How to simulate real torque of induction motor?
Duc Le
on 18 May 2022
Latest activity Reply by Loc Minh Duc LE
on 19 May 2022
I am trying to simulate a 3 phase induction motor on Simulink by using its dynamic model. This is the information in its datasheet:
It also has 4 poles and J = 0.0117 kgm², star connection. After calculate parameters, I get these values:
Lm = 0.176 H; Rr=Rs = 0.613Ω; Llr=Lls = 0.00669 H.
Also in the datasheet, they said that the output torque = 705 Nm at speed 35 rpm, but the model cannot reach that value. The simulation motor run backward if applied torque greater than 70 Nm ( at about 1200 rpm). But in reality, this motor can withstand 705 Nm, I have seen it run.
Can someone help me with this problem? Thank you
4 Comments
Time Descending
You can start with some of these examples:
you can use an asynchronous machine to model an induction motor.
Regards,
Joel
Thank you for your reply.
I get the same result with asynchronous machine from simscape, motor cannot turn with the torque greater than rated torque (19.7 Nm)
can you share the model that you put together and how you calculated the parameters?
I get the dynamic from this link:
And asynchronous machine from this link:
Ablout the calculation:
Motor is star connection Vs=400/√3 = 230V; Is=6.4A
- No load test:
cos(phi) = 0.76 -> sin(phi) = 0.65
Xm = Vs/(Is*sin(phi)) = 230/(6.4*0.65) = 55.288
Lm = Xm/(2*pi*50) = 0.176 H
- Lock rotor test:
Ia/In = 8.2 -> Ia=8.2*6.4=52.48 A
Ma/Mn = 3.4 = Pa/Pn -> Pa=3.4*3000 = 10200 kW
Z = Vs/Ia = 230/52.48 = 4.38 ohm
R = P/(Ia^2) = (10200/3)/(52.48^2) = 1.23 ohm
X = √((Z^2)-(R^2)) = 4.2 ohm
Rs=Rr = R/2 = 0.615 ohm
Llr = Lls = X/(2*(2*pi*50))=0.00669 H
Sign in to participate
