Induction Motor Simulation

Induction Motor Simulation-Synchronous Reference Frame Model Formulation Ref. Material :Induction machine simulation for the given Example [ 5.6]- Section 5.8 ,in the book by R.Krishnan : Electric Motor drives- Modeling , Analysis and control

1- Mr. Srinivas G. - Dr. Sastry V Vedula Induction Motor Simulation-Synchronous Reference Frame Model formulation Induction machine simulation for the given Example [ 5.6]- Section 5.8 ,in the book by R.Krishnan : Electric Motor drives- Modeling , Analysis and control is as follows:

Given data:200v, 4-pole, 3-ph, 60 Hz, Y connected, Rs=0.183 Ohms, Rr=0.277 Ohms, Lm=0.0538H, Ls=0.0553H, Lr=0.056H, B=0, Load torque = Tl=0 N-m, J=0.0165kg-m2, Base Power is 5hp.

70.7% of rated voltage is applied.

Va = 115.5sin(wt), Vb = 115.5sin(wt-120), Vc = 115.5sin(wt+120).[ These are supposed to phase to neutral voltages: 0.707*200 * sqrt (2)/ sqrt (3) shall be the peak phase voltage]Tm is load torque set to zero

2Induction motor simulation S R F Model3

Induction Motor Parameters in Per Unit ( P.U.) SystemThe model considered above is in p.u. The following are considered as base quantities:Wb = 2f = 377 = base frequency.Vb3(line-line) = 200v, Vb3(phase) = 115.5v, Vb2(phase) = sqrt(2) x Vb3(phase)=163.3v, Pb = 5HP=5x746 = 3730W, Base current, Ib3(line-line)=Pb/(sqrt(3) x Vb3(line)) = 3730/(sqrt(3) x 200) = 10.767AIb3(phase) = Ib3(line)=10.76 ( as it is star connected)Ib2(phase) = sqrt(2) x 10.76 = 15.21693793AZb = Vb2 (line)/Ib2 (line) = 163.3/15.21694=10.7342ohms

The p.u quantities are calculated as follows:Rsn = Rs/Zb = 0.183/10.7342= 0.017048314(p.u)Rrn = Rr/Zb = 0.277/10.7342= 0.025805371 (p.u)Wb x Lls = Wb x (Ls Lm) / Zb = 377x(0.0015)/10.7342 = 0.052682 (p.u)Wb x Llr = Wb x (Lr Lm) / Zb = 377x(0.0018)/10.7342 = 0.063218502 (p.u)Wb x Lm = Wb x (Lm) / Zb =377x(0.0538)/10.7342 = 1.88953 (p.u)Inertia constant (seconds), H = J x (Wb)^2/(2 x Pb x (p/2)^2) = 0.0165x377^2/(2x3730x4) =0.07819 seconds.

4Induction Motor Configuration : S R F Model5

Induction Motor Parameters taken6

7Simulation Configuration parameters considered:

Stator 3-Phase Phase Voltages / CurrentsVabcIabc8

Stator 3-Phase Phase Voltages / Currents under steady state 9

VabcIabcStator dq voltages and currents w.r.t time VqsVdsIqsIds10

Stator dq voltages and currents under steady state11VqsVdsIqsIds

Rotor d, q Currents IqrIdr12

Rotor d, q currents under steady state13

IqrIdrSpeed and Torque CurvesSpeedTorque14

Speed and Torque under steady state15

SpeedTorqueTorque vs Speed Charateristics16

Speed rad/sTorque N-mComparison of resultsSimulation: For stator voltages Vas, Vbs, Vcs are exactly same as book result. Book result: voltages are in purely sine wave. Simulation: Vqs is approximately zero. Vds is obtained as dc quantity with value 0.7073 p.u.Book result: Vqs is zero. Vds value is 0.7 p.u.Simulation results of Iqs and Ids currents have same oscillations as compared with results of book. Simulation results of Iqr and Idr currents have same oscillations as compared with results of book. Simulation results of speed and torque have same oscillations as compared with results of book. Speed settles at 1 p.u and torque at 0 which are same as book.

17ConclusionsBy simulating Induction motor model considering synchronous reference frame Vqs and Vds are DC quantities.

The above curves resemble similar curves given as solution for the problem in the book.

A similar speed and torque curves will be achieved if simulated the other two reference frames. i.e., with rotor reference frame and stator reference frame.

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