Brilliant explanation and depending upon your point of view, it may or may not have answered one of the pertinent questions, which was :-

"Why is the current reduced in Star format?"

The reason is that from phase to phase in star, the current must flow through two windings, = double the impedance, = cheaply said (not exactly true, but near enough for Government work!) = Half the current.

In delta format, each winding is connected phase to phase, so there is only the impedance of one winding, therefore a higher current will flow.....

If you regard for simplicities sake impedance as resistance, it may make it easier to understand.

Resistance is a terminology for a non capacitive and non inductive resistance, it reacts basically in the same way for both AC and DC currents. (except that some resistors are internally wire wound and can have an inductive component, so be careful here!, carbon resistors have basically no reactiveness)

Impedance is a term for a component that either has capacitive or inductive or both and reacts differently to AC than to DC, in fact it "impedes" changes in the voltage/current....which AC does 50 or 60 times a second in most of the world....

I hope that this explains that last question for you exactly.

All confusion started with the theory exits for star -Delta connection. When connected in star, Line current & Phase current are same I Ph=I-L & Vph= V-Line/root 3 It means full line current has to flow through phases while phase voltage reduce by 58%of that Line voltage . So by theory there is no reduction of current in star connection.

However in delta ,line voltage & Phase volage are same while phase current reduce by 58% of lne current.

Inspite of all theory ,actually current reduce in star.Any way your explaination is also logical.

To reduce the confusion, consider the voltages across each winding.

In star, the voltage across each winding is equal to the lin voltae divided by rt(3). On a 400V supply, that results in a voltage across each winding or 230 volts.

In Delta, the voltage across each winding is equal to the line voltage, so on a 400V supply, the voltage across each winding is 400V.
As the voltage across each winding is rt(3) higher in delta than in star, the current through each winding is higher by rt(3) also.

In star connection, the total current in the line is equal to the current of the star connected winding. In delta, the line current is the vector sum of the currents from two windings with the currents having a phase offset.
The net result is that the line current in delta is three times the line current in star.

Best regards,
Mark Empson

http://www.LMPhotonics.com | http://www.LMPForum.com

Thank you for the prompt reply.You are very logical & correct.Thanks again.

Regards

You know, if you are going to plagiarize, you should at least give credit to the original author!

Original source of his entire response! Click here.

Hello PeterRyan

Thank you for posting the contents of my web page in response to this question, it would be helpful if you provide a link back to the original source for others to follow.

Best regards,

Mark Empson

http://www.LMPhotonics.com | http://www.LMPForum.com

Thank You Sir For Replying

The Star-Delta starter also comes with other variations.

There is slip-ring and resistance combination starters, and the soft-starters, with the latest introduction of the Inverter system.

Modern electronic components make the motor starter system more refined and "Power-grid" friendly. What you saw in the previous postings by different individuals are the step function control, or the on-off control. But the soft starter and the Inverter controls are proportional control and also PID and FUZZY LOGIC control, making our world a better place to live in.

To know what I am talking about, do a Google on the key words "PID" or Proportional, Integrative and Derivative Control, and Inverter control, and Fuzzy logic control.

In star the voltage is 1/root3 or57%of rated volts of Delta connection. So if a motor starts in star the staring current is limited to 57%of rated current.As motor accelerates and we change to Delta at 75-80 % of rated RPM full voltage is applied .What is achieved is actually the duration for which the starting current persists as at the time of Change over to delta there will be a current pulse which is for a short duration.Further this can be used only when the motor is designed to Run in Delta.Starting current duration also depends on Load Inertia.In high inertia loads it persists till motor attains the full rated RPM or in other words on the acceleration time.The value of starting current is determined by the design while the duration by voltage applied and the load inertia.

So the idea is to start it with high impedance star windings, then, when the speed is increased enough to allow some back emf, you can switch it over to delta windings. (I know...they are the same windings, just connected differently....grin!)

Is that right, or have I missed the whole point?

Spot on!

Hello Yusef1

Not quite true, it is not about back emf, that is a DC motor thing. It is about the impedance of the motor. The impedance of most standard induction motors only begons to rise significantly when the rotor speed is greater than 80% speed, so the transition to delta should only occur at speeds close to full speed.

If you step to delta at less than 80% full speed, you will draw a very high resynchronising transient and then a high current (close to Locked Rotor Current) until the motor is close to full speed.

You may find the pages at http://www.lmphotonics.com/m_control.htm and http://www.lmphotonics.com/m_start.htm may help to give a better picture. (I wont copy and past the whole pages!!)

Best regards,

Mark Empson

http://www.LMPhotonics.com | http://www.LMPForum.com

Right. I think I got it. I tend to think of "back emf" as the result of the motor acting as a generator, and the faster it goes, the higher the voltage it creates in opposition to the incomiming voltage. And I think of impedance as the choking of current (modified by lead and lag and copper/iron losses as the case may be) caused by collapsing magnetic fields in coils which result in "choking" of the incoming current...current limiting as opposed to voltage limiting. From a technician's point of view, I don't see much difference in a constant cps input motor curcuit.

On the other hand, I am just the technician who has to work with the blessed things, not the engineer who has to design them...but it helps to know these things when your employee has a bright idea...like putting a rheostat onto the input of the motor..to explain to the twit why this is not a good idea.

It depends, for what reason you ask the question? Is the reason to understand the theory? or torque problems? or cost? or mechanical? In practise it is not the preferred option, as it has dis-advantages. With this option the electrical equipment is increased and labour costs. The perceive advantage, within a single installation is that this method of starting a motor reduces the starting current or "in rush" by more then half, but it also reduces the applied torque, so where is the real advantage? Their is many other ways to solve this problem, ether mechanically, electrical or a combination of both. Electrically this can be solved by a electronic soft starter, or a electronic variable speed drive. Mechanically this can be solved by a fluid coupling, ratio of pulley's, and drive options. But everything comes back to cost, capital and running/ maintenance costs? And of course at the end of the day the process that the motor is required for? So this needs more debate on the application rather then the method?