AC Induction motor startup current

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Friend gave me this task he had back in school to calculate the necessary amperage for a fuse that is put in series with one of the phases for a 3 phase induction motor.
I could calculate the working current of the motor from the rated kW and cos voltage/current angle, but I couldn't figure out how to calculate the starting current.

The parameters are as such , motor is put in a 3 phase delta (triangle) connection, supply voltage 380v, motor rated power 9.4Kw, power coefficient is 0,91cos.
From this I took 9.4kW and expressed that as 9400w which I divided by the power coefficient times voltage, I got 27.18 amps, the total current consumed, divide that by 3 as each phase uses one third of the current and I got 9.06 amps as the working current for a phase.
Obviously I could crudely take those 9 amps and multiply them by about 7 (times the nominal current for an induction motor at startup) but the task gives additional data, so my question then is using this additional data is it possible to determine the starting current and how?

additional data: mechanic efficiency = 0.98
Kp=Ip/In = 5 (really don't know what this means)
and lastly lower case alpha α = 2.0


thanks.
 

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  • #2
jim hardy
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There are published guidelines for motor fusing , available at national electrical code and at fuse manufacturers sites

eg
https://m.littelfuse.com/~/media/electrical/application-notes/littelfuse-motor-protection-guide.pdf
http://www.cooperindustries.com/con.../BUS_Ele_Tech_Lib_Motor_Protection_Tables.pdf
As to starting current
someplace on the motor nameplate should be an entry called "NEMA KVA Code" or similar
it's a single letter code telling you what is the starting current
you look up the value corresponding to the one letter code in a table
it's expressed in kva per horsepower


here's the top half of the table, rest of it is in the link
244337


https://www.electricalengineeringtoolbox.com/2015/12/nema-locked-rotor-kva-for-three-phase.html
old jim
 
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If we take R&M technical data resource for a 12.5 HP/9.3 kW induction motor at 380 V 50 Hz the rated current it is about 20 A[from other source the efficacy it is only 88% and power factor only 0.8]. However, in your case I=9400/SQRT(3)/0.98/0.91=16 A.
Ip/In could be the start current divided by rated then the start current it is about 80 A. In my opinion α it could be Tα/Tn the starting torque to rated torque ratio. Usually the start current it is between 6 to 8 times the rated. However, for 16 poles [750 rpm] this could be 5.
Still, in my opinion, some data are not entirely real. For instance efficacy of 98% is difficult to achieve.
244406
 
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Well this is a crude school task so I guess the numbers aren't meant to add up to real life performance.
@Babadag
so you are dividing the rated power in watts with the square root of (mechanical efficiency divided by cosine) ? Then why the you put (3) after sqrt, I don't understand this one?

I am trying to solve the equation you gave but I can't , I'm getting different numbers
 
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Irated[A]=P[kW]*1000/[√3*cosfi*eff*V]
[only 3 is under sqrt[√]
Irated=9400/[SQRT(3)*0.98*0.91*380]=16 A
 
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@Babadag ok and if I had a single phase induction motor , would I then have to use sqrt of 2 or any sqrt or none at all?
 
  • #7
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If "none at all" it means 1 then it is correct.
 

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