 kw. If PF= 0.8, this comes to 394 kVA. Unless you are starting all the motors at once, what is the problem?

I certainly wouldn't be wasting money on SD starters for such small drives, DOL will be fine.

Hi,

since 500 kVA transformer full load current = 721.7 A

total of 18 no of IM current = (full load current of 17 kW)*18=441.7 A

for DOL starting (10 times FLC expected)=4417 Amps.

My question is from where will this current be supplied - as 500 kVA transformer full load current is only 721.7 Amps.

Regs,

http://www.idc-online.com/conferences/PAPERMichaelHamilton.pdf

appears, short time overloading will be there for transformer using DOL starting, which can reduce the life of the transformer.

Regs,

No. Transformers can handle brief overloading, up to a point, without damage or even serious voltage drop. You will not be anywhere near that point.

You have two options:

A/ Employ an engineer that knows what he's talking about.
B/ Buy a bigger transformer.

A/ Will cost less.

Anonymous : I am sure you are kidding.

As suggested there won't be any issue UNLESS you start all the motor simultaneouly.

The question is if you want to start all of them Together.

If not why bother???

Concen comes if FLA of all the motors except the biggest one + starting inrush of biggest motor overshoot, even the overloading capacity of transformer. Is that the case? Lest dont worry. As starting inrush is for few secs (max 30) this must not be an issue.

If there was a power failure while all the motors were on and running, and then after a minute the power returns, they'd all be starting at once.

Not if the system is designed correctly. The individual starters should de-latch on power loss.

Mind you with the OP's comments so far it could be a distinct possibility.

with minimum of 4% short circuit impedance, the s.c. current is 18,042 Amps.

But including all 18 motors simultaneous DOL starting uses: 4416.7 Amps.

Again, full load current of transformer is : 441.7 Amps.

By this calculation, transformer is overloaded more than 6 times.

Regs,

Then fix your motor (re)starting sequence with suitable delays.

But this way, if motor no 5 is to be used, then all the motors no from 1 to 4 should be started before motor 5 starts and run. I need to start and run just motor no 5 only for a defined period. (This applies to all individual motors apart from motor no 1 because for motor 1 no delay is needed).

Regs,

A simple and cheap answer, purchase a book of pro-forma written warning forms that you can give to your plant operators if they break the rules you set. If they win three written warnings, dismiss them.

The slightly more expensive method for you personally:
Find a new job as you're clearly incompetent!

Regs,

Your concerns appear to be from an incomplete application of theory and experience. A motor's inrush lasts long enough for the motor to get close to its operating speed.

• You have not told us the type of load the motor(s) supply, so you have seen a conservative starting time of 15-30 seconds (appropriate for a large centrifugal load with a lot of inertia). Most loads start much sooner--within 1-2 seconds.
• You are looking at a 10x size inrush, which is typical for very energy-efficient motors; the typical "rule of thumb" is to use 6x to 8x.
• You appear to be equating transient overloads with long-term overloads. During a transient overload the components will be subjected to significant amounts of heating. However, this then dissipates by conduction, convection, and radiation for the many minutes to hours before there is another significant transient overload. Your transformer has a very comfortable margin between its present load and its design capacity. So, any transient overload for the starting of one or more motors is not going to overheat it.
• Motors that meet NEMA standards (for the USA) have a code letter marked on their nameplate. This gives the actual inrush kVA as a multiplier of the motor's nameplate output power.

Use a scope meter or fast-acting multimeter and measure the actual line voltage while you are starting that questioned motor. I suspect the voltage drop at the transformer's output will be less than 10%.

--JMM

1. the type of load is barrage gates (which are very heavy) of an irrigation systems. The induction motors are intended to raise or lower it by use of hydaulic system.

I am unaware how much time is temporary overload. Since it is 6 times overloading, its pretty high. Only 150% to 200% overload may be preferable for temporary overloads, that too, for very little time. (I rdo not exact % of overload the dry type 500 kVA transformer can handle for specified time, I could not find it anywhere. Only I know that for oil filled transformer they talk upto 200% load for 30 min overloading.)

2. Thanks, corrected.

3. I am aware that this is just a transient overload. But I have read in IEC standards that any significant overload for specified duration can reduce the life of transformer. The long term overload or continuous overload, you must referring to type S1 duty, which gives continuous overload of 10%.

4. I am not aware of such inrush kVA rating, in south-asia.

Regs,

Sir,

Thanks for the reply.

Even though the gates may be very heavy, your motor is operating a hydraulic pump that then moves the gates. That pump, if you observe it, will get up to full speed in only a few seconds or less (probably 2-4 seconds at most), because the pump has very little inertial load.

Your wiring normally has two different types of protection:

• Against short circuits, which is provided by fuses or circuit breakers, and
• Against overloads, which is normally provided by an overload sensing device in the motor controller, such as an overload relay. Fuses sized very close to the motor's nameplate running current can also be used to provide this overload protection, but that approach is seldom used.

The running overload protection typically comes in different time classes (IEC classes are 10, 20, & 30), which tell how much time (approximately in seconds) the motor will be given to start and get up to speed before any overload is acted-on. The sensing elements are usually small resistors that produce heat in the overload relay to mimic the heat produced inside the motor. (Electronic overload relays have the same purpose but do it differently.) The setting of this overload relay usually is to the nameplate current of the motor.

If you have this type of overload protection for your motor, and it is working, then all your worries about inrush heating and overloading of the transformer and other wiring can be set aside. The electrical codes have been written, based on over 100 years of experience with equipment in daily use. The people who have written them include manufacturers, electricians, insurers, engineers, and many other types of people. Check the motor's data and the controller's setting. Put an ammeter on it and see how long it takes to start the motor, until its current drops down to approximately the value on the nameplate.

Come back and post this information so we can all see it.

--John M.

Induction motors Sq cage draw 6times the rated current at starting.So the voltage drop at starting needs to be calculated for which you have use the Impedance of the Trf.Moto can accelerate normally on low inertia loads at 80% of rated terminal voltage and will stall below this.Also the drop will depend upon loading on transformer when you start the motor.When you use a star/ delta starter at the point of changing over from star to Delta 6 times the current will flow.In other words the use of the star /delta starter reduces the duration for which the starting current persists.Starting current is drawn till motor reaches full rated RPM which in turn depends upon the Inertia of the load.If loads of soft ,low inertia equipment like centrifugal pumps the acceleration time is short.You have to do a lot of calculation .Best for you will be to use soft starters.