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Discussion Starter · #1 · (Edited)
we have a 22kw motor 37amps that smoked really bad.tried to check the control to see if there was a problem with the overload settings bt attached is the circuit that I saw(that grey wire on F4 is for resetting
I hope the pic is clear.if you see the relay on your far left written F4. that's where the overload is. and the input to the contactor relay is S1,S1,S1 from the three current transformers for the three phase and the output is S2,S2,S2 which goes back to the CT. and the overload was. set at 0.9. I have never come across such type of motor protection.does anyone know about this.its from an old panel from india

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What is the CT ratio and what is the motor full load current? Larger motor starters often use CTs for the overload relay, but that would be very unusual for a motor that small.
 

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Discussion Starter · #3 ·
What is the CT ratio and what is the motor full load current? Larger motor starters often use CTs for the overload relay, but that would be very unusual for a motor that small.
5/50A. rated current 37.5A. please explain to me how this principle works and why its set. at 0.9

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5/50A. rated current 37.5A. please explain to me how this principle works and why its set. at 0.9
From my understanding, If you have a 50:5 ct ratio.... This means when the the load is drawing 50A, the relay will be supplied 5A. If only 40A is being drawn by the motor, then you will have 4A flowing to the relay. Typically current transformers are used on large loads as it is not practical to have 50A on the control side to operate control relays, and metering. Using the ct's allows you to step down that current to a lower level, for controls like in your picture. Your relays and equipment is much cheaper when it is rated to handle 5A as opposed to 50A. The setting of 0.9 is the setting where at 0.9 x 5A (4.5A) the normally closed overload will open. This means when the motor for example draws 45A the overload relay will have 4.5A flow to it and it will open. If it was set at 0.5x the motor would trip out at 25A. Hopefully that helps. Not the most technical explanation but maybe it helps you out. I'm sure Jraef and big john will chime in to correct me.
 

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The ratio of the CT (5 divided by 50 = .1) is the ratio of your FLA to the setting of the overload. 37 FLA x .1 = OL setting at 3.7 amps. :thumbsup:
 

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Discussion Starter · #7 ·
From my understanding, If you have a 50:5 ct ratio.... This means when the the load is drawing 50A, the relay will be supplied 5A. If only 40A is being drawn by the motor, then you will have 4A flowing to the relay. Typically current transformers are used on large loads as it is not practical to have 50A on the control side to operate control relays, and metering. Using the ct's allows you to step down that current to a lower level, for controls like in your picture. Your relays and equipment is much cheaper when it is rated to handle 5A as opposed to 50A. The setting of 0.9 is the setting where at 0.9 x 5A (4.5A) the normally closed overload will open. This means when the motor for example draws 45A the overload relay will have 4.5A flow to it and it will open. If it was set at 0.5x the motor would trip out at 25A. Hopefully that helps. Not the most technical explanation but maybe it helps you out. I'm sure Jraef and big john will chime in to correct me.
ok I get most of that. thanks
since am at work now I went to.check on it the load is drawing 31 amps. so per our CT ratio at the relay its 3.1 right? now the O.L setting is at 0.9
thus 0.9×3.1=2.79A
this means the overload will only open at(0.9×5=4.5A) right? this seems to be on a higher side so I have reduced it to 0.7×5A=3.5A
this means when the load rises to 35 amps the overload should open. (it should be able to with stand the inrush current when the motor is first made ON. the motor is star connected)

btwn
we have AC motors as high as 85kw which use the normal overloads not the CT type.are there any advantages to it?

lets imagine a worst case scenario all the three CT fail at once so we wont have a reference input to the relay will the motor stop running? will it give an over voltage alarm? thanks


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Typically the setting on the overload relay is in amps and with a 50 to 5 ratio, 0.9 would be way too low to let the motor run. If it is in percentage (0.9 setting = 90%) of the 0-5 amp range then it would be set to trip when the motor is pulling over 45 amps. With a 37 amp motor that would be just short of 125% of the full load current. 125% is permitted by the NEC for motors with a service factor of 1.15 or greater. If the service factor is less, than the maximum setting of the overload protection would be 115% of the nameplate full load current.
 

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Typically the setting on the overload relay is in amps and with a 50 to 5 ratio, 0.9 would be way too low to let the motor run. If it is in percentage (0.9 setting = 90%) of the 0-5 amp range then it would be set to trip when the motor is pulling over 45 amps. With a 37 amp motor that would be just short of 125% of the full load current. 125% is permitted by the NEC for motors with a service factor of 1.15 or greater. If the service factor is less, than the maximum setting of the overload protection would be 115% of the nameplate full load current.
I agree, with most IEC bimetal OL relays, the value on the dial is Amps, not percentage.

If that is the case, then it may be that your unit was designed for a 50:1 CT ratio, but someone put in a 50:5 CT, which means the OL relay would almost instantly trip. If you look, it also appears that someone may have put in a jumper across the aux contacts, which means it might as well not even be there and the motor would burn up. Given that the motor DID burn up, I would start with that assumption and work from there to confirm.

But to be sure, get the manufacturer and model number of that OL relay and check it out.
 

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test your o/l, trip it manually and your motor should stop (stop the contactor) if contact from o/l is bypass, it will never stop motor (maybe overload is burned also), check for current on the overlaod itself to see if ct are ok
 

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Sorry, I have a better monitor now, I can see it is a Siemens overload relay from the old "World Series" line from the 1980s, so probably a 3UA... series. The dial numbers on that device are Amps, not percent. So your setting is 0.9A and on a motor current draw of 37A at Full Load, that 50:5 CT ratio is sending it 3.7A, over 400% of the setting. Most likely it was in a constant state of being tripped and would immediately trip on every start after being reset, so rather than someone trying to figure out why, they put a jumper across the Trip contacts that were supposed to shut down the motor starter. So if you never overloaded the motor, you never noticed, but the first time there was a problem, you had ZERO protection and the motor fried.
 

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Discussion Starter · #12 · (Edited)
Sorry, I have a better monitor now, I can see it is a Siemens overload relay from the old "World Series" line from the 1980s, so probably a 3UA... series. The dial numbers on that device are Amps, not percent. So your setting is 0.9A and on a motor current draw of 37A at Full Load, that 50:5 CT ratio is sending it 3.7A, over 400% of the setting. Most likely it was in a constant state of being tripped and would immediately trip on every start after being reset, so rather than someone trying to figure out why, they put a jumper across the Trip contacts that were supposed to shut down the motor starter. So if you never overloaded the motor, you never noticed, but the first time there was a problem, you had ZERO protection and the motor fried.
haha! jraef nothing wrong with your monitor. actually I took that picture with ma phone which is not of so good quality bt it helps me learn and get updates from this forum (China make)

hey I even wonder how you saw that loop with that quality bt that was good.
now that loop is actually from point 95 giving power to point 97 and 98 (NO) contact. for trip indication purpose but point 95 and 96 is intact.

btwn.
if all CT were to fail at one go would the motor still run?

by the way we had to put that motor on a drive bt still retaining that circuit. incase of drive fail through a kind of changeover as shown in pics. View attachment 34141 View attachment 34141 View attachment 34141
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