D.O.L starter overload settings

[TomAng]

Hi everyone
ive got my AM2 coming up and im unsure about the over load settings for the D.O.L, ive seen 2 different equations and im trying to work out which is right.
so say my motor is 750w would my calculation be 750/400/1.732/0.8 = 1.35
or
750/400 x 1.732 x 0.8 = 2.59
i bought a course to learn and they use the second equation and some how got the first equations answer?
thanks in advance.

 

[brianmoooore]

750W on a motor usually refers to the output power, not the input, which will be higher. FLA is the figure you want off of the plate.

 

[telectrix]

i.e if the rating plate says 10A you set the overload at 10A. although it used to be common practice to set at 10% over FLC.

 

[plugsandsparks]

Or use thicker oil.?????

 

[TomAng]

i.e if the rating plate says 10A you set the overload at 10A. although it used to be common practice to set at 10% over FLC.

ok have you done the AM2 in the uk? just both my tutor and training video said this is the formula.
only problem is i cant remember whether he said to divide or times for the formula and the training video has both.
but if you have done the AM2 in the uk then your answer could be right.

 

[Kyle_Trainee]

i.e if the rating plate says 10A you set the overload at 10A. although it used to be common practice to set at 10% over FLC.

In the AM2 you only get the output power so what would the equation be to determine the FLC, would it be I= P/Vx1.73x0.8 also where does the 0.8 value come from

ive also attached basically what you get in the AM2 bay this is from a training rig at college

 

[WallaceP]

In the AM2 you only get the output power so what would the equation be to determine the FLC, would it be I= P/Vx1.73x0.8 also where does the 0.8 value come from

ive also attached basically what you get in the AM2 bay this is from a training rig at college

I think 0.8 is the value of the power factor

 

[TomAng]

Another question about the DOL. when testing IR. should I disconnect cables from the contactor? Or will it be able to take the 500v

 

[Lucien Nunes]

would my calculation be 750/400/1.732/0.8 = 1.35
or
750/400 x 1.732 x 0.8 = 2.59

Think about what each term means and you will see that the second version has a simple mistake: it has been written with the brackets missing that show the order of calculation. If you put the brackets in, the two formulas are the same and give the same result.

The aim here is to calculate the line current from the the power P, by dividing it by the line voltage V (line), also dividing it by sqrt (3), and also dividing it by the efficiency / power factor term (0.8).

I (line) = P / V (line) / sqrt(3) / 0.8
= 750 / 400 / 1.732 / 0.8
= 1.35A

Logically the second version must be wrong because it's implying three phases carry less power than one phase per line amp and that the motor has an impossible over-unity power factor. But re-write it with the brackets and all is good:

750 / (400 * 1.732 * 0.8)

Multiply what is inside the brackets first, then divide the power by that intermediate result, and you have the correct answer of 1.35A again.

would it be I= P/Vx1.73x0.8

And you're being careless with brackets too!

 

[Darkwood]

In response to tel, you're showing your age, we don't recommend setting any overload beyond the rated capacity of the motor, if the motor is running continuously at just over full load then you can get a runaway effect and overheat the motor without the overload operating in time, your advice relies on a comfort margin and a short lived overcurrent which means the overload won't trip as designed on some cases, might be rare but it still is a factor of motor failure.