Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science Electric Machines

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1 Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science Electric Machines Problem Set solutions September 8, 3 e dλ dt Problem : Torque is, for this system, T = Mi i sin θ and voltage in coil is v = = d dt M cos θi.. if i = constant, then the answers are trivial: e T v = MI I sin ωt = ωmi sin ωt These are plotted in Figure Problem, Constant Secondary Current Torque, Nm Terminal Voltage Time, s Figure : Problem : Fixed Current in Coil. if λ =, M i = I cos ωt L Torque is straightforward: Flux in coil is: e (MI ) (MI ) T = sin ωt cos ωt = sinωt L L M M I M I λ = M cos ωt I cos ωt = cos ωt = ( + cosωt) L L L

2 Problem, Zero Secondary Flux.5 Terminal Voltage Torque, Nm Time, s Figure : Problem : Zero flux in Coil Differentiating this we have These are plotted in Figure M I v = sinωt 3. Connected to a resistor, this is most conveniently worked in complex amplitudes: or which gives us where Torque is: voltage in coil is v = L V = RI = jωl I + jωmi jωm I = I R + jωl i = I cos (ωt + φ) ωm I = J I R + (ωl ) π ωl φ = tan R e T = MI I sinωt cos(ωt + φ) MI I = (sin(ωt + φ) sin φ) dλ which can be evaluated to be: dt v di = ωm sinωti + M cos ωt dt = ωm I (sin ωt cos(ωt + φ) + cos ωt sin(ωt + φ)) = ωm I sin(ωt + φ)

3 .5 Problem, Secondary shorted through resistor Terminal Voltage Torque, Nm Time, s Figure 3: Problem : Coil connected to Ohm Resistor These are plotted in Figure 3 4. The last part of this is simply taking the average value of torque and plotting over a range of speeds. This is plotted in Figure 4. 3

4 .384 Problem, Average Torque N m RPM Figure 4: Problem : Average Torque 4

5 Problem :. using the principal of virtual work, W m = Li + λi We don t know L but if it doesn t vary it doesn t matter, and force is just f e = m W x λ = i x From the voltage test described in the problem statement: dλ v = = dt dλ dx dx dt and if u = dx dt = 5m/s, and we can also use t =, so that: x u and then force is: dλ A α = xe u dx u x dλ f e = i dx = (4e)xe x Problem Set, part.3.. N m Figure 5: Force vs. position. Energy required to lift the magnet is W = mgh = =.49J. This is also force integrated over distance: x W = I 4exe dx. Using αx xe dx = α (αx ) e αx 5

6 where α =, Then required current is: x. xe dx = 4 (. + ) e = 4. e.49 I = 6.5A 8 4 6

7 Problem 3:. The rail width is such that, when the yoke is centered, each side has an overlap of w. Area is square meter, so force per unit area must be: Then flux density across the gap must be: Required current is: F 5, B P = = = 6 Pa = A.5 µ F B = µ.58t A B NI = g 37,7A-T. To accommodate lateral motion, note that total permeance of the system is: where µ P = P + P ( w ) µ l x P = g ( ) w µ l + x P = g then ( ) w µ l x P = g w Suspension Force is: Or, inverting, required ampere-turns is: This is plotted in the top part of Figure. (NI) P F = g 8Fwg NI = µ (w 4x ) 3. Lateral force is found by differentiating the permeance with respect to lateral position x: (NI) P (NI) x µ l F x = = x w g 4. Since we have already calculated NI(x), we may combine that expression and the derivative of permeance with respect to x to get: 8gFx F x = w 4x This is plotted in the lower half of Figure. 7

8 x Problem Set, Problem Amp Turns x 4 Restoring, N Offset, m Figure 6: Required Ampere-Turns and Lateral Force 8

9 % Problem Set, Problem % Parameters L =.; L =.; M =.8; I = 5; R = ; om = *pi*6; t =-/6:/6:/6; % Part : Constant Current I = 5; Tm = M*I*I; Vm = om*m*i; T_e = -M*I*I.* sin(om.* t); V = -om*m*i.* sin(om.* t); fprintf( Problem Set, Problem \n ) fprintf( Part : Torque Magnitude = %g\n, Tm) fprintf( Part : Voltage Magnitude = %g\n, Vm) figure() subplot plot(t, T_e) title( Problem, Constant Secondary Current ) ylabel( Torque, Nm ) subplot plot(t, V); ylabel( Terminal Voltage ) xlabel( Time, s ) % Part : Rotor shorted Tm = (M*I)^/(*L); Vm = om*m^*i/l; T_e = (M^/(*L))*I^.* sin(*om.*t); V = (om*m^*i/l).* sin(*om.*t); fprintf( Part : Torque Magnitude = %g\n, Tm) fprintf( Part : Voltage Magnitude = %g\n,vm) figure() subplot plot(t, T_e) title( Problem, Zero Secondary Flux ) ylabel( Torque, Nm ) subplot 9

10 plot(t, V); ylabel( Terminal Voltage ) xlabel( Time, s ) % Part 3: Rotor shorted through a resistor Im = om*m*i/sqrt((om*l)^+r^); phi = atan(om*l/r); theta = pi/ - phi; T_e3 = -(M*I*Im/).* (sin(theta) - sin(*om.* t + theta)); V3 = om*m*im.* sin(*om.* t + theta); figure(3) subplot plot(t, T_e3) title( Problem, Secondary shorted through resistor ) ylabel( Torque, Nm ) subplot plot(t, V3); ylabel( Terminal Voltage ) xlabel( Time, s ) % part 4: Average torque over a speed range RPM = 5:5; Om = (*pi/6).* RPM; I_ = I.* Om.* M./ sqrt(r^ + (Om.* L).^); Theta = pi/ - atan(om.* L./ R); Tav = -.5*M*I.* I_.* sin(theta); figure(4) plot(rpm, Tav) title( Problem, Average Torque ); ylabel( N-m ) xlabel( RPM )

11 % Problem set, problem, 3 a = 5; A = *exp(); %x =.:.:; t = -.5:.:.5; f = A.* t.* exp(-a.* abs(t)); %g = exp(-a.* abs(x)).* (-a.* abs(x) -)./ a^; figure() %subplot() plot(t, f) %subplot() %plot(x, g) title( Coil Voltage ) xlabel( time, ms ) ylabel( Volts ) axis([ ]) grid on % now we can plot force vs. position u = 5; x = -.5:.:.5; D = (A/u^).* x.* exp(-(a/u).* abs(x)); figure() plot(x, D) title( d flux/dx ) ylabel( Wb/m ) xlabel( distance from center, m ) axis([ ]) grid on % now get energy stored B = A/u^; b = a/u; x = :.:.5; W_ = -(B/b).* x.* exp(-b.* x) + (B/b^).* (-exp(-b.* x)); % energy with respe figure(4) plot(x, W_) title( Energy from center of coil ) ylabel( Joules ) xlabel( Distance from coil center, m ) x =.; xx =.:.:.5; W = (B/b).* (x * exp(-b * x) - xx M =.5; v = sqrt((/m).* W); % to get energy with respect to this position.* exp(-b.* xx)) + (B/b^).* (exp(-b*x) - exp(-b.* xx

12 x =.; W = (B/b) *(x *exp(-b*x) - x*exp(-b*x)) + (B/b^) * (exp(-b*x) - exp(-b*x)); v = sqrt((/m) * W); fprintf( Velocity = %g\n, v) figure(3) plot(xx, v) title( Velocity vs. x ) ylabel( m/s ) xlabel( position, m )

13 % 6.685, Fall, 3 problem set, problem 3 F = 5; % required force muzero = pi*4e-7; W =.; l = ; g =.5; A = W*l; B = sqrt(*muzero*f/a); NIz = *g*b/muzero; % width % length % gap % resulting area fprintf( Required Supporting Flux Density %g T\n, B) fprintf( Required Coil Current %g A-t\n,NIz) d=w/4; % increment of x x = -W/4:d:W/4; % over this range of x NIsq = 8*F*W*g^./(muzero*l*(W^- 4.*x.^)); NI =sqrt(nisq); Fx = - 8*g*F.*x./(W^ - 4.*x.^); figure() subplot plot(x, NI) title( Problem Set, Problem 3 ) ylabel( Amp Turns ) grid on subplot plot(x, Fx) ylabel( Restoring, N ) xlabel( Offset, m ) grid on 3

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Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science Electric Machines

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines Problem Set Solutions September 17, 5 Problem 1: First, note that λ 1 = L 1 i 1 +