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Trying to model a 3 phase induction motor circuit for a certain production line

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Deepak
Deepak 2023 年 4 月 7 日
コメント済み: Deepak 2023 年 4 月 11 日
We have a setup where we are monitoring the current supplied to the a 3 phase induction motor in a manufacturing plant. The current is read using CT sensors attached to the supply .
We are trying to identify current patterns during various fault situation but don't have enough data to do so. Hence, I am trying to model similar circuit in matlab simulink.
Model attached.
The nameplate spec of the motor are as follows :
RPM = 1380, A = 1.05, PF = 0.74, V = 415+-10% ,Hz = 50+-5%, kW = 0.37, HP = 0.50, EFF = 66.0, Frame = 71 , AMB = 50 ,IP = 55, InCl = 'F', Duty = 87, Emcl = 'TEFC'
Am from data science background and having trouble here regarding the parameters to be set. This mode made with the help of a few youtube videos. I have used parameters for the model from an example model on simulink except the First 3 [pow,volt,freq]. I have also set the supply voltage phases with a difference of 120 deg, freq = 50Hz and the supply voltage as Supply Voltage = (sqrt(2) * Vrms) / sqrt(3) Where Vrms is the given voltage on motor nameplate, which comes to be as 338.85 for this motor.
The problem I am facing is that the readings are supposed to be in ranges of 0.6 (as in the real world reading we are getting) as the motor is of max 1.04 Amps, but here the readings are of greater values. I thought of using Gain block before scope to scale these but one of the electrical guys said I should look into setting the parameters right.
The whole project is on me now and I don't have any mentor for this domain. Any help in setting up the simulation right is appreciated. Thank you!!.

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Joel Van Sickel
Joel Van Sickel 2023 年 4 月 11 日
Try using the motor parameter estimation tool on the block. Here are some parameter to help you out.
Nominal torque = 2.58
synchrnoous speed = 1500
Nominal speed = 1380
leave the ratios alone for now.
I don't know how close this will get you to your lab results, but it does bring the current down to something a bit more believable.
Regards,
JOel
  4 件のコメント
Deepak
Deepak 2023 年 4 月 11 日
Oh I see, that makes things much easier. Thanks. Also I found out the the 3 ratios from the motor's spec sheet along with a bunch of other values
Torque(kgm) = 0.26, Efficiency % [ FL,3/4L,1/2L] = [66.0 65.0 58.0], Power Factor [ FL,3/4L,1/2L] = [0.74 0.64 0.53], GD^2 (kgm2) = 0.0022, Wt.(kg) = 20 , I(ST)/I(N) = 3.5, T(ST)/T(N) = 1.9, T(PO)/T(N) = 2.1.
Setting the following parameters changed the motor type to double cage too.
The Speed (RPM) under constant full load (2.56) comes as around 13 and under no load is 30 which is far cry from the rated speed. Am assuming its due to those resistances being so high which I might have to look into it a bit further. Anyways, thanks for your inputs. Much appreciated.
Deepak
Deepak 2023 年 4 月 11 日
Setting up the supply voltage appropritately helped. The peak supply voltage for each phase would be 415V * sqrt(2) = 587V

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