Motor current in star and delta connection

[tariqzia]

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

Thanks Ingenieur, It makes clear.
just more clarification plz,
From eq i=p/1.73*380*pf i=20.9 A This will be delta current. Yes?
so i think we cannot use above equation directly to calculate star current.First we calculate delta current from above equation then we divide by root 3 to get star current.
Am i right?

 

[Sahib]

Thanks Ingenieur, It makes clear.
just more clarification plz,
From eq i=p/1.73*380*pf i=20.9 A This will be delta current. Yes?
so i think we cannot use above equation directly to calculate star current.First we calculate delta current from above equation then we divide by root 3 to get star current.
Am i right?

No. You should divide the result by 3, not by root 3.

 

[winnie]

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

In what operating state (stalled, accelerating, full load, no load, etc) is the motor operating in where you see current proportional to the square of the supply voltage?

Thanks
Jon

 

[Sahib]

In what operating state (stalled, accelerating, full load, no load, etc) is the motor operating in where you see current proportional to the square of the supply voltage?

Thanks
Jon

Is he stating or implying that ie current proportional to square of voltage?!

 

[winnie]

Is he stating or implying that ie current proportional to square of voltage?!

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

He is saying that the connection change from delta to star (which effectively changes the voltage across the coils by sqrt(3) ) that the current will change by a factor of 3, in other words current proportional to the square of the applied voltage.

I am simply asking which motor state gives this proportionality.

I don't disagree that it is a possibility, but it clearly is not true in _all_ motor operating states.

-Jon

 

[Sahib]

He is saying that the connection change from delta to star (which effectively changes the voltage across the coils by sqrt(3) ) that the current will change by a factor of 3, in other words current proportional to the square of the applied voltage.

I am simply asking which motor state gives this proportionality.

I don't disagree that it is a possibility, but it clearly is not true in _all_ motor operating states.

-Jon

Suppose at any one condition such as no load, accelerating etc the motor phase winding impedance 'z'. Let line voltage V. Then star line current Is=(1/1.732)*V/z. The delta phase current Id=V/z. Then delta line current Idl=1.732*V/z. So ( Idl/Is)=(1.732*V/z)/(1/1.732)*V/z=3. So Is=Idl/3.

 

[Ingenieur]

Suppose at any one condition such as no load, accelerating etc the motor phase winding impedance 'z'. Let line voltage V. Then star line current Is=(1/1.732)*V/z. The delta phase current Id=V/z. Then delta line current Idl=1.732*V/z. So ( Idl/Is)=(1.732*V/z)/(1/1.732)*V/z=3. So Is=Idl/3.

bingo
it is calculated at steady state full load but should be ~ proportional for starting and various loading

when you look at info for wye-delta starters they tell you iy ~ 1/3 id
but do not elaborate
the above is a good explanation

 

[winnie]

Okay, there is a point that I was missing:

In the locked rotor state:
The voltage across a coil is reduced by sqrt(3) in the star connection, so the current through the coil is reduced by sqrt(3).

The current in the delta configuration puts pairs of coils together at each terminal, thus by changing to the star configuration and adjusting voltage to maintain the _same_ coil current the star configuration reduced current by sqrt(3).

The combination of these two effects reduces the current by a factor of 3.

Thanks
Jon

 

[Ingenieur]

Okay, there is a point that I was missing:

In the locked rotor state:
The voltage across a coil is reduced by sqrt(3) in the star connection, so the current through the coil is reduced by sqrt(3).

The current in the delta configuration puts pairs of coils together at each terminal, thus by changing to the star configuration and adjusting voltage to maintain the _same_ coil current the star configuration reduced current by sqrt(3).

The combination of these two effects reduces the current by a factor of 3.

Thanks
Jon

yes sir
1/sqrt3 x the voltage (same v over 2 coils, each coil sees ~ line-neut v) and 1/sqrt3 x the current (due to 2 coils, twice the z for the same v, thus reduced current)

 

[Besoeker]

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

Most likely delta. The 380V is most likely to be the delta configuration.

 

[GeorgeB]

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.

I 100% agree with Jraef. Let me give another example.

I was called from 200 miles away to troubleshoot an injection molding machine that had been misbehaving for 2 years since being commissioned. The OEM had replaced the motor, pump, motor starter and verified all wiring.

Machine would start and run ok for hours, days or weeks, then trip the overloads.

Turns out the OEM had designed for use in applications where wye-delta starting was needed, but this plant ordered for across-the-line start. Only 3 wires from the motor ... connected in wye. The pump was unloaded for start, so probably only 15% or so of full load torque was needed to start.

We never learned why it sometimes tripped; my money is on either operator settings or intermittent slightly low voltage. I made them open the motor terminal box ... they were SURE it was correct ... swapped the jumpers on the motor (what I call European style with 6 binding posts and either 2 or 3 jumpers) to delta and all was fine; they said it seemed to start too fast<g>.

I was asked how I knew ... told the truth, just logical troubleshooting, didn't know. OEM fumed at being back charged for downtime and multiple service calls. My employer was asked to send me in 3 more times to provide training.

My philosophy ... keep an open mind, and everyone lies even when they don't know they are lying.

 

[Ingenieur]

P = sqrt3 v i (pf eff)
s = slip
n = asynch motor speed rps = synch speed x (1-s)
= (2 f /p(oles)) (1-s)
v = i Z and Z = 2 Pi f L or v = i 2 Pi f L
T = P 88/(motor speed rps) ... converted speed to rps 5252/60=88
v = 460/3

substituting
convert P from W to HP ie divide by 746
T = (P/746) 88 / [(2 f /p)(1-s)]
= (sqrt3 (i 2 Pi f L) i (pf eff) 88/746) / [(2 f / p)(1-s)]
assume
pf eff ~ 0.8
4 pole
plug and simplify
T = (2 i^2 L) / f(1-s)

assume 480/3, 100 HP, 120 A so Z = 460/120 = 3.83 Ohm ignore R
L = 0.61/f
T = 1.24 i^2 /(f(1-s))
assume 2% slip at flc
T = 1.24 (120^2) /(60(1-0.02)) = 304 lb ft
check
T = 100 5252 / ((1-0.02)1800) = 298 lb ft
close enough

so for a 4 pole motor 100 HP 460/3 at 60 Hz
T = 0.02 (i^2) / (1-s)
break down torque at 35% slip = 445 lb ft ~ 150% of rated T
can be done for other HP and poles by changing Z/L