# Star Delta Starting

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Dear All,

As we know that any induction motor draws a peak current from supply when it starts. So the contactors should be able to pass on such a large starting current. But we have Japanese machines which star contactors have less rating (<30 - 40%) as compare with delta contactors in their star delta starters. some one can explain it why star contactors have less rating?

If we want to start, for example, a 37KW, 4poles, 400VAC, 50HZ motor on star delta mode, what should be the rating of star delta contactor?

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Line and Delta contactors need to be rated at 58% motor nameplate FLC. The Star point contactor needs to be rated at 33% motor nameplate FLC. Contactor utilisation category AC3 also applies.

Regards,

GGOSS

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Line and Delta contactors need to be rated at 58% motor nameplate FLC. The Star point contactor needs to be rated at 33% motor nameplate FLC. Contactor utilisation category AC3 also applies.

Regards,

GGOSS

Dear GGOSS,

Are you sure about your answer? If we consider it right, it means that a 37KW 380V 70.9A motor can start through (70.9*0.58 = 41A) 41A rated Line and Delta contactor and 23A rated star contactor. I think these are too de-rated. We have a same motor which Line and Delta contactor are 70A rating and Star contactor is 50A rating. Please revise your answer as I have really confused.

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The answer provided to you is correct. If you are not prepared to accept that, you should not hesitate to contact any manufacturer or supplier of switchgear products for confirmation.

Regards,

GGOSS

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Rzn,

At a given voltage, the current in Star is the current in Delta divided by the square root of 3 (1.732) which works out to be 58%. So the Starting contactor (K3 on Marke's diagram) need only be rated for that current. When switched into Run, technically the Run contactor (K2) need only be rated for the remaining 42% current, but most manufacturers just use the same size as the Starting contactor. The Star contactor (K1) is only going to carry 58% of the Starting current, and .58 x .58 = roughly 33% of the FLA.

The fact that some supplier has chosen to provide larger contactors is probably based on some economic benefit they get from it, i.e. they buy a larger quantity of the same sizes and use them for smaller motors rather than carry more inventory.

"He's not dead, he's just pinin' for the fjords!"
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Jraef,

Could you please shed some more light of the 42% issue. As far as I can see; when the motor is connected in Delta (Run) both contactors are effectively connected in series within the delta loop of the motor and hence they would see the same current which is of-course equal to line current/rt3. Am I missing something?

Regards,

GGOSS

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When the motor is star connected, the current through the motor windings is one third of the rated current of the delta connected motor. => the star contactor is rated at 33% of the motor rating.

When the motor is connected in delta, the current through the motor windings is 1/rt3 of the line current for the delta connected motor. => The two other contactors are rated at 1/rt3 * the motor rating = 58% of the motor rating.

Best regards,

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Dear All,

Thanks a lot for very useful informations.

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• 1 month later...

The Star contactor (K1) is only going to carry 58% of the Starting current, and .58 x .58 = roughly 33% of the FLA.

Dear Mr. Jraef,

I was reading that topic and your answer, I am not cleared about ".58 x .58 = roughly 33%". Can you explain it more in detail?

"Don't assume any thing, always check/ask and clear yourself".

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1/rt3 = 0.58 (approx)

1/rt3 x 1/rt3 = 1/3

Best regards,

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Thanks Mr. marke.

"Don't assume any thing, always check/ask and clear yourself".

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• 1 month later...

Rizwan,

The star contactor's job is tie the three phases ( that's why you see the 2 loops at the top of the star contactor) and as the current is reduced during star (Y) the contactor rating is lesser than line and delta contactors.

However there are certain applications like a stone grinder or pulveriser, where the starting current is more than a regular application and hence the contactors are oversized.

Centrifuges used in Food processing industry have a very starting current (because of inertia) but running current is low and hence the regular sizing formula is not applicable.

Special Horses for special races...

You may review your application and then size the contactors.

If it is regular application then sizing can be done as suggested in earlier posts.

Chaterpilar

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Hello Chaterpilar

The actual start current drawn through the star contactor of a star delta starter is dependant on the rotor design of the motor, rotor speed and line voltage only. It is independant of the connected load.

The connected load will determine how long the motor takes to get to full speed and the application will determine the number of starts per hour.

The contactors on a star delta starter are sometimes oversized, due to extended starting times, or start frequencies, but not the start current magnitude.

Best regards,

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Hello Chaterpilar

The actual start current drawn through the star contactor of a star delta starter is dependant on the rotor design of the motor, rotor speed and line voltage only. It is independent of the connected load.

The connected load will determine how long the motor takes to get to full speed and the application will determine the number of starts per hour.

The contactors on a star delta starter are sometimes over sized, due to extended starting times, or start frequencies, but not the start current magnitude.

Best regards,

Side story:

I once worked on a fish oil separator project (50HP Alpha Laval) started with Star-Delta which took 7.5 minutes to accelerate in star before switching to delta. The motors were specially designed with a stator core twice as large as normal and a rotor from a 150HP motor. The star current was still the same percentage, but the accel time was so long that they severely over sized all of the contactors. Even so, they didn't last long because the switching transient was horrible; it would occasionally cause the circuit breakers to fail. We replaced them with soft starters using 150HP frames programmed for the 50HP motors. The soft starters accelerated them in 3 minutes at 300% current limit

"He's not dead, he's just pinin' for the fjords!"
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Marke,

The No-load amps of say a 100 HP motor ( during Y start) will be same as the same motor connected to a pulveriser...? the motor will not see the load..? during Y ...?

respectfully,

Chaterpilar

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Hello Chaterpilar

Yes that is correct.

Under high slip conditions, (speed less than 95% speed, the current drawn by the motor is a function of the rotor design, the terminal voltage and the rotor slip. The rotor can not see the driven load, just the frequency of the rotor current.

If you take a motor and connect it to a high inertia flywheel and start it DOL, it will draw Locked rotor current and the current will only drop significantly as the rotor reaches full speed. This could take say 30 seconds and an ameter will show the high current, somewhere in the order of 600 - 750% of rated current.

Now apply the same motor to say a pump with a very low inertia and DOL start it. The motor will initially draw Locked rotor current and the current will only drop significantlyas the motor reaches full speed. Exactly the same current profile as with the high inertia load only it is now starting is 1 - 2 seconds. An ameter will show a lower current because of the time for the ammeter to respond. A high speed chart recorder would show the same magnitude of current.

Next start the same motor open shaft, and the current profile will be exactly the same, but for a shorter period of time again.

If we connect the motor in star, the same thing happens except that the current drawn is one third of DOL current. (The start torque is one third of DOL torque)

Have a look at http://www.lmphotonics.com/m_control.htm

Best regards,

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Good work Mr. Marke.

As you said that the rating of star contactor defines according to starting current of motor and acceleration time of rotor up to 95% speed. Now, for example, if the rotors of three 10HP motors accelerate up to 95% speed in 5sec (Eddy current motors), 1 minute and 5 minutes then how can we determine the rating of star contactor?

And what about main contactor? It is often used as 58% of the motor rating. What would be the rating of this contactor for above mentioned applications?

"Don't assume any thing, always check/ask and clear yourself".

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Hello AB2005

This quite an involved topic, but let me see if I can give you some idea.

Firstly, there are a number of different ratings that apply to contactors and these ratings are essentially thermally determined.

When the contactor is closed, and current flows through the contacts, there is power dissipated in the contacts which causes the contacts to heat up. If the contacts get too hot, the contacts will degrade and eventually fail.

The rate of degradation is a function of the maximum contact temperature. The temperature rise of the contacts is related to the power dissipated in those contacts and that is proportional to the square of the current. The temperature of the contacts does take time to rise, so if the peak power dissipation occurs for a short time, it will only elevate the temperature by a small amount.

The contactor is given a thermal current rating, usually designated AC1. This is the maximum continuous current of the contactor when used at it's rated ambient temperature. If the ambient temperature is increased, the contact temperature will increase and the life of the contactor will reduce. The contactor can be derated such that the maximum contact temperature does not rise above the maximum and this will result in the design life being maintained. Reducing the ambient temperature will enable a small increase in current without compromising the contactor life.

Under AC1 ratings, there must be no overload current beyond the AC1 rating.

Motor starters use AC3 rated contactors. AC3 rated contactors are rated for an high overload for a short time. This overload current is sufficient to DOL start a motor and would typically be in the order of 6 times the motor rating. This results in 36 times the power dissipated in the contacts during start and will cause a rapid elevation in temperature. Provided that the length of time this overload occurs for is restricted, the contacts are rated at a lower current and the frequency of the overloads is restricted, the maximum temperature of the contacts are kept below the maximum temperature and the life of the contacts is OK.

If the overload current is increased, the overload time must be reduced, and/or the continuous rating must be reduced.

If the overload current is reduced, the overload time and/or the continuous current can be increased.

If the overload time is increased, the overload current and/or the continuous current must be reduced.

With all the above in mind, now look at the star contactor.

The star current is one third of the delta current, so the overload current for a contactor rated at one third of the motor rating will equate to the same power dissipated during the star connection as it would be if it was starting a motor of one third of the size, however, in the case of the star contactor, there is no continuous current flow which allows the contacts to cool. This means that it is possible to push the contactor a little harder under normal conditions, but most do not do this.

If the start time is increased beyond the design time of the contactor, then the contactor should be derated, so a larger contactor needs to be used.

The derating of contactors under extended starting times is a function of the design parameters of that contactor and this can only be defined by the contactor manufacturer. Some manufacturers publish curves to enable you to do this.

If you double the start time, you double the total power dissipated in the contacts. If you reduce the current to 71%, you will halve the total power dissipated in a given time. I would suggest that everytime you double the start time, you should reduce the current to 71%. At four times the time, you should halve the current.

If a 30 second start time was acceptable for the contacts, the at 1 minute start time, you should increase the contactor rating to 1/0.7071 = 1.414 times the current rating. At two minites, the current rating should be 2 times and a 4 minutes the current rating should be 2 x 1.414 = 2.828 times the current rating.

For extended times, you could use contactors with an AC1 rating equal to the start current and then you would have no issues. This would require the star contactor to be AC1 rated at around twice the motor rating.

The same theory applies to the main contactor, and as the delta contactor will be subjected to similar overloads if he motor does not reach full speed before the changeover, it should be rated the same as the main contactor.

Best regards,

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Well done Mr. Marke,

What a great explanation.

Thanks a lot for spending your valuable time for posting such a lengthy and accurate answer.

"Don't assume any thing, always check/ask and clear yourself".

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Thanks Marke for your insight into Y part of Y-D starting.

regards,

Chaterpilar

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• 1 month later...

Good Day to everybody.

Can anybody help me with this? I have a 7.5HP motor connected to DOL cobtrol and its been working for 5yrs. Is there any advantage if i will connect it to Star-Delta control?

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It changes nothing in the operating mode, but it could allow starting with less flicker if that is a problem. I have never seen that to be a problem on 7.5HP however. I would just leave it alone.

What were you thinking it could do for you?

"He's not dead, he's just pinin' for the fjords!"
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DOL is the most efficient way of starting a motor/machine. The move to any form of reduced voltage starting (star/delta, primary resistance, auto-transformer or soft starter) is influenced by the need to reduce starting current or to reduce mechanical stress.

The type of reduced voltage starter you select depends very much upon the starting characteristics of the motor and the torque required by the load. Without that information it is difficult to advise what advantages are possible.

What type of machine is the motor driving?

Are you experiencing voltage fluctuation related issues at moment? If so, describe them.

Are you experiencing any mechanical issues at the moment? If so, describe them.

Answering the above questions may provide members the additional insight needed before they can advise you further.

Regards,

GGOSS

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I am using this 7.5HP motor on AHU and I never experienced any voltage fluctuation nor any mechanical issues. But just curious, coz based on what I had read in the LM Photonics Ltd. (http:/lmphotonics.com/star_delta.htm) that "Traditionally in many supply regions, there has been a requirement to fit a reduced voltage starter on all motors greater than 5HP (4 KW)."

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