Star-delta overload protection

Hello everyone,
here I have one problem, which I can not understand, can you please help me?
In attachment you will find a basic star-delta diagram, with unreal calculation (Z is not real value, it's just for calculation and understanding the princip of star-delta). According to these current values, I dont't know what should be the values of my overload relays F1 and F2. Can you please help me?

Thank you.



Insights Author
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Your photo is very dim and hard to read. Can you post a better one? Use a better light source.
I can read it. I looked at how carefully the layout might be made from a blank page left to right and the jump-over lines hand made, possibly by starting line segments and on the way, from several attempts, at the outside wire on the left clarified as utilizing that its then the outside wire on the right. I can't assist but it's a good example of presenting all the qualities forum requests of to discuss the problem. I have trouble with my room overhead light in the center of the ceiling casting the camera's shadow every direction I turn. I've managed better illumination sometimes, yet another rewarding common subset topic. There is effort in this. My compliments, the programmers that wrote visio have a feature that unoverlaps lines, teaching a dumb lump of sand to just do that is hard. Neat example. I can't help on the values, but I'd like to deflect the deflation on an unintended roadblock when I get sick frustrated I ask questions, I criticize unthinkingly because I have to burp and forgot. This is an interesting question, I got curious if the relays unzip or throw at the same time? But that's not his question either. Try putting the pic in paint or gimp to lighten it, my phone was clear enough. Lol open it on your phone instead of your laptop and move the laptop lid around for viewing angle. copy it to paper so you can read it in your own hand with your own materials. Sometimes this is archaeology! Turn the lights out.

People sometimes have to brace the phone because not they shake, but breathing itself moves the phone and we play phone tag with the autorefocusser! Is it set for daylight or is there a big flash glare on one side of the page....take pic from an angle?
Let's laugh and call these rich people's problems in the hands of poor people using their phone with clunky interfaces. The handwriting is quite legible. Now I check if the phone retyped something with spellfixer.
Does M values matter? Is that an M? Even locating, centering, and labeling parts of drawings is a set of routines.

I'm sorry I didn't just know the formulas but I've been taking pictures of printouts while my computer is down, or a book page example with flash glare. Even sharing experience with that as my classroom memory goes is good. Perpetual phone tag with the auto focusser while breathing vs. its processing delay refocussing is like, free drinks....when you don't want them, or if you do. um, unreadability is off topic, but I assert my phone and my mk1 eyeballs with reader glasses worked. Nice drawing and the eqns fit on the same page/pic too. Space between lines on drawing is very good. If we're going to twitter about it. My computer being down keeps me from scanning napkin drawings into jpg from the out of ink all in one printer, too. Looks like same! Hm, maybe that's why they need an overload relay. Hee ;) get the ventilator fan running too..oh not anymore... hang in there help is on the way. Reason for deletion, I'm a blithering idiot blithering? That took some work with clarity.
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I've never seen a wye-delta starter with a second overload relay 'F2'. 'F2' detects the same current that flows through overload relay 'F1', and I don't see the reason for adding it.

Either overload relays were:
  • placed in the mains line and set to motor full load current. The down side is it won't properly detect an overloaded winding until after the transition timer switches the star relay open, and closes 'K2'.
  • or "in the delta" (as 'F1' is as shown in your drawing), in which case it would be set to 58% of motor full load current, (1/√3) * FLA.

wye-delta starter.jpg
I’ll try to explain more the phenomenon:
The actual schematic diagram of an induction motor it is similar with a transformer.

The difference is the rotor side impedance and output voltage depends on s[slip].Slip is related to the difference between the magnetic field rotation-equal to the synchronous speed-a bit more than no-load speed- and the actual rotor speed. It changes all the time- from start =1 to 0.02-0.05 at full load:
s=(synchronous speed-actual speed)/synchronous speed.
The reactance is changing more since depends also directly on slip since the rotor current frequency f=frated*s
That means for 60 Hz at s=1 [start] f=60 and for full load it could be 0.05*60=3 Hz
If the rotor side of the diagram will be divided by nst/nr[stator number of turns/rotor number of turns] and by s then we may attach this to the stator side:

Let's say the rotor is a squirrel cage short-circuit rotor then V'r=0
At start the Io[the magnetic core equivalent current] is negligible we have this:


For start this formula is close to the actual. For other slip this it is only informative.
What we know then we have an impedance at start [this is the minimum] and other impedance at other s [and maximum at the no-load steady].
Let’s take an induction motor of 200 kW at 440 V A Ilockrot=6*Irated
Irated=381 A[line current] but the current in the transformer windings connected in delta will be 381/sqrt(3)=220.1 A. and cos(φ)=0.86 Z=440/220=2
So let's say 1.72+j1.02 [Z=sqrt(1.72^2+1.02^2)=2] it is the impedance at full load [when R/s could be significant]
At start D.O.L-across the line- [in delta line current ] IstD=Ilockrot=6*Irated=2288.7 A and p.f.[cos(φ)]=0.3 then Zst= 0.1+ j0.32 at start [Z=0.333].
At start, connected in star, the voltage will be only 440/sqrt(3)=254 V per phase and the current at start will be 0.33*IstD=762.9 A[254/.333=763.7A]
Post #5 doesn't clarify.

In post 1 you used a Z of 2.0 for both star and delta calculations, but in post #5, Z=0.333 in the star example although it remained 2.0 in the delta example. Further, a delta line current of 380A is in a believable range, but in the star example in post #1, a current of 75.15 amps seems too low, while the revised current of 763A in post #5 is impossibly high for a 200 kW motor.

Consider a real-life example, a WEG, W22 series, 200 kW, 4 pole, 1487 RPM/50 Hz motor (product code 12962587) is rated 380V/360 amps full load, 99A no-load (magnetization current), 2160A locked rotor, with a maximum locked rotor on-time of 37 seconds (cold winding) and 21 seconds (hot winding).
  • 360 amp full load ("outside the delta"; mains current)
  • 380V/sqrt(3) = 220V
  • 220/380 is a ratio of 0.58. This can also be expressed as 1/sqrt(3).
  • 0.58 * 360 = 208A line amps "in the delta".

The same reasoning applies to contactor sizing. This FAQ from Schneider Electric may be of assistance.

According to these current values, I don't know what should be the values of my overload relays F1 and F2.
F1 and F2 ratings would be identical (although the only purpose I can see for F2 would be redundancy in case F1 failed to function).

In this WEG motor example it would be set to 208 amps.

My experience with across-the-line started AC motors taps out at 100 horsepower, thermal-magnetic overload relays in this size range are already quite large, and I don't know if they are available into the 200 kW size range. The only open-transition star-delta system I've worked on was on a 300 HP motor, and used 500:5A output current transformers fed through an adjustable thermal-mag relay (if memory serves, adjustable from 3.7A to 5.5A) for overload protection.
In my opinion, this schematic diagram it is for starting purpose. In my opinion also it is nonsense to work in star a motor destined for delta since the torque will decrease at 58% only. The advantage of the star starting is the line current will be 33% of the delta starting current.
For your task will be the next calculation:
for star connection:
line current Il=Ip=Vp/Zp=Vl/(1.73Zp)=440/(1.73x2)=127 A,
total power S=1.73xIlxVl=1.73x127x440= 97 kW
for delta connection:
line current Il=1.73Ip=1.73Vp/Zp=1.73Ul/Zp=1.73x440/2=380 A,
total power S = 1.73xIlxVl = 1.73x380x440 = 289 kW

Read more here.

For your diagram through F1 and F2 will pass equal current, but in case of delta connection current will be more than in case of star connection.

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