Star/Delta AC Motors

I do believed that you don't really understand the principle of the motor and it's applications. These normal motors are designed to be connected in either Star Connection or in Delta Connection. Some of the motor are design to have both type of connection available. Either, the motor connected in Star or in Delta connection, each of them have their own specialty in driving the load that the motor is connected to.

The horse power or are the same regardless of what type of the motor connection is. If the motor are rated at 10HP it will remain at 10 HP either it is connected in Star Connection or in Delta Connection. For these motor that have dual voltages, either you connect in delta connection (for low voltage) or in Star connection (high voltage), the final power of these motor are the same as proven by the formula as follows;

P=V*I*1.732*PF

For Delta connection ( Lower Voltage), your input voltage is low but your ampere will be high as base on the above formula. You still get the same motor power, and that goes to the Star Connection.

In IEE regulation stated that "any motor that higher than 13HP must be provided with a Star/Delta starters" that why you have star/delta starter for the motor. It is not that simple just to fix a bigger motor to the load or just start with one motor with bigger horse power.

Do you understand the starting current curve of these induction motor and do you know what is the maximum starting current of these motor in the DOL Starter?

For a 10HP motor with direct on line starter (DOL), your motor staring current is about 122 amps and if you put bigger motor to run your machine with DOL starter, you are going to trip off the entire power supply line from the national grid system.

Star-Delta Starter provide with a lower starting current and when the speed is up to 75% of it normal rated speed or the torque, it then change to delta connection for final speed and torque. This Star-Delta Starter prevent your power supply system from sudden voltage dip and it also help to prevent your machine damages during starting of these big motors.

Inrush current has quite a bit to do with it, but I've always been told that the primary reason for a star/delta setup for a big motor is heat. You keep a big motor in delta all the time you're going to burn it up real quickly as it's going to draw a lot of current at first, only some of which is initially transferred to motion. The vast majority of it will become heat energy, and it will melt things.

My advice? Go with a VFD. :)

Your theory is somewhat accurate. However, the purpose of star delta starting is to decrease the torque where certain applications could do serious damage to the connected load. For example, star connected motors (as most as in the US) put the voltage at 1.732 and I don't think that requires explination, therefore reducing LR and torque initially to 33%. Subsequently, motors used in the USA are, in fact, soft start motors. Wye-Delta is very common in the rest of Europe, even on large HP motors.

I think you have it a little backwards. The starting voltage of a wye connected motor is line/1.732 subsequently causing low voltage starting reducing the torque and LR current to 33%. When transition is complete wheter open or closed, the delta will see maximimum torque and voltage.

There is no magic on Star Delta motors and starting.

Consider a simple 3 winding 3-phase motor with one winding having two leads for each winding, an in and an out. Lets also say that the design voltage for each winding is 480 VAC.

If you connect the three windings in a triangle configuration, called delta, each winding has the full 480 VAC across it and experiences full load current during operation and 6-7 x full load current during starting for a NEMA design B induction motor, which most industrial motors are today.

If you connect the motor with one lead of each winding tied together and the other lead from each winding tied to one of the 3-phase power leads, the current has to pass through two windings betwen each phase and the voltage across each winding is 480 VAC / 1.732 or 277 VAC. This reduced voltage across the leads results in a softer start of the motor and less current drawn from the 3-phase power source.

In a WYE-DELTA motor starter there are two contactor arrangements, one for STAR and one for DELTA and a timer that initially powers the motor in a STAR configuration and after a time delay to allow the motor to spin up to near running speed switches the motor connections in an open transition to DELTA configutration which gives the motor full power, speed and horsepower available to the load.

Notes:

1) The motor is not run at full nameplate horsepower while in the STAR connection.

2) About 20 years ago I worked on a machine for installation at Boeing in Everett Washington. Our machine had a 300 HP main drive motor. It required a NEMA size 7 STAR-DELTA motor starter that occupied an enclosure 84 inches high x 36 inches wide x 24 inches deep. This was an excellent application for STAR-DELTA reduced voltage starting of the 300 HP motor that kept the motor's starting current low enough to protect the plant's power system.

Thanks,
Greg

Hi All

A side topic on the star/delta AC motors, which is what happens when the incoming voltage drops?

A: ampere drawn by the motor goes up to compensate for the loss in voltage?

OR,

B: Because of V=IR, ampere drawn drops?

at the instant of starting, machine acts as an L-R load and therefore you can calculate inrush current from V = X * I, where X is the equivalent transient impedance.

Quote" at the instant of starting, machine acts as an L-R load and therefore you can calculate inrush current from V = X * I, where X is the equivalent transient impedance."

Not at the starting, but when the machine is operating and experience a voltage drop.

Does this sounds like enenergy conservation where Power = I*V, where I goes up to compensate for the decrease in V?

But I am unable to explain what happens to V=IR, Ohm's Law, the fundamental of electricity?

Anyone with practical or theoretical experience and know-how can help?

Thank you!

if you drive the motor from utility without any control (open loop or closed loop), then motor works depending on the internal circuit of the motor and the mechanical equations (torque, speed, shaft power etc.) Moreover, motor can be modeled simply as L-R-emf circuit, where emf is proportional to shaft speed. Therefore, without any control action, if input voltage drops, the current drops from the formula I = (Vinput - Vemf) / X, where X is the impedance of the circuit. At that time, electrical shaft torque is reduced that is proportional to input current and shaft speed reduces. With the reduction of shaft speed, back emf reduces and input current increases that means increased electrical torque. This natural balancing mechanism forces the system to an equilibirium point at which eletrical torque is equal to mechanical torque (friction, load torque etc.) and shaft speed is constant at a lower value. However, in the case of controlled-motor drive, controller tries to stabilize the output power and sinks more current for reduction of input voltage typically.

If voltage is lower then the starting current will, of course, be lower, motor's torque will become lower too, but the starting time will be longer. If by that the motor with its reduced torque turns out to be unable at all to cope with the load at its shaft, then at best the protection turns the voltage off, at worst - the motor burns.

If load is constant on the motor or not cut off, it will try to draw more current since it is moving towards locked rotor status.

The question of low current due to low voltage is not applicable in this particular case.

DHAYANANDHAN.S,

CR4 MEMBER,

INDIA.

The incoming voltage does not drop. It's simply applied voltage in a star connected motor is line voltage/1.732. When it transistions into delta the line voltage is then applied voltage. Say you have a 480 volt wye-delta motor. When it first starts 480v/1.732=277 starting volts. When it switches to delta you then have 480 volts, line voltage applied that will give you close to full torque.

The OP has mixed together two different items: dual-voltage (e.g. 480/240) motors, and star-delta (480/277).

A typical dual-voltage motor has 6 windings and 9 leads. For high voltage, the windings are connected in series (3 pairs) For low (1/2) voltage, they are connected in parallel (again, 3 pairs).

A three-winding star-delta motor would have 6 leads. The motor would be rated at one voltage (full delta volts). When started in star mode, the voltage on each winding is 1/sqrt 3 times the full voltage. This is only momentary, however, and is not intended to be run for long or to have a HP rating.

A 6-winding dual-voltage star-delta motor would have 12 leads, which can be connected in series for full voltage or parallel for half voltage. In addition, at either voltage, it can be further connected in star for starting, delta for running.

Simon Wan mentions an IEE (IEC?) requirement for >= 13 HP (10 KW) motors to have star-delta capability. This sounds like a European thing, and it doesn't seem to make much sense. After all, reduced-voltage starting can also be done by resistive/reactive starters, autotransformer starters, or variable frequency drives.

Of the responses so far, Greg's is most correct (GA).

The wye-delta reduced voltage is very common in Europe. Many shipboard electrical systems that I design or in fact wye-delta starting, small to large HP.

hi,

i saw some three phase motors ''american made'' 75hp with no star /delta operation. it is operated only on star connection all the time, and it is working fine.

also i saw some three phase motors ''german made'' 75hp but operated with the star/delta mode.

both are used to run the same loads ''water pumps''

why is that? whitch one is right? which one is better? or its some thing to do with the manufacturer design?

Star-delta motors are special, e.g., if not transition to delta in a short period of time will burn up.

For the same HP a star connected motor will produce 3 times more heat (when compared to a delta connected motor) as the windings will carry the line current in star connection. So, at low voltages it is preferred to have delta connection.

But, if the motor is a high voltage motor, if you connect in delta to get the lesser heat generation advantage, then you may have to rate the winding to body insulation for full line voltage, whereas in a similar case, with star connection you have to insulate only for the phase voltage. At higher voltages, teh insulation costs are considerable.So, HV motors are connected in star. Whereas at low voltages, there is not much cost difference between phase voltage insulation and line voltage insulation.

hi,

the american star connected motor is 480v 3phase, the german star/delta motor is 380v 3phase.

so you mean the 380v motor ''german'' is better design as the heat will be less three times from the 480v motor?

i mean why we use star/delta way?

You kindly recall that in Star connection Line voltage is 1.732 times the pahse voltage and line current is equal to phase current. In delta connection, line voltage is equal to phase voltage and line current is 1.732 times the phase current.

In a motor, line current is what is measured at the terminal connecting cables and phase current is that carried by the windings.

Obviously, in a delta connected motor, the winding current or the phase current is only 1/1.732 times or only about 56% of the line current (i.e.) the current that would flow in the windings in a star connected motor.

So, for the given HP rating, definitely, in a star connected motor the heat generated in the winding will be 3 times that in a delta connected winding.

Only because of this advantage of less heat produced in the windings, do all LV motors are preferred to have a delta connected winding.

If your interstest in German motors - you can find it here http://www.motor-mauer.de