Download pdf - EE 410/510: Electromechanical - UAH 410 Chapt 6.pdf · Example 6.2 • Calculate and plot the electromagnetic torque if the balanced current set is clear all; % Calculation of the

EE 410/510: Electromechanical SystemsChapter 6

• Chapter 6: Synchronous Machines• Radial Topology Synchronous Reluctance Motorsad a opo ogy Sy c o ous e ucta ce oto s

• Single Phase

• Three Phase

• Radial Topology Permanent Magnet Synchronous MotorsSynchronous Motors

• Two Phase

• Three Phase

• Mathematical Models

• Axial topology PM Synchronous Machines• Axial topology PM Synchronous Machines

• Conventional 3 Phase Machines

Note: We will be skipping multiple sections of this chapter in attempt to provide a clear introduction to thein attempt to provide a clear introduction to the material and allow us to move onto other equally important topics

All figures taken from primary textbook unless otherwise cited.5/21/2010 1

Single Phase Synchronous R l t M tReluctance Motors

• Unlike Induction machines Synchronous motors use permanentSynchronous motors use permanent magnet rotors.

• Thus there is not mismatch between the induction coupling of a squirrel cage and the stator coils leading to slipcage and the stator coils leading to slip

• The electromagnetic and rotational torques are equal

h h di f i l• The phase diagram for rotational velocity and torque in a PM synchronous motor looks much more like a DC motor phase diagram


• Assume that the motor drawn here has initial conditionshas initial conditions

Note:Direct axis aligns the N and S poles

N

• And that

• The magnetizing inductance on the rotor is

pof the magnet

Quadrature axis is perpendicular to the pole

S

• Which varies in maximum and minimum values as

to the pole direction


• Assume that the motor drawn here has initial conditionshas initial conditions

• And that

• The magnetizing inductance on the rotor is

• Which varies in maximum and minimum values as

• The energy and electric torque generated are


• Electromagnetic torque is not developed in h if l DC i li dthe system if only DC current is supplied

• Thus the applied current must be sinusoidal in nature to provide a constant driving force to the rotor

• Yielding the following torque equation

• The average torque generated by the magnetic circuit is

• Furthermore, one finds that the phase current required to maximize Te and eliminate

• Thus ideally providing

eadditional ripples in the wr is • However it is impossible to use this value

for i, thereby insuring that such systems will always have some oscillation in wr


SIMULINK of Single Phase Radial SR MotorSIMULINK of Single Phase Radial SR Motor

Radial Single Phase SR Motor ExampleRadial Single Phase SR Motor Example

3 Phase SR Motors3 Phase SR Motors

SS

N


• Using the circuit described

• Where as, bs, and cs represent the , , pindividual stator windings

• Inductance along the quadrature axis is labeled mq

• Inductance along the pole direction• Inductance along the pole direction (direct axis) of the magnet is md

• Magnetizing inductance is m average

• Leakage inductance is ls

• Flux linkage between as, bs, and cs is labeled abcs

• Mutual inductance between as, bs, and cs is labeled abcsand cs is labeled abcs




22 MM iPLT 2 MMe iPLT

Example 6.2

• Calculate and plot the electromagnetic torque if theelectromagnetic torque if the balanced current set is

clear all;% Calculation of the Developed Electromagnetic Torqueth=0:0.01:4*pi; % angular rotor displacementphi=0.3245; % phase current angleIM 10 P 4 LDm 0 05

• Let iM = 10 A, P=4, LM=0.05H

Th T 28 2842

IM=10; P=4; LDm=0.05;% Balanced three-phase current setIas=sqrt(2)*IM*sin(th+phi*pi/3); % current in the as windingIbs=sqrt(2)*IM*sin(th-(2-phi)*pi/3); % current in the bs windingIcs=sqrt(2)*IM*sin(th+(2+phi)*pi/3); % current in the cs windingA = sin(2*th);B = sin(2*th+2*pi/3);

22 MMe iPLT

• Thus Te = 28.2842B sin(2 th 2 pi/3);C = sin(2*th-2*pi/3);D = sin(2*th+4*pi/3); E = sin(2*th-4*pi/3);% Calculation of the electromagnetic torque developedTe =(P*LDm/2)*[Ias.*Ias.*A + 2.*Ias.*Ibs.*C + 2.*Ias.*Ics.*B + Ibs.*Ibs.*E + 2.*Ibs.*Ics.*A + Ics.*Ics.*D];%Pl h li d b i di

No ripple in applied torque%Plot the currents applied to abc windingshold on;plot (th,Ias,'-',th,Ibs, '--',th,Ics,'-.');% axis([0,4*pi,-10,10]);xlabel ('Angular Displacement,\theta_r [rad]','FontSize',14);ylabel ('Phase Currents','FontSize',14);title ('Phase Currents, i_a_s, i_b_s and i_c_s [A]','FontSize',14);% Plot of the torque developed versus the angular displacement3 phase voltage % Plot of the torque developed versus the angular displacementTep = sqrt(2)*P*LDm*IM*IM;plot (th, Tep);plot (th,Te); %axis ([0,4*pi,0,15]);xlabel ('Angular Displacement,\theta_r [rad]','FontSize',14);ylabel ('Electromagnetic Torque','FontSize',14);title ('Electromagnetic Torque, T_e [N-m]','FontSize',14);

3 phase voltage

Model for 3 phase PM SR Motor

• From

• We can write the following differential equations


• Cont.


• Lyshevski then provides a simpler version of these equations that are derived using the MATLAB symbolic toolbox.

• The following notations are required to address his simplified set of equations


• Lyshevski then provides a simpler version of these equations that are derived using the MATLAB symbolic toolbox.

• The following code is developed

L=sym('[Lls+Lbm-Ldm*Cl,-Lbm/2-Ldm*C2,-Lbm/2-Ldm*C3,0;-Lbm/2-Ldm*C2,Lls+Lbm-Ldm*C4,-Lbm/2-Ldm*C5,0;-Lbm/2-Ldm*C3,-Lbm/2-Ldm*C5,Lls+Lbm-Ldm C4, Lbm/2 Ldm C5,0; Lbm/2 Ldm C3, Lbm/2 Ldm C5,Lls LbmLdm*C6,0;0,0,0,2*J/P]');R=sym('[-rs, 0, 0, 0;0, -rs, 0, 0;0,0, -rs, 0; 0, 0, 0, -2*Bm/P]');I=sym ('[Ias; Ibs; Ics; Wr]');V=sym('[vas;vbs;ves;-TL]');K ('[Ld *2*W *(Sl*I +S2*Ib +S3*I ) Ld *2*W *(S2*I +S4*Ib +S5*I ) Ld *2*K=sym('[Ldm*2*Wr*(Sl*Ias+S2*Ibs+S3*Ics);Ldm*2*Wr*(S2*Ias+S4*Ibs+S5*Ics);Ldm*2*Wr*(s3*Ias+S5*Ibs+S6*Ics);Te]');L1=inv(L); L2=simplify(L1);FS1=L2*R*I; FS2=simplify(FS1)FS3=L2*V; FS4=simplify(FS3)

• Loading MATLAB one can easily recognizes that the “simplified” equations thought to reduce

p y( )FS5=L2*K; FS6=simplify(FS5)FS7=FS2+FS4-FS6; FS=simplify(FS7)

• Loading MATLAB, one can easily recognizes that the simplified equations thought to reduce the calculation time required are enormous.

• However there is said to be enough symmetry in this problem such that


• However there is said to be enough symmetry in the equations above that they can be reduced to the following generic form

• Note that this form is still highly nonlinear, and quite complex


• Cont.


• Cont.


• It turns out the most efficient manner by which one can develop a working model requires one simple assumption

F S h t th l t i d t ti l l l iti th ! ( li )– For a Synchronous motor the electric and rotational angular velocities are the same! (no slip)

• One can therefore change the reference frame from the stator to that of the spinning rotor, and thereby remove the extra rotational degree of freedom and significantly simplify the problem

• We can examine how this is achieved using a transformation matrix that was poorly introduced in Section 5 2 2Section 5.2.2

• Examining the system from the potential and induction currents generated in the quadrature and direct axis of the magnets, one can write

Park transformation matrix

• The beauty of this transformation matrix, is that it removes the complexity of the phase and t ti l l f th t bl di tlrotational angle from the motor problem directly


• The resulting current and torque relations may be written as:

• Because we=wr, the solution to these equations must match that of the stator reference framed equations we have used until this point.equations we have used until this point.

• Thus, we have a much simpler version of the differential equations in terms of the constants nominally provided for the motor!!!


• The problem is now solved by applying direct “Park” transformations

• Let us solve the following example. Consider the following 3‐phase PM SR Motor

• 4 pole, 110V, 400 rad/sec, 40 KW

• Rs =0.01 Ohm, Lmd =0.0012 H, Lmq = 0.0002 H, J =0.6 kg‐m2, and Bm = 0.003 N‐m‐s/rad