Motor current in star and delta connection

[jumper]

Patience Tariq.

We are 10 plus hours behind you, so it is midnight or so here.

Not as many members are participating this late, you will get responses eventually.

We are glad to have you here, just give us a bit of time to respond.

 

[Besoeker]

So, according to the link I posted the power formulas are the same, because the differences in the voltage and amperage equations basically balance each other out.

Not really. The voltage on the windings is reduced, the current is reduced. That's the whole point of star delta starting. The raison d'être.
It is not a simple linear linear relationship which is why there is not a simple answer to the OP's question. Treat it that way and you'd maybe get 3.7kW in the star configuration if the 11kW pertains to the delta connection. There are a lot of ifs.
And what I have been trying to explain only to have my posts deleted. This one too probably will be seen as argumentative but I try to stick to facts when others don't. Perhaps you may now understand why I got very irked about the deletions.

 

[jumper]

Not really. The voltage on the windings is reduced, the current is reduced. That's the whole point of star delta starting. The raison d'être.
It is not a simple linear linear relationship which is why there is not a simple answer to the OP's question. Treat it that way and you'd maybe get 3.7kW in the star configuration if the 11kW pertains to the delta connection. There are a lot of ifs.
And what I have been trying to explain only to have my posts deleted. This one too probably will be seen as argumentative but I try to stick to facts when others don't. Perhaps you may now understand why I got very irked about the deletions.

Well, the OP is online now, so the way I see it the best thing we should do is address his questions as best we can now. An active audience is a good thing IMO.

 

[Besoeker]

Well, the OP is online now, so the way I see it the best thing we should do is address his questions as best we can now. An active audience is a good thing IMO.

Thank you for the very polite response.

 

[Besoeker]

If i use the power equation Iline=11kw/1.73*380*0.90*eff,, I=19 A

If this is line current of delta, then phase current will be 19/1.73 =10.74A
So now what will be current in star?
Can we use above equation to calculate star current directly?

You could I suppose. Whether it will be correct is another matter. In star it won't produce the rated power or torque. It may not reach rated speed so operate at high slip.
Stated efficiency and power factor go out the window. You can't simply pro rata the values.

 

[mivey]

You could I suppose. Whether it will be correct is another matter. In star it won't produce the rated power or torque. It may not reach rated speed so operate at high slip.
Stated efficiency and power factor go out the window. You can't simply pro rata the values.

If it is the same motor. The OP may be asking about comparing winding currents on an 11 kW motor #1 designed for delta connection with motor #2 having an 11 kW rating designed to be wye connected.

Tariqzia, is this scenario comparing two motors wound two different ways or for one motor connected two different ways?

 

[jumper]

One for the mods.

The mods ask that we all please just answer the OPs questions.

 

[topgone]

I am looking at this as if he has a 6 lead motor and a wye-delta transition starter
regardless, makes little difference in the math

But remember that most small motors are wired delta. Configuring a motor designed for delta and wiring it in wye, supplying it with the same voltage sure lets out holy smokes!

 

[Jraef]

Thanks for your reply.
iline=11kw/1.732*380*.90
I=19A

What will be current if this motor is only connected in delta? VL-L 380
And what will be current with same voltage if motor is connected in star? VL-L 380
Please explain

Before this can be answered, you need to KNOW FIRST how the motor was wound from the factory as it pertains to the nameplate ratings. Since you are describing a 380V motor, we can assume this is an IEC design, and typically in a motor like that, the 380V connection already IS the Star pattern. If that's the case, the discussion is over. But if by chance the motor is wound as in Delta for 380V, the reconnecting it in Star would indeed reduce the effective winding voltage to 57% of normal or 220V, so the torque and current capability of the motor would be reduced to 33% (57% squared). Capability is an important concept here as you will see.

Now you must discuss what happens with the LOAD connected to that motor. If the load remains the same and required all 11kW of mechanical power from that motor, the loss of 67% of its torque will result in the slip becoming very high and the motor pulling very high current yet likely not developing sufficient torque to fully accelerate it. The end result then is a stall condition. If on the other hand your 11kW motor was only using 3kW to move your load, that load is sufficiently low enough to be handled by the newly configured motor without stalling.

But here's the thing; if your load only required 3kW, why was an 11kW motor used in the first place? It makes no sense to have done that. Still, I see people do this sort of thing when they have an old system that was built for higher production and over time, becomes under utilized. So rather than just going out and buying a smaller motor, they use what they have by changing them from Delta to Star connection. There is actually an argument to make for this because the one thing it can do for you (again, only because the LOAD is lower now), is to reduce a small amount of the losses in the motor windings. A portion of the Magnetizing current losses are tied to the voltage across the windings so by reducing that voltage, you reduce those losses. It's a fraction of a fraction however so not really worth doing for its own sake.

 

[jim dungar]

My question is how can we calculate motor current in delta and star.

Why do you want to make this calculation?
Is this a homework (theoretical) problem or an actual installation?

Most of the previous posts are guessing at the actual conditions, and therefore guessing that their discussions are proper answers.

 

[tariqzia]

If it is the same motor. The OP may be asking about comparing winding currents on an 11 kW motor #1 designed for delta connection with motor #2 having an 11 kW rating designed to be wye connected.

Tariqzia, is this scenario comparing two motors wound two different ways or for one motor connected two different ways?

Consider it a single motor, have option to connect it in star and delta. If i connect it in star it should have less starting current as compared to delta starting.
Now my question is if i want to calculate starting current,how can i calculate it.?
if i use eq i=11kw/1.73*380*pf 21A (will this current is for delta) this value will be less
if i use eq i=11kw/1.73*220*pf 36A (will this current is for star) this value will be high

 

[tariqzia]

Why do you want to make this calculation?
Is this a homework (theoretical) problem or an actual installation?

Most of the previous posts are guessing at the actual conditions, and therefore guessing that their discussions are proper answers.

Just theoretical problem.

 

[Besoeker]

Hello every one

I have a confusion in calculating motor current for delta and star with same voltage.
My question is how can we calculate motor current in delta and star.
Suppose we have a motor rating 11kw, 380 volt PF 0.90,
How can i calculate line current if motor is connected in delta?
if i use the power eq Iline=11kw/1.73*380*0.90*eff, this calculated current is for star or delta.

Your answer will be much appreciated

Tariq

Hi Tariq and welcome.

It isn't the same voltage though. In delta each winding get 380V. In star it is about 220V and the current greatly reduced.
That's why we have star delta starters. Though not commonly at powers as low as 11kW.

 

[tariqzia]

Hi Tariq and welcome.

It isn't the same voltage though. In delta each winding get 380V. In star it is about 220V and the current greatly reduced.
That's why we have star delta starters. Though not commonly at powers as low as 11kW.

Thanks for your response, my question is how to calculate current for star and delta connected motor. Eq I=KW/1.73*380*pf will give current for star or delta? Please explain

 

[winnie]

The point has been raised several times, that this is not a simple calculation, because it depends not only on the characteristics of the motor, but also the characteristics of the mechanical load connected to the motor.

As Besoeker has already stated, when you have the same applied voltage to the motor terminals, the voltage across the windings changes by a factor of sqrt(3) when you change between delta and star. The star connection puts lower voltage across the windings. So if you have 380V across the coils in delta you have 220V across the coils when connected in delta.

If the motor can successfully supply the mechanical load in both situations, and if the mechanical load stays the same, then the motor will behave as an approximately constant power electrical load. In this specific case the reduced voltage requires greater current to maintain constant power. In this case the star connection will require sqrt(3) greater current than the delta connection.

Far more common is the use of the star connection to _start_ the motor, using contactors to switch to the delta connection as the motor comes up to speed. In this case the mechanical case to consider is the _locked_ condition when the rotor is stationary. When the motor is at zero speed, if connected at full voltage it will draw something in the neighborhood of 6x full load current (the exact amount depends on the design of the motor). By reducing the applied voltage the motor the locked rotor current drawn is similarly reduced. In this particular case the _locked rotor_ current is _reduced_ by a factor of sqrt(3) in the star connection versus the delta connection.

So you see that with 2 different mechanical states the change from delta to star could _increase_ or _decrease_ the current. Without knowing the mechanical conditions for the motor you are asking about we cannot give you a definitive answer.

Note that star/delta starting is applicable to mechanical loads where the torque requirement is reduced when the speed is reduced (think of something like a fan). When the load is stationary the load torque is small, so the reduced starting torque is not a problem. If the torque requirement of the load is constant then star/delta starting likely means that the motor cannot accelerate at all.

-Jon

 

[Sahib]

If the torque requirement of the load is constant then star/delta starting likely means that the motor cannot accelerate at all.

-Jon

If the load torque, whether it is constant or variable, is lower than starting torque, the motor would accelerate irrespective of whether it has star or delta winding at start.

 

[Ingenieur]

i wye = (1/sqrt3 x 1/sqrt3) x i delta
= 1/3 x i delta
both line currents

assume 11 kw, pf 0.8, 380/3 delta ratings
delta line i = 20.9 A
wye line i = 20.9/3 = 7 A

I posted the math but it was deleted
I'll try to find it

 

[Ingenieur]

delta starting torque 150% x rated torque
wye ~ 1/3 of that or 50%
if torque required to start > 50% of rated fl torque motor may stall
this may occur with some machines, conveyors, hoists, elevators

 

[Jraef]

delta starting torque 150% x rated torque
wye ~ 1/3 of that or 50%
if torque required to start > 50% of rated fl torque motor may stall
this may occur with some machines, conveyors, hoists, elevators

There's the answer I think you were looking for tariqzia.

Current and torque are basically the same in terms of percentages and vary by the square of the voltage difference. So if you you use Start-Delta starting, or just permanently connect a motor in Star when it was supposed to be Delta, the effect is the same. The EFFECTIVE voltage across the coils in Star will be lower by the sq. root of 3, so divided by 1.732 (or multiplied by the inverse, 57.7%). Applying the torque formula to that then means the torque and thereby current, becomes .577 squared or 33.3% (1/3) of the torque/current at the designed voltage. That carries through to the starting torque/current as Ingeniuer pointed out.

So yes, you will reduce your current to 1/3, but you ALSO reduce your peak torque capability to 1/3 as well. Peak torque is what your motor uses to accelerate and also RE-accelerate a motor after a change in load. Without the same torque available from the motor, if your LOAD connected to the motor is not also reduced to 1/3, then the slip increases, the motor draws more current, and if it cannot accelerate or even run the load with that increased torque/current, it will overload and go off line (hopefully) or burn up.

Side note that I'm thinking MIGHT be the basis of your question:
If you DO happen to have a motor that is 3x the size necessary for the load connected to it, then you can reconnect it in Star permanently. This is done in a lot of developing countries where old large industrial facilities are repurposed and the new purpose is significantly lower. I see this happen a lot in India for example (through connections I have there, I've never been there) so I can imagine that Pakistan has similar issues going on. The reason they do this is because in a motor, there is a certain amount of fixed energy used in just turning the coils and steel into magnets, and a portion of that portion of the energy used on that is lost due to things like eddy currents in the steel. These are called the "core losses" and vary directly with the applied voltage. So by reconnecting a motor in Star where it becomes 57.7% of the effective coil voltage, the core losses are reduced to 57.7% as well. Again, this represents a fraction of a fraction of a fraction of the total motor energy used so the risk/reward ratio is very low. But in some places they feel it's worth it.

 

[kwired]

Consider it a single motor, have option to connect it in star and delta. If i connect it in star it should have less starting current as compared to delta starting.
Now my question is if i want to calculate starting current,how can i calculate it.?
if i use eq i=11kw/1.73*380*pf 21A (will this current is for delta) this value will be less
if i use eq i=11kw/1.73*220*pf 36A (will this current is for star) this value will be high

When you have that option the wye connection is intended for 1.732 times the input voltage as when delta connected. The individual coils see same voltage across them either way.

Take a motor with coils individually rated 220 volts and connect them in delta to a 220 volt supply - each coil sees 220 volts.
Take same coils and connect them in wye configuration and supply it with 380 volts - it will be 220 volts across each coil to the center of the wye.

Now take a set of motor windings designed to run in delta at 480 volts, but connect them in wye to 480 volts, those coils designed for 480 are only going to see 277 across them, the motor will develop less torque as a result connected in this way - but we do that to reduce the surge level of current during starting and don't leave them connected in wye for long at all in this starting scheme.