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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