Motor current in star and delta connection

[tariqzia]

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

 

[Ingenieur]

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

that is delta i
wye/star will be i/sqrt3
this assumes it runs delta at 380/3

 

[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

For a start, it will not be 11kW for both connections.

 

[kwired]

For a start, it will not be 11kW for both connections.

presuming same input voltage both ways.

 

[Besoeker]

presuming same input voltage both ways.

Well, only 380V is mentioned.

 

[jumper]

The following link shows the formulas and such for the two types of connections.

I find it a nice little comparison, easy to follow in my opinion.

https://www.electricaltechnology.or...arison-between-Star-and-Delta-Connections.jpg

That jpg is a bit hard to read, here is the webpage URL.

https://www.electricaltechnology.org/2014/09/comparison-between-star-and-delta-connections.html

 

[Ingenieur]

The following link shows the formulas and such for the two types of connections.

I find it a nice little comparison, easy to follow in my opinion.

https://www.electricaltechnology.or...arison-between-Star-and-Delta-Connections.jpg

That jpg is a bit hard to read, here is the webpage URL.

https://www.electricaltechnology.org/2014/09/comparison-between-star-and-delta-connections.html

good chart
since line I is reduced by 1/sqrt3 so would power since it is proportional

 

[Besoeker]

good chart
since line I is reduced by 1/sqrt3 so would power since it is proportional

That's simply not correct. The voltage is also reduced.
And it isn't a linear relationship.

 

[Ingenieur]

That's simply not correct. The voltage is also reduced.
And it isn't a linear relationship.

the v is not reduced
it is distributed over 2 coils (2 times the Z) reducing the current
y P = sqrt3 v I
d P = sqrt3 v I/sqrt3
Py = 0.577 Pd

 

[jumper]

Let's take notice of the fact the the OP is new and this is first post.

Please, no tangents and keep it simple.

Thanks

 

[Besoeker]

the v is not reduced

Per winding it is.

 

[Ingenieur]

Per winding it is.

that is what I said but you edited my post
same v over 1 vs 2 coil

but for the power calc S = sqrt3 v i, v is the same, only i is reduced by (1-1/sqrt)

 

[Besoeker]

that is what I said but you edited my post
same v over 1 vs 2 coil

It isn't. But can we quit now?

 

[jumper]

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.

Yes, I know what I am saying is simplistic, but could say that it is essentially correct?

 

[Ingenieur]

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.

Yes, I know what I am saying is simplistic, but could say that it is essentially correct?

not really, but close
you transition from y to d
current in y is reduced vs d since you have the same v, but in y it is across 2 coils, and only one in d, in essence double (actually 1.732) times the impedence
this reduces i in y mode since Z increases and v is constant
this is the whole reason for doing it: to reduce starting current/power
once spinning transition to d full current/power
ol' timey soft start lol

so the equations are the same except
line i in y is 1/sqrt3 of that in d in this case, y to d transition
meaning current/power in y are ~0.577 (1/sqrt3) of the d value

 

[jumper]

Ah, okay I see what you mean.

I reworked the equations substituting values to understand my mistake better.

 

[tariqzia]

The following link shows the formulas and such for the two types of connections.

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?

 

[tariqzia]

not really, but close
you transition from y to d
current in y is reduced vs d since you have the same v, but in y it is across 2 coils, and only one in d, in essence double (actually 1.732) times the impedence
this reduces i in y mode since Z increases and v is constant
this is the whole reason for doing it: to reduce starting current/power
once spinning transition to d full current/power
ol' timey soft start lol

so the equations are the same except
line i in y is 1/sqrt3 of that in d in this case, y to d transition
meaning current/power in y are ~0.577 (1/sqrt3) of the d value

So you mean power in star will be 0.577*11kw?

So calculated current by the eq iline=11kw/1.73*380*0.90 =19A will be delta line current.
line current in star will be iline=11kw*0.577/1.73*380*0.90 =10.72A will be star line current. but i think this should be the phase current of delta?

 

[tariqzia]

that is delta i
wye/star will be i/sqrt3
this assumes it runs delta at 380/3

Thanks for your response.

So calculated current by the eq iline=11kw/1.73*380*0.90 =19A will be delta line current.
line current in star will be iline=11kw*0.577/1.73*380*0.90 =10.72A will be star line current. but i think this should be the phase current of delta?

 

[tariqzia]

For a start, it will not be 11kW for both connections.

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