Induction motor DOL starting characteristics 1

[yorgi]

Good afternoon , A colleague and I have been asked to install a compressor with a 3.3kV 1MW motor as its prime mover. We have been chasing the manufacturer for the typical Starting characteristic. We have been assured that the compressor can only start in an unloaded condition. That would suggest that the starting current would be between 3 and 5 times full load current but for a short time maybe 1 to 2 seconds. The issue being the potential Volt drop that the starting current may induce. Is the suggested starting time accurate or are we potentialy looking at an extended starting period.

 

[waross]

I would model the starting current at 6 times for 5 seconds. (Canadian code stipulates 6 times in the absence of other information.) If you can live with those results you are probably safe. If The voltage drop is more than you can accept, then try harder to get exact information from the manufacturer. The acceptable voltage drop depends a lot on the location. What may be acceptable on a dedicated industrial feeder may be unacceptable in an area where power is shared with residential and or commercial loads. Who are you neighbours, electrically speaking.

Bill
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"Why not the best?"
Jimmy Carter

 

[7anoter4]

If it is an IEC standard motor the time could be less.
The starting time depends on the acceleration torque, motor rpm and inertia moment of the motor and the load. See-for instance:

http://www05.abb.com/global/scot/sc...6fc6003e4e05/$File/Motor guide GB 02_2005.pdf

section 4.3.3 Starting characteristics
ts=2*pi()*rpm/60*(JM+J'L)/Tacc ts =starting time [sec]
JM=motor inertia moment [kg*m^2 ] JL=load inertia moment [kg*m^2]
J'L=JL*(rpmLoad/rpmMotor)^2 if there is here a gearbox.
Tacc=Tmotor-Tload acceleration torque[Nm]
Tmotor=0.45*(Ts+Tmax) approx. Ts=start torque Tmax =maximum torque
Tload=kl*Tratedload Tratedload=Tnmot=rated motor torque [usually].
Kl=1/3 for centrifugal compressor or pump.
Let's take an ABB motor HV Cast iron motors 3 KV 1000 KW 600 rpm .See:

http://www05.abb.com/global/scot/sc...hnical_IEC_catalog_FINAL_EN_092011_lowres.pdf

JM=140.6 kg.m^2 Tmax/Tn=1.9 Ts/Tn=0.7 Tn=16006 Nm
Tmotor=0.45*(1.9+0.7)*16006=18727 Nm Tload=16006/3=5335 Nm
Tacc=18727-5335=13392 Nm
ts=2*pi()*600/60*(2*140.6)/13392=1.32 sec.

 

[waross]

After reading 7anoter4's post, I accept that my starting time was high. I will accept under 2 seconds start time for a low inertia load.
However, the voltage drop remains the same value but for a lessor time. Here, by code, I must use 6 times normal current unless I can justify a lower starting current. I see that the starting current ratio for the ABB motors shown in one of 7anoter4's post run from 5 times to 5.5 times normal current. Nice information, 7anoter4, thanks.
I am not shy about suggesting to a sales rep that if such information is not immediately available then possibly I am dealing with the wrong company. I started with no patience and I don't have much of it left. grin

Bill
--------------------
"Why not the best?"
Jimmy Carter